1000 Electronic Devices & Circuits Mcqs

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1000 Electronic Devices & Circuits MCQs Set by ltm3542 2

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Our 1000+ Electronic Devices and Circuits questions and answers focuses on all areas of Electronic Devices and Circuits subject covering 100+ topics in Electronic Devices and Circuits. These topics are chosen from a collection of most authoritative and best reference books on Electronic Devices and Circuits. One should spend 1 hour daily for 2-3 months to learn and assimilate Electronic Devices and Circuits comprehensively. This way of systematic learning will prepare anyone easily towards Electronic Devices and Circuits interviews, online tests, examinations and certifications. Highlights – 1000+ Multiple Choice Questions & Answers in Electronic Devices and Circuits with explanations – Every MCQ set focuses on a specific topic in Electronic Devices and Circuits Subject Who should Practice these Electronic Devices and Circuits Questions? – Anyone wishing to sharpen their knowledge of Electronic Devices and Circuits Subject – Anyone preparing for aptitude test in Electronic Devices and Circuits – Anyone preparing for interviews (campus/off-campus interviews, walk-in interview and company interviews) – Anyone preparing for entrance examinations and other competitive examinations – All – Experienced, Freshers and Students

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Here’s list of Questions & Answers on Electronic Devices and Circuits Subject covering 100+ topics: 1. Questions & Answers on EDC Overview The section contains questions and answers on basics of electronic devices and circuits, methods, frequency responses, time signals, network theorems, analysis of circuits, modulation, transmission and coupled circuits. Basic Concepts Method of Analysis Frequency Response – 1 Frequency Response – 2 Two Port Network Continous Time Signals Random Process

Noise Amplitde Modulation Digital Transmission Semicondunctor Physics Network Theroms Circuit analysis in S domain Magneticaly coupled circuits

2. Questions on Conduction in Semiconductors The section contains questions and answers on semiconductor electrons, holes and conductivity, donor and acceptor impurities, fermi level semiconductor, charge densities, diffusion, carrier lifetime, continuity equation and hall effect. Electrons and Holes in Semiconductor Conductivity of a Semiconductor Donor and Acceptor impurities Charge Densities in a Semiconductor impurities Fermi Level in a Semiconductor having Impurities

Diffusion Carrier Life Time The Continuity Equation The Hall Effect

3. Questions & Answers on Semiconductor-Diode Characteristics The section contains questions on pn junction qualitative theory, p-n junction diode, band structure of open circuited p-n junction, components in p-n junction diode, volt ampere

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characteristics, diode resistance and capacitances, pn diode switching times, breakdown and tunnel diodes, point contact diode, p-i-n diode and its characteristics. Qualitative Theory of the p-n junction The P-N Junction as a Diode Band Structure of an Open-Circuited p-n junction The Current Components in a P-N junction diode Quantitative Theory of the P-N Diode Currents The Volt Ampere Characteristics The Temperature Dependence of P-N Characteristics

Diode Resistance Diode Capacitances PN Diode Switching Times Breakdown Diodes Tunnel Diodes and its Characteristics p-i-n Diode and its Characteristics The Point Contact Diode

4. Questions on Diodes The section contains questions on basics of diode, types of diodes which includes zener diode and others, limiting and clamping circuits, rectifiers and characteristics of junction diode and diode forward characteristic modelling. The Ideal Diode Modelling the Diode Forward Characteristic Zener Diode

Rectifiers Limiting and Clamping Circuits and Some Special Types of diodes Diode Circuits

5. Questions & Answers on Application of Diodes The section contains questions and answers on half-wave and full-wave rectifier, bridge and voltmeter rectifier, inductor and capacitor filters, l-section filters, clc filters and voltage regulation. Half-Wave Rectifier Full-wave Rectifier Bridge Rectifier Inductor Filters Capacitor Filters 6. Questions on Transistor Characteristics

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L-Section Filter CLC Filter Voltage Regulation Using Zener Diode The Rectifier Voltmeter

The section contains questions on transistor amplifier, transistor construction, junction transistor, common base, emitter and collector configuration, ce and cb characteristics, dc load lines, transistor switch and switching times. The Junction Transistor The Transistor as an Amplifier Transistor Construction The Common Base Configuration The Common Emitter Configuration The Common Collector Configuration

The CE Characteristics The CB Characteristics DC Load Lines Transistor as a Switch Transistor Switching Times

7. Questions & Answers on Transistor Biasing and Thermal Stabilization The section contains questions and answers on operating point, bias stability, collector-to-base and collector-emitter bias, self bias, bias compensation, thermistor, thermal runway and stability. Collector-to-Base Bias Self-Bias Bias Compensation Thermal Runaway Thermal Stability The Operating Point

Bias Stability Emitter Feedback Bias Collector-Emitter Feedback Bias Stabilization against Variations in VBE and Beta for Self Bias Circuit Thermistor and Sensistor Compensation

8. Questions on Signals and Amplifiers The section contains questions and answers on basics of signals, amplifiers, sinusoidal steady state analysis, amplifier circuit models and frequency response. Signals Amplifiers Circuit Models for Amplifier

Frequency Response of Amplifier Sinusoidal Steady State Analysis

9. Questions & Answers on Operational Amplifiers The section contains questions and answers on ideal operational amplifiers, inverting and noninverting configuration, differentiators and differential amplifiers, operational amplifiers, finite open loop gain effect, circuit performance bandwidth and large signal operations. The Ideal Operational Amplifiers The Inverting Configuration The Non Inverting Configuration 6

Integrators and Differentiators DC Imperfections in Operational Amplifiers Effect of Finite Open-Loop gain and Bandwidth

Difference Amplifiers

on Circuit Perfomance Large Signal Operations on operational Amplifiers

10. Questions on MOS Field Effect Transistors (MOSFETs) The section contains questions and answers on basics of MOSFET, device structure, physical and small signal operation of MOSFET, basics of MOSFET configurations and circuit biasing, body effect and discrete MOSFET circuits. MOSFETs Device Strucuture and Physical Operation MOSFETs Current-Voltage Characterisitcs MOSFETs Circuits at DC MOSFET in Amplfier Design MOSFET in Small Signal Operation

Basic MOSFET Amplifier Configurations Biasing in MOS Amplifier Circuit Discrete-Circuit MOS Amplifiers The Body Effect

11. Questions & Answers on Bipolar Junction Triodes (BJTs) The section contains questions on basics of BJT, device structures and physical operations, circuits, current-voltage properties, amplifier design, signal operations and circuit configuration and biasing. BJTs Device Strucutres and Physical Operations BJTs Current-Voltage Characteristics BJT Circuits at DC BJT in Amplifier Design

Small Signal Operations and Model Basic BJT Amplifier Configuration Biasing in BJT Amplifier Circuits Spread Spectrum

12. Questions on Small-Signal Low-Frequency AC models of Transistors The section contains questions on ac models and analysis, transistor amplifier, biasing parameters, two port devices and hybrid model, transistor hybrid model, h-parameters and its measurement, cb transistor physical model, hybrid model in ce, cb and cc, ac and dc analysis problems, transistor circuit analysis and millers theorem. The AC Analysis of a Small-Signal LowFrequency Common Emitter Transistor Biasing Parameters Problems on AC and DC Analysis Two-Port Devices and Hybrid Model 7

Conversion Formulas for the Parameters of the Three-Transistor Configurations Analysis of Transistor Amplifier Circuit using h-parameters Comparision of Transistor Amplifier

Transistor Hybrid Model Determination of the h-parameters Measurement of h-parameters

Linear Analysis of a Transistor Circuit Physical Model of a CB Transistor Approximate Hybrid model and its use in CE,CB,CC Millers Theorem

13. Questions & Answers on Field-Effect Transistors The section contains questions and answers on junction field effect transistor, pinch off voltage, insulated gate, fet small signal model, common source and drain amplifier, fet biasing, fet amplifier and unijunction transistors. The Junction Field-Effect Transistor The Pinch off Voltage Vp The JFET Volt-Ampere Characteristics The FET Small-Signal Model The Insulated-Gate FET(MOSFET)

The Common-Source Amplifier The Common-Drain Amplifier Biasing the FET A Generalized FET Amplifier The Unijunction Transistors

14. Questions on Large Signal Amplifiers The section contains questions on class a large signal amplifiers, second and higher order harmonic distortion. Class A Large Signal Amplifiers Second Harmonic Distortion

Higher-Order Harmonic Distortion

15. Questions & Answers on Low Frequency Transistor Amplifier Circuit The section contains questions and answers on cascading amplifiers, decibel, cc and cb configurations, ce amplifier, emitter follower, high input resistance transistor circuit, cascode transistor and amplifiers. Cascading Transistor Amplifiers N-Stage Cascading Amplifiers The Decibel Simplified CE Hybrid Model Simplified Calculations for the CC Configuration Simplified Calculations for the CB Configuration

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CE Amplifier with an Emitter Resistance The Emitter Follower High Input Resistance Transistor Circuit The Cascode Transistor Configuration Difference Amplifiers

16. Questions on High Frequency Transistor The section contains questions on high frequency t model, hybrid pi models, alpha cutoff frequency, ce short circuit current frequency response, resistive load and transistor amplifier response. The High Frequency T Model The CB Short Circuit Current Frequency Response The Hybrid PI CE Transistor Model Hybrid PI Conductances In Low Frequency H Parameters The CE Short Circuit Gain Obtained With The Hybrid PI Model

The Alpha Cutoff Frequency The CE Short Circuit Current Frequency Response Current Gain With Resistive Load Transistor Amplifier Response, Taking Source Resistance Into Account

17. Questions & Answers on Frequency Response The section contains questions on low and high frequency responses on cs and ce amplifiers, high frequency models of bjt and mosfet, millers theorem, high frequency response of source and emitter followers, differential amplifiers, mos and bipolar cascode amplifiers, cd-cs, cc-ce, cd-ce, cc-cb and cd-cg configurations, pole splitting, frequency and miller compensation.

Low-Frequency Response of the CS Amplifiers Low-Frequency Response of the CE Amplifiers High-Frequency Model of the BJT High-Frequency Model of the MOSFET High-Frequency Response of the CS Amplifiers High-Frequency Response of the CE Amplifiers The High Frequency Gain Function Determining the 3-dB Frequency fH 9 Approximate Determination of fH Miller’s Theorm High-Frequency Response of the CG

High-Frequency Response of the MOS Cascode Amplifier High-Frequency Response of the Bipolar Cascode Amplifier High Frequency Response of the Source Followers High-Frequency Response of the Emitter Followers High Frequency Response of the Differential Amplifiers The CD-CS, CC-CE and CD-CE Configurations The CC-CB and CD-CG Configurations High Frequency Response of Multistage Amplfiers Frequency Compensation Miller Compensation and Pole Splitting

18. Questions on Differential and Multistage Amplifiers 9

The section contains questions and answers on input bias, voltage, current, multistage and differential amplifiers, feedback structure and negative feedback properties, transconductance and transresistance amplifiers, a, b and ab output stages, feedback effects, power bjts, heat sinks and variations in ab classification. Input Bias and Offset Currents of the Bipolar Differential Amplifier Differential Amplifier with Active Load Multistage Amplifier The General Feedback Structure Properties of Negative Feedback Voltage Amplifiers Current Amplifiers Tranconductance Amplifiers Transresistance Amplifiers Feedback Voltage Amplifier (Series-Shunt) Determining the Loop Gain The Transfer Function of the Feedback Amplifier

Effect of Feedback on the Amplifier Poles (The Nyquist Plot) Effect of Feedback on the Amplifier Poles Stability Study using Bode Plots Classification of Output Stages Class A output Stage Class B Output Stage Calss AB Output Stage Biasing the Class AB Circuit Power BJTs Transistor case and Heat Sinks Variations on Class AB Configuration

1. Questions & Answers on EDC Overview The section contains questions and answers on basics of electronic devices and circuits, methods, frequency responses, time signals, network theorems, analysis of circuits, modulation, transmission and coupled circuits. Basic Concepts Method of Analysis Frequency Response – 1 10

Noise Amplitde Modulation Digital Transmission

Frequency Response – 2 Two Port Network Continous Time Signals Random Process

Semicondunctor Physics Network Theroms Circuit analysis in S domain Magneticaly coupled circuits

Questions & Answers (MCQs) focuses on “Basic Concepts”. 1. A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is 11

a) 160.2 C b) -160.2 C c) 16.02 C d) -16.02 C View Answer Answer: c Explanation: n 1020, Q = ne = e 1020 = 16.02 C. Charge on sphere will be positive. 2. A lightning bolt carrying 15,000 A lasts for 100 s. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is a) 13.33 C b) 75 C c) 1500 C d) 1.5 C View Answer Answer: d Explanation: dQ = i dt = 15000 x 100µ = 1.5 C. 3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is a) 0.5 A b) 2 A c) 3.33 mA d) 0.3 mA View Answer Answer: b Explanation: i = dQ/dt = 120/60 = 2A. 4. The energy required to move 120 coulomb through 3 V is a) 25 mJ b) 360 J c) 40 J d) 2.78 mJ View Answer Answer: b Explanation: W = Qv = 360 J. 12

5. Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is

a) 30 V b) 25 V c) 20 V d) 15 V View Answer Answer: d Explanation: 100 = 65 + V2 => V2 = 35 V V3 – 30 = V2 => V3 = 65 V 105 – V3 + V4 – 65 = 0 => V4 = 25 V V4 + 15 – 55 + V1 = 0 => V1 = 15 V. 6. Req = ?

a) 11.86 ohm b) 10 ohm c) 25 ohm d) 11.18 ohm View Answer Answer: d Explanation: Req – 5 = 10(Req + 5)/(10 + 5 +Req). Solving for Req we have Req = 11.18 ohm. 13

7. In the circuit the dependent source

a) supplies 16 W b) absorbs 16 W c) supplies 32 W d) absorbs 32 W View Answer Answer: d Explanation: P = VIx = 2Ix Ix = 2 x 16 or 32 watt (absorb). 8. Twelve 6 resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is (in ohm) a) 5/6 b) 6/5 c) 5 d) 6 View Answer

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Answer: c Explanation:

9. The energy required to charge a 10 µF capacitor to 100 V is a) 0.1 J b) 0.05 J c) 5 x 10(-9) J d) 10 x 10(-9) J View Answer Answer: b Explanation: Energy provided is equal to 0.5 CVxV. 10. A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec 15

interval. The value of capacitance is a) 0.75 mF b) 1.33 mF c) 0.6 mF d) 1.67 mF View Answer Answer: d Explanation: Voltage gain = 1/C x i(t2 – t1), where t2 – t1 = 10s. Hence C = 1.67 mF.

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Questions & Answers (MCQs) focuses on “Method of Analysis”. 1. V1 = ?

a) 0.4 Vg b) 1.5 Vg c) 0.67 Vg d) 2.5 Vg View Answer Answer: b Explanation: Apply nodal analysis, V1 = Vs(4/6 + 1/3)/(1/6 + 1/3 + 1/6) = 1.5 Vs. 2. Va = ?

a) -11 V 17

b) 11 V c) 3 V d) -3 V View Answer Answer: b Explanation: Va = 2(3+1) + 3 = 11 V. 3. V1 = ?

a) 120 V b) -120 V c) 90 V d) -90 V View Answer Answer: d Explanation: -V1/60 – V1/60 + 6 = 9 or V1 = -90 V.

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4. Va = ?

a) 4.33 V b) 4.09 V c) 8.67 V d) 8.18 V View Answer Answer: c Explanation: (Va – 10)/4 + Va/2 = 4 or Va = 8.67 V. 5. V2 = ?

a) 0.5 V b) 1.0 V c) 1.5 V d) 2.0 V View Answer Answer: d Explanation: V2/20 + V2+10/30 = 0.5 V or V2 = 2.0 V.

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6. Ib = ?

a) 0.6 A b) 0.5 A c) 0.4 A d) 0.3 A View Answer Answer: b

Explanation: 20

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7. i1 = ?

a) 0.6 A b) 2.1 A c) 1.7 A d) 1.1 A View Answer Answer: a Explanation: i1 = 10/36+64 + 0.5 => i1 = 0.6 A. 8. i1 = ?

a) 1 mA b) 1.5 mA c) 2 mA d) 2.5 mA View Answer

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Answer: b Explanation: 75 = 90k (i1 – 7.5m) => 150 = 100ki1 => i = 1.5 mA. 9. i1 = ?

a) 4 A b) 3 A c) 6 A d) 5 A View Answer Answer: b Explanation: 3 – 5i1 – 12 => i1 = 3A. 10. i1 = ?

a) 20 mA b) 15 mA c) 10 mA d) 5 mA View Answer

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Answer: b Explanation: 45 = 2ki1 + 500(i1 + 15m) => i1 = 15 mA.

Questions & Answers (MCQs) focuses on “Frequency Response – 1”. (Q.1-Q.3) A parallel resonant circuit has a resistance of 2k ohm and half power frequencies of 86 kHz and 90 kHz. 1. The value of capacitor is a) 6 µF b) 20 nF c) 2 nF d) 60 µF View Answer

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Answer: b Explanation: BW = 2p(90-86)k = 1/RC or C = 19.89 nF. 2. The value of inductor is a) 4.3 mH b) 43 mH c) 0.16 mH d) 1.6 mH View Answer Answer: c Explanation: w = (86 + 90)k/2 = 88 = (1/LC)(0.5) or L = 0.16 mH. 3. The quality factor is a) 22 b) 100 c) 48 d) 200 View Answer Answer: a Explanation: Q = w/BW = 176pK/8pk = 22. (Q.4-Q.5) A parallel resonant circuit has a midband admittance of 25 X 10(-3) S, quality factor of 80 and a resonant frequency of 200 krad s. 4. The value of R (in ohm) is a) 40 b) 56.57 c) 80 d) 28.28 View Answer Answer: a Explanation: At mid-band frequency Y = 1/R or R = 1000/25 or 40 ohm. 5. The value of C is a) 2 µF b) 28.1 µF c) 10 µF d) 14.14 µF View Answer 24

Answer: c Explanation: Q = wRC or C = 80/(200 x 1000 x 40) or 10 µF. 6. A parallel RLC circuit has R 1 k and C 1 F. The quality factor at resonance is 200. The value of inductor is a) 35.4 H b) 25 H c) 17.7 H d) 50 H View Answer Answer: b Explanation: Use Q = R (L/C)0.5. 7. A parallel circuit has R = 1k ohm , C = 50 µF and L = 10mH. The quality factor at resonance is a) 100 b) 90.86 c) 70.7 d) None of the above View Answer Answer: c Explanation: Use Q = R (L/C)0.5. 8. A series resonant circuit has L = 1 mH and C = 10 F. The required R (in ohm) for the BW = 15 9 . Hz is a) 0.1 b) 0.2 c) 0.0159 d) 500 View Answer Answer: a Explanation: Use BW = R/L. 9. For the RLC parallel resonant circuit when R = 8k, L = 40 mH and C = 0.25 F, the quality factor Q is a) 40 b) 20 c) 30 25

d) 10 View Answer Answer: b Explanation: use Q = R (C/L)0.5. 10. A series resonant circuit has an inductor L = 10 mH. The resonant frequency w = 10^6 rad/s and bandwidth is BW = 103 rad/s. The value of R and C will be a) 100 F, 10 ohm b) 100 pF, 10 ohm c) 100 pF, 10 Mega-ohm d) 100 µF, 10 Meg-ohm View Answer Answer: b Explanation: Use wxw = 1/LC and BW = R/L.

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Questions and Answers focuses on “Frequency Response – 2”. 1. The maximum voltage across capacitor would be

a) 3200 V b) 3 V c) 3 V d) 1600 V View Answer

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Answer: a

Explanation: 2. Find the resonnant frequency for the circuit.

a) 346 kHz b) 55 kHz c) 196 kHz d) 286 kHz View Answer

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Answer: b

Explanation: 3. Determine the resonant frequency of the circuit.

a) 12 9 . kHz b) 12.9 MHz c) 2.05 MHz d) 2.05 kHz View Answer

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Answer: c

Explanation: 4. The value of C and A for the given network function is

a) 10 µF, 6 b) 5 µF, 10 c) 5 µF, 6 d) 10 µF, 10 View Answer

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Answer: c

Explanation: 5. H(w) = Vo/Vi = ?

a) 0.6 / jw(1 + 0.2jw) b) 0.6 / jw(5 + jw) c) 3 / jw(1 + jw) d) 3 / jw(20 + 4jw) View Answer

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Answer: a

Explanation: 6. H(w) = Vo/Vi = ?

a) 1 / jw(5 + j20w) b) 1 / jw(5 + j4w) c) 1 / jw(5 + j30w) d) 5 / jw(5 + j6w) View Answer Answer: a

Explanation: 32

7. The value of input frequency is required to cause a gain equal to 1.5. The value is

a) 20 rad/s b) 20 Hz c) 10 rad/s d) No such value exists View Answer Answer: d

Explanation: 8. In the circuit shown phase shift equal to 45 degrees and is required at frequency w = 20 rad/s. The value of R (in kilo-ohm) is

a) 200 b) 150 33

c) 100 d) 50 View Answer Answer: d

Explanation: 9. For the circuit shown the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is

a) 20 rad/s b) 20 Hz c) 40 rad/s d) 40 Hz View Answer

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Answer: a

Explanation: 10. The bode diagram for the Vo/Vs for the given circuit is

a)

b)

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c)

d) View Answer Answer: d

Explanation:

Questions & Answers (MCQs) focuses on “Two Port Network”.

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1. [T] = ?

View Answer

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Answer: d

Explanation: 2. [h] = ?

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View Answer Answer: a Explanation: V2 = 2I2 + 4I1 and I1 = 0.5I2 + 0.5(V1 – 2V2) => V1 = 4I1 + 1.5V2.

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3. [y] = ?

View Answer Answer: a Explanation: I1 = V1(Ya + Yab) – V2(Yab) and I2 = -V1(Yab) + V2(Yb + Yab).

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4.The [y] parameter of a two port network is given by

View Answer 41

Answer: b Explanation: The new parameter will be the sum of the previous network and the resistor parameter. 5. The [y] parameter for a 2-port network and the network itself are given below.

The value of Vo/vs is _______ a) 3/32 b) 1/16 c) 2/33 d) 1/17 View Answer

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Answer: a

Explanation: 6. [y] = ?

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View Answer

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Answer: b

Explanation: 7. A 2-port resistive network satisfy the condition A = D = 3/2B = 4/3C. The z11 of the network is a) 4/3 b) 3/4 c) 2/3 d) 3/2 View Answer Answer: a Explanation: z11 = A/C = 4/3. 8. A 2-port network is driven by a source Vs =100 V in series with 5 ohm, and terminated in a 25 ohm resistor. The impedance parameters are

The Thevenin equivalent circuit presented to the 25 ohm resistor is a) 80 V, 2.8 ohm 45

b) 160 V, 6.8 ohm c) 100 V, 2.4 ohm d) 120 V, 6.4 ohm View Answer Answer: b Explanation: 100 = 25I1 + 2I2, V2 = 40I1 + 10I2 Also V2 = 160I1 + 6.8I2 using above equations we get, Vth = 160 V and Rth = 6.8 ohm. 9. Find V1 and V2.

a) -68.6 V, 114.3 V b) 68.6 V, -114.3 V c) 114.3 V, -68.6 V d) -114.3 V, 68.6 V View Answer Answer: b Explanation: 800 = 10V1 – V2 and 3V2 = 5V1 = 0.

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Questions & Answers (MCQs) focuses on “Continous Time Signals”. (Q.1-Q.2) The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b=2. 1. The probability that more than 3 cars will arrive during any 10 min period is a) 0.249 b) 0.143 c) 0.346 d) 0.543 View Answer Answer: b Explanation: Evaluate 1 – P(x = 0) – P(x = 1) – P(x = 2) – P(x = 3). 2. The probability that no car will arrive is a) 0.516 b) 0.459 c) 0.246 d) 0.135 View Answer Answer: d Explanation: Evaluate P(x = 0). (Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution. 3. The probability that there will be five or more murder in a given week is a) 0.1847 b) 0.2461 c) 0.3927 47

d) 0.4167 View Answer Answer: a Explanation: P(5 or more) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 0.1847. 4. On the average, how many weeks a year can Delhi expect to have no murders ? a) 1.4 b) 1.9 c) 2.6 d) 3.4 View Answer Answer: c Explanation: P(0) = 0.0498. Hence average number of weeks per year with no murder is 52 x P(0) = 2.5889 week. 5. How many weeds per year (average) can the Delhi expect the number of murders per week to equal or exceed the average number per week? a) 15 b) 20 c) 25 d) 30 View Answer Answer: d Explanation: P(3 or more) = 1 – P(0) – P (1) – P(2) = 0.5768. Therefore average number of weeks per year = 52 x 0.5768 or 29.994 weeks. (Q.6-Q.8) The random variable X is defined by the density f(x) = 0.5u(x) e(0.5x) 6. The expect value of g(x) = X3 is a) 48 b) 192 c) 36 d) 72 View Answer Answer: a Explanation: Solve E[g(x)] = E[X3].

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7. The mean of the random variable x is a) 1/4 b) 1/6 c) 1/3 d) 1/5 View Answer Answer: a Explanation: Solve integral (x f(x) dx) from negative infinity to x. 8. The variance of the random variable x is a) 1/10 b) 3/80 c) 5/16 d) 3/16 View Answer Answer: b Explanation: Variance is given by E[X230] – 1/16. (Q.9-Q.10) A joint sample space for two random variable X and Y has four elements (1,1), (2,2), (3,3) and (4,4). Probabilities of these elements are 0.1, 0.35, 0.05 and 0.5 respectively. 9. The probability of the event{X 2.5, Y 6} is a) 0.45 b) 0.50 c) 0.55 d) 0.60 View Answer Answer: a Explanation: The required answer is given by Fxy(2.5, 6.0) = 0.1 + 0.35 = 0.45. 10. The probability of the event that X is less than three is a) 0.45 b) 0.50 c) 0.55 d) 0.60 View Answer

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Answer: b Explanation: The required answer is given by Fx(3.0) = Fxy(3.0, infinity) = 0.1 + 0.35 + 0.05 = 0.50.

Questions & Answers (MCQs) focuses on “Random Process”. 1. For random process X = 6 and Rxx (t, t+t) = 36 + 25 exp(|t|). Consider following statements: (i) X(t) is first order stationary. (ii) X(t) has total average power of 36 W. (iii) X(t) is a wide sense stationary. (iv) X(t) has a periodic component. Which of the following is true? a) 1, 2, and 4 b) 2, 3, and 4 c) 2 and 3 d) only 3 View Answer Answer: c Explanation: X Constant and Rxx() is not a function of t, so X(t) is a wide sense stationary. So (i) is false & (iii) is true. Pxx = Rxx(0) 36+25 = 61. Thus (ii) is false if X(t) has a periodic component, then RXX(t) will have a periodic component with the same period. Thus (iv) is false. 50

2. White noise with power density No/2 = 6 microW/Hz is applied to an ideal filter of gain 1 and bandwidth W rad/s. If the output’s average noise power is 15 watts, the bandwidth W is a) 2.5 x 10 (-6) b) 2.5p x 10 (-6) c) 5 x 10 (-6) d) p5 x 10 (-6) View Answer Answer: b Explanation: Pyy = 1/2p Integral(?xx(w) |H(w)|^2 dw ) from plus infinity to minus infinity. Hence solve for W. (Q.3-Q.4) The two-level semi-random binary process is defined by X(t) A or -A where (n 1)T < t < nt and the levels A and -A occur with equal probability. T is a positive constant and n = 0, ±1, ±2. 3. The mean value E[X(t)] is a) 1/2 b) 1/4 c) 1 d) 0 View Answer Answer: d Explanation: E[X(t)] = A P(A) – (-A)P(-A) which is zero. 4. The auto correlation Rxx(t1 = 0.5T, t2 = 0.7T) will be a) 1 b) 0 c) A x A d) 0.5 (A x A) View Answer Answer: c Explanation: Here Rxx is AxA if both t1 and t2 are different and zero if they are same. Hence the answer is AxA. 5. Air craft of Jet Airways at Ahmedabad airport arrive according to a Poisson process at a rate of 12 per hour. All aircraft are handled by one air traffic controller. If the controller takes a 2 – minute coffee break, what is the probability that he will miss one or more arriving aircraft? a) 0.33 b) 0.44 51

c) 0.55 d) 0.66 View Answer Answer: a Explanation: P (miss/or more aircraft) = 1 – P(miss 0) = 1 – P(0 arrive). 6. A stationary random process X(t) is applied to the input of a system for which h(t) = u(t) t2 e(8t) . If E[X(t)] = 2, the mean value of the system’s response Y(t) is a) 1/128 b) 1/64 c) 3/128 d) 1/32 View Answer Answer: c Explanation: The mean value of Y(t) is integral of h(t)dt over negative infinity to positive infinity which gives the value equal to 3/128. 7. A random process is defined by X(t) + A where A is continuous random variable uniformly distributed on (0,1). The auto correlation function and mean of the process is a) 1/2 & 1/3 b) 1/3 & 1/2 c) 1 & 1/2 d) 1/2 & 1 View Answer Answer: b Explanation: E[X(t)X(t+t)] = 1/3 and E[X(t)] = 1/2 respectively. (Q.8-Q.9) The auto correlation function of a stationary ergodic random process is shown below.

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8. The mean value E[X(t)] is a) 50 b) sqrt(50) c) 20 d) sqrt(20) View Answer Answer: d Explanation: Lim |t| tends to infinity, Rxx(t) = 20 = X2. hence X is sqrt(20). 9. The E[X2(t)] is a) 10 b) sqrt(10) c) 50 d) sqrt(50) View Answer Answer: c Explanation: Rxx(0) = X2 = 50. 10. The variance is a) 20 b) 50 c) 70 d) 30 View Answer Answer: d Explanation: Here X = 0, y = 0, Rxx(0) = 5, Ryy(0) = 10. The only value that satisfies all the given conditions is 30.

Questions & Answers (MCQs) focuses on “Noise”. 1. In a receiver the input signal is 100 V, while the internal noise at the input is 10 V. With amplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of 53

receiver is a) 2 b) 0.5 c) 0.2 d) None of the mentioned View Answer Answer: a Explanation: NF = (100/10)/(2/0.4) or 2. 2. A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is a) 1.8 µV b) 8.4 µV c) 4.3 µV d) 12.6 µV View Answer Answer: a Explanation: v2 = 4kBTR where symbols have their usual meanings, hence v = 1.8 µV. 3. A resistor R 1 k is maintained at 17C. The rms noise voltage generated in a bandwidth of 10 kHz is a) 16 x 10(-14) V b) 0.4 µV c) 4 µV d) 16 x 10(-18) V View Answer Answer: b Explanation: v2 = 4kBTR where symbols have their usual meanings, hence v = 0.4 µV. 4. A mixer stage has a noise figure of 20 dB. This mixer stage is preceded by an amplifier which has a noise figure of 9 dB and an available power gain of 15 dB. The overall noise figure referred to the input is a) 11.07 b) 18.23 c) 56.43 d) 97.38 View Answer

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Answer: a Explanation: The required answer is 7.94 + (100-1)/31.62 = 11.07. 5. A system has three stage cascaded amplifier each stage having a power gain of 10 dB and noise figure of 6 dB. the overall noise figure is a) 1.38 b) 6.8 c) 4.33 d) 10.43 View Answer Answer: c Explanation: The gain at each stage is 10db, hence the required answer is 4 + (4-1)/10 + (41)/100 or 4.33. (Q.6-Q.8) An amplifier when used with a source of average noise temperature 60 K, has an average operating noise figure of 5. 6. The Te is a) 70 K b) 110 K c) 149 K d) 240 K View Answer Answer: d Explanation: The required answer is 60 (5-1) or 240K. 7. If the amplifier is sold to engineering public, the noise figure that would be quoted in a catalog is a) 0.46 b) 0.94 c) 1.83 d) 2.93 View Answer Answer: c Explanation: The required answer is 1 + (240/290). 8. What average operating noise figure results when the amplifier is used with an antenna of temperature 30 K? a) 9.54 db 55

b) 10.96 db c) 11.23 db d) 12.96 db View Answer Answer: a Explanation: The required answer is 1 + 240/30 = 9 or 9.54 db. 9. What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7? a) 203 K b) 215 K c) 235 K d) 255 K View Answer Answer: a Explanation: The required answer is 290 x (1.7-1) or 203K. 10. If a matched attenuator with a loss of 3.2 dB is placed between the source and the amplifier’s input, what is the operating spot noise figure of the attenuator amplifier cascade if the attenuator’s physical temperature is 290 K? a) 9 db b) 11.3 db c) 10.4 db d) 13.3 db View Answer Answer: d Explanation: Here Te is 290[(20.89-1) + (2.089)(7.98-1)] or 4544.4K. Hence the required answer is 1 + 4544.4/290 = 212 or 13.3 db.

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Questions & Answers (MCQs) focuses on “Amplitde Modulation”. (Q.1-Q.3) An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V. 1. The modulation index is a) 20 b) 4 c) 0.2 d) 10 View Answer Answer: c Explanation: 20 + 4sin(500πt) = 20(1 + 0.2sin(500πt)), hence the modulation index is 0.2. 2. Total signal power is a) 208 W b) 204 W c) 408 W d) 416 W View Answer Answer: b Explantion: Pc = 20×20/2 or 200 W. Pt = Pc(1 + 0.2×0.2/4) or 204 W. 3. Total sideband power is a) 4 W b) 8 W c) 16 W d) 2 W View Answer Answer: a Explanation: Pt – Pc = 204 – 200 = 4W. 57

4. An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is a) 12.12 kW b) 31.12 kW c) 6.42 kW d) none of the mentioned View Answer Answer: a Explanation: Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW. 5. The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is a) 0.68 b) 0.73 c) 0.89 d) 0.95 View Answer Answer: a Explanation: 400/326 = 1 + (α2)/2, therefore α is 0.68. 6. A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be a) 5 kW b) 8.46 kW c) 12.5 kW d) 6.25 kW View Answer Answer: c Explanation: Pi = Pt -Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW. 7. A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is a) 2.23 kW b) 2.36 Kw c) 1.18 kW d) 1.26 kW View Answer

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Answer: a Explanation:Pt = Pc (1 + 0.49/2). 8. A carrier is simultaneously modulated by two sine waves with modulation indices of 0.4 and 0.3. The resultant modulation index will be a) 1.0 b) 0.7 c) 0.5 d) 0.35 View Answer Answer: c Explanation: α2 = 0.32 + 0.42 = 0.52 or α = 0.5. 9. In a DSB-SC system with 100% modulation, the power saving is a) 100% b) 55% c) 75% d) 100% View Answer Answer: b Explanation: This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%. 10. A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power is a) 11.25 kW b) 12.5 kW c) 15 kW d) 17 kW View Answer Answer: a Explanation: The required answer is 1 (1 + 0.42 + 0.32) or 11.25 kW.

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Questions & Answers (MCQs) focuses on “Digital Transmission”. 1. Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a binary PCM signal based on this quantization scheme will be a) 5 W b) 10 W c) 20 W d) None of the mentioned View Answer Answer: b Explanation: The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since v 10, we have BW = 10 W. 2. In PCM system, if the quantization levels are increased form 2 to 8, the relative bandwidth requirement will a) Remain same b) Be doubled c) Be tripled d) Become four times View Answer

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Answer: c Explanation: If L = 2, then 2 = 2n or n = 1 ND. If L = 8, then 8 = 2n or n = 3. So relative bandwidth will be tripled. 3. A speech signal occupying the bandwidth of 300 Hz to 3 kHz is converted into PCM format for use in digital communication. If the sampling frequency is8 kHz and each sample is quantized into 256 levels, then the output bit the rate will be a) 3 kb/s b) 8 kb/s c) 64 kb/s d) 256 kb/s View Answer Answer: c Explanation: fs = 8 kHz, 2n = 256 or n = 8. Bit Rate = 8 x 8k = 64 kb/s. 4. Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated? a) 120 kbps b) 200 kbps c) 240 kbps d) 180 kbps View Answer Answer: d Explanation: Nyquist Rate = 2 x 30k = 60 kHz 2n should be greater than or equal to 6. Thus n 3, Bit Rate = 60×3 = 18 kHz. 5. Four voice signals. each limited to 4 kHz and sampled at Nyquist rate are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be a) 8 kbps b) 64 kbps c) 256 kbps d) 512 kbps View Answer Answer: c Explanation: Nyquist Rate 2 x 4k = 8 kHz Total sample 4 x 8 = 32 k sample/sec 256 = 28, so that 8 bits are required Bit Rate 32k x 8 = 256 kbps. 61

6. A TDM link has 20 signal channels and each channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is a) 1180 K bits/sec b) 1280 K bits/sec c) 1180 M bits/sec d) 1280 M bits/sec View Answer Answer: b Explanation: Total sample 8000 x 20 = 160 k sample/sec Bit for each sample 7 + 1 = 8 Bit Rate = 160k x 8 = 1280 kilobits/sec. 7. Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal to a) 5 kHz b) 20 kHz c) 40 kHz d) 80 kHz View Answer Answer: d Explanation: fm = 5 kHz, Nyquist Rate 2 x 5 = 10 kHz Since signal are sampled at twice the Nyquist rate so sampling rate 2 x 10 = 20 kHz. Total transmission bandwidth 4 x 20 = 80 kHz. 8. A sinusoidal massage signal m(t) is transmitted by binary PCM without compression. If the signal to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be a) 8 b) 10 c) 12 d) 14 View Answer Answer: a Explanation: 3(L2)/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 28, So 8 bits per sample is required. 9. A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40 dB. The minimum storage 62

capacity needed to accommodate this signal is a) 1.12 KBytes b) 140 KBytes c) 168 KBytes d) None of the mentioned View Answer Answer: b Explanation: (SNR)q = 1.76 + 6.02(n) = 40 dB, n = 6.35 We take the n = 7. Capacity = 20 x 8k x 7 = 1.12 Mbits = 140 Kbytes. 10. A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is a) 2.04 V b) 1.08 V c) 4.08 V d) 2.16 V View Answer Answer: b Explanation: Amax = (0.1 x 68k)/(2000p) or 1.08V.

Questions & Answers (MCQs) focuses on “Semiconductor Physics”. In the problems assume the parameter given in following table. Use the temperature T= 300 K unless otherwise stated.

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1. In germanium semiconductor material at T 400 K the intrinsic concentration is (x 10^14 per cc) a) 26.8 b) 18.4 c) 8.5 d) 3.6 View Answer Answer: c

Explanation: 2. The intrinsic carrier concentration in silicon is to be no greater than ni = 1 x 10^12 cc. The maximum temperature allowed for the silicon is ( Eg = 1.12 eV) a) 300 K 64

b) 360 K c) 382 K d) 364 K View Answer Answer: c

Explanation: 3. Two semiconductor material have exactly the same properties except that material A has a bandgap of 1.0 eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of material B is a) 2016 b) 47.5 c) 58.23 d) 1048 View Answer Answer: b

Explanation: 4. In silicon at T = 300 K the thermal-equilibrium concentration of electron is n0 = 5 x 10^4 cc. The hole concentration is a) 4.5 x 1015 cc b) 4.5 x 1015 m3 c) 0.3 x 10-6 cc d) 0.3 x 10-6 m3 View Answer Answer: a Explanation: ni x ni = no x po. 5. In silicon at T = 300 K if the Fermi energy is 0.22 eV above the valence band energy, the value of p0 is a) 2 x 1015 cm3 65

b) 1015 cm3 c) 3 x 1015 cm3 d) 4 x 1015 cm3 View Answer Answer: a

Explanation: 6. The thermal-equilibrium concentration of hole p0 in silicon at T = 300 K is 1015 cm3. The value of n0 is a) 3.8 x 108 cm3 b) 4.4 x 104 cm3 c) 2.6 x 104 cm3 d) 4.3 x 108 cm3 View Answer Answer: b

Explanation:

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7. In germanium semiconductor at T 300 K, the acceptor concentrations is Na 1013 cm3 and donor concentration is Nd 0. The thermal equilibrium concentration p0 is a) 2.97 x 109 cm3 b) 2.68 x 1012 cm3 c) 2.95 x 1013 cm3 d) 2.4 cm3 View Answer Answer: c

Explanation: 8. A silicon sample doped n type at 10^18 cm3 have a resistance of 10 ohm. The sample has an area of 10^(-6) cm2 and a length of 10 µm . The doping efficiency of the sample is (µn = 800 cm2/V-s ) a) 43.2% b) 78.1% c) 96.3% d) 54.3% View Answer Answer: b

Explanation: 9. Six volts is applied across a 2 cm long semiconductor bar. The average drift velocity is 104 cms. The electron mobility is a) 4396 cm2/V-s 2 b) 3 x 104 cm2/V-s c) 6 x 104 cm2V-s 67

d) 3333 cm2/V-s View Answer Answer: d

Explanation: 10. A particular intrinsic semiconductor has a resistivity of 50 (ohm-cm) at T = 300 K and 5 (ohm-cm) at T = 330 K. If change in mobility with temperature is neglected, the bandgap energy of the semiconductor is a) 1.9 eV b) 1.3 eV c) 2.6 eV d) 0.64 eV View Answer Answer: b

Explanation:

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Questions & Answers (MCQs) focuses on “Network Theroms”. Circuit for Q.1-Q.2

1. Vth = ? a) 1 V b) 2 V c) 3 V d) 4 V View Answer Answer: d Explanation: Vth = (6)(6)/(6 + 3) or 4 V. 2. Rth = ? (in ohm) a) 2 69

b) 3 c) 4 d) 5 View Answer Answer: c Explanation: Rth = (3||6) + 2 or 4 ohm. 3. A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 ohm-1, the power dissipated by the bulb is a) 80 W b) 1800 W c) 112.5 W d) 228 W View Answer Answer: c Explanation: r = Voc/Isc = 1.2 ohm Power used by bulb = (24 x 24) x 2/(1.2 + 2) x (1.2 + 2) or 11.5 Watt. Circuit for Q.4-Q-5

4. In = ? a) 1.5 A b) 3 A c) 6 A d) 10 A View Answer Answer: b Explanation: V1 = (15/2)/(1/4 + 1/2 + 1/2) In = Isc = V1/2 = 3 A.

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5. Rn = ? (in ohm) a) 10/3 b) 4 c) 6 d) 10 View Answer Answer: a Explanation: Rn = (2||4) + 2. Circuit for (Q.6-Q.7)

6. Vth = ? a) -2 V b) -1 V c) 1 v d) 2 V View Answer Answer: c Explanation: Vth = (2)(3)(1)/3+3 = 1V. 7. Rth = ? (in ohm) a) 5/6 b) 6/5 c) 5/3 d) 3/5 View Answer Answer: a Explanation: Rth = 1||5 or 5/6 ohm.

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8. The equivalent to the given circuit is

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View Answer Answer: b Explanation: After killing all source equivalent resistance is R. Open circuit voltage is v1. 9. V1 = ?

a) 6 V b) 7 V c) 8 V d) 10 V View Answer

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Answer: a

Explanation: 10. i1 = ?

a) 3 A b) 0.75 mA c) 2 mA d) 1.75 mA View Answer

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Answer: b

Explanation:

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Questions & Answers (MCQs) focuses on “Circuit analysis in S domain”. 1. The network function (s2 + 4s)/(s + 1)(s + 2)(s + 3) represents a) RC impedance b) RL impedance c) LC impedance d) None of the mentioned View Answer Answer: d Explanation: Poles and zero does not interlace on negative real axis so it is not a immittance function. 2. The network function (3s2 + 8s)/(s + 1)(s + 3) represents a) RC impedance b) RL impedance c) LC impedance d) None of the mentioned View Answer Answer: c Explanation: The singularity nearest to origin is a zero. So it may be RL impedance or RC admittance function. Because of (D) option it is required to check that it is a valid RC admittance function. The poles and zeros interlace along the negative real axis. The residue of Yrc(s)/s is positive. 3. The network function (s + 1)(s + 4)/s(s + 2)(s + 5) represents a) RC impedance b) RL impedance c) LC impedance d) All of the mentioned View Answer Answer: b Explanation: The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function.

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4. The network function s^2 + 10s + 24/s2 + 8s + 15 represents a) RC impedance b) RL impedance c) LC impedance d) None of the mentioned View Answer Answer: d Explanation: The singularity is near to origin is pole. So it may be RC impedance or RL admittance function. 5. A valid immittance function is a) (s + 4)(s + 8)/(s + 2)(s – 5) b) s(s + 1)/(s + 2)(s + 5) c) s(s + 2)(s + 3)/(s + 1)(s + 4) d) s(s + 2)(s + 6)/(s + 1)(s + 4) View Answer Answer: d Explanation: a) pole lie on positive real axis b) poles and zero does not interlace on axis. c) poles and zero does not interlace on axis. d) is a valid immittance function. 6. The network function (s2 + 8s +15)/(s2 + 6s + 8) is a) RL admittance b) RC admittance c) LC admittance d) All of the mentioned View Answer Answer: a Explanation: The singularity nearest to origin is a pole. So it may be a RL admittance or RC impedance function. 7. The voltage response of a network to a unit step input is Vo(s) = 10/s(s2 + 8s + 16). The response is a) under damped b) over damped c) critically damped

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d) can’t be determined View Answer Answer: c Explanation: The characteristic equation has real and repeated roots (-4, -4). Hence it is critically damped. 8. The current response of a network to a unit step input is Io(s) = 10(s + 2)/s(s2 + 11s + 30). The response is a) Under damped b) Over damped c) Critically damped d) None of the mentioned View Answer Answer: b Explanation: The roots are real and unequal (-6, -5) for the characteristic equation. Hence it is over damped. Circuit for q.9-Q.10

9. The ratio of the transfer function Io/Is is a) s(s + 4)/(s2 + 3s + 4) b) s(s + 4)/(s + 1)(s + 3) c) (s2 + 3s + 4)/s(s + 4) d) (s + 1)(s + 3)/s(s + 4) View Answer Answer: b Explanation: Io/Is = (s + 4)/ (s + 4 + 3/s) = s(s + 4)/(s + 1)(s + 3). 10. The response is a) Over damped 78

b) Under damped c) Critically damped d) can’t be determined View Answer Answer: a Explanation: The characteristic roots are real and unequal (-1, -3), therefore it is over damped.

Questions & Answers (MCQs) focuses on “Magneticaly coupled circuits”. (Q.1-Q.2) For the circuit given below i1 = 4 sin(2t) and i2 = 0

1. v1 = ? a) -16 cos 2t V b) 16 cos 2t V c) 4 cos 2t V d) -4 cos 2t V View Answer Answer: b Explanation: v1 = 2 d(i1)/dt + d(i2)/dt. 2. v2 = ? a) 2 cos 2t V b) -2 cos 2t V c) 8 cos 2t V 79

d) -8 cos 2t V View Answer Answer: c Explanation: v2 = d(i1)/dt + d(i2)/dt. Circuit for Q.3-Q.4

3. If i1 = 0 and i2 = 2 sin(4t), the voltage v1 is a) -24 cos(4t) V b) 24 cos(4t) V c) 1.5 cos(4t) V d) -1.5 cos(4t) V View Answer Answer: b Explanation: v1 = 3 d(i1)/dt – 3 d(i2)/dt. 4. If i1 = e(-2t) and i2 = 0, the voltage v1 is a) -6e(-2t) V b) 6e(-2t) V c) 1.5e(-2t) V d) -1.5e(-2t) V View Answer Answer: c Explanation: v1 = 4 d(i1)/dt – 3 d(i2)/dt.

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Circuit for Q.5-Q.8

5. If i1 = 3 cos(4t) and and i2 = 0. Find v1. a) -24 sin(4t) V b) 24 sin(4t) V c) 1.5 sin(4t) V d) -1.5 sin(4t) V View Answer Answer: a Explanation: v1 = 2 d(i1)/dt – 2 d(i2)/dt. 6. If i1 = 3 cos(4t) and and i2 = 0. Find v2. a) -24 sin(4t) V b) -36 sin(4t) V c) sin(4t) V d) -sin(4t) V View Answer Answer: a Explanation: v2 = -3 d(i1)/dt + 2 d(i2)/dt. 7. If i1 = 4 cos(3t) and and i1 = 0. Find v2. a) 12 cos(3t) V b) -12 cos(3t) V c) -24 cos(3t) V d) 24 cos(3t) V View Answer Answer: c Explanation: v1 = 2 d(i1)/dt – 2 d(i2)/dt. 8. If i2 = 4 cos(3t) and and i1 = 0. Find v2. a) -12 cos(3t) V b) -24 cos(3t) V 81

c) -36 cos(3t) V d) -48 cos(3t) V View Answer Answer: c Explanation: v2 = 3 d(i1)/dt + 2 d(i2)/dt. 9. Leq = ?

a) 4 H b) 6 H c) 7 H d) 0 H View Answer Answer: c Explanation: Leq = L1 + L2 – 2M. 10. Leq = ?

a) 2 H b) 4 H c) 6 H d) 8 H View Answer 82

Answer: a Explanation: Leq = L1 + L2 – 2M.

2. Questions on Conduction in Semiconductors The section contains questions and answers on semiconductor electrons, holes and conductivity, donor and acceptor impurities, fermi level semiconductor, charge densities, diffusion, carrier lifetime, continuity equation and hall effect. Electrons and Holes in Semiconductor Conductivity of a Semiconductor Donor and Acceptor impurities Charge Densities in a Semiconductor impurities 83

Diffusion Carrier Life Time The Continuity Equation The Hall Effect

Fermi Level in a Semiconductor having Impurities

Questions & Answers (MCQs) focuses on “Electrons and Holes in Semiconductor”.

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1. Which of the following expressions represents the correct distribution of the electrons in the conduction band? (gc(E)=density of quantum states, fF(E)=Fermi dirac probability a) n(E)=gc(E)*fF(E) b) n(E)=gc(-E)*fF(E) c) n(E)=gc(E)*fF(-E) d) n(E)= gc(-E)*fF(-E) View Answer Answer: a Explanation: The distribution of the electrons in the conduction band is given by the product of the density into Fermi-dirac distribution. 2. What is the value of the effective density of states function in the conduction band at 300k? a) 3*1019 cm-3 b) 0.4*10-19 cm-3 c) 2.5*1019 cm-3 d) 2.5*10-19 cm-3 View Answer Answer : c

Explanation : Substituting the values of mn=m0 ,h=6.626*10-34J/s ,k=1.38*10-23 and T=300K, we get Nc=2.5*1019 cm-3. 3. In a semiconductor which of the following carries can contribute to the current? a) Electrons b) Holes c) Both d) None View Answer Answer : c Explanation : In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current. 4. Which of the following expressions represent the Fermi probability function? a) fF(E)=exp(-[E-EF]/KT) b) fF(E)=exp(-[EF-E]/KT) c) fF(E)=exp([E-EF]/KT)

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d) fF(E)=exp(-[EF-E]/KT) View Answer Answer : b Explanation : It is the correct formula for the Fermi probability function. 5. Electrons from valence band rises to conduction band when the temperature is greater than 0 k. Is it True or False? a) True b) False View Answer Answer : a Explanation : As the temperature rises above 0 k, the electrons gain energy and rises to the conduction band from the valence band. 6. What is the intrinsic electrons concentration at T=300K in Silicon? a) 1.5*1010cm-3 b) 1.5*10-10cm-3 c) 2.5*1019cm-3 d) 2.5*10-19cm-3 View Answer Answer : a Explanation : Using the formula, We get, ni=1.5*1010cm-3. 7. The intrinsic Fermi level of a semiconductor depends on which of the following things? a) Emidgap b) mp* c) mn* d) All of the mentioned View Answer Answer : d Explanation : From the above formula, Efidepends on all of the options given.

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8. What is the difference between the practical value and theoretical value of ni? a) Factor of 1 b) Factor of 2 c) Factor of 3 d) Factor of 4 View Answer Answer : b Explanation : This is practically proved. 9. The thermal equilibrium concentration of the electrons in the conduction band and the holes in the valence band depends upon? a) Effective density of states b) Fermi energy level c) Both A and B d) Neither A nor B View Answer Answer : c Explanation : The electrons and holes depends upon the effective density of the states and the Fermi energy level given by the formula, . 10. In which of the following semiconductor, the concentration of the holes and electrons is equal? a) Intrinsic b) Extrinsic c) Compound d) Elemental View Answer Answer : a Explanation : In the intrinsic semiconductor, ni=pi that is the number of the electrons is equal to the number of the holes. Whereas in the extrinsic conductor ni is not equal to pi.

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Questions & Answers (MCQs) focuses on “Conductivity of a Semiconductor”. 1. What is the SI unit of conductivity? a) Ωm b) (Ωm)-1 c) Ω d) m View Answer Answer : b Explanation: The formula of the conductivity is the σ=1/ρ. So, the unit of resistivity is Ωm. Now, the unit of conductivity becomes the inverse of resistivity. 2. Which of the following expressions doesn’t represent the correct formula for Drift current density? a) J=σE b) J=qnµE c) J=µE d) None View Answer Answer : c Explanation : The following formulae represent the correct expression for drift current density, J=σE And J=qnµE. 3. Does a semiconductor satisfy the ohm’s law? a) True b) False View Answer

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Answer : a Explanation : V=IR J=σE I/A=σ(V/L) V=(L/ σA)*I=(ρL)*I/A=IR Thus, above equation satisfies Ohm’s law. 4. In which range of temperature, freeze out point begins to occur? a) Higher range b) Lower range c) Middle range d) None View Answer Answer : b Explanation : At lower range of temperature, the concentration and conductivity decreases with lowering of the temperature. 5. Which of the following expression represents the correct formula for the conductivity in an intrinsic material? a) ρ=e(μn+μp )ni b) σ=e(μn+μp )ni c) σ=1/(e(μn+μp )ni) d) ρ=1/(e(μn+μp )ni) View Answer Answer : b Explanation: Option b is the correct formula. 6. What is the voltage difference if the current is 1mA and length and area is 2cm and 4cm2 respectively?(ρ=2Ωm) a) 0.025V b) 25V c) 0.25V d) None View Answer Answer : d Explanation : V=IR R=ρl/(A)=2*2/4=100Ω V=1mA*100 =0.1V. 89

7. Is resistivity is a function of temperature? a)True b)False View Answer Answer : a Explanation : Resistance depends on the temperature and the resistivity depends on the resistance, so now the resistivity depends on the temperature. 8. What is the electric field when the voltage applied is 5V and the length is 100cm? a) 0.5V/m b) 5V/m c) 50V/m d) None View Answer Answer : c Explanation : E=V/L=5/100cm=5V/m. 9. Calculate the average random thermal energy at T=300K? a) 0.038eV b) 3.8eV c) 38eV d) 0.38eV View Answer Answer : a Explanation : Average random thermal energy=3/2*k*T=0.038eV.

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10.

In the above figure, a semiconductor having an area ‘A’ and length ‘L’ and carrying current ‘I’ applied a voltage of ‘V’ volts across it. Calculate the relation between V and A? a) V= ((ρ*L)/A)*I b) V= ((ρ*A)/L)*I c) V= ((ρ*I)/(A*L)) d) V=((ρ*I*A*L) View Answer Answer : a Explanation: Option A, satisfies the Ohm’s law which is V=IR where R=(ρl)/A.

Questions & Answers (MCQs) focuses on “Donor and Acceptor impurities”. 1. At what temperature the donor states are completely ionized? a) 0 K b) ROOM c) 300K d) 900K View Answer 91

Answer: b Explanation: At room temperature, the donors have donated their electrons to the conduction band. 2. The opposite of ionization takes place at which temperature? a) 0 K b) ROOM c) 300 K d) 900K View Answer Answer: a Explanation: AT 0 K, all the donors and acceptors are in their lowest energy levels. 3. What do you mean by the tem ‘FREEZE-OUT’? a) All the electrons are frozen at room temperature b) None of the electrons are thermally elevated to the conduction band c) All the electrons are in the conduction band d) All the holes are in the valence band View Answer Answer: b Explanation: Freeze out means none of the electrons are transmitted to the conduction band. 4. Which of the following expressions represent the correct formula for the density of electrons occupying the donor level? a) nd=Nd-Nd+ b) nd=Nd-Nd– c) nd=Nd+Nd+ d) nd=Nd+Nd– View Answer Answer: a Explanation: The density of the electrons is equal to the electrons present in the substrate minus the number of donors present. 5. Which of the following band is just above the intrinsic Fermi level for n-type semiconductor? a) Donor band b) Valence band c) Acceptor band d) Conduction band View Answer 92

Answer: a Explanation: For n-type semiconductors, the donor band is just above the intrinsic Fermi level. 6. At absolute zero temperature, which level is above the Fermi energy level in the case of donors? a) Donor energy level b) Acceptor energy level c) Conduction Band d) Valence Band View Answer Answer: c Explanation: At T=0 K, the tem exp(-∞)=0 in the expression of

Thus, EF>ED So, only conduction band lies above the Fermi energy level. 7. At T=0 K, the location of Fermi level with respect to the Ec and Ed for the n type material is? a) Above than conduction band b) Midway c) Lower than Ed d) Greater than Ed View Answer Answer: b Explanation: At T=0 K, the Fermi level of n band lies between the midway of Ec and Ed as intrinsic Fermi level always lies between the Ec and Ev. 8. At absolute zero temperature, which level is below the Fermi energy level in the case of acceptors? a) Donor energy level b) Valence Band c) Conduction band d) Acceptor energy level View Answer Answer: b Explanation: At T=0 K, the tem exp(-∞)=0 in the expression of

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So, only valence band lies below the Fermi energy level of the acceptors. 9.

For the above n-type semiconductor, what is B knows as? a) Valence Band b) Conduction Band c) Donor Energy level d) Acceptor energy level View Answer Answer: c Explanation: For n-type semiconductors, the donor energy level is always greater than the Fermi level energy. 10.

For the above given figure, identify the correct option for satisfying the above semiconductor figure? a) P type, A-Conduction band, B-donor energy band, C- Valence band b) P type, A-Conduction band, B-acceptor energy band, C- Valence band c) n type, A-Conduction band, B-donor energy band, C- Valence band d) n type, A-Conduction band, B-acceptor energy band, C- Valence band View Answer

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Answer: c Explanation: The given figure has B band below the intrinsic Fermi level, so that would be acceptor energy band and will be a p-type semiconductor. Questions and Answers focuses on “Charge Densities in a Semiconductor impurities”. 1. Is n/p=ni2 is a correct formula? a) True b) False View Answer Answer: b Explanation: The correct formula is n*p=ni2. 2. Calculate the number of electrons is the number of holes are 15*1010? a) 15*1010 b) 1.5*108 c) 1.5*109 d) 1.5*1010 View Answer Answer: c Explanation: n*p=(1.5*1010)2 n*15*1010=1.5*1.5*1010*1010 n=1.5*109 electrons. 3. For which type of material, the number of free electron concentration is equal to the number of donor atoms? a) P type semiconductor b) Metal c) N-type semiconductor d) Insulator View Answer Answer: c Explanation: The n-type semiconductor has equal concentration of free electron and donor atoms. 4. Identify the correct condition for a semiconductor to be electrically neutral. a) Nd+p=Na+n b) Nd-p=Na+n 95

c) Nd+p=Na-n d) Nd-p=Na-n View Answer Answer: a Explanation: The sum of the number of donors and the holes is equal to the sum of the number of the acceptors and the electrons. 5. Do the Fermi energy level changes in a semiconductor? a) True b) False View Answer Answer: a Explanation: The Fermi energy level changes as the electron and hole concentrations change because of the formula which defines the position of the Fermi level depending on the concentration of holes and electrons. 6. Consider a silicon wafer having Nc=2.8*1019cm-3 and the Fermi energy is .25eV below the conduction band. Calculate the equilibrium concentrations of electrons at T=300K? a) 18*1016cm-3 b) 1.8*1016cm-3 c) 1.8*1014cm-3 d) 180*1016cm-3 View Answer Answer: a Explanation: n0=Nc*exp(-Eg/KT)=2.8*1019*exp(-0.25/0.0259) =18*1016cm-3. 7. If Ef>Efi, then what is the type of the semiconductor? a) n-type b) P-type c) Elemental d) Compound View Answer Answer: a Explanation: For n-type, the Fermi energy level is greater than the intrinsic Fermi energy level because in an energy band, Fermi level of donors is always greater than that of the acceptors.

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8. The 1-fF (E) increases in which of the following band for n type semiconductor? a) Conduction band b) Donor band c) Acceptor band d) Valence band View Answer Answer: d Explanation: For an n-type semiconductor, the probability of fF (E) decreases in the valence band. The probability of finding the electron in the conduction band is more. 9. The fF (E) decreases in which of the following band for p-type semiconductor? a) Conduction band b) Donor band c) Acceptor band d) Valence band View Answer Answer: a Explanation: The probability of finding the electron in the conduction band decreases for a ptype semiconductor because in a p-type semiconductor, the holes will be in conduction band rather than the electrons. 10. Do the intrinsic Fermi energy level changes with the addition of dopants and acceptors? a) True b) False View Answer Answer: b Explanation: The intrinsic Fermi energy level always remains constant because it is an imaginary level taken to distinguish between the Fermi level of the types of semiconductor.

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Questions & Answers (MCQs) focuses on “Fermi Level in a Semiconductor having Impurities”. 1. Which states get filled in the conduction band when the donor-type impurity is added to a crystal? a) Na b) Nd c) N d) P View Answer Answer: b Explanation: When the donor-type impurity is added to a crystal, first Nd states get filled because it is of the highest energy. 2. Which of the following expression represent the correct formulae for calculating the exact position of the Fermi level for p-type material? a) EF = EV + kTln(ND / NA ) b) EF = -EV + kTln(ND / NA ) c) EF = EV – kTln(ND / NA ) d) EF = -EV – kTln(ND / NA ) View Answer

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Answer: a Explanation: The correct position of the Fermi level is found with the formula in the ‘a’ option. 3. Where will be the position of the Fermi level of the n-type material when ND=NA? a) Ec b) Ev c) Ef d) Efi View Answer Answer: a Explanation: When ND=NA, kTln(ND/NA )=0 So, Ef=Ec. 4. When the temperature of either n-type or p-type increases, determine the movement of the position of the Fermi energy level? a) Towards up of energy gap b) Towards down of energy gap c) Towards centre of energy gap d) Towards out of page View Answer Answer: c Explanation: whenever the temperature increases, the Fermi energy level tends to move at the centre of the energy gap. 5. Is it true, when the temperature rises, the electrons in the conduction band becomes greater than the donor atoms? a) True b) False View Answer Answer: a Explanation: When the temperature increases, there is an increase in the electron-hole pairs and all the donor atoms get ionized, so now the thermally generated electrons will be greater than the donor atoms. 6. If the excess carriers are created in the semiconductor, then identify the correct energy level diagram.

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a)

b)

c)

d) View Answer Answer: a Explanation: The diagram A refers the most suitable energy level diagrams because Efp>Ef>Efi>Efp>Ev.

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7. If excess charge carriers are created in the semiconductor then the new Fermi level is known as Quasi-Fermi level. Is it true? a) True b) False View Answer Answer: a Explanation: Quasi-fermi level is defined as the change in the level of the Fermi level when the excess chare carriers are added to the semiconductor. 8. Ef lies in the middle of the energy level indicates the unequal concentration of the holes and the electrons? a) True b) False View Answer Answer: b Explanation: When the Ef is in the middle of the energy level, it indicates the equal concentration of the holes and electrons. 9. Consider a bar of silicon having carrier concentration n0=1015 cm-3 and ni=1010cm-3. Assume the excess carrier concentrations to be n=1013cm-3, calculate the quasi-fermi energy level at T=300K? a) 0.2982 eV b) 0.2984 eV c) 0.5971 eV d) 1Ev View Answer Answer: b Explanation: =1.38*10-23*300*ln(1013+1015/1013) =0.2984 eV. 10. From the above equation, assuming the same values for the for ni, n= p and T. Given that p0=105cm-3. Calculate the quasi-fermi energy level in eV? a) 0.1985 b) 0.15 c) 0.1792

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d) 0.1 View Answer Answer: c Explanation: Using the same equation,

Substituting the respective values, EFi – EFp=0.1792 eV.

Questions & Answers (MCQs) focuses on “Diffusion”. 1. What is the SI unit of electron diffusion constant? a) cm2/s b) m2/s c) m/s 102

d) none View Answer Answer: b Explanation: J=eDdn/dx So D=q*m/ (q*s*(1/m)) =m2/s. 2. Calculate the diffusion current density when the concentration of electron varies from the 1*1018 to 7*1017 cm-3 over a distance of 0.10 cm.D=225cm2/s a) 100 A/cm2 b) 108 A/cm2 c) 0.01A/cm2 d) None View Answer Answer: b Explanation: J=eDdn/dx J=1.6*10-19*225*(1018-(7*1017))/0.1 =108A/cm2. 3. Which is the correct formula for the Jp? a) Jp=qDdp/dx b) Jp=pDdn/dx c) Jp=-qDdp/dx d) None View Answer Answer: c Explanation: Jp is negative for the p type of semiconductors. 4. What is the direction of the electron diffusion current density relative to the electron flux? a) Same direction b) Opposite to each other c) Perpendicular to each other d) At 270 degrees to each other View Answer Answer: b Explanation: From the graph between electron concentration and the distance, we can see that the direction of the electron diffusion current density is opposite to the electron flux.

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5. In diffusion, the particles flow from a region of _______ to region of ___________ a) High, low b) Low , high c) High , medium d) Low, medium View Answer Answer: a Explanation: Diffusion is the process of flow of particles form the region of the high concentration to a region of low concentration. 6. Which of the following parameter describes the best movement of the electrons inside a semiconductor? a) Velocity gradient b) Diffusion c) Mobility d) Density gradient View Answer Answer: c Explanation: Mobility is defined as the movement of the electrons inside a semiconductor. On the other hand, velocity gradient is the ratio of velocity to distance. 7. Which of the following term isn’t a part of the total current density in a semiconductor? a) Temperature b) µ c) e d) E View Answer Answer: a Explanation: J=enµE+epµE+eDdn/dx-eDdp/dx So, temperature isn’t a part of the equation. 8. What does dn/dx represent? a) Velocity gradient b) Volume gradient c) Density gradient d) None View Answer

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Answer: c Explanation: dn/dx represent velocity gradient. 9. Calculate the diffusion constant for the holes when the mobility of the holes is 400cm2/V-s and temperature is 300K? a) 1.035m m2/s b) 0.035m m2/s c) 1.5m m2/s d) 1.9m m2/s View Answer Answer: a Explanation: Dp=VT*μn = (1.38*10-23*300*400*10-2)/ (1.6*10-19) = 1.035m m2/s. 10. Calculate the diffusion constant for the electrons when the mobility of the electrons is 325cm2/V-s and temperature is 300K? a) 0.85 m2/s b) 0.084 m2/s c) 0.58 m2/s d) 0.95 m2/s View Answer Answer: c Explanation: Dn=VT*μn = (1.38*10-23*300*325*10-2)/ (1.6*10-19) = 0.084 m2/s.

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Questions and Answers for Freshers focuses on “Carrier Life Time”. 1. What is the range of the carrier lifetime? a) Nanoseconds to microseconds b) Nanoseconds to hundreds of microseconds c) Nanoseconds to tens of microseconds d) Nanoseconds to milliseconds View Answer Answer: b Explanation: Carrier lifetime is defined as the existence of any carrier for τ seconds. Carrier lifetime ranges from nanoseconds to hundreds of microseconds. 2. What is the process number of Schokley-Read-Hall Theory processes? Process-‘ The capture of an electron from the conduction band by an initially neutral empty trap’ a) Process1 b) Process2 c) Process3 d) Process4 View Answer Answer: a Explanation: This is the first process of Schokley-Read-Hall theory of Recombination. 3. Calculate the recombination rate if the excess carrier concentration is 1014cm-3 and the carrier lifetime is 1µseconds. a) 108 b) 1010 c) 1020 d) 1014 View Answer Answer: c Explanation: The recombination rate, R=δn/τ. So, R=1014 / 10-6 R=1020.

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4. Calculate the capture rate where Cn=10, Nt=1010cm-3, n=1020 and fF (Et)=0.4. a) 6*1030 b) 5*1030 c) 36*1030 d) 1.66*1029 View Answer Answer: a Explanation: Rcn=Cn*N_t*(1-fF (Et))*n Substituting the values, Rcn=6*1030. 5. Calculate the emission rate where En=2.5, Nt=1010cm-3 and fF (Et)=0.6 . a) 15*1010 b) 1.5*1010 c) 15*1011 d) 1.5*1011 View Answer Answer: b Explanation: Ren=En*Nt*(fF (Et)) Substituting the values, Ren=1.5*1010. 6. At what condition, the rate of electron capture from the conduction band and the rate of the electron emission back into the conduction band must be equal? a) Thermal equilibrium b) At room temperature c) T=250K d) At boiling temperature View Answer Answer: a Explanation: At thermal equilibrium, the electron capture rate and the emission rate will be same in the conduction band. 7. Calculate the carrier lifetime when Cp=5 and Nt=1010cm-3. a) 2*1011 b) 2*10-11 c) 20*10-11 d) 20*1011 View Answer 107

Answer: b Explanation: τp=1/(Cp*Nt ) =1/(5*1010) =2*10-11. 8. The number of majority carriers that are available for recombining with excess minority carriers decreases as the excess semiconductor becomes intrinsic. Is it true? a) True b) False View Answer Answer: a Explanation: With the increase in the number of the majority carriers, the carriers for the recombination will be decreasing with the excess minority carriers and will finally become intrinsic as the concentrations will be same. 9. Which of the following is used as the recombination agent by semiconductor device manufactures? a) Silver b) Gold c) Platinum d) Aluminium View Answer Answer: b Explanation: Gold is used as a recombination agent because of its chemical properties as it was used in the Bohr’s experiment. Thus the device designer can obtain the desired carrier lifetimes by introducing gold into silicon under controlled conditions. 10. The rate of change of the excess density is proportional to the density. Is it true of false? a) True b) False View Answer Answer: a Explanation: The rate of change of the excess density depends on the density of the semiconductor and the rate with respect to time is also dependent on it.

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Questions & Answers (MCQs) focuses on “The Continuity Equation”. 1. What does p/τ represent? a) holes b) time c) holes per second lost d) p per unit time View Answer Answer: c Explanation: Option (c) represents the holes per second lost by recombination per unit volume. 2. Which of the following is the Taylor’s expression? a) b) c) d) View Answer Answer: a Explanation: Option a represents the correct formula for the Taylor’s expression.

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3. Calculate the number of coulombs per second if the area is 4cm2, recombination rate of hole is 1000 cm-3/s and the differential length is 2mm. a) 1.28*10-23 b) 1.28*10-22 c) 1.28*10-21 d) 1.28*10-20 View Answer Answer: b Explanation: number of coulombs per second= eAdxp/τ =1.6*10-19*4*10-4*2*10-3*1000 =1.28*10-22. 4. The current entering the volume at x is I and leaving is I+Δi , the number of coulombs per second will be equal to δI. Is it true or false? a) True b) False View Answer Answer: a Explanation: Coulombs per second is known as the current. The differential current will be the current through the semiconductor. 5. The change in the carrier density is due to a) Flow of incoming flux b) Flow of outgoing flux c) Difference of flow between incoming and outgoing flux d) Difference of flow between incoming and outgoing flux plus generation and minus recombination View Answer Answer: d Explanation: The change in the carrier density describes the continuity equation which is equal to the difference between the incoming and outgoing flux plus generation and minus recombination. 6. What of the following conditions satisfies when the number of holes which are thermally generated is equal to the holes lost by recombination? a) I≠0 b) dp/dt≠0 c) g=p/τ

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d) g≠p/τ View Answer Answer: c Explanation: Under the equilibrium conditions, I will be zero and then the dp/dt will aso be equal to zero in the continuity equation. Then, g= p/τ is left which is option c. 7. What is the diffusion length for holes when Dp=25cm2/s and τp=25s? a) 25cm b) 1cm c) 0.04cm d) 50cm View Answer Answer: a Explanation: Lp=√(Dp*τp) =√(25*25) =25cm. 8. Which of the following represents the continuity equation? a) dp/dt=-(p-p0)/τp+Dp(d2p/dx2)-µpd(ρϵ)/dx b) dp/dt=-(p-p0)/τp-Dp(d2p/dx2)-µpd(ρϵ)/dx c) dp/dt=-(p-p0)/τp+Dp(d2p/dx2)+µpd(ρϵ)/dx d) dp/dt=(p-p0)/τp-Dp(d2p/dx2)-µpd(ρϵ)/dx View Answer Answer: a Explanation: Option a represents the correct equation of the continuity equation for holes. 9. What is the diffusion length for electrons when Dn=10cm2/s and τn=40s? a) 50cm b) 25cm c) 20cm d) 15cm View Answer Answer: c Explanation: Ln=√(Dn*τn) =√(10*40) =20cm.

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10. Which of the following represents the best definition for the diffusion length for holes? a) Average distance which an electron is injected travels before recombining with an electron b) Average distance which a hole is injected travels before recombining with an electron c) Average distance which a hole is injected travels before recombining with a hole d) Average distance which an electron is injected before recombining with a hole View Answer Answer: b Explanation: Diffusion length for holes is represented as the average distance which a hole is injected travels before recombining with an electron. It is the distance into the semiconductor at which the injected concentration falls to 1/ϵ of its value at x=0.

Questions & Answers (MCQs) focuses on “The Hall Effect”. 1. In the Hall Effect, the directions of electric field and magnetic field are parallel to each other. The above statement is a) True b) False View Answer Answer: b Explanation: To make Lorentz force into the effect, the electric field and magnetic field should be perpendicular to each other.

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2. Which of the following parameters can’t be found with Hall Effect? a) Polarity b) Conductivity c) Carrier concentration d) Area of the device View Answer Answer: d Explanation: The Hall Effect is used for finding the whether the semiconductor is of n-type or ptype, mobility, conductivity and the carrier concentration. 3. In the Hall Effect, the electric field is in x direction and the velocity is in y direction. What is the direction of the magnetic field? a) X b) Y c) Z d) XY plane View Answer Answer: c Explanation: The Hall Effect satisfies the Lorentz’s Force which is E=vxB So, the direction of the velocity, electric field and magnetic field should be perpendicular to each other. 4. What is the velocity when the electric field is 5V/m and the magnetic field is 5A/m? a) 1m/s b) 25m/s c) 0.2m/s d) 0.125m/s View Answer Answer: a Explanation: E=vxB v=E/B =5/5 =1m/s. 5. Calculate the hall voltage when the Electric Field is 5V/m and height of the semiconductor is 2cm. a) 10V b) 1V 113

c) 0.1V d) 0.01V View Answer Answer: c Explanation: Vh=E*d =5*2/100 =0.1V. 6. Which of the following formulae doesn’t account for correct expression for J? a) ρv b) I/wd c) σE d) µH View Answer Answer: d Explanation: B=µH So, option d is correct option. 7. Calculate the Hall voltage when B=5A/m, I=2A, w=5cm and n=1020. a) 3.125V b) 0.3125V c) 0.02V d) 0.002V View Answer Answer: d Explanation: Vh=BI/wρ =5=2/ (5*10-2*105*1.6*10-19) =0.002V. 8. Calculate the Hall Effect coefficient when number of electrons in a semiconductor is 1020. a) 0.625 b) 0.0625 c) 6.25 d) 62.5 View Answer Answer: b Explanation: R=1/ρ

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=1/(1.6*10-19*1020) =0.0625. 9. What is the conductivity when the Hall Effect coefficient is 5 and mobility is 5cm2 /s. a) 100 S/m b) 10 S/m c) 0.0001S/m d) 0.01 S/m View Answer Answer: c Explanation: µ=σR σ =µ/R =5*10-4/5 =0.0001 S/m. 10. In Hall Effect, the electric field applied is perpendicular to both current and magnetic field? a) True b) False View Answer Answer: a Explanation: In Hall Effect, the electric field is perpendicular to both current and magnetic field so that the force due to magnetic field can be balanced by the electric field or vice versa.

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3. Questions & Answers on Semiconductor-Diode Characteristics The section contains questions on pn junction qualitative theory, p-n junction diode, band structure of open circuited p-n junction, components in p-n junction diode, volt ampere characteristics, diode resistance and capacitances, pn diode switching times, breakdown and tunnel diodes, point contact diode, p-i-n diode and its characteristics. Qualitative Theory of the p-n junction The P-N Junction as a Diode Band Structure of an Open-Circuited p-n junction The Current Components in a P-N junction diode Quantitative Theory of the P-N Diode Currents The Volt Ampere Characteristics The Temperature Dependence of P-N Characteristics

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Diode Resistance Diode Capacitances PN Diode Switching Times Breakdown Diodes Tunnel Diodes and its Characteristics p-i-n Diode and its Characteristics The Point Contact Diode

Questions and Answers for freshers focuses on “Qualitative Theory of the p-n junction”. 1. The donor ions is represented by a positive plus sign. Is it True or False? a) True b) False View Answer Answer: a Explanation: The donor atom donates the extra ion to the semiconductor. Therefore, it is represented by the positive plus sign. 2. Initially, the p-type carriers are located to the____________of the semiconductor. a) Right b) Left c) Middle d) Top View Answer Answer: b Explanation: The p-type carriers are nominally located to the left of the junction and n-type carriers are to the right. 3. The displacement of the charges results in a) Magnetic field b) Electric field c) Rust d) Hall effect View Answer Answer: The flow of carriers in a semiconductor results in the electric field across the junction. The electric field thus makes the current flow in the device. 4. What is the value of 1 micron? a) 10-6cm b) 10-5cm 117

c) 10-4cm d) 10-3cm View Answer Answer: c Explanation: 1 micron=10-4cm=10-6cm. 5. Which of the following results when the equilibrium established in a semiconductor? a) Restrain the process of diffusion b) Electric field becomes very high c) Both a and b d) None of these View Answer Answer: c Explanation: As the electric field is very high, the flow of the carries will be restricted and the equilibrium will be obtained. 6. Which of the following options doesn’t defined for the necessity for the existence of the potential barrier? a) Contact b) Potential c) Diffusion d) Fermi dirac View Answer Answer: d Explanation: The potential barrier is formed at the junction of the semiconductor. It’s necessity of the potential barrier is known as the contact, potential or diffusion. 7. Under the open-circuited conditions the net hole current must be zero. Is this statement is True or false? a) True b) False View Answer Answer: a Explanation: The net hole current is zero because if this wasn’t true, the hole density at one end of the semiconductor would continue to increase indefinitely with time, a situation which is obviously physically impossible. 118

8. The un-neutralised ions in the neighbourhood of the junction are known as a) Depletion charges b) Uncovered charges c) Mobile ions d) Counter ions View Answer Answer: b Explanation: The un-neutralised ions in the neighbourhood of the junction are known as uncovered ions because they are not mobile. 9. Which of the following doesn’t defines for the junction which is depleted of mobile charges? a) Depletion region b) Uncovered region c) Space charge region d) Transition region View Answer Answer: b Explanation: The junction which is depleted of mobile charges is known as depletion region or space charge region and transition region. 10. Convert 10 micron to meters. a) 10-5m b) 107m c) 10-6m d) 10-4m View Answer Answer: a Explanation: Since, 1 micron=10-6m 10 micron =10-5m.

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Questions & Answers (MCQs) focuses on “The P-N Junction as a Diode”. 1. How many junction/s do a diode consist? a) 0 b) 1 c) 2 d) 3 View Answer Answer: b Explanation: Diode is a one junction semiconductor device which has one cathode and anode. The junction is of p-n type. 2. If the positive terminal of the battery is connected to the anode of the diode, then it is known as a) Forward biased b) Reverse biased c) Equilibrium d) Schottky barrier View Answer Answer: a Explanation: When a positive terminal is connected to the anode, the diode is forward biased which lets the flow of the current in the circuit. 120

3. During reverse bias, a small current develops known as a) Forward current b) Reverse current c) Reverse saturation current d) Active current View Answer Answer: c Explanation: When the diode is reverse biased, a small current flows between the p-n junction which is of the order of the Pico ampere. This current is known as reverse saturation current. 4. If the voltage of the potential barrier is V0. A voltage V is applied to the input, at what moment will the barrier disappear? a) V< V0 b) V= V0 c) V> V0 d) V<< V0 View Answer Answer: b Explanation: When the voltage will be same that of the potential barrier, the potential barrier disappears resulting in flow of current. 5. During the reverse biased of the diode, the back resistance decrease with the increase of the temperature. Is it true or false? a) True b) False View Answer Answer: a Explanation: Due to the increase in the reverse saturation current due to the increase in the temperature, the back resistance decrease with the increasing temperature. 6. What is the maximum electric field when Vbi=2V , VR=5V and width of the semiconductor is 7cm? a) -100V/m b) -200V/m c) 100V/m d) 200V/m View Answer

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Answer: b Explanation: Emax=-2(Vbi+VR)/W =-2(2+5)/ (7*10-2) =-200V/m. 7. When the diode is reverse biased with a voltage of 6V and Vbi=0.63V. Calculate the total potential. a) 6V b) 6.63V c) 5.27V d) 0.63V View Answer Answer: b Explanation: Vt=Vbi+VR =0.63+6 =6V. 8. It is possible to measure the voltage across the potential barrier through a voltmeter? a) True b) False View Answer Answer: b Explanation: The contacts of the voltmeter have some resistance which will not accurately measure the voltage across the potential barrier. Thus, it is not possible to measure the voltage across the potential barrier. 9. What will be the output of the following circuit? (Assume 0.7V drop across the diode)

a) 12V b) 12.7V c) 11.3V d) 0V View Answer

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Answer: c Explanation: V=12-0.7 =11.3V. 10. Which of the following formula represents the correct formula for width of the depletion region? a) b) c) d) View Answer Answer: a Explanation: Option a is the correct formula.

Questions and Answers for Experienced people focuses on “Band Structure of an Open-Circuited p-n junction”. 1. The conduction band edge in the p material is not at the same level to that of conduction band edge in the n material. Is it true or false? a) True b) False View Answer Answer: a Explanation: In a p-n junction diode, the energy levels of the p material and n material will not be at same level. They will be different. So, the conduction band edge as well as the valence band edge of the p material will not be same to that of the n material.

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2. Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction? a) Eo = Ecn – Ecp b) Eo = Ecp – Ecn c) Eo = Ecp + Ecn d) Eo = -Ecp – Ecn View Answer Answer: b Explanation: The shift in the energy of the energy level will be the difference of the conduction band edge of the p material and conduction band edge of n material. In the energy level diagram, the conduction band edge of p material is higher than that of the n material. 3. Calculate the Eo given that Nd=1.5*1010cm-3, Na=1.5*1010cm-3 at temperature 300K? a) 1.5*1010eV b) 0.256eV c) 0eV d) 4.14*10-21eV View Answer Answer: c Explanation: Eo=kTln((Nd*Na)/(ni)2) Substituting k=1.38*10-23/K, T=300k and the values ofNd,Naand ni, We get Eo=0eV. 4. In a p-n junction, the valence band edge of the p material is greater than which of the following band? a) Conduction band edge of n material b) Valence band edge of n material c) Conduction band edge of p material d) Fermi level of p material View Answer Answer: b Explanation: When the p-n junction is formed, the energy levels of the p- material go higher than the n material. That’s why the valence band of the p material will be greater than that of the n material. 5. Which of the following equations represent the correct expression for the band diagram of the p-n junction? (E1=difference between the fermi level of material and conduction band of n material and E2=difference between the conduction band of n material and fermi level of n material) 124

a) Ecn – E f = (1/2)*EG – E1 b) Ecn – E f = (1/2)*EG – E2 c) Ef – Ecp = (1/2)*EG – E1 d) Ecn – Ef = (1/2)*EG + E1 View Answer Answer: a Explanation: From the energy band diagram of the p-n junction, the option ‘a’ satisfies that band diagram. 6. Calculate the value of Eo when pno=104cm-3 and ppo=1016cm-3 at T=300K. a) 1meV b) 0.7meV c) 0.7eV d) 0.1meV View Answer Answer: c Explanation: Eo=kTln(ppo/pno) Substituting the values, we get Eo=0.7eV. 7. Calculate the value of Dp when µp=400cm/s and VT=25mV. a) 1 b) 0.01 c) 0.1 d) 10 View Answer Answer: c Explanation: Dp= µp*VT =400*10-2*25*10-3 =0.1. 8. What is the value of kT at room temperature? a) 0.0256eV b) 0.25eV c) 25eV d) 0.0025eV View Answer 125

Answer: a Explanation: kT=1.38*10-23*300K =4.14*10-21/ (1.6*10-19) =0.0256eV. 9. Is Vo depends only on the equilibrium concentrations. Is it true or false? a) True b) False View Answer Answer: a Explanation: Vo is the contact potential of the junction when the junction is in equilibrium. If, the junction is not in the equilibrium, Vo can’t be calculated. 10. Calculate Vo when ppo=1016cm-3, pno=104cm-3 and Vt=25mV. a) 69V b) 6.9V c) 0.69V d) 0.069V View Answer Answer: c Explanation: Vo=VT ln⁡(ppo/pno ) =25*10-3*ln(1016/104) =0.69V.

Questions and Answers for Experienced people focuses on “The Current Components in a P-N junction diode”. 126

1. When a forward biased is applied to a diode, the electrons enter to which region of the diode? a) P-region b) N-region c) P-n junction d) Metal side View Answer Answer: a Explanation: When the forward biased is applied, the electrons enter to the p-region and the holes enter to the n-region so that holes can flow from p-region to n-region. Whereas, the electrons can travel from n-region to p-region. 2. The number of injected minority carriers falls off linearly with the increase in the distance from the junction. Is it true or false? a) True b) False View Answer Answer: b Explanation: The number of minority carriers fall off exponentially rather than linearly with the increase in the distance from the junction. 3. What is the total current in a diode when x=0? a) I = Ipn (0) – Inp (0) b) I = Ipn (0) + Inp (0) c) I = -Ipn (0) – Inp (0) d) I = -Ipn (0) + Inp (0) View Answer Answer: b Explanation: At junction, the total current is equal to the minority hole current plus the minority electron current. 4. The current in the diode is 1. Unipolar 2. Bipolar a) I only b) II only c) I and II both

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d) Neither I nor II View Answer Answer: b Explanation: The current in the diode consists of both the electrons and holes. So, it is bipolar. 5. The current is constant throughout the device. Is it true or false? a) True b) False View Answer Answer: a Explanation: The current in the device is constant but the proportion due to the electrons and holes varies with distance. 6. Which of the following statements is correct under forward biased p-n diode? a) current enters n side as hole current and leaves p side as electron current b) current enters n side as electron current and leaves p side as hole current c) current enters p side as hole current and leaves n side as electron current d) current enters p side as hole current and leaves p side as electron current View Answer Answer: c Explanation: When the current flows in a p-n diode, the current enters p side as hole current and leaves n side as electron current. 7. Calculate the total current when Ipn (0)=1mA and Inp (0)=2mA. a) 1mA b) -1mA c) 0 d) 3mA View Answer Answer: d Explanation: I=Ipn (0)+Inp (0) =1mA+2mA =3mA. 8. What is the hole current in the p region of the diode? a) Ipp (x)=I-Inp (x) b) Ipp (x)=I+Inp (x) 128

c) Ipp (x)=-I-Inp (x) d) Ipp (x)=-I+Inp (x) View Answer Answer: a Explanation: The hole current in the p region is equal to the total current minus the minority electrons in the p region. 9. What does Inp represent? a) Hole current in n region b) Hole current in p region c) Electron current in n region d) Electron current in p region View Answer Answer: d Explanation: Inp constitutes of the electron current in the p region. It is the minority electron carrier in the p region. 10. Deep into the p side the current is a drift current Ipp of holes sustained by the small electric field in the semiconductor. Is the statement true or false? a) True b) False View Answer Answer: a Explanation: In the p region, the drift current is sustained into the p region by the small electric field which is formed at the junction in the semiconductor. So, the above statement is true.

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Quantitative Theory of the P-N diode Currents”. 1. What is the thickness of ‘space charge region’ or ‘transition region’ in P-N junction diode? a) 1 micron b) 5 micron c) 10 micron d) 2.876 micron View Answer Answer: a Explanation: The region of the junction is depleted by mobile charges, hence it is called space charge region or depletion region or transition region which is 10-4 cm = 10-6 m= 1 micron. 2. If what of the following is doped into a semiconductor say germanium a P-N junction is formed. a) Electrons and Protons b) Protons and Neutrons c) Neutrons and Electrons d) Gallium and Phosphorus View Answer Answer: d Explanation: A P-N junction is formed only when a donor impurities and acceptor impurities are added to either side of a semiconductor like silicon and germanium. 3. Which of the factors doesn’t change the diode current. a) Temperature b) External voltage applied to the diode c) Boltzmann‘s constant d) Resistance View Answer Answer: d Explanation: I = Io [e(v/nVt) -1], as shown in this equation the diode current is dependent on temperature , 130

voltage applied on the diode , Boltzmann’s constant but diode current is not dependent on resistance as it is independent of resistance. 4. The product of mobility of the charge carriers and applied Electric field intensity is known as a) Drain velocity b) Drift velocity c) Push velocity d) Pull velocity View Answer Answer: b Explanation: When the semiconductors like silicon and germanium is implied by an electric field the charge carriers when get drifted by certain velocity known as drift velocity. Drift velocity is product of mobility of charge carriers and field intensity. 5. The tendency of charge carriers to move from a region of heavily concentrated charges to region of less concentrated charge is known as. a) Depletion current b) Drain current c) Diffusion current d) Saturation current View Answer Answer: c Explanation: In a semiconductor the charge will always have a tendency to move from higher concentrated area to less concentrated area to maintain equilibrium this movement of charges will result in diffusion current. 6. If the drift current is 100mA and diffusion current is 1A what is the total current in the semiconductor diode. a) 1.01 A b) 1.1 A c) 900m A d) 10 A View Answer Answer: b Explanation: We know that the total current in a semiconductor is equal to sum of both drift current and diffusion current. Total current = 1A + 100mA =1.1A.

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7. Which of the following is reverse biased?

a) A) b) B) c) C) d) D) View Answer Answer: c Explanation: The P-N junction diode is forward bias when the voltage applied to p type is greater than the n type and vice versa, since the voltage applied to p type is less in C) it is the answer. 8. The drift velocity is 5V and the applied electric field intensity 20v/m what will be the mobility of charge carriers. a) 100 m2/ (vs) b) 4 m2/ (vs) c) 15 m2/ (vs) d) 0.25 m2/ (vs) View Answer Answer: d Explanation: We know that mobility of charge carriers is drift velocity divide by applied electric field intensity. Mobility = drift velocity / field intensity. 9. When there is an open circuit what will be the net hole current. a) 5A 132

b) 0.05A c) 0.5A d) 0A View Answer Answer: d Explanation: If there is any current present under open circuit there will be an indefinite growth of holes at one end of the semiconductor which is practically not possible hence zero amperes. 10. Rate of change of concentration per unit length in a semiconductor is called as. a) Concentration change b) Concentration mixture c) Concentration gradient d) Concentration variant View Answer Answer: c Explanation: In a semiconductor the holes as well as electrons which are the charge carriers is not equally concentrated on all regions of the semiconductor the change in their rate is referred as concentration gradient.

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“The Volt Ampere Characteristics”. 1. The voltage equivalent of temperature (Vt) in a P-N junctions is given by. a) T/1000 volts b) T/300 volts c) T/1600 volts d) T/11600 volts View Answer Answer: d Explanation: We know that the P-N junction is temperature dependent, it varies with the change in temperature the measure of change that is the voltage equivalent of temperature is given by Vt = T/11600 volts. 2. At room temperature what will be voltage equivalent of temperature. a) 10 mV b) 4.576 mV c) 26 mV d) 98 V View Answer Answer: c Explanation: Room temperature is 27o C = 300 k .We know that Vt= T/11600 volts by substituting the value of T we get 300/11600 = 26mV. 3. In a P-N junction the positive voltage at which the diode starts to conduct consequently is called. a) Cut off voltage b) Saturation voltage c) Knee voltage d) Breakdown voltage View Answer Answer: c Explanation: At a certain critical voltage, a large reverse current flows and the diode is said to be in breakdown region, at this region the diode will be forward biased and starts to conduct consequently.

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4. In volt ampere characteristics the current increases with voltage _________ a) Exponentially b) Equally c) Sinusoidal d) Unequally View Answer Answer: a Explanation: The current in the volt ampere characteristics increases exponentially with respect to voltage I(t) = eV(t). 5. The cut off voltage for diode of silicon semiconductor and germanium semiconductor is ____ volts. a) 0.5 and 0.1 b) 0.7 and 0.3 c) 1 and 0.5 d) 0.5 and 1 View Answer Answer: b Explanation: The cut off voltage is the voltage only after which the semiconductors conduct, the cut off voltage for silicon is 0.7V in the sense the silicon diode will conduct only when voltage is more than 0.7V and 0.3 for germanium. 6. What would be the current and voltage when there is no external voltage applied on the diode? a) 0 b) 0.7 c) 0.3 d) 1 View Answer Answer: a Explanation: When there is no external voltage applied on the circuit it acts as an open circuit and there will be no flow of charges hence the current and voltage will be zero. 7. In P-N junction V-I characteristics during forward biased, at what region the current increase is very low. a) Saturation b) Depletion c) Cut off d) Breakdown View Answer

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Answer: b Explanation: In the V-I characteristics the change in the current with respect to voltage is very less in depletion region due to the large resistance in the circuit as the resistance deceases by a certain value the current increases exponentially with voltage. 8. In a P-N junction diode during forward bias if the current increases more than the value that is rated will destroy the diode. a) True b) False View Answer Answer: a Explanation: If the current in the P-N junction diode during forward bias increases beyond the value rated on it will destroy the diode because voltage is directly proportional to current so extreme voltage will burn the diode down. 9. When the P-N junction diode is forward bias the current in circuit is controlled by. a) External voltage b) Capacitance c) Resistance d) Internal voltage View Answer Answer: c Explanation: When the P-N junction is in forward bias that is the p side connected to the positive terminal of voltage source the current in the circuit can be varied by varying the resistance, the current flow decreases as the resistance increases and vice versa. 10. The P-N junction diode conducts in which direction. a) Reverse direction b) Forward direction c) Both Forward and Reverse direction d) Neither Forward nor Reverse direction View Answer Answer: b Explanation: The P-N junction diode conducts only in forward direction, it will not conduct in reverse direction so only Zener Diode was introduced as it conducts in both forward and reverse direction.

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The Temperature Dependence of P-N Characteristics”. 1. The magnitude of the electric charge (e) is given by ____________ a) -1.6*10-19 C b) 1.6*10-19 C c) 9.11*10-31 C d) 1.637*10-37 C View Answer Answer: a Explanation: The charge of the electron is the magnitude of electric force that an electron exerts on other particles which is equal to -1.6*10-19 C, the negative sign indicates the direction of force. 2. What is the forbidden gap voltage for silicon material? a) 1.46 V b) 1.56 V c) 10 V d) 1.21 V View Answer 137

Answer: d Explanation: The forbidden gap voltage of a material is numerically equal to forbidden gap energy of the material which is 1.21 joules for silicon so forbidden gap voltage will be 1.21 V. 3. Which of the following parameters of P-N junction diode increases with temperature. a) Cut in voltage b) Reverse saturation current. c) Ideality factor d) Resistance View Answer Answer: b Explanation: Reverse saturation current at temperature T2 is 2[(T2 –T1)/10] times greater than reverse saturation current at temperature T1 where T2 is greater than T1. 4. Which of the following diodes do not exhibits a constant reverse saturation current with the change in reverse saturation voltage. a) 1N909 b) 1N405 c) 1N207 d) 1N676 View Answer Answer: c Explanation: 1N207 is the germanium diode for which the reverse saturation current is not constant which the change in voltage due to the leakage in the surface of the diode and due to the generation of new current carriers. 5. Which of these P-N junction characteristics are not dependent on temperature. a) Junction resistance b) Reverse saturation current c) Bias current d) Barrier voltage View Answer Answer: a Explanation: As the temperature of the P-N junction increases the current increases and the voltage decreases so the barrier voltage, reverse saturation current, bias current changes with temperature but junction resistance is independent of temperature.

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6. As the temperature to the P-N junction increases the current increases due to? a) Leakage in bias region b) Electron-hole pair c) Leakage in P region d) Leakage in N region View Answer Answer: b Explanation: As the temperature to the P-N junction increases the mobility of charges increases thus increases the electron-hole pair which proportionally increases the current in the P-N junction diode. 7. By what percentage the reverse saturation current increases with 10 C rise in the temperature. a) 25% b) 12.5% c) 50% d) 7% View Answer Answer: d Explanation: As the temperature to the P-N junction diode increases the mobility of charges increases thus increasing the current, the reverse saturation current increases by 7% with 10C rise in temperature and doubles with every 100C rise in temperature. 8. What will be the decrease of barrier voltage with the rise in 10C in temperature? a) 10V b) 1mV c) 10mV d) 2mV View Answer Answer: d Explanation: As the temperature to the P-N junction diode increases the voltage across the junction decreases and the current increases with every degree rise in temperature the barrier voltage increases by 2mV. 9. What will be the reverse saturation current in the junction when the voltage across the junction is 0? a) 0.3A b) 0.7A c) 0A

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d) 1.24A View Answer Answer: c Explanation: When the voltage across the junction is zero in the sense there will be potential difference between the junctions hence there will be no movement of electrons and holes, hence the current will be 0. 10. The breakdown voltage of the P-N junction diode decreases due to the increase in. a) Reverse saturation current b) Reverse leakage current c) Bias voltage d) Barrier voltage View Answer Answer: b Explanation: Breakdown voltage of the diode is inversely proportional to the reverse leakage current so it decreases with the increase in reverse leakage current.

Questions & Answers (MCQs) focuses on “Diode Resistance”. 1. The static resistance R of the diode is given by __________ a) V/I b) V*I c) V+I

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d) V-I View Answer Answer: a Explanation: According to Ohms law the electric current in the circuit is directly proportion to voltage and inversely proportional to resistance so, R=V/I. 2. In the volt ampere characteristics of the diode, the slope of the line joining the operating point to the origin at any point is equal to reciprocal of the _________ a) resistance b) conductance c) voltage d) current View Answer Answer: a Explanation: In the diode’s volt ampere characteristics, the line joining the operating point and the origin, at any point of the line is equal to the conductance so, it is reciprocal of the resistance. 3. At room temperature (VT = 26) what will be the approximate value of r when n=1 and I=100mA? a) 26 ohms b) 2.6 ohms c) 260 ohms d) 2600 ohms View Answer Answer: c Explanation: We know that R= (n*VT) /I, by substituting the value of n, VT, I we get R= 260 ohms, (1*26)/100*10-3 = 260 ohms. 4. In the diode volt ampere characteristics what will be the resistance if a slope is drawn between the voltages 50 to 100 and corresponding current 5 to 10? a) 5 ohms b) 10 ohms c) 50 ohms d) 100 ohms View Answer Answer: b Explanation: We know that, in volt ampere characteristics the resistance is equal to the reciprocal

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of the line joining the origin and operating point, R = dV/dI, by substituting the value of dV and dI we get R= 10ohms. 5. In piecewise linear characteristics what will be the RF value if the slope is 0.5? a) 25 m ohms b) 50 m ohms c) 2 ohms d) 10 ohms View Answer Answer: c Explanation: In piecewise linear characteristics the forward resistance will be equal to reciprocal of the slope so, RF = 1/slope, RF = 1/0.5 which is equal to 2 ohms. 6. A diode will behave as an open circuit if the voltage in the circuit is less than __________ a) cut off voltage b) saturation voltage c) leakage voltage d) threshold voltage View Answer Answer: d Explanation: The diode made up of semiconductor has a certain threshold voltage only after which it behave as closed circuit in the sense it performs some operation if the threshold voltage is greater than the voltage in circuit. 7. What will be the approximate value of thermal voltage of diode? a) 25mV at 300K b) 30mV at 180K c) 25mV at 180K d) 30mV at 300K View Answer Answer: a Explanation: We know that the thermal voltage of diode is approximately equal to room temperature which is 300K then for all practical purpose the thermal voltage of diode is taken as 25mV so it will be 25mV at 300K. 8. What will be the thermal voltage of the diode if the temperature is 300K? a) 25.8 mV b) 50 mV c) 50V 142

d) 19.627 mV View Answer Answer: a Explanation: The thermal voltage of the diode is given by, VT = KT/q, by substituting the values of T, K which is Boltzmann constant and q which is the charge of the electron we get VT = (300*1.38*10-23)/ (1.602*10-19), VT= 25.8mV. 9. What will be the diode resistance if the current in the circuit is zero? a) 0 ohms b) 0.7 ohms c) 0.3 ohms d) 1 ohms View Answer Answer: a Explanation: When the current in the circuit is zero there will be no flow of charges to resist hence the diode resistance will be zero. 10. Which of these following is not a characteristic of an ideal diode? a) Perfect conductor when forward bias b) Zero voltage across it when forward bias c) Perfect insulator when reverse bias d) Zero current through it when forward bias View Answer Answer: d Explanation: The diode acts as an ideal diode when it is a perfect conductor and has zero voltage across it during forward bias, a perfect insulator and zero current through it during reverse bias.

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Questions & Answers (MCQs) focuses on “Diode Capacitances”. 1. Compared to a PN junction with NA=1014/CM3, which one of the following is true for NA=ND= 1020/CM3? a) depletion capacitance decreases b) depletion capacitance increases c) depletion capacitance remains same d) depletion capacitance can’t be predicted View Answer Answer: b Explanation: We know, CT=Aε/W and W ∝ (1/NA+1/ND) 1/2. So, CT ∝ (1/NA+1/ND)-1/2 So when NA and ND increases, depletion capacitance CT increases. 2. If CT is the transition capacitance, which of the following are true? 1) in forward bias, CT dominates 2) in reverse bias, CT dominates 3) in forward bias, diffusion capacitance dominates 4) in reverse bias, diffusion capacitance dominates a) 1 only b) 2only c) 2 and 3 d) 3 only View Answer Answer: c Explanation: In reverse bias condition, depletion region increases and acts as an insulator or dielectric medium. So, the transition capacitance increases. In forward bias condition, due to stored charge of minority carriers, diffusion capacitance increases. 3. For an abrupt PN junction diode, small signal capacitance is 1nF/cm2 at zero bias condition.If the built in voltage, Vbi is 1V, the capacitance at reverse bias of 99V is? a) 0.1nF/cm2 144

b) 1nF/cm2 c) 1.5nF/cm2 d) 2nF/cm2 View Answer Answer: a Explanation: Cjo is the capacitance at zero bias, that is VR=0V, Cjo=Cj for VR=0V. We know, Cj = Cjo/(1+(VR/Vbi))m , m=1/2 for abrupt. So, putting Cj=0.1nF/cm2 where, VR=99V and Vbi=1V we get, Cjo= 0.1(1+99)1/2 = 0.1nF/cm2. 4. The built in capacitance V0 for a step graded PN junction is 0.75V. Junction capacitance Cj at reverse bias when VR=1.25V is 5pF. The value of Cj when VR=7.25V is? a) 0.1pF b) 1.7pF c) 1pF d) 2.5Pf View Answer Answer: d Explanation: We know, Cj1/ Cj2=[(V0+VR2)/(V0+VR2)]1/2 So, Cj2=Cj1/ {(0.75+7.25)/(0.75+1.25)}1/2 we get Cj2=Cj1 /2 =5/2=2.5Pf. 5. Consider an abrupt PN junction. Let V0 be the built in potential of this junction and VR be the reverse bias voltage applied. If the junction capacitance Cj is 1pF for V0+VR =1V, then for V0+VR =4V what will be the value of Cj? a) 0.1pF b) 1.7pF c) 1pF d) 0.5Pf View Answer Answer: d Explanation: We know, Cj1/ Cj2=[(V0+VR2)/(V0+VR1)]1/2 Cj2=Cj1(1/4)1/2=1/2 . We get Cj2=1/2=0.5pF. 6. A silicon PN junction diode under revers bias has depletion width of 10µm, relative permittivity is 11.7 and permittivity, ε0 =8.85×10-12F/m. Then depletion capacitance /m2 =? a) 0.1µF/m2 b) 1.7µF/m2 c) 10µF/m2

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d) 0.5µF/m2 View Answer Answer: c Explanation: We know, CT =Aε0εr /W CT/A= (8.85×10-12)(11.7)/10 =10 By putting the values we get 10µF/m2. 7. The transition capacitance, CT of a PN junction having uniform doping in both sides, varies with junction voltage as ________ a) (VB )1/2 b) (VB )-1/2 c) (VB )1/4 d) (VB )-1/4 View Answer Answer: b Explanation: CT = K/(V0+VB)1/2 As it’s having uniform doping on both sides, the voltage V0 will be zero. So, CT=K/(VB)1/2. The variation of transition capacitance with built in capacitance is (VB )-1/2. 8. The CT for an abrupt PN junction diode is ________ a) CT = K/(V0+VB)1/2 b) CT = K/(V0+VB)-1/2 c) CT = K/(V0+VB)1/3 d) CT = K/(V0+VB)-1/3 View Answer Answer: a Explanation: For an abrupt PN junction diode, CT = K/(V0+VB)n. Here, n=1/2 for abrupt PN junction diode and 1/3 for linear PN junction diode. When the doping concentration of a diode varies within a small scale of area, then the diode is called as an abrupt diode. 9. The diffusion capacitance of a PN junction _______ a) decreases with increasing current and increasing temperature b) decreases with decreasing current and increasing temperature c) increasing with increasing current and increasing temperature d) doesnot depend on current and temperature View Answer

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Answer: b Explanation: CD =τ I /n0 VT Where, I is the current and VT is temperature factor. The diffusion capacitance is directly proportional to current and indirectly proportional to the temperature. 10. Transition capacitance is also called as _______ a) diffusion capacitance b) depletion capacitance c) conductance capacitance d) resistive capacitance View Answer Answer: b Explanation: Transition capacitance occurs in reverse bias. We obtain a depletion layer in that case. Hence it’s also called as depletion capacitance. The diffusion capacitance occurs in forward bias.

Questions & Answers (MCQs) focuses on “PN Diode Switching Times”. 1. Diode acts as a short circuit when switched from forward to reverse bias for some time due to______ a) Accumulation of minority charge carriers when it’s in forward bias b) Accumulation of majority charge carriers when it’s in forward bias c) Accumulation of minority charge carriers when it’s in reverse bias d) Accumulation of majority charge carrier when it’s in reverse bias View Answer Answer: a Explanation: When a diode is switched suddenly, it persists the conducting property for a short time in its reverse bias also. This leads to excess minority charge carrier settlement at potential barrier. Hence acts as a short circuit. 2. Reverse recovery time for a diode is? a) Time taken to eliminate excess minority charge carriers b) Sum of storage time (TS) and transition time (TT) 147

c) Time taken to eliminate excess majority charge carriers d) Time elapsed to return to non conduction state View Answer Answer: a Explanation: The time period for which diode remains in conduction state even in reverse direction is called storage time. The time elapsed to return the non conduction state is called transition time. Their sum is called reverse recovery time. 3. Switching speed of P+ junction depends on. a) Mobility of minority carriers in P junction b) Life time of minority carriers in P junction c) Mobility of majority carriers in N junction d) Life time of minority carriers in N junction View Answer Answer: d Explanation: Switching leads to move holes in P region to N region as minority carriers. Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively. 4. Time taken for a diode to reach 90% of its final value when switched from steady state is______ a) 2.3*time constant b) 2.2*time constant c) 1.5*time constant d) equals the time constant View Answer Answer: b Explanation: Time constant = RC. To reach 90% of the final value, time taken is 2.2 of RC. Time constant is the time required to discharge the capacitor, through the resistor, by 36.8%. 5. Which of the following are true? 1) In reverse bias, the diode undergoes stages of storage and transition times 2) Minority charge carriers accumulation makes the diode as a short circuit 3) Storage time is the sum of recovery and transition times a) 1 only b) 2 and 3 c) 3 only d) 1 and 2 only View Answer 148

Answer: d Explanation: When a diode is switched from forward to reverse bias, storage and transition times takes place. The accumulation time or the life time of minority carriers makes it a short circuit. The conduction property is holds for a short period of time in reverse bias also. 6. In a circuit below, the switch is at position 1 at t<0 and at position 2 when t=0. Assume diode has zero voltage drop and storage time. For 0
a) 5V b) -5V c) 0v d) 10V View Answer Answer: b Explanation: At position ‘1’ when connected to +5V, the diode is forward biased and acts as a short circuit. So, VR is 5V. For 0
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a) 1mA b) 2mA c) -2mA d) -1mA View Answer Answer: c Explanation: Initially, the diode is in forward bias. When suddenly switched to reverse bias, upto a storage time limit, it conducts during storage time period. We know that, current I=V/R=-20/10K=-2mA. 8. A PN junction diode with 100Ω resistor is forward biased such that 100A current flows. If voltage across this combination is instantaneously reversed to 10V at t=0, the reverse current that flows through diode at t=0 is?

a) 10mA 150

b) 100mA c) -100mA d) -10mA View Answer Answer: b Explanation: At t=0, V=-10V. During storage time, current still flows. We know that, current I=V/R=10/100Ω=100mA from N to P region. 9. The delay in switching between the ON and OFF states is due to _________ a) The time required to change amount of excess minority carriers stored in quasi-neutral regions b) The time required to change amount of excess majority carriers stored in quasi-neutral regions c) The conduction between storage time and recovery time d) The exponential increase in carriers in N region View Answer Answer: a Explanation: When switched instantaneously it stays in a quasi state i.e.., temporary state which stores charges. The delay is produced due to this charge settlement. The diode needs to discharge these excess carriers in order to return the non conduction stage. 10. The delay time can be reduced by? a) decreasing lifetime and increasing ratio of reverse to forward current b) increasing lifetime and decreasing ratio of reverse to forward current c) increasing lifetime and increasing ratio of reverse to forward current d) decreasing lifetime and decreasing ratio of reverse to forward current View Answer Answer: a Explanation: When the current increases the depletion layer decreases and the storage and transition time decreases. A decreased depletion layer can easily discharge the excess carrier and thereby lessens the delay time.

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Questions & Answers (MCQs) focuses on “Breakdown Diodes”. 1. A zener diode works on the principle of_________ a) tunneling of charge carriers across the junction b) thermionic emission c) diffusion of charge carriers across the junction d) hopping of charge carriers across the junction View Answer Answer: a Explanation: Due to zener effect in reverse bias under high electric field strength, electron quantum tunneling occurs. It’s a mechanical effect in which a tunneling current occurs through a barrier. They usually cannot move through that barrier. 2. Which of the following are true about a zener diode? 1) it allows current flow in reverse direction also 2) it’s used as a shunt regulator 3) it operates in forward bias condition a) 3 only b) 1 and 2 c) 2 and 3 d) 2 only View Answer Answer: b Explanation: The operation of a zener diode is made in reverse bias when breakdown occurs. So, it allows currnt in reverse direction. The most important application of a zener diode is voltage or shunt regulator. 3. When the voltage across the zener diode increases_________ a) temperature remains constant and crystal ions vibrate with large amplitudes b) temperature increases and crystal ions vibrate with large amplitudes c) temperature remains constant and crystal ions vibrate with smaller amplitudes 152

d) temperature decreases and crystal ions vibrate with large amplitudes View Answer Answer: b Explanation: When voltage is increased, the tunnelling at reverse bias increases. The voltage rises temperature. The crystal ions with greater thermal energy tend to vibrate with larger amplitudes. 4. For the zener diode shown in the figure, the zener voltage at knee is 7V, the knee current is negligible and the zener dynamic resistance is 10Ω. If the input voltage (Vi) ranges from 10 to 16 volts, the output voltage (Vo) ranges from?

a) 7 to 7.29V b) 6 to 7V c) 7.14 to 7.43V d) 7.2 to 8V View Answer Answer: c Explanation: If i is the current flowing, then V0=10i+7 i=(VI-7)/210. By substituting, if VI=10V then i=1/70 and V0=(1/7)+7=7.14V if VI =16V then i=3/70 and V0=(3/7)+7=7.43V. 5. In the circuit below, the knee current of ideal zener diode is 10mA. To maintain 5V across the RL, the minimum value of RL is?

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a) 120 b) 125 c) 250 d) 100 View Answer Answer: b Explanation: Here, IKNEE=10mA, VZ=5V. I=IL+IZ. I= (10-5)/100=50mA Now, 50=10+ILMAX . ILMAX=40mA. RLMIN=5/40mA=125 Ω. 6. The zener diode in the circuit has a zener voltage of 5.8V and knee current of 0.5mA. The maximum load current drawn with proper function over input voltage range between 20 and 30V is?

a) 23.7mA b) 20mA c) 26mA d) 48.3mA View Answer Answer: a Explanation: Here, I1MAX=IZMIN+ILMAX. 154

IZMIN =0.5mA, I1MAX =(V1MAX-VZ )/RS . Putting the values we get , I1MAX =24.2mA. So, 24.2-0.5=23.7mA. 7. In the given limiter circuit, an input voltage Vi=10sin100πt is applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______

a) 6.1V,-0.7V b) 0.7V,-7.5V c) 7.5V,-0.7V d) 7.5V,-7.5V View Answer Answer: c Explanation: With VI= 10V when maximum, D1 is forward biased, D2 is reverse biased. Zener is in breakdown region. VOMAX=sum of breakdown voltage and diode drop=6.8+0.7=7.5V. VOMIN=negative of voltage drop=-0.7V. There will be no breakdown voltage here. 8. The 6V Zener diode shown has zener resistance and a knee current of 5mA. The minimum value of R so that the voltage does not drop below 6V is?

a) 1.2Ω b) 80 Ω c) 50 Ω d) 70 Ω View Answer 155

Answer: b Explanation: Here, Vz =6V, IZMIN=5mA.IS=IZMIN+ILMAX. 80=5+ILMAX . ILMAX=75Ma.RLMIN=VI/ILMAX=6/75mA =80 Ω. 9. Avalanche breakdown in zener diode is ______ a) electric current multiplication takes place b) phenomenon of voltage multiplication takes place c) electrons are decelerated for a period of time d) sudden rise in voltage takes place. View Answer Answer: a Explanation: The carriers in transition region are accelerated by electric field to energies. That energies are sufficient to create electron current multiplication. A single carrier that is energized will collide with another by gaining energy. Thus an avalanche multiplication takes place. 10. The zener diode is heavily doped because______ a) to have low breakdown voltage b) to have high breakdown voltage c) to have high current variations d) to maintain perfect quiescent point View Answer Answer: a Explanation: The value of reverse breakdown voltage at which zener breakdown occurs is controlled by amount of doping. If the amount of doping is high, the value of voltage at which breakdown occurs will decrease. Better doping gives a sooner breakdown voltage. Questions & Answers (MCQs) focuses on “Tunnel Diodes and its Characteristics”. 1. If ‘X’ corresponds to a tunnel diode and ‘Y’ to an avalanche diode, then__________ a) X operates in reverse bias and Y operates in forward bias b) X operates in reverse bias and Y operates in reverse bias c) X operates in forward bias and Y operates in forward bias d) X operates in forward bias and Y operates in reverse bias View Answer Answer: d Explanation: In forward bias, negative resistance helps for tunnel diode to operate. Here, the current decreases with increase in voltage. If they are used in reverse bias, they are called as back diodes. Avalanche diode operates in reverse bias at breakdown region. 156

2. The range of tunnel diode voltage VD, for which slope of its V-I characteristics is negative would be? (The VP is the peak voltage and VV is the valley voltage).

a) VD > 0 b) 0 VD > VP d) VD > VV View Answer
depicted below: 3. Tunnel diode has a very fast operation in__________ a) gamma frequency region b) ultraviolet frequency region c) microwave frequency region d) radio frequency region View Answer 157

Answer: c Explanation: Tunnel diode is a type of semiconductor which works on tunneling effect of electrons in microwave region. So, tunnel diode has a very fast operation in microwave region. 4. Which of the following are true about a tunnel diode? 1) it uses negative conductance property 2) it operates at high frequency 3) fermilevel of p side becomes higher than the n side in forward bias a) 1 only b) 1 and 2 c) 3 only d) 2 and 3 View Answer Answer: b Explanation: The negative resistance property helps in the operation of tunnel diode. As the tunnel diode works at high frequency, its applications are mostly in that range. High frequency oscillators are based on the resonant tunneling diode. 5. The depletion layer of tunnel diode is very small beacause______ a) its abrupt and has high dopants b) uses positive conductance property 158

c) its used for high frequency ranges d) tunneling effect View Answer Answer: a Explanation: When the P and N regions are very highly doped, the depletion layer comes closer. The tunnel diode is also highly doped. Its doping concentration varies within a small scale. So it’s an abrupt diode. For these reasons, the depletion region is small. 6. With interments of reverse bias, the tunnel current also increases because________ a) electrons move from balance band of pside to conduction band of nside b) fermi level of pside becomes higher than that of nside c) junction currrent decreases d) unequality of n and p bandedge View Answer Answer: a Explanation: When the forward bias is increased, the tunnel current is also increased upto a certain limit. This happens when the electron movement takes place from P to N side. 7. The tunnneling involves_______ a) acceleration of electrons in p side 159

b) movement of electrons from n side conduction band to p side valance band c) charge distribution managementin both the bands d) positive slope characteristics of diode View Answer Answer: b Explanation: Tunneling means a direct flow of electrons across small depletion region from N side conduction band to P side valance band. The electrons begin to accelerate in the N side of the semiconductor. 8. Tunnel diodes are made up of________ a) Germanium and silicon materials b) AlGaAs c) AlGaInP d) ZnTe View Answer Answer: a Explanation: Germanium and silicon materials have low band gaps and flexibility. That matches tunnel diode requirements. The remaining materials emits the energy in terms of light or heat. 9. For a tunnel diode, when ‘p’ is probability that carrier crosses the barrier, ’e’ is energy,’w’ is width. (-A*e*w) a) p ∝ e 160

(-A*e*w) b) p ∝ 1/ e (A*e*w) c) p ∝ e (A*e*w) d) p ∝ 1/e View Answer Answer: a Explanation: The carrier jump occurs without any loss of energy due to small depletion layer. The probability of the carrier to jump across a barrier depends on the energy and width of the band. This variess exponentially for a given carrier. 10. In the construction of tunnnel diode,why is the pellet soldered to anode contact and a tindot to cathode contact via a mesh screen? a) for better conduction and reduce inductance respectively b) for heat dissipation and increase conduction respectively c) for heat dissipation and reduce induction respectively d) for better conduction and reduce inductance respectively View Answer Answer: c Explanation: Anode goes through better heat dissipation. So the pellet is used for the purpose. The tindot via mesh screen resists inductive effects caused at the cathode. Conduction is an independent factor which can’t be controlled.

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11. What happens to a tunnel diode when the reverse bias effect goes beyond the valley point? a) it behaves as a normal diode b) it attains increased negative slope effects c) reverse saturation current increases d) beacomes independent of temperature View Answer Answer: a Explanation: After the valley point is crossed, the tunnel diode obtains positive slope resistance. That is similar to the characteristics of a normal diode. So it behaves like a normal diode after beyond valley point.

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Questions & Answers (MCQs) focuses on “p-i-n Diode and its Characteristics”. 1. PIN diode is a photosensitive diode because of _______ a) large current flow in p and n region b) depletion layer increases giving a larger surface area c) stronger covalent bonds d) low carrier storage View Answer Answer: b Explanation: An intrinsic layer that is sandwiched between p and n layers. This gives a larger surface area making it compatible for photosensitivity. Reverse bias causes an increased depleted region in a PIN diode. 2. During forward bias, the PIN diode acts as _______ a) a variable resistor b) a variable capacitor c) a switch d) an LED View Answer Answer: a Explanation: In forward bias, the forward resistance decreases and acts as a variable resistor. The low frequency model of a PIN diode neglects the input capacitive values. 3. During reverse bias, the PIN diode acts as _______ a) Variable resistor b) Switch c) Variable capacitor d) LED View Answer Answer: c Explanation: In reverse bias, the intrinsic layer is completely covered by depletion layer. The stored charges vanishes acting like a variable capacitor. The high frequency model of a PIN diode neglects the input resistances.

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4. When the p and n regions are used for high resistivity, the depletion region at the respective places is called _________ a) Q and ϒ regions b) ϒ and π regions c) Q and π regions d) π and ϒ regions View Answer Answer: d Explanation: When p region is used for high resistance, the depletion layer is high at p side.When n side is used the depletion layer is high at n side. They are called as π and ϒ regions respectively. 5. The applications for PIN diode are __________ a) Microwave switch b) LED c) Voltage regulator d) Amplifier View Answer Answer: a Explanation: Being employed at 300Hz, the swept voltage is attained at π region.Then it’s used as a microwave switch. Swept voltage is nothing but, the voltage at which the complete intrinsic layer is swept out as a depleted one. 6. In high frequency model, the values of resistance ‘R’ and capacitance ‘C’ are _______ a) 0.1 to 10KΩ and 0.02 to 2pF respectively b) 1 to 10KΩ and 0.02 to 2pF respectively c) 10 to 100KΩ and 0.02 to 2pF respectively d) 0.1 to 10KΩ and 2 to 20pF respectively View Answer Answer: a Explanation: At high frequency, the applied values for resistance and capacitance is 0.1 to 10KΩ and 0.02 to 2pF respectively. At high frequencies, it almost acts as a perfect resistor. 7. What happens in PIN diode for low frequency model? a) reactance decreases b) conductance increases c) resistance increases d) reactance increases View Answer 164

Answer: d Explanation: In a low frequency model, the resistance decreases and reactance increases.Here the variable resistance is neglected. At low frequencies, the charge can be removed and the diode can be turned off. 8. Which of the following is true about a PIN diode? a) it’s photosensitive in reverse bias b) it offers low resistance and low capacitance c) it has a decreased reversed breakdown voltage d) carrier storage is low View Answer Answer: a Explanation: Due to increased depletion region, the covalent bonds break and increase the surface area for photosensitivity. This property is used in fields of light sensors, image scanners, artificial retina systems. 9. In the application of frequency models, the value of forward current is _____ a) IF = A(µPP + µNN)q b) IF = A(µPN + µNP)q c) IF = A(µPP – µNN)q d) IF = A(µPN – µNP)q View Answer Answer: a Explanation: The forward current depends on mobility and carrier concentration. In frequency models, the value of forward current is IF = A*(µPP + µNN)q. Where, µP and µN are the mobility of p and n type charge carriers respectively. 10. The forward resistance for a PIN diode is given by ________ a) RF = W/σP b) RF = W/σN c) RF = WσP d) RF = WσN View Answer Answer: b Explanation: Forward resistance for a PIN diode depends on the width, current density and positive carrier concentration of the diode. No diode is perfectly ideal. In practise, a diode offers a small resistance in forward bias which is called as forward resistance.

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Questions & Answers (MCQs) focuses on “The Point Contact Diode”. 1. The materials that are used in the construction of point contact diode are _________ a) Silicon b) SnTe or Bi2Te3 c) GaS or CdS d) HgI View Answer Answer: b Explanation: The diode base of SnTe or Bi2Te3 is highly detection sensitive. They are mechanically stable over long periods of use either as harmonic generators or mixers. They are emphasized in the 2-200THz region. 2. Which of the following are true? 1) point contact diode has a metal whisker to make pressure contact during its operation 2) it has high voltage rating 3) its V-I characteristics are constant 4) it has low breakdown voltage a) 1 and 2 b) 3 only c) 1 and 4 d) 2 only View Answer

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Answer: c Explanation: The diode contains a die of a semiconductor material on which an epitaxial layer is deposited. It uses the metal whisker to make pressure contact against that layer. It has a low breakdown voltage. 3. In the forward bias condition, the resistance of point contact diode is_________ a) less than that of a general PN diode b) greater than that of a general PN diode c) equal to that of a general PN diode d) varies exponentially than that of a general PN diode View Answer Answer: a Explanation: The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. This characteristic of contact diode makes its capacitance high at high frequency. A small capacitive current flows in the circuit. 4. The barrier layer capacitance of a point contact diode is_________ a) 0.1pF to 1pF b) 5pF to 50pF c) 0.2pF to 2pF d) 0.008µF to 20µF View Answer Answer: a Explanation: The barrier capacitance at the point is very low about 0.1pF to 1pF. The capacitance between the cat whisker and crystal is less compared to junction diode capacitance between both sides of the diode. For a general PN diode is 0.008µF to 20µF. 5. The cat whisker wire present in the contact diode is used for_________ a) for heat dissipation b) for charge transfer between sections c) maintaining the pressure between sections d) preventing current flow View Answer Answer: c Explanation: The operation of a contact diode depends on the pressure of contact between semiconductor crystal and point. The cat whisker wire presses against the crystal to form a section and the section allows the current flow. This is similar to the behaviour of PN diode.

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6. The semiconductor junctions those are present in a contact diode_________ a) beryllium-copper and bronze-phosphor b) beryllium-phosphor and bronze-copper c) mercury-iodine d) tin-tungsten View Answer Answer: a Explanation: The diode contains two sections having a small rectangular crystal of N type silicon and a fine beryllium-copper and bronze-phosphor. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other. 7. The application of a contact diode is_________ a) Clampers and clippers b) Voltage multipliers c) Rectifiers d) AM detectors View Answer Answer: d Explanation: The point contact diodes are the oldest microwave semiconductor devices. They were developed during world war 2. They have excessive applications in microwave fields and used as receivers and detectors. 8. The operating frequencies of the point contact diode is_________ a) 30KHz or above b) 10GHz or above c) 30GHz or above d) 10KHz or above View Answer Answer: b Explanation: It’s used in high frequency conversions and circuits in the order of 10KHz or above. The reactance due to capacitance is high and at high frequency a very small capacitive current flows. 9. During the manufacture of point contact diode, why is a relatively large current passed from cat whisker to silicon crystal? a) to control the amount of current flow b) to form small region of p type material c) to allow mechanical support for the sections

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d) to form anode and cathode regions View Answer Answer: b Explanation: The behaviour of a contact diode is similar to that of a PN diode. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other. 10. What is the capacitive reactance across the point contact diode when compared to normal PN junction diode a) lower b) higher c) equal d) cannot be determined View Answer Answer: a Explanation: The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. A small capacitive current flows in the circuit.

4. Questions on Diodes The section contains questions on basics of diode, types of diodes which includes zener diode and others, limiting and clamping circuits, rectifiers and characteristics of junction diode and diode forward characteristic modelling. The Ideal Diode Modelling the Diode Forward Characteristic

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Rectifiers Limiting and Clamping Circuits and Some Special

Zener Diode

Types of diodes Diode Circuits

Questions & Answers (MCQs) focuses on “The Ideal Diode”.

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1. For an ideal diode which of the following is true? a) It allows the passage of current in the forward bias with zero potential drop across the diode b) It does not allow the flow of current in reverse bias c) All of the mentioned d) None of the mentioned View Answer Answer: c Explanation: Both of the facts mentioned hold true for an ideal operational amplifier. 2. Consider an AC sine wave voltage signal being used to connect a diode and a resistor as shown in the figure. The variation of the voltage across the diode (Vd) with respect to time (t) is given by

a)

b)

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c) d) None of the mentioned View Answer Answer: c Explanation: Vd = Vi – Vo. 3. The figure below shows a circuit for charging a 12-V battery. If Vs is a sinusoid with 24-V peak amplitude, the fraction of each cycle during which the diode conducts is

a) One quarter of a cycle b) One-third of a cycle c) One half of the cycle d) Three quarters of a cycle View Answer Answer: b Explanation: For one half of the cycle the diode conducts when 24 cos(Θ) = 12 or Θ is 600. Hence for a complete cycle the diode does not conducts for 2Θ or 1200. Hence the diode does not conducts for a third of a cycle. 4. Diodes can be used in the making of a) Rectifiers b) LED lamps c) Logic gates d) All of the mentioned View Answer

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Answer: d Explanation: Diodes are used to make rectifiers (full wave or half wave rectifiers are the most common examples), LED lamps uses diodes (diodes are generally doped for their use in this purpose) and logic gates can also be made using diodes using the fact that diodes are conducting only in the forward biased configuration. 5. For the connections shown below, the equivalent logic gate is

a) OR gate b) AND gate c) XOR gate d) NAND gate View Answer Answer: a Explanation: The following circuit behaves as a OR logic gate. 6. For the connections shown below, the equivalent logic gate is

a) OR gate b) AND gate c) XOR gate d) NAND gate View Answer Answer: b Explanation: The circuit shown behaves as an AND logic gate. 173

7. The figure below shows a circuit for an AC voltmeter. It utilizes a moving-coil meter that gives a full-scale reading when the average current flowing through it is 1 mA. The moving-coil meter has a 50-Ω resistance. The value of R that results in the meter indicating a full-scale reading when the input sine-wave voltage VI is 20 V peak-to-peak is

a) 3.183 kΩ b) 3.133 kΩ c) 3.183 Ω d) 56.183 Ω View Answer Answer: b Explanation: Average value of input voltage= 20/π or 6.366 V. Since there is diode half of the time there is no voltage across the voltmeter hence the average value of the input voltage is 10/π or 3.183V. The total resistance required to generate 1mA current is 3.183 kΩ out of which 50Ω is provided by the moving coil. Therefore, a resistance of 3.133 kΩ is required. 8. The value of I and V for the circuit shown below are

a) 2A and 5V respectively b) -2A and 5V respectively c) -2A and -5V respectively d) 2A and 0V respectively View Answer Answer: d Explanation: I = V/R or 5/2.5 or 2A For an ideal diode when it is conducting the potential drop across its terminal is zero. 174

9. The units frequently used to measure the forward bias and reverse bias current of a diode are a) µA and µA respectively b) µA and mA respectively c) mA and µA respectively d) mA and mA respectively View Answer Answer: c Explanation: The currents in forward bias is generally in MA and in the reverse bias it is very small and is generally measured in µA. 10. A diode a) Is the simplest of the semiconductor devices b) Has a characteristic that closely follows that of a switch c) Is two terminal device d) All of the mentioned View Answer Answer: d Explanation: Both the statements are true for a diode.

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Questions & Answers (MCQs) focuses on “Modelling the Diode Forward Characteristic”. 1. Which of the following is method to model a diode’s forward characteristics? a) Iteration method b) Graphical method c) Constant-voltage drop model d) All of the mentioned View Answer Answer: d Explanation: All of the mentioned are different methods used to model a diode’s forward characteristics. 2. A voltage regulator needs to provide a constant voltage in spite of the fact that there may be a) Change in the load current drawn from the terminals of the regulator b) Change of the DC power supply that feeds the regulator circuit c) None of the mentioned d) All of the mentioned View Answer Answer: d Explanation: Voltage regulators are required in both of the situations mentioned.

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(Q.3-Q.4) For the circuit shown below calculate

3. Calculate the %age change in the regulated voltage caused by a change of ±10% in the input voltage. (RL is not connected to the circuit) a) ± 0.5% b) ± 1% c) ± 5% d) ± 10% View Answer Answer: a Explanation: Current I in the circuit = (10 – 2.1) / 1 or 7.9 mA Incremental resistance of each diode is rd = 25mV/7.9mA at room temperature or 3.2Ω Total resistance of the diode connection = r = 9.6Ω The series combination of R and r acts as a voltage divider, hence ▲V0 = ±r/r+R or 9.5 mA or ± 0.5%. 4. Calculate the change in the voltage when RL is connected as shown a) -10 mA b) -15 mA c) -20 mA d) -25 mA View Answer Answer: c Explanation: When RL is connected it draws approx. 2.1v/1000 A current or 2.1 mA. Hence the current in the diode decreases by approximately 2.1 mA. The voltage change across the diode is thus -2.1mA * 9.6 Ω or -20mA.

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5. The value of the diode small-signal resistance rd at bias currents of 0.1 mA is a) 250 Ω b) 25 Ω c) 2.5 Ω d) 0.25 Ω View Answer Answer: a Explanation: Thermal voltage at room temperature is 25mV and the current is 0.1mA. rd = 25mV/0.1mA or 250 Ω. 6. In many applications, a conducting diode is modeled as having a constant voltage drop, usually approximately a) 1 V b) 0.1 V c) 0.7 V d) 7V View Answer Answer: c Explanation: Normally a silicon diode has a voltage drop of 0.7V. 7. The graphical method of modeling a diode characteristics is based on a) Iteration method b) Constant voltage drop method c) Small signal approximation d) Exponential method View Answer Answer: d Explanation: Graphical method uses equations obtained in the exponential method and then the Answer of these equations is obtained by plotting them on a graph. 8. For small-signal operation around the dc bias point, the diode is modeled by a resistance equal to the a) Slope of the i-v characteristics of the diode at the bias point b) Square root of the slope of the i-v characteristics of the diode at the bias point c) Inverse of the slope of the i-v characteristics of the diode at the bias point d) Square root of the inverse of the slope of the i-v characteristics of the diode at the bias point View Answer

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Answer: d Explanation: It is a required characteristic for small-ground operations. 9. The other name for bias point is a) Quiescent point b) Node point c) Terminal point d) Static point View Answer Answer: a Explanation: Quiescent point is also called base point. It is due to this name that is also called Q– point. 10. In iteration method for modelling a diode the answer obtained by each subsequent iteration is a) Is close to the true value b) Is close to the value obtained by exponential method c) All of the mentioned d) None of the mentioned View Answer Answer: c Explanation: True value is obtained by the exponential method and therefore both of the statements are true.

Questions & Answers (MCQs) focuses on “Zener Diode”. 1. Zener diodes are also known as a) Voltage regulators b) Forward bias diode c) Breakdown diode d) None of the mentioned View Answer

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Answer: c Explanation: Zener diodes are used as voltage regulators but they aren’t called voltage regulators. They are called breakdown diodes since they operate in breakdown region. 2. Which of the following is true about the resistance of a Zener diode? a) It has an incremental resistance b) It has dynamic resistance c) The value of the resistance is the inverse of the slope of the i-v characteristics of the Zener diode d) All of the mentioned View Answer Answer: d Explanation: All of the statements are true for the resistance of the zener diode. 3. Which of the following is true about the temperature coefficient or TC of the Zener diode? a) For Zener voltage less than 5V, TC is negative b) For Zener voltage around 5V, TC can be made zero c) For higher values of Zener voltage, TC is positive d) All of the mentioned View Answer Answer: d Explanation: All of the mentioned are true for the TC of a zener diode. 4. Which of the following can be used in series with a Zener diode so that combination has almost zero temperature coefficient? a) Diode b) Resistor c) Transistor d) MOSFET View Answer Answer: a Explanation: If a Zener diode of TC of about -2mV is connected with a forward diode (which has a TC of about +2mV) in series, the combination can be used to obtain a very low (close to zero) TC. 5. In Zener diode, for currents greater than the knee current, the v-i curve is almost a) Almost a straight line parallel to y-axis b) Almost a straight line parallel to x-axis c) Equally inclined to both the axes with a positive slope 180

d) Equally inclined to both the axes with a negative slope View Answer Answer: b Explanation: Note that the curve is v-I curve and not an i-v curve. 6. Zener diodes can be effectively used in voltage regulator. However, they are these days being replaced by more efficient a) Operational Amplifier b) MOSFET c) Integrated Circuits d) None of the mentioned View Answer Answer: c Explanation: ICs have been widely adapted by the industries over conventional zener diodes as their better replacements for a voltage regulators. 7. A 9.1-V zener diode exhibits its nominal voltage at a test current of 28 mA. At this current the incremental resistance is specified as 5 Ω. Find VZ0 of the Zener model. a) 8.96V b) 9.03V c) 9.17V d) 9.24V View Answer Answer: b Explanation: VZ = VZo + MZ IZT 9.1 = VZo + 5 * 28 * 10-3 VZo = 8.96v VZ = VZo + 5IZ = 8.96 * 5IZ. 8. A shunt regulator utilizing a zener diode with an incremental resistance of 5 Ω is fed through an 82-Ω resistor. If the raw supply changes by 1.0 V, what is the corresponding change in the regulated output voltage? a) 72.7 mV b) 73.7 mV c) 74.7 mV d) 75.7 mV View Answer

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Answer: c

Explanation: 9. A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rz of 10 Ω and 5.1-V devices with rz of 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R. a) -30mV/mA and 120mV/mA respectively b) 30mV/mA and 60mV/mA respectively c) -60mV/mA and +60mV/mA respectively d) -30mV/mA and -120mV/mA respectively View Answer Answer: d Explanation: Three 6.8v zeners provide 3*6.8 = 20.4v with 3 * 10 =30Ω Resistance, neglecting R, we have load Regulation = -30mV/mA. For 5.1 Zeners we need 4 diodes to provide 20.4v with 4 * 30 =120Ω Resistance. load Regulation = -120mV/mA . 10. Partial specifications of a Zener diode is provided. VZ = 10.0 V, VZK = 9.6 V, and IZT = 50 mA. Assuming that the power rating of a breakdown diode is established at about twice the specified Zener current (IZT), what is the power rating of each of the diodes described above? a) 1.04 W b) 0.104 W c) 10.4 mW d) 1.04 mW View Answer

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Answer: a

Explanation:

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Questions & Answers (MCQs) focuses on “Rectifiers”. 1. Which of the following isn’t a type of rectifier? a) Precision Half-wave Rectifier b) Bridge Rectifier c) Peak Rectifier d) None of the mentioned View Answer Answer: d Explanation: All of the mentioned are different types of a rectifier. 2. For a half wave or full wave rectifier the Peak Inverse Voltage of the rectifier is always a) Greater than the input voltage b) Smaller than the input voltage c) Equal to the input voltage d) Greater than the input voltage for full wave rectifier and smaller for the half wave rectifier View Answer Answer: b Explanation: The peak input voltage is smaller than the input voltage due to the presence of diode(s). A single diode reduces the output voltage by approximately 0.7V. 3. For a half-wave rectifier having diode voltage VD and supply input of VI, the diode conducts for π – 2Θ, where Θ is given by a) tan -1 VD/VI b) tan-1 VD/VI – VI c) sin-1 VD/VI d) sin-1 VD/VI – VI View Answer Answer: c Explanation: The diode doesn’t conducts when VD ≥VI . Hence Θ = sin-1 (D/VI). 4. Bridge rectifier is an alternative for a) Full wave rectifier 184

b) Peak rectifier c) Half wave rectifier d) None of the mentioned View Answer Answer: a Explanation: Bridge rectifier is a better alternative for a full wave rectifier. 5. Which of the following is true for a bridge rectifier? a) The peak inverse voltage or PIV for the bridge rectifier is lower when compared to an identical center tapped rectifier b) The output voltage for the center tapped rectifier is lower than the identical bridge rectifier c) A transistor of higher number of coil is required for center tapped rectifier than the identical bridge rectifier d) All of the mentioned View Answer Answer: d Explanation: All of the given statements are true for a bridge rectifier. 6. The diode rectifier works well enough if the supply voltage is much than greater than 0.7V. For smaller voltage (of few hundreds of millivolt) input which of the following can be used? a) Superdiode b) Peak rectifier c) Precision rectifier d) None of the mentioned View Answer Answer: a Explanation: For the supply voltages less than 0.7V super diodes are used. 7. A simple diode rectifier has ‘ripples’ in the output wave which makes it unsuitable as a DC source. To overcome this one can use a) A capacitor in series with a the load resistance b) A capacitor in parallel to the load resistance c) Both of the mentioned situations will work d) None of the mentioned situations will work View Answer Answer: b Explanation: A capacitor is parallel with a resistor can only makes ripples go away. Series connection will become equal to an open circuit once the capacitor is fully charged. 185

8. Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting a) 1.06 % b) 2.12 % c) 3.18% d) 4.24% View Answer Answer: c Explanation: w Δt ~ √(2Vr/Vp) Θ = √(2 X 2/100) Θ = 0.2 rad or 3.18% of the cycle (Q.9-Q.10) The op amp in the precision rectifier circuit is ideal with output saturation levels of ±12 V. Assume that when conducting the diode exhibits a constant voltage drop of 0.7 V.

9. Find V– when VI is -1V. a) 0V b) 0.7V c) 1V d) 1.7V View Answer Answer: a Explanation: VI = -1v Vo = 1v VA = 1.7v V– = 0v Virtual gnd as negative feedback is closed through R. VI > 0 D1 conducts D2 cutoff. VI < 0 D2 conducts D1 cutoff.

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V 0

⁄ VI = -1.

10. Find V0 when VI is 2V. a) 0V b) 0.7V c) 1V d) 1.7V View Answer Answer: a Explanation: VI = 2v Vo = 0v VA = -0.7v V– = 0v Virtual gnd as negative feedback is closed through R. VI > 0 D1 conducts D2 cutoff. VI < 0 D2 conducts D1 cutoff. V 0 ⁄ VI = -1.

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“Limiting and Clamping Circuits and Some Special Types of diodes”. 1. Which of the following circuits can be used as limiter or clamper or both?

a)

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b)

c) d) All of the mentioned View Answer Answer: d Explanation: Each of the circuit can be either as a clamper or limiter or both. 2. The V0 vs VI curve for the below circuit is

a) 189

b)

c)

d) View Answer Answer: d Explanation: Only this curve represents the characteristics correctly. 3. For the circuit and the input signal shown which of the following is true for the output voltage?

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a) The output waveform is a square wave with lowest peak clamped to 0V b) The output wave is a square wave with lowest peak clamped to -6V c) The output waveform is a square wave with highest peak clamped to 4V d) The output waveform is a straight line with the value of output voltage equal to 10V View Answer Answer: a Explanation: It is a clamped capacitor circuit. 4. Which of the following is not true for the duty cycle of a waveform? a) Duty cycles can be used to describe the percent time of an active signal in an electrical device b) Duty cycle can be used to determine the percentage of the time a signal is active c) 60% duty cycle means that the waveform is active for 40% of the total time d) 50% duty cycle means that the waveform is non-active for 50% of the total time View Answer Answer: c Explanation: 60% duty time means that a signal is active for 60% of the total time. 5. The maximum and the minimum voltage across the diode D1 respectively is

a) 2Vp and 0V respectively b) 0V and -2Vp respectively c) Vp and -Vp respectively 191

d) 2Vp and Vp respectively View Answer Answer: b Explanation: It is a voltage doubler circuit. 6. Limiting and clamping circuits are employed in a) FM transmitters b) Television receivers and transmitter c) Production of various signal waveforms such as trapezoidal, square or rectangular waves d) All of the mentioned View Answer Answer: d Explanation: All of the given are uses of limiting and clamping circuits. 7. Which of the following is not true for a Schottky-Barrier Diode (SBD)? a) It is formed by bringing metal into contact with a moderately doped n-type semiconductor material. b) In the SBD, current is conducted by mainly by minority carriers. c) The forward voltage drop of a conducting SBD is lower than that of a pn-junction diode. d) SBD are used in the design of a special form of bipolar-transistor logic circuits View Answer Answer: d Explanation: Current is conducted largely by electrons which are majority current carriers. 8. Photodiodes are part of a) Optoelectronics b) High Intensity Discharge c) Low pressure Discharge d) None of the mentioned View Answer Answer: a Explanation: Optoelectronics are also called photodiodes. 9. Which of the following is true for LEDs a) The light emitted by a LED is inversely proportional to the current flowing through the diode b) LED operate in a manner opposite to the working of the optoelectronics or photodiodes c) LED cannot be used to generate coherent source of light

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d) None of the mentioned View Answer Answer: b Explanation: The working of the optoelectronics is opposite to that of the LEDs. 10. Optoisolator is a) A combination of LED and varactor in the same package b) A combination of a varactor and photodiode in the same package c) A combination of LED and a photodiode in the same package d) A combination of photodiode and Schottky-Barrier Diode in the same package View Answer Answer: c Explanation: By definition, an optoisolator is a package containing both LED as well as a photodiode.

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Questions & Answers (MCQs) focuses on “Diode Circuits”. (Q.1-Q.4) For the given circuits and input waveform determine the output waveform

194

1.

View Answer

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Answer: d Explanation: Diode is off when Vi < 5 and for Vi > 5, Vo = 5V.

2.

View Answer

196

Answer: c Explanation: Diode is off when Vi + 2 > 0.

3.

View Answer 197

Answer: d Explanation: For V1 < 4 the diode is on otherwise its off.

4.

198

View Answer Answer: c Explanation: During +ve half cycle when Vi < 8 both the diodes are off. For V1 > 8, D1 is on. Similarly for the negative half cycle. 5. The maximum load current that can be drawn is

a) 1.4 mA b) 2.3 mA c) 1.8 mA d) 2.5 mA View Answer Answer: a Explanation: At regulated poer supply, Is = 30-9/15k or 1.4 mA. The load current will remain less than this current.

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6. For the circuit shown diode cutting voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is (in kilo ohm)

a) 6.25 b) 12.5 c) 25 d) 30 View Answer Answer: a

Explanation: 7. The circuit inside the box in fig. P3.1.31. contains only resistor and diodes. The terminal voltage vo is connected to some point in the circuit inside the box. The largest and smallest possible value of vo most nearly to is respectively

a) 15 V, 6 V b) 24 V, 0 V c) 24 V, 6 V d) 15 V, 9 V View Answer Answer: d Explanation: The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage. 200

8. The Q-point of the zener diode in the circuit shown below is

a) (0.34 mA, 4 V) b) (0.34 mA, 4.93 V) c) (0.94 mA, 4 V) d) (0.94 mA, 4.93 V) View Answer Answer: a

Explanation:

5. Questions & Answers on Application of Diodes The section contains questions and answers on half-wave and full-wave rectifier, bridge and voltmeter rectifier, inductor and capacitor filters, l-section filters, clc filters and voltage regulation. Half-Wave Rectifier Full-wave Rectifier Bridge Rectifier Inductor Filters Capacitor Filters 201

L-Section Filter CLC Filter Voltage Regulation Using Zener Diode The Rectifier Voltmeter

Questions & Answers (MCQs) focuses on “Half-Wave Rectifier”. 1. The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________ a) Vm/√2RL b) Vm/(RF+RL)π c) 2Vm/√π 202

d) Vm/RL View Answer Answer: b Explanation: For a half wave rectifier, the IDC=IAVG=Im/π I= Vmsinωt/(RF+RL)=Imsinωt Im =Vm/ RF+RL So, IDC=Im/π=Vm/(RF+RL). 2. The below figure arrives to a conclusion that _________

a) for Vi > 0, V0=-(R2/R1)Vi b) for Vi > 0, V0=0 c) Vi < 0, V0=-(R2/R1)Vi d) Vi < 0, V0=0 View Answer Answer: b Explanation: The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now negative. So, the diode 1 is reverse biased and diode 2 is forward biased. Then output is clearly zero. 3. What is the output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (Di=0)

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a) 0 b) -Vi c) Vi d) 2Vi View Answer Answer: c Explanation: When the input of the inverted mode op-amp is positive, the output is negative. The diode is reverse biased. The input appears at the output. 4. In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be? a) 3.99V b) 2.5V c) 5.97V d) 6.77V View Answer Answer: b Explanation: The standard form of a sine wave is Vmsinωt. BY comparing the given information with this equation, Vm =50. Power=Vm2/RL=50*50/1000=2.5V. 5. In a half wave rectifier, the sine wave input is 200sin300t. The average value of output voltage is? a) 57.876V b) 67.453V c) 63.694V d) 76.987V View Answer

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Answer: c Explanation: Comparing with the standard equation, Vm=200V. Average value is given by, Vavg=Vm/π. So, 200/π=63.694. 6. Efficiency of a half wave rectifier is a) 50% b) 60% c) 40.6% d) 46% View Answer Answer: c Explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. For half wave it’s 40.6%. It’s given by, Vout/Vin*100. 7. If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be? a) 5V b) 4.9V c) 4.3V d) 6.7V View Answer Answer: c Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=Vm-Vd=5-0.7=4.3V. 8. Transformer utilisation factor of a half wave rectifier is _________ a) 0.234 b) 0.279 c) 0.287 d) 0.453 View Answer Answer: c Explanation: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.

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9. If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________ a) 150Hz b) 200Hz c) 100Hz d) 300Hz View Answer Answer: c Explanation: The ripple frequency of the output and input is same. This is because, one half cycle of input is passed and other half cycle is seized. So, effectively the frequency is the same. 10. Ripple factor of a half wave rectifier is_________(Im is the peak current and RL is load resistance) a) 1.414 b) 1.21 c) 1.4 d) 0.48 View Answer Answer: b Explanation: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.

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Questions & Answers (MCQs) focuses on “Full-wave Rectifier”. 1. Efficiency of a centre tapped full wave rectifier is _________ a) 50% b) 46% c) 70% d) 81.2% View Answer Answer: d Explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. It’s usually expressed in percentage. For centre tapped full wave rectifier, it’s 81.2%. 2. A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage? a) 0.562V b) 0.785V c) 0.954V d) 0.344V View Answer Answer: c Explanation: The ripple voltage is (Vϒ)RMS=ϒVDC /100. VDC=0.636*VRMS* √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482. Hence, (Vϒ)RMS=0.482*198 /100=0.954V. 3. A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load. 207

a) 2V b) 5V c) 4V d) 1V View Answer Answer: a Explanation: We know that, PDC=VDC2/RL. So, VDC=(PDC*RL)1/2=100001/2=100V. Here, ϒ=0.02 ϒ=VAC/VDC=VAC/100.So, VAC=0.02*100=2V. 4. A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input. a) 368.98mW b) 275.2mW c) 145.76mW d) 456.78mW View Answer Answer: b Explanation: The AC power input PIN=IRMS2(RF+Rr). IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA So, PIN=(13.5*10-3)2*(1500+10)=275.2mW. 5. In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________

a) 54% b) 48% c) 26% d) 81% View Answer Answer: b Explanation: The ripple factor ϒ= [(IRMS/IAVG)2 – 1]1/2. IRMS =Im 208

/√2=Vm/(Rf+RL)√2=200/1.01=198. (Secondary line to line voltage is 800/2=400. Due to centre tap Vm=400/2=200) IRMS=198/√2=140mA, IAVG=2*198/π=126mA. ϒ=[(140/126)2-1]1/2=0.48. So, ϒ=48%. 6. If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________ a) 100Hz b) 50Hz c) 25Hz d) 500Hz View Answer Answer: a Explanation: In the output of the centre tapped rectifier, one of the half cycle is repeated. The frequency will be twice as that of input frequency. So, it’s 100Hz. 7. Transformer utilization factor of a centre tapped full wave rectifier is_________ a) 0.623 b) 0.678 c) 0.693 d) 0.625 View Answer Answer: c Explanation: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.693. 8. In the circuits given below, the correct full wave rectifier is _________

a)

209

b)

c)

d) View Answer Answer: c Explanation: When the input is applied, a full wave rectifier should have a current flow. The 210

flow should be in the same direction for both positive and negative half cycles. Only the third circuit satisfies the above condition. 9. If the peak voltage on a centre tapped full wave rectifier circuit is 5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is_________ a) 4.3V b) 9.3V c) 5.7V d) 10.7V View Answer Answer: b Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=2Vm-Vd = 10-0.7 = 9.3V. 10. In a centre tapped full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be _________ a) 50Hz b) 100Hz c) 25Hz d) 200Hz View Answer Answer: b Explanation: The equation of sine wave is in the form Vmsinωt. So, by comparing we get ω=100. Frequency, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, fout = 100Hz. Questions & Answers (MCQs) focuses on “Bridge Rectifier”. 1. DC average current of a bridge full wave rectifier (where Im is the maximum peak current of input). a) 2Im b) Im c) Im/2 d) 1.414Im View Answer Answer: b Explanation: Average DC current of half wave rectifier is Im. Since output of half wave rectifier 211

contains only one half of the input. The average value is the half of the area of one half cycle of sine wave with peak Im. This is equal to Im. 2. DC power output of bridge full wave rectifier is equal to (Im is the peak current and RL is the load resistance). a) 2 Im2RL b) 4 Im2RL c) Im2RL d) Im2 RL/2 View Answer Answer: b Explanation: DC output power is the power output of the rectifier. We know VDC for a bridge rectifier is 2Vm and IDC for a bridge rectifier is 2Im. We also know VDC=IDC/RL. Hence output power is 4Im2RL. 3. Ripple factor of bridge full wave rectifier is? a) 1.414 b) 1.212 c) 0.482 d) 1.321 View Answer Answer: c Explanation: Ripple factor of a rectifier measures the ripples or AC content in the output. It’s obtained by dividing AC rms output with DC output. For full wave bridge rectifier it is 0.482. 4. If input frequency is 50Hz then ripple frequency of bridge full wave rectifier will be equal to_________ a) 200Hz b) 50Hz c) 45Hz d) 100Hz View Answer Answer: d Explanation: Since in the output of bridge rectifier one half cycle is repeated, the frequency will be twice as that of input frequency. So, f=100Hz. 5. Transformer utilization factor of bridge full wave rectifier _________ a) 0.623 b) 0.812 212

c) 0.693 d) 0.825 View Answer Answer: b Explanation: Transformer utilization factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For bridge full wave rectifier it’s equal to 0.693. 6. If peak voltage on a bridge full wave rectifier circuit is 5V and diode cut in voltage os 0.7, then the peak inverse voltage on diode will be_________ a) 4.3V b) 9.3V c) 8.6V d) 3.6V View Answer Answer: d Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the circuit. If PIV rating of diode is less than this value breakdown of diode may occur.. Therefore, PIV rating of diode should be greater than PIV in the circuit, For bridge rectifier PIV is Vm-VD = 5-1.4=3.6. 7. Efficiency of bridge full wave rectifier is_________ a) 81.2% b) 50% c) 40.6% d) 45.33% View Answer Answer: a Explanation: It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. Efficiency of a rectifier is the effectiveness of rectifier to convert AC to DC. It’s usually expressed inn percentage. For bridge full wave rectifier, it’s 81.2%. 8. In a bridge full wave rectifier, the input sine wave is 40sin100t. The average output voltage is_________ a) 22.73V b) 16.93V c) 25.47V d) 33.23V View Answer 213

Answer: c Explanation: The equation of sine wave is in the form Emsinωt. Therefore, Em=40. Hence output voltage is 2Em=80V. 9. Number of diodes used in a full wave bridge rectifier is_________ a) 1 b) 2 c) 3 d) 4 View Answer Answer: d Explanation: The model of a bridge rectifier is same as Wein Bridge. It needs 4 resistors. Bridge rectifier needs 4 diodes while centre tap configuration requires only one. 10. In a bridge full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be_________ a) 50Hz b) 200Hz c) 100Hz d) 25Hz View Answer Answer: c Explanation: The equation of sine wave is in the form of Emsinωt. So, ω=100 and frequency (f)=ω/2=50Hz. Since output of bridge rectifier have double the frequency of input, f=100Hz.

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Questions & Answers (MCQs) focuses on “Inductor Filters”. 1. What is the effect of an inductor filter on a multi frequency signal? a) Dampens the AC signal b) Dampens the DC signal c) To reduce ripples d) To change the current View Answer Answer: a Explanation: Presence of inductor usually dampens the AC signal. Due to self induction induces opposing EMF or changes in the current. 2. The ripple factor (ϒ) of inductor filter is_________ a) ϒ = RZ3/√2ωL b) ϒ = RZ/3√2ωL c) ϒ = RZ3√2/ωL d) ϒ = RZ3/√2ωL View Answer Answer: b Explanation: Ripple factor will decrease when L is increased and RL. Inductor has a higher dc resistance. It depends on property of opposing the change of direction of current. 3. The inductor filter gives a smooth output because_________ a) It offers infinite resistance to ac components b) It offers infinite resistance to dc components c) Pulsating dc signal is allowed d) The ac signal is amplified View Answer Answer: a Explanation: The inductor does not allow the ac components to pass through the filter. The main purpose of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output. 4. A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the dc load current. a) 0.7mA b) 17mA c) 10.6mA 215

d) 20mA View Answer Answer: c Explanation: For a rectifier with an inductor filter, VDC=2Vm/π, Idc=VDC/RL=2Vm/RLπ IDC=2*250/(3.14*15*103)=10.6mA. 5. The output of a rectifier is pulsating because_________ a) It has a pulse variations b) It gives a dc output c) It contains both dc and ac components d) It gives only ac components View Answer Answer: c Explanation: For any electronic devices, a steady dc output is required. The filter is used for this purpose. The ac components are removed by using a filter. 6. A dc voltage of 380V with a peak ripple voltage not exceeding 7V is required to supply a 500Ω load. Find out the inductance required. a) 10.8henry b) 30.7henry c) 28.8henry d) 15.4henry View Answer Answer: c Explanation: Given the ripple voltage is 7V. So, 7=1.414VRMS ϒ=VRMS/VDC=4.95/380=0.0130. ϒ=1/3√2(RL/Lω) So, L=28.8henry. 7. The inductor filter should be used when RL is consistently small because_________ a) Effective filtering takes place when load current is high b) Effective filtering takes place when load current is low c) Current lags behind voltage d) Current leads voltage View Answer Answer: a Explanation: When RLis infinite, the ripple factor is 0.471. This value is close to that of a rectifier. So, the resistance should be small. 216

8. The output voltage VDC for a rectifier with inductor filter is given by_________ a) (2Vm/π)-IDCR b) (2Vm/π)+IDCR c) (2Vmπ)-IDCR d) (2Vmπ)+IDCR View Answer Answer: a Explanation: The inductor with high resistance can cause poor voltage regulation. The choke resistance, the resistance of half of transformer secondary is not negligible. 9. What causes to decrease the sudden rise in the current for a rectifier? a) the electrical energy b) The ripple factor c) The magnetic energy d) Infinite resistance View Answer Answer: c Explanation: When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also helps to prevent the current to fall down too much. 10. A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the ripple factor (ϒ). a) 0.1 b) 0.6 c) 0.5 d) 0.4 View Answer Answer: d Explanation: ϒ=IAC/IDC, IAC=2√2Vm/3π(RL2+4ω2L2)1/2 By putting the values, IAC=4.24Ma. VDC=2Vm/π, IDC=VDC/RL=2Vm/RL π IDC=2*250/(3.14*15*103)=10.6mA. ϒ=4.24/10.6=0.4.

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Questions & Answers (MCQs) focuses on “Capacitor Filters”. 1. In a shunt capacitor filter, the mechanism that helps the removal of ripples is_________ a) The current passing through the capacitor b) The property of capacitor to store electrical energy c) The voltage variations produced by shunting the capacitor d) Uniform charge flow through the rectifier View Answer Answer: b Explanation: Filtering is frequently done by shunting the load with capacitor. It depends on the fact that a capacitor stores energy when conducting and delivers energy during non conduction. Throughout this process, the ripples are eliminated. 2. The cut-in point of a capacitor filter is_________ a) The instant at which the conduction starts b) The instant at which the conduction stops c) The time after which the output is not filtered d) The time during which the output is perfectly filtered View Answer Answer: a Explanation: The capacitor charges when the diode is in ON state and discharges during the OFF state of the diode. The instant at which the conduction starts is called cut-in point. The instant at which the conduction stops is called cut-out point. 3. The rectifier current is a short duration pulses which cause the diode to act as a_________ a) Voltage regulator b) Mixer c) Switch 218

d) Oscillator View Answer Answer: c Explanation: The diode permits charge to flow in capacitor when the transformer voltage exceeds the capacitor voltage. It disconnects the power source when the transformer voltage falls below that of a capacitor. 4. A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5? a) 114.87Ω b) 167.98Ω c) 115.47Ω d) 451.35Ω View Answer Answer: c Explanation: For a half wave filter, ϒ=1/2√3 fCRL=1/2√3*50*10-3*RL RL=103/5√3=115.47Ω. 5. A 100µF capacitor when used as a filter has 15V ac across it with a load resistor of 2.5KΩ. If the filter is the full wave and supply frequency is 50Hz, what is the percentage of ripple frequency in the output? a) 2.456% b) 1.154% c) 3.785% d) 3.675% View Answer Answer: b Explanation: For a full wave rectifier, ϒ=1/4√3 fCRL =1/4√3*50*10-3*2.5 =0.01154. So, ripple is 1.154%. 6. A full wave rectifier uses a capacitor filter with 500µF capacitor and provides a load current of 200mA at 8% ripple. Calculate the dc voltage. a) 15.56V b) 20.43V c) 11.98V

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d) 14.43V View Answer Answer: d Explanation: The ripple factor ϒ=IL/ 4√3 fCVDC VDC=200*10-3/ 4√3 *50*500*8 =14.43. 7. The charge (q) lost by the capacitor during the discharge time for shunt capacitor filter. a) IDC*T b) IDC/T c) IDC*2T d) IDC/2T View Answer Answer: a Explanation: The ‘T’ is the total non conducting time of capacitor. The charge per unit time will give the current flow. 8. Which of the following are true about capacitor filter? a) It is also called as capacitor output filter b) It is electrolytic c) It is connected in parallel to load d) It helps in storing the magnetic energy View Answer Answer: b Explanation: The rectifier may be full wave or half wave. The capacitors are usually electrolytic even though they are large in size. 9. The rms ripple voltage (Vrms) of a shunt filter is_________ a) IDC/2√3 b) IDC2√3 c) IDC/√3 d) IDC√3 View Answer Answer: a Explanation: The ripple waveform will be triangular in nature. The rms value of this wave is independent of slopes or lengths of straight lines. It depends only on the peak value.

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10. A shunt capacitor of value 500µF fed rectifier circuit. The dc voltage is 14.43V. The dc current flowing is 200mA. It is operating at a frequency of 50Hz. What will be the value of peak rectified voltage? a) 18.67V b) 16.43V c) 15.98V d) 11.43V View Answer Answer: b Explanation: We know, Vm=Vdc+Idc/4fC =14.43+ {200/ (200*500)} 103 =14.43+2=16.43V.

Questions & Answers (MCQs) focuses on “L-Section Filter”. 1. An L section filter with L=2henry and C= 49µF is used in the output of a full wave single phase rectifier that is fed from a 40-0-40 V peak transformer. The load current is 0.2A. Calculate the output dc voltage. a) 20.76V b) 24.46V c) 34.78V d) 12.67V View Answer Answer: b Explanation: Given, VL= 40V. VDC=2/π*VL=2/π*40=25.46V. 221

2. Calculate LC for a full wave rectifier which provides 10V dc at 100mA with a maximum ripple of 2%. Input ac frequency is 50Hz. a) 40*10-6 b) 10*10-6 c) 30*10-6 d) 90*10-6 View Answer Answer: a Explanation: LC=1/6√2ω2ϒ ω=2πf=314 By putting the values, LC=1/6√2(314)2 0.02=40*10-6. 3. The value of inductance at which the current in a choke filter does not fall to zero is_________ a) peak inductance b) cut-in inductance c) critical inductance d) damping inductance View Answer Answer: c Explanation: When the value of inductance is increased, a value is reached where the diodes supplies current continuously. This value of inductance is called critical inductance. 4. The condition for the regulation curve in a choke filter is_________ a) LC≥RL/3ω b) LC≤RL/3ω c) L≥RL/3ω d) LC≥RL3ω View Answer Answer: a Explanation: IDC should not exceed the negative peak of ac component. So, the regulation curve in direct output voltage against load current for a filter is given the relation LC≥R L/3ω. 5. The ripple factor for an l section filter is_______ a) ϒ= 1/6√2ω2LC b) ϒ= 6√2ω2LC c) ϒ= 6√3ω2LC d) ϒ= 1/6√3ω2LC View Answer 222

Answer: a Explanation: The ripple factor is the ratio of root mean square (rms) value of ripple voltage to absolute value of dc component. It can also be expressed as the peak to peak value. 6. The output dc voltage of an LC filter is_______ a) VDC=2Vm/π + IDCR b) VDC=Vm/π – IDCR c) VDC=2Vm/π – 2IDCR d) VDC=2Vm/π – IDCR View Answer Answer: d Explanation: The value for VDC is same as that of inductor filter. If inductor has no dc resistance, then VDC=2Vm/π. If R is the series resistance of transformer, then VDC=2Vm/π – IDCR. 7. The rms value of ripple current for an L section filter is_______ a) IRMS=√2/3*XL*VDC b) IRMS=√2/3*XL*VDC c) IRMS=√2/3*XL *VDC d) IRMS=√2/3*XL*VDC View Answer Answer: a Explanation: The ac current through L is determined primarily by XL=2ωL. It is directly proportional to voltage produced and indirectly proportional to the reactance. 8. What makes the load in a choke filter to bypass harmonic components? a) capacitor b) inductor c) resistor d) diodes View Answer Answer: a Explanation: When the capacitor is shunted across the load, it bypasses the harmonic components. This is because it offers low reactance to ac ripple component. In another hand the inductor offers high impedance to harmonic terms. 9. The ripple to heavy loads by a capacitor is_______ a) high b) depends on temperature c) low 223

d) no ripple at all View Answer Answer: c Explanation: Ripple factor is inversely proportional to load resistance for a capacitor filter. So, the ripples that are produced are low when the load is high. 10. In a choke l section filter_______ a) the inductor and capacitor are connected across the load b) the inductor is connected in series and capacitor is connected across the load c) the inductor is connected across and capacitor is connected in series to the load d) the inductor and capacitor are connected in series View Answer Answer: b Explanation: The choke filter is sometimes also called as L section filter because the inductor and capacitor are connected in the shape ’L’ or inverted manner. But several sections of this choke filter are employed to smooth the output curve and make it free from ripples.

Questions & Answers (MCQs) focuses on “CLC Filter”. 224

1. What is the number of capacitors and inductors used in a CLC filter? a) 1, 2 respectively b) 2, 1 respectively c) 1, 1 respectively d) 2, 2 respectively View Answer Answer: b Explanation: A very smooth output can be obtained by a filter consisting of one inductor and two capacitors connected across each other. They are arranged in the form of letter ‘pi’. So, these are also called as pi filters. 2. Major part of the filtering is done by the first capacitor in a CLC filter because _________ a) The capacitor offers a very low reactance to the ripple frequency b) The capacitor offers a very high reactance to the ripple frequency c) The inductor offers a very low reactance to the ripple frequency d) The inductor offers a very high reactance to the ripple frequency View Answer Answer: a Explanation: The CLC filters are used when high voltage and low ripple frequency is needed than L section filters. The capacitor in a CLC filter offers very low reactance to the ripple frequency. So, maximum of the filtering is done by the first capacitor across the L section part. 3. At f=50Hz, the ripple factor of CLC filter is_________ a) ϒ=5700RL / (LC1C2) b) ϒ=5700/ (LC1C2RL) c) ϒ=5700LC1/ (C2RL) d) ϒ=5700C1C2/ (LRL) View Answer Answer: b Explanation: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 5700 / (LC1C2RL) at 50Hz. 4. A single phase full wave rectifier makes use of pi section filter with 10µF capacitors and a choke of 10henry. The secondary voltage is 280V and the load current is 100mA. Determine the dc output voltage when f=50Hz. a) 345V b) 521V c) 243V 225

d) 346V View Answer Answer: d Explanation: Given, VRMS=280V So, V¬m = 1.414*280=396V. From theory of capacitor filter, VDC = Vm –IDC/4fC=396-0.1/ (4*50*10*10-6)=346V. 5. For a given CLC filter, the operating frequency is 50Hz and 10µF capacitors used. The load resistance is 3460Ω with an inductance of 10henry. Calculate the ripple factor. a) 0.165% b) 0.142% c) 0.178% d) 0.321% View Answer Answer: a Explanation: We have, ϒ=5700 / (LC1C2RL) =5700 / (10*100*10-12*3460) =0.165%. 6. The inductor is placed in the L section filter because_________ a) It offers zero resistance to DC component b) It offers infinite resistance to DC component c) It bypasses the DC component d) It bypasses the AC component View Answer Answer: a Explanation: The inductor offers high reactance to ac component and zero resistance to dc component. So, it blocks the ac component which cannot be bypassed by the capacitors. 7. The voltage in case of a full wave rectifier in a CLC filter is_________ a) Vϒ = IDC/2fC b) Vϒ = IDC fC c) Vϒ = IDC/fC d) Vϒ = 2IDCfC View Answer Answer: a Explanation: T he filter circuit is a combination of capacitors and inductors. The RMS value depends on the peak value of charging and discharging magnitude, VPEAK. 226

8. The advantages of a pi-filter is_________ a) low output voltage b) low PIV c) low ripple factor d) high voltage regulation View Answer Answer: c Explanation: Due to the use of two capacitors with an inductor, an improved filtering action is provided. This leads to decrement in ripple factor. A low ripple factor signifies regulated and ripple free DC voltage. 9. What is the relation between time constant and load resistance? a) They don’t depend on each other b) They are directly proportional c) They are inversely proportional d) Cannot be predicted View Answer Answer: c Explanation: If the load resistance value is large, the discharge time constant will be of a high value. Thus the capacitors time to discharge will get over soon. This lowers the amount of ripples in the output and increases the output voltage. If the load resistance is small, the discharge time constant will be more with decrease in output voltage. 10. The output waveform of CLC filter is superimposed by a waveform referred to as_________ a) Square wave b) Triangular wave c) Saw tooth wave d) Sine wave View Answer Answer: c Explanation: Since the rectifier conducts current only in the forward direction, any energy discharged by the capacitor will flow into the load. This result in a DC voltage upon which is superimposed a waveform referred to as a saw tooth wave.

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Questions and Answers for Entrance exams focuses on “Voltage Regulation Using Zener Diode”. 1. The percentage voltage regulation (VL) is given by_________ a) (VNL-VL)/VNL*100 b) (VNL+VL)/VNL*100 c) (VNL-VL)/VL*100 d) (VNL+VL)/VL*100 View Answer Answer: a Explanation: The change in the output voltage from no load to full load condition is called as voltage regulation, where VNL is the voltage at no load condition. It is used to maintain a nearly constant output voltage. If the regulation is high, the output voltage is stable. 2. The limiting value of the current resistor used in a Zener diode (when used as a regulator) a) (R)min=[(Vin)max + VZ/R b) (R)min=[(Vin)max-VZ]/R c) (R)min=[(Vin)max-VZ]R d) (R)min=[(Vin)max+ VZ]R View Answer Answer: b Explanation: When the input voltage is maximum, the load current is minimum, the Zener current should not increase the maximum rated value. Therefore there should be a minimum value of resistor. 3. When the regulation by a Zener diode is with a varying input voltage, what happens to the voltage drop across the resistance? a) Decreases b) Has no effect on voltage 228

c) Increases d) The variations depend on temperature View Answer Answer: c Explanation: When the input voltage varies, the input current also varies. This makes more current to flow in the diode. This increase in the current should balance a change in the load current. Hence the voltage drop increases across the resistor. 4. In the given limiter circuit, an input voltage Vi=10sin100πt is applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______

a) 6.1V,-0.7V b) 0.7V,-7.5V c) 7.5V,-0.7V d) 7.5V,-7.5V View Answer Answer: c Explanation: With VI= 10V when maximum, D1 is forward biased, D2 is reverse biased. Zener is in breakdown region. VOMAX=sum of breakdown voltage and diode drop=6.8+0.7=7.5V. VOMIN=negative of voltage drop=-0.7V. There will be no breakdown voltage here. 5. Determine the maximum and minimum values of load current for which the Zener diode shunt regulator will maintain regulation when VIN=24V and R=500Ω. The Zener diode has a VZ=12V and (IZ)MAX=90mA. a) 40mA, 0mA respectively b) 36mA, 5mA respectively c) 10mA, 6mA respectively d) 21mA, 0mA respectively View Answer

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Answer: d Explanation: The current through the resistance R is given by, I=(VIN-VZ)/R= (2412)/500=24mA. (IL)MAX=I-(IZ)MIN=24-3=21mA .This current is less than (IZ)MAX. So, we assume that all the input current flows through the Zener diode. Under this condition, (IL)MIN is 0mA. 6. Determine the minimum value of load resistance that can be used in the circuit with (IZ)Min=3mA. The input voltage is 10V and the resistance R is 500Ω. The Zener diode has a VZ=6V 0and (IZ)MAX=90mA. a) 1KΩ b) 2.4KΩ c) 1.2KΩ d) 3.6KΩ View Answer Answer: c Explanation: The I=(VIN-VZ)/R=(10-6)/500=8mA. (IL)MAX=I-(IZ)MIN=8-3=5mA. (RL)MIN=VZ/(IL)MAX=6/5m=1.2KΩ. 7. A Zener regulator has to handle supply voltage variation from 19.5V to 22.5V. Find the magnitude of regulating resistance, if the load resistance is 6KΩ. The Zener diode has the following specifications: breakdown voltage =18V, (IZ)Min=2µA, maximum power dissipation=60mW and Zener resistance =20Ω. a) 0 < R < 500Ω b) 77.8 < R < 500Ω c) 77.8 < R < 100Ω d) 18 < R < 500Ω View Answer Answer: b Explanation: (PZ)MAX/rZ=(IZ)MAX2 . So, (IZ)MAX =60m/20=54.8µA. IL=VO/RL=18/6000=3mA. RMAX=(VMin-VZ)/[( IZ)Min+( IL)MAX]=(19.5-18)/(2µ+3m)=500Ω. RMin=(VMAX-VZ)/[( IZ)MAX+( IL)Min]=(22.5-18)/(54.8m+3m)=77.8Ω. 8. A transistor series regulator has the following specifications: VIN=15V, VZ=8.3V, β=100, R=1.8KΩ, RL=2KΩ. What will be the Zener current in the regulator circuit? a) 4.56mA b) 3.26mA c) 4.56mA d) 3.68mA View Answer

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Answer: d Explanation: We know, VO=VZ-VBE=8.3-0.7=7.6V. VCE=VIN-V0=15-7.6=7.4V. So, IR=(VINVZ)/R=(15-8.3)/1.8m=3.72mA. IL=VO/RL=7.6/2000=3.8mA. IB=IL/ β=3.8mA/100=0.038mA. Finally, IZ=IR-IB=3.72-0.038=3.682mA. 9. When is a regulator used? a) when there are small variations in load current and input voltage b) when there are large variations in load current and input voltage c) when there are no variations in load current and input voltage d) when there are small variations in load current and large variations in input voltage View Answer Answer: a Explanation: The regulator has following limitations: 1.It has low efficiency for heavy load currents 2. The output voltage changes slightly due to Zener impedance. Hence, it is used when there are small variations in load current and input voltage. 10. A transistor in a series voltage regulator acts like a variable resistor. The value of its resistance is determined by _______ a) emitter current b) base current c) collector current d) it is not controlled by the transistor terminals View Answer Answer: b Explanation: The principle is based on the fact that a large fraction of the increase in input voltage appears across the transistor so that the output voltage remains to be constant. When input voltage is increased, the output voltage also increases which biases the transistor towards less current.

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6. Questions on Transistor Characteristics The section contains questions on transistor amplifier, transistor construction, junction transistor, common base, emitter and collector configuration, ce and cb characteristics, dc load lines, transistor switch and switching times. The Junction Transistor The Transistor as an Amplifier Transistor Construction The Common Base Configuration The Common Emitter Configuration The Common Collector Configuration

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The CE Characteristics The CB Characteristics DC Load Lines Transistor as a Switch Transistor Switching Times

Questions & Answers (MCQs) focuses on “The Junction Transistor”. 1. The advantages over the vacuum triode for a junction transistor is_________ a) high power consumption b) high efficiency c) large size d) less doping View Answer

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Answer: b Explanation: A junction transistor is an analogous to a vacuum triode. The main difference between them is that a transistor is a current device while a vacuum triode is a voltage device. The advantages of a transistor over a vacuum triode are long life, high efficiency, light weight, smaller in size, less power consumption. 2. What is the left hand section of a junction transistor called? a) base b) collector c) emitter d) depletion region View Answer Answer: c Explanation: The main function of this section is to supply majority charge carriers to the base. Hence it is more heavily doped in comparison to other regions. This forms the left hand section of the transistor. 3. In an NPN transistor, the arrow is pointed towards_________ a) the collector b) the base c) depends on the configuration d) the emitter View Answer Answer: d Explanation: As regards to the symbols, the arrow head is always at the emitter. The direction indicates the conventional direction of current flow. In case of PNP transistor, it is from base to emitter. 4. Which of the following is true in construction of a transistor? a) the collector dissipates lesser power b) the emitter supplies minority carriers c) the collector is made physically larger than the emitter region d) the collector collects minority charge carriers View Answer Answer: c Explanation: In most of the transistors, the collector is made larger than emitter region. This is due to the fact that collector has to dissipate much greater power. The collector and emitter cannot be interchanged.

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5. In the operation of an NPN transistor, the electrons cross which region? a) emitter region b) the region where there is high depletion c) the region where there is low depletion d) P type base region View Answer Answer: d Explanation: The electrons in the emitter region are repelled by the negative terminal of the battery towards the emitter junction. The potential barrier at the junction is reduced due to forward bias and base region is very thin and lightly doped, electrons cross the P type base region. 6. Which of the following are true for a PNP transistor? a) the emitter current is less than the collector current b) the collector current is less than the emitter current c) the electrons are majority charge carriers d) the holes are the minority charge carriers View Answer Answer: b Explanation: The 2 – 5% of holes is lost in recombination with electrons in the base region. The majority charge carriers are holes for a PNP transistor. Thus the collector current is slightly less than the emitter current. 7. In the saturated region, the transistor acts like a_________ a) poor transistor b) amplifier c) open switch d) closed switch View Answer Answer: d Explanation: In saturated mode, both emitter and collector are forward biased. The negative of the battery is connected to emitter and similarly the positive terminals of batteries are connected to the base. The transistor now acts like a closed switch. 8. When does the transistor act like an open switch? a) cut off region b) inverted region c) saturated region

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d) active region View Answer Answer: a Explanation: In cut off region, both the junctions are reverse biased. The transistor has practically zero current because the emitter does not emit charge carriers to the base. So, the transistor acts as open switch. 9. If the emitter-base junction is forward biased and the collector-base junction is reverse biased, what will be the region of operation for a transistor? a) cut off region b) saturated region c) inverted region d) active region View Answer Answer: d Explanation: When the emitter-base junction is forward biased and the collector-base junction is reverse biased, the transistor is used for amplification. A battery is connected to collector base circuit. The positive terminal is connected to the collector while the negative is connected to the base. 10. The transfer of a signal in a transistor is_________ a) low to high resistance b) high to low resistance c) collector to base junction d) emitter to base junction View Answer Answer: a Explanation: A forward biased emitter base junction has a low resistance path. A reversed biased junction has a high resistance path. The weak signal is introduced in a low resistance circuit and the output is taken from the high resistance circuit.

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Questions and Answers for Aptitude test focuses on “The Transistor as an Amplifier”. 1. The emitter current consist of_________ a) electrons passing from collector to emitter b) holes crossing from base to collector c) electron current Ine constituted by electrons d) immobile charge carriers View Answer Answer: c Explanation: The emitter current consists of two parts. It consists of hole current IpE constituted by holes. The other part is that it consists the electron current InE constituted by electrons. 2. The total emitter current (IE) is given by_________ a) IE = IpE * InE b) IE = IpE – InE c) IE = IpE / InE d) IE = IpE + InE View Answer Answer: d Explanation: The total emitter current is the sum of InE and IpE. In commercial transistors, the doping of emitter region is made much heavier than base. Hence current by majority charge carriers InE is negligible when compared to current by minority charge carriers IpE. 3. A common base transistor amplifier has an input resistance of 20Ω and output resistance of 100kΩ. If a signal of 400mV is applied between emitter and base, find the voltage amplification. Assume αac to be one. a) 20 b) 50 c) 30 d) 25 View Answer Answer: b Explanation: IE = V/R=400M/20=20mA 237

IC=αIE= 1*20mA=20mA. VO=IC*RL=20m*1k=20V Amplification, A= VO/signal voltage=20V/400m=50. 4. The amplification factor for a transistor is given by_________ a) A=αRL/re b) A=αRLre c) A=re/ αRL d) A=RL/reα View Answer Answer: a Explanation: One of the most important application of a transistor is an amplifier. A small change in signal voltage produces an appreciable change in emitter current because the input circuit has low resistance (α=∆IC/IE). 5. Why is the silicon mostly chosen when compared to germanium? a) low power consumption b) high efficiency c) greater working temperature d) large ICBO View Answer Answer: c Explanation: The normal working temperature of germanium is approximately 70°C .The normal working temperature of silicon is approximately 150°C. The other advantages of using a silicon material are, it has a smaller ICBO and its variations are smaller with temperature. 6. The change in output voltage across the load resistor for a transistor during amplification is_________ a) RL *α*∆IE b) RL *∆IE/α c) RL *α2*∆IE d) RL *α1/2*∆IE View Answer Answer: a Explanation: A small change of voltage ∆Vi between emitter and base causes a relatively large emitter current change ∆IE. We define by the symbol α that fraction of this current change which is collected and passes through RL. 7. A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc? a) 0.565 238

b) 0.754 c) 1.24 d) 0.995 View Answer Answer: d Explanation: Emitter current IE=IC+IB=100+0.5=100.5mA. αdc=IC/IE=100/100.5=0.995. 8. A germanium transistor used as an amplifier has a collector cut off current ICBO=10µA at a temperature 27°C and β=50. What is the collector current when the base current is 0.25mA? a) 10.76mA b) 13.01mA c) 15.67mA d) 11.88Ma View Answer Answer: b Explanation: IC=βIB+(1+β)ICBO IC=50*0.25/1000+51/100000=13.01mA. 9. In a PNP germanium transistor, the cut in voltage is about_________ a) -0.1V b) -0.01V c) -0.05V d) -0.07V View Answer Answer: a Explanation: The cut in voltage of germanium is lower than that of silicon. If both germanium and silicon are in parallel, Ge starts conducting earlier and stops silicon from conducting. 10. In a PNP transistor operating in active region, the main stream of current is_________ a) drift of holes b) drift of electrons c) diffusion of holes d) diffusion of electrons View Answer Answer: c Explanation: The emitter-base junction is forward biased while collector-base junction is

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reversed biased. The transistor now operates in active region. Here, it can be used for amplification purpose.

Questions & Answers (MCQs) focuses on “Transistor Construction”. 1. Which gas is used to fill the chamber in the grown junction type transistor construction? a) helium b) boron c) nitrogen d) oxygen View Answer Answer: c Explanation: In the process of transistor construction, a crucible is placed in the chamber. This chamber consists of hydrogen or nitrogen. These gases help in the prevention of oxidation. It also contains purified Ge or Si at a temperature few degrees above its melting point. 2. In a grown junction type construction, the method used form a junction transistor is_________ a) alloy type diffusion b) mesa type c) speed variation method d) fused junction type View Answer

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Answer: c Explanation: The grown junction may be formed by suddenly varying the rate of pulling the seed crystal from the melt. This method is based on the fact that proportion in which N and P type impurities crystallise i.e.., enter the grown crystal depends on the rate of pulling. 3. Which of the following methods take impurity variation method for transistor construction? a) alloy type diffusion b) grown junction type c) epitaxial type d) mesa type View Answer Answer: b Explanation: In impurity variation method, the impurity content of the semiconductor is altered in its type as well as the quantity. For example, in making NPN germanium grown junction transistor, a small type of N type impurity is added to molten germanium and the crystal growth is started. 4. Which of the following is true about grown junction type construction? a) N type impurity is added to P type impurity b) Boron helps in the prevention of oxidation c) The seed is pulled to a large distance for a correct growth d) Slow pulling leads to the formation of P type crystal View Answer Answer: d Explanation: This method is based on the fact that proportion in which N and P type impurities crystallise i.e.., enter the grown crystal depends on the rate of pulling. If the pulling rate is small, a P type crystal is grown. If the pulling rate is fast, an N type crystal is grown. 5. What is the melting point of indium in alloy type transistors? a) 300°C b) 200°C c) 155°C d) 100°C View Answer Answer: c Explanation: This is similar to soldering and PNP transistor is generally is made by this process. In this method, first of all N type germanium is obtained. The N type wafer and indium dots are placed in a furnace and heated to about 500°C.

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6. The non rectifying base contact is made from_________ a) welding a strip b) germanium c) indium d) graphite View Answer Answer: a Explanation: Leads for emitter and collector are soldered to the dots making non rectifying contacts. Further, non rectifying base contact is usually made from a welding a strip or loop of gold plated wire to the base plate. 7. What is the thickness of wafer in the alloy type transistors? a) 1-2m inch b) 3-5m inch c) 5-6m inch d) 4-7m inch View Answer Answer: b Explanation: The wafer of crystal has a 3-5m inch thickness and 80m inch square. This is placed in a graphite jig with a dot of prepared indium. One dot of an indium is 3 times larger than the other. 8. The larger dot of the indium is used as_________ a) base b) emitter c) control pin d) collector View Answer Answer: d Explanation: The wafer is placed in a graphite jig with a dot of prepared indium. One dot of an indium is 3 times larger than the other. Finally the larger dot is used as collector. The smaller dot is used as emitter. 9. The electrical properties of a transistor in alloy type construction is determined by_________ a) space between the junctions in the wafer b) proportions of N and P type impurities c) the pulling rate of crystal d) uniformity of the crystal lattice View Answer 242

Answer: a Explanation: Large area collector junction helps in collecting most of the holes emitted from the emitter ensuring that the collector current almost equals the emitter current. The spacing between two junctions inside germanium wafer is very small and determines the electrical properties. 10. The grown junction type transistors is generally used for_________ a) PNP transistors b) NPN transistors c) Both transistors d) Depends on the material used View Answer Answer: b Explanation: Grown junction type transistors are manufactured through growing single large crystal which is slowly pulled from the melt in crystal growing furnace. This is generally used for NPN transistors.

Questions & Answers (MCQs) focuses on “The Common Base Configuration”. 1. The AC current gain in a common base configuration is_________ a) -∆IC/∆IE b) ∆IC/∆IE c) ∆IE/∆IC d) -∆IE/∆IC View Answer

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Answer: a Explanation: The AC current gain is denoted by αac. The ratio of change in collector current to the change in emitter current at constant collector base voltage is defined as current amplification factor. 2. The value of αac for all practical purposes, for commercial transistors range from_________ a) 0.5-0.6 b) 0.7-0.77 c) 0.8-0.88 d) 0.9-0.99 View Answer Answer: d Explanation: For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current. 3. A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc? a) 0.787 b) 0.995 c) 0.543 d) 0.659 View Answer Answer: b Explanation: Emitter current IE=IC+IB =100+0.5=100.5mA αdc=IC/IE=100/100.5=0.995. 4. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current. a) 0.01mA b) 0.07mA c) 0.02mA d) 0.05mA View Answer Answer: c Explanation: Here, IC=4.9/5K=0.98mA α = IC/IE .So, IE=IC/α=0.98/0.98=1mA. IB=IE-IC=1-0.98=0.02Ma.

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5. The emitter current IE in a transistor is 3mA. If the leakage current ICBO is 5µA and α=0.98, calculate the collector and base current. a) 3.64mA and 35µA b) 2.945mA and 55µA c) 3.64mA and 33µA d) 5.89mA and 65µA View Answer Answer: b Explanation: IC=αIE + ICBO =0.98*3+0.005=2.945mA. IE=IC+IB . So, IB=3-2.495=0.055mA=55µA. 6. Determine the value of emitter current and collector current of a transistor having α=0.98 and collector to base leakage current ICBO=4µA. The base current is 50µA. a) 1.5mA b) 3.7mA c) 2.7mA d) 4.5mA View Answer Answer: c Explanation: Given, IB=50µA=0.05mA ICBO=4µA=0.004Ma IC=α/(1- α)IB+1/(1- α)ICBO=2.45+0.2=2.65Ma IE=IC+IB=2.65+0.05=2.7mA. 7. The negative sign in the formula of amplification factor indicates_________ a) that IE flows into transistor while IC flows out it b) that IC flows into transistor while IE flows out it c) that IB flows into transistor while IC flows out it d) that IC flows into transistor while IB flows out it View Answer Answer: a Explanation: When no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. αdc=-IC/IE. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current. 8. The relation between α and β is _________ a) β=α/(1-α) 245

b) α= β/(1+β) c) β=α/(1+α) d) α= β/(1- β) View Answer Answer: b Explanation: β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β)ICBO. 9. A transistor has an IE of 0.9mA and amplification factor of 0.98. What will be the IC? a) 0.745mA b) 0.564mA c) 0.236mA d) 0.882mA View Answer Answer: d Explanation: Given, IE = 0.9mA, α=0.98 We know, α= IC/IE So, IC=0.98*0.9=0.882mA. 10. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current? a) 3mA and 55µA b) 2.945mA and 55µA c) 3.64mA and 33µA d) 5.89mA and 65µA View Answer Answer: a Explanation: (IC – ICBO)/α=IE = (2.945-0.002)/0.98=3mA. IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.

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Questions & Answers (MCQs) focuses on “The Common Emitter Configuration”. 1. The base current amplification factor β is given by_________ a) IC/IB b) IB/IC c) IE/IB d) IB/IE View Answer Answer: a Explanation: The current amplification factor (β) is given by IC//IB. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor. 2. In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO. a) 10µA b) 100µA c) 90µA d) 500µA View Answer Answer: b Explanation: IC=α IE +ICBO =0.995*10mA+0.5µA=9.9505mA. IB=IE-IC=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199 ICEO=9.9505-199*0.0495=0.1mA==100µA. 3. A germanium transistor with α=0.98 gives a reverse saturation current ICBO=10µA in a CB configuration. When it is used in CE configuration with a base current of 0.22µA, calculate the collector current. a) 0.9867mA b) 0.7654mA c) 0.51078mA d) 0.23456mA View Answer Answer: c Explanation: Given, ICBO=10µA, α=0.98 and IB =0.22µA. IC=α/ (1-α) IB+ 1/(1-α) ICBO 0.01078+0.5=0.51078mA.

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4. In CE configuration, if the voltage drop across 5kΩ resistor connected in the collector circuit is 5V. Find the value of IB when β=50. a) 0.01mA b) 0.25mA c) 0.03mA d) 0.02mA View Answer Answer: d Explanation: IC=V across RL/RL=5V/5KΩ=1mA. IB=IC/β=1/50=0.02mA. 5. A transistor is connected in CE configuration. Collector supply voltage Vcc=10V, RL=800Ω, voltage drop across RL=0.8V, α=0.96. What is base current? a) 41.97µA b) 56.78µA c) 67.67µA d) 78.54µA View Answer Answer: a Explanation: Here, IC=0.8/800=1mA β= α/ (1-α)=0.96/1-0.96=24. Now, IB=IC/ β=1/24=41.67µA. 6. The collector supply voltage for a CE configured transistor is 10V. The resistance R L=800Ω. The voltage drop across RL is 0.8V. Find the value of collector emitter voltage. a) 3.7V b) 9.2V c) 6.5V d) 9.8V View Answer Answer: b Explanation: Here, IC=0.8/800=1mA. We know, VCE=VCC-ICRL =10-0.8=9.2V. 7. The relation between α and β is_________ a) β = α/ (1-α) b) α = β/(1+β) c) β = α/ (1+α) 248

d) α = β/(1- β) View Answer Answer: b Explanation: β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β) ICBO. 8. In ICEO, wt does the subscript ‘CEO’ mean? a) collector to base emitter open b) emitter to base collector open c) collector to emitter base open d) emitter to collector base open View Answer Answer: c Explanation: The subscript ‘CEO’ means that it is collector to emitter base open. It is called as the leakage current. It occurs in a reverse bias in PNP transistor. The total current can be calculated by IC=βIB+IC. 9. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________ a) dc current gain b) base current amplification factor c) emitter current amplification factor d) ac current gain View Answer Answer: d Explanation: The ac current gain is given by β=∆IC/∆IB. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain. 10. The range of β is _________ a) 20 to 500 b) 50 to 300 c) 30 to 400 d) 10 to 20 View Answer Answer: a Explanation: Almost in all the transistors, the base current is less than 5% of the emitter current. Due to this fact, it is generally greater than 20. Usually it ranges from 20 to 500. Hence this

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configuration is frequently used when appreciable current gain as well as voltage gain is required.

Questions & Answers (MCQs) focuses on “The Common Collector Configuration”. 1. The current amplification factor ϒdc is given by_________ a) IE/IB b) IB/IE c) IC/IE d) IE/IC View Answer Answer: a Explanation: When no signal is applied, then the ratio of emitter current to base current is called as ϒdc of the transistor. As the collector is common to both input and output circuits, hence the name common collector configuration. 2. The relation between α and β is given by _________ a) 1/(1-α)=1- β b) 1/(1+α)=1+ β c) 1/(1-α)=1+ β d) 1/(1+α)=1- β View Answer Answer: c Explanation: The current amplification factor (β) is given by IC//IB. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor. β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β) ICBO.

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3. The CC configuration has an input resistance_________ a) 500kΩ b) 750kΩ c) 600kΩ d) 400kΩ View Answer Answer: b Explanation: It has a high input resistance and very low output resistance so the voltage gain is always less one. It is used for driving a low impedance load from a high impedance source. 4. The application of a CC configured transistor is_________ a) voltage multiplier b) level shifter c) rectification d) impedance matching View Answer Answer: d Explanation: The most important use of CC transistor is an impedance matching device. It is seldom used for amplification purposes. The current gain is same as that of CE configured transistor. 5. What is the output resistance of CC transistor? a) 25 Ω b) 50 Ω c) 100 Ω d) 150 Ω View Answer Answer: a Explanation: The CC transistor has a very low value of output resistance of 25 Ω. The voltage gain is always less one. It is used for driving a low impedance load from a high impedance source. 6. Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance. a) 20kΩ b) 10kΩ c) 50kΩ d) 60kΩ View Answer 251

Answer: b Explanation: ro=∆VCE/∆IC =3/0.3m=10kΩ. 7. A change in 300mV in base emitter voltage causes a change of 100µA in the base current. Determine the dynamic input resistance. a) 20kΩ b) 10kΩ c) 30kΩ d) 60kΩ View Answer Answer: c Explanation: ro=∆VBE/∆IB =300m/100µ=30kΩ. 8. The point on the DC load line which is represented by ‘Q’ is called _________ a) cut off point b) cut in point c) breakdown point d) operating point View Answer Answer: d Explanation: The point which represents the values of IC and VCE that exist in a transistor circuit when no signal is applied is called as operating point. This is also called as working point or quiescent point. 9. When is the transistor said to be saturated? a) when VCE is very low b) when VCE is very high c) when VBE is very low d) when VBE is very high View Answer Answer: a Explanation: When VCE is very low, the transistor said to be saturated and it operates in saturated region of characteristic. The change in base current IB does not produce a corresponding change in the collector voltage IC. 10. The input resistance is given by _________ a) ∆VCE/∆IB 252

b) ∆VBE/∆IB c) ∆VBE/∆IC d) ∆VBE/∆IE View Answer Answer: b Explanation: The ratio of change in base emitter voltage (∆VBE) to resulting change in base current (∆IB) at constant collector emitter voltage (VCE) is defined as input resistance. This is denoted by ri.

Questions & Answers (MCQs) focuses on “The CE Characteristics”. 1. The input characteristics of a CE transistor is_________

a)

b) 253

c)

d) View Answer Answer: b Explanation: A graph of IB against VBE is drawn. The curve so obtained is known as input characteristics. The collector emitter voltage (VCE) is kept constant. 2. The input resistance is given by _________ a) ∆VCE/∆IB b) ∆VBE/∆IB c) ∆VBE/∆IC d) ∆VBE/∆IE View Answer Answer: b Explanation: The ratio of change in base emitter voltage (∆VBE) to resulting change in base current (∆IB) at constant collector emitter voltage (VCE) is defined as input resistance. This is denoted by ri. 3. Which of the following depicts the output characteristics of a CE transistor?

a) 254

b)

c)

d) View Answer Answer: d Explanation: A graph of IC against VCE is drawn. The curve so obtained is known as output characteristics. The base current (IB) is kept constant. 4. The output resistance is given by _________ a) ∆VCE/∆IB b) ∆VBE/∆IB c) ∆VBE/∆IC 255

d) ∆VCE/∆IC View Answer Answer: d Explanation: The ratio of change in collector emitter voltage (∆VCE) to resulting change in collector current (∆IC) at constant base current (IB) is defined as output resistance. This is denoted by ro. 5. Which of the following cases damage the transistor? a) when VCE is increased too far b) when VCE is decreased too far c) when VBE is increased too far d) when VBE is decreased too far View Answer Answer: a Explanation: When VCE is increased too far, collector base junction completely breaks down and due to this avalanche breakdown, collector current increases rapidly. This is not shown in the characteristic. In this case, the transistor is damaged. 6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________ a) inverted region b) active region c) cut off region d) cut in region View Answer Answer: b Explanation: In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors. 7. The small amount of current which flows even when base current IB=0 is called_________ a) IBEO b) ICBO c) ICEO d) IC View Answer Answer: c Explanation: In the cut off region, a small amount of collector current flows even when base 256

current IB is zero. This is called ICEO. Since the main current is also zero, the transistor is said to be cut off. 8. A change in 700mV in base emitter voltage causes a change of 200µA in the base current. Determine the dynamic input resistance. a) 2kΩ b) 10kΩ c) 3kΩ d) 3.5kΩ View Answer Answer: c Explanation: ro=∆VBE/∆IB =700m/200µ=3.5kΩ. 9. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance. a) 20kΩ b) 10kΩ c) 50kΩ d) 60kΩ View Answer Answer: b Explanation: ro=∆VCE/∆IC =3/0.3m=10kΩ. 10. Which of the following points locates the quiescent point? a) (IC, VCB) b) (IE, VCE) c) (IE, VCB) d) (IC, VCE) View Answer Answer: a Explanation: The quiescent point is best located between the cut off and saturation point. IE= VEE/RE, VCB=VCC-ICRL. It is denoted by ‘Q’.

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Questions & Answers (MCQs) focuses on “The CB Characteristics”. 1. The input resistance in a CB transistor is given by _________ a) ∆VCE/∆IB b) ∆VBE/∆IB c) ∆VBE/∆IC d) ∆VEB/∆IE View Answer Answer: d Explanation: The ratio of change in emitter base voltage (∆VEB) to resulting change in emitter current (∆IE) at constant collector base voltage (VCB) is defined as input resistance. This is denoted by ri. 2. The output resistance of CB transistor is given by _________ a) ∆VCB/∆IC b) ∆VBE/∆IB c) ∆VBE/∆IC d) ∆VEB/∆IE View Answer Answer: a Explanation: The ratio of change in collector base voltage (∆VCB) to resulting change in collector current (∆IC) at constant emitter current (IE)¬ is defined as output resistance. This is denoted by ro. 3. Which one of the following depicts the output characteristics for a CB transistor?

a)

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b)

c)

d) View Answer

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Answer: b Explanation: A graph of IC against VCB is drawn. The curve so obtained is known as output characteristics. The emitter current (IE) is kept constant. 4. The input characteristics of a CE transistor is_________

a)

b)

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c)

d) View Answer Answer: c Explanation: A graph of IE against VEB is drawn. The curve so obtained is known as input characteristics. The collector base voltage (VBC) is kept constant. 5. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance? a) 30Ω b) 60Ω c) 40Ω d) 50Ω View Answer

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Answer: c Explanation: The ratio of change in emitter base voltage (∆VEB) to resulting change in emitter current (∆IE) at constant collector base voltage (VCB) is defined as input resistance. This is denoted by ri. We know, ∆VEB/∆IE=ri =200/5=40Ω. 6. When the collector junction is reverse biased and emitter junction is forward biased, the operating region of the transistor is called_________ a) inverted region b) active region c) cut off region d) cut in region View Answer Answer: b Explanation: In the active region, for small values of base current, the effect of collector voltage over collector current is small while for large base currents this effect increases. The shape of characteristic here is same as that of CB transistors. 7. Which of the following corresponds to the output circuit of a CB transistor? a) VBE b) IB c) VCB d) VCE View Answer Answer: c Explanation: Here, the quantity collector to base voltage corresponds to the output circuit of a CB transistor. The complete electrical behaviour of a transistor can be described by stating the relation between these quantities. 8. The input of a CB transistor is given between_________ a) collector and emitter terminals b) base ad collector terminals c) ground and emitter terminals d) emitter and base terminals View Answer Answer: d Explanation: The name of the CB transistor says that it’s a common based one. The input is

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given between the emitter and base terminals and the output is taken between collector and base terminals. 9. The current gain of the CB transistor is_________ a) less than or equal to unity b) equal to unity c) greater than unity d) remains same View Answer Answer: a Explanation: The input current flowing into the emitter terminal must be higher than the base current and collector current to operate the transistor. Therefore the output collector current is less than the input emitter current. 10. The input characteristics of a CB transistor resembles_________ a) Forward biased diode b) Illuminated photo diode c) LED d) Zener diode View Answer Answer: b Explanation: The input characteristics resemble the illuminated photo diode and the output characteristics resemble the forward biased diode. This transistor has low input impedance and high output impedance.

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Questions & Answers (MCQs) focuses on “DC Load Lines”.

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1. Which of the following depicts the DC load line?

a)

b)

c)

d) View Answer

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Answer: a Explanation: In transistor circuit analysis, sometimes it is required to know the collector currents for various collector emitter voltages. The one way is to draw its load line. We require the cut off and saturation points. 2. For the circuit shown, find the quiescent point.

a) (10V, 4mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, IE=VEE/RE=30/10kΩ=3mA IC=α IE =IE =3mA VCB=VCC-ICRL=25-15=10V. So, quiescent point is (10V, 3mA). 3. Which of the following depicts the load line for the circuit shown below?

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a)

b)

c)

d) View Answer

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Answer: d Explanation: We know, IE=VEE/RE=15/5kΩ=3mA IC=α IE =IE =3mA VCB=VCC-ICRL=20-15=5V. So, quiescent point is (5V, 3mA). 4. For the circuit shown, find the quiescent point.

a) (6V, 1mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, VCE=12V (IC)SAT =VCC/RL=12/6K=2mA. IB=10V/0.5M=20µA. IC= βIB=1mA. I VCE=VCC-ICRL=12-1*6=6V. So, quiescent point is (6V, 1mA).

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5. Which of the following depicts the load line for the given circuit?

a)

b)

269

c)

d) View Answer Answer: d Explanation: We know, VCE=6V (IC)SAT =VCC/RL=10/2K=5mA. IB=10V/0.5M=20µA. IC= βIB=1mA. I VCE=VCC-ICRL=10-1*2=8V. So, quiescent point is (8V, 1mA). 6. The DC equivalent circuit for an NPN common base circuit is.

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View Answer Answer: a Explanation: In the common base circuit, the emitter diode acts like a forward biased ideal diode, while collector diode acts as a current source due to transistor action. Thus an ideal transistor may be regarded as a rectifier diode in the emitter and a current source at collector. 271

7. The DC equivalent circuit for an NPN common emitter circuit is.

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View Answer Answer: b Explanation: In the common emitter circuit, the ideal transistor may be regarded as a rectifier diode in the base circuit and a current source in the collector circuit. In the current source, the direction of arrow points in direction of conventional current. 8. What is the other representation of the given PNP transistor connected in common emitter configuration?

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View Answer Answer: d Explanation: The emitter junction is forward biased with the help of battery VEE by which, negative of the battery is connected to the emitter while positive is connected to base. RE is the emitter resistance. The collector junction is reversed biased.

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9. What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode? a) IC = βIB b) IC > βIB c) IC >> βIB d) IC < βIB View Answer Answer: d Explanation: When in a transistor is driven into saturation, we use VCE(SAT) as another linear parameter. In, addition when a transistor is biased in saturation mode, we have IC < βIB. This characteristic used to prove that the transistor is indeed biased in saturation mode. 10. For the circuit shown, find the quiescent point.

a) (10V, 4mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, IE=VEE/RE=10/5kΩ=2mA IC=α IE =IE =2mA VCB=VCC-ICRL=20-10=10V. So, quiescent point is (10V, 2mA).

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Questions & Answers (MCQs) focuses on “Transistor as a Switch”. 1. In which region a transistor acts as an open switch? a) cut off region b) inverted region c) active region d) saturated region View Answer Answer: a Explanation: In this mode, both the junctions are reverse biased. The transistor has practically zero current because the emitter does not emit charge carriers to the base. There is negligibility current due to minority carriers. In this mode the transistor acts as an open switch. 2. In which region a transistor acts as a closed switch? a) cut off region b) inverted region c) active region d) saturated region View Answer Answer: d Explanation: In this mode, both the junctions are forward biased. The negative terminal of the battery is connected to the emitter. The collector current becomes independent of base current. In this mode the transistor acts as a closed switch.

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3. Which of the following circuits act as a switch?

a)

b)

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c)

d) View Answer Answer: b Explanation: This is an inverter, in which the transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device. 4. The current which is helpful for LED to turn on is_________ a) emitter current b) base current c) collector current d) depends on bias View Answer 278

Answer: c Explanation: Depending on the type of load, a collector current is induced that would turn on the motor or LED. The transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device. 5. Which of the following statements is true? a) Solid state switches are applications for an AC output b) LED’s can be driven by transistor logics c) Only NPN transistor can be used as a switch d) Transistor operates as a switch only in active region View Answer Answer: b Explanation: Output devices like LED’s only require a few milliamps at logic level DC voltages and can therefore be driven directly by the output of a logic gate. However, high power devices such as motors or lamps require more power than that supplied by an ordinary logic gate so transistor switches are used. 6. The base emitter voltage in a cut off region is_________ a) greater than 0.7V b) equal to 0.7V c) less than 0.7V d) cannot be predicted View Answer Answer: c Explanation: From the cut off characteristics, the base emitter voltage (VBE) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, VBE > 0.7 and IC=0. For a PNP transistor, the emitter potential must be negative with respect to the base. 7. In saturation region, the depletion layer_________ a) increases linearly with carrier concentration b) decreases linearly with carrier concentration c) increases by increasing the emitter current d) decreases by decreasing the emitter voltage drop View Answer Answer: d Explanation: Here, the transistor will be biased so that maximum amount of base current is applied, resulting in maximum collector current resulting in minimum emitter voltage drop which results in depletion layer as small as possible and maximum current flows through the transistor. 279

8. The base emitter voltage in a saturation region is_________ a) greater than 0.7V b) equal to 0.7V c) less than 0.7V d) cannot be predicted View Answer Answer: d Explanation: From the saturation mode characteristics, the transistor acts as a single pole single throw solid state switch. A zero collector current flows. With a positive signal applied to the base of transistor it turns on like a closed switch. 9. The switching of power with a PNP transistor is called_________ a) sourcing current b) sinking current c) forward sourcing d) reverse sinking View Answer Answer: a Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it. 10. The switching of power with a NPN transistor is called_________ a) sourcing current b) sinking current c) forward sourcing d) reverse sinking View Answer Answer: b Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to supply and the transistor switches the power to it.

Questions & Answers (MCQs) focuses on “Transistor Switching Times”.

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1. The collector current will not reach the steady state value instantaneously because of_________ a) stray capacitances b) resistances c) input blocking capacitances d) coupling capacitance View Answer Answer: a Explanation: When a pulse is given, the collector current will not reach the steady state value instantaneously because of stray capacitances. The charging and discharging of capacitance makes the current to reach a steady state value after a given time constant. 2. For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?

a) 20ms b) 50ms c) 60ms d) 70ms View Answer Answer: b Explanation: At t < 0, the BJT is OFF in cut off region. IB=0 as β=∞, so IC=IE. When t > 0, switch opens and BJT is ON. The voltage across capacitor increases. From the input loop, -5VBE-I(4.3K)+10=0 and gives I=1mA. IC1=1-0.5=0.5mA. VC1=0.7+4.3+10=-5V. IC1=C1dVC1/dt. From this equation, we get t=50ms. 3. The technique used to quickly switch off a transistor is by_________ a) reverse biasing its emitter to collector junction b) reverse biasing its base to collector junction 281

c) reverse biasing its base to emitter junction d) reverse biasing any junction View Answer Answer: c Explanation: The technique used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The advantage of this circuit is that it is not necessary to have high voltage control signal. 4. The disadvantage of using the method of reverse biasing base emitter junction is_________ a) high voltage control signal b) low voltage control signal c) output swing d) incomplete switching of output View Answer Answer: d Explanation: This method is used to quickly switch off a transistor is by reverse biasing its base to collector junction. It is demonstrated in a high voltage switching circuit. The disadvantage of using the method of reverse biasing base emitter junction is that the output does not switch completely to GND due to forward voltage drop of the diode. 5. Which of the following circuits helps in the applications of switching times?

a)

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b)

c)

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d) View Answer Answer: b Explanation: This is an inverter, in which the transistor in the circuit is switched between cut off and saturation. The load, for example, can be a motor or a light emitting diode or any other electrical device. 6. Which of the following helps in reducing the switching time of a transistor? a) a resistor connected from base to ground b) a resistor connected from emitter to ground c) a capacitor connected from base to ground d) a capacitor connected from emitter to ground View Answer Answer: a Explanation: Connecting a resistor connected from base of a transistor to ground/negative voltage helps in reducing the switching the switching time of the transistor. When transistor saturate, there is stored charge in the base that must be removed before it turns off. 7. The time taken for a transistor to turn from saturation to cut off is _________ a) inversely proportional to charge carriers b) directly proportional to charge carriers c) charging time of the capacitor

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d) discharging time of the capacitor View Answer Answer: b Explanation: When sufficient charge carriers exist, the transistor goes into saturation. When the switch is turned off, in order to go into cut off, the charge carriers in the base region need to leave. The longer it takes to leave, the longer it takes for a transistor to turn from saturation to cut off. 8. The switching of power with a PNP transistor is called _________ a) sourcing current b) sinking current c) forward sourcing d) reverse sinking View Answer Answer: a Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it. 9. The base emitter voltage in a cut off region is _________ a) greater than 0.7V b) equal to 0.7V c) less than 0.7V d) cannot be predicted View Answer Answer: c Explanation: From the cut off characteristics, the base emitter voltage (VBE) in a cut off region is less than 0.7V. The cut off region can be considered as ‘off mode’. Here, VBE < 0.7 and IC=0. For a PNP transistor, the emitter potential must be negative with respect to the base. 10. Switching speed of P+ junction depends on _________ a) Mobility of minority carriers in P junction b) Life time of minority carriers in P junction c) Mobility of majority carriers in N junction d) Life time of minority carriers in N junction View Answer Answer: d Explanation: Switching leads to move holes in P region to N region as minority carriers. 285

Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively.

7. Questions & Answers on Transistor Biasing and Thermal Stabilization The section contains questions and answers on operating point, bias stability, collector-to-base and collector-emitter bias, self bias, bias compensation, thermistor, thermal runway and stability. Collector-to-Base Bias Self-Bias Bias Compensation Thermal Runaway Thermal Stability The Operating Point

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Bias Stability Emitter Feedback Bias Collector-Emitter Feedback Bias Stabilization against Variations in VBE and Beta for Self Bias Circuit Thermistor and Sensistor Compensation

Questions & Answers (MCQs) focuses on “Collector-to-Base Bias”. 1. The collector current (IC) that is obtained in a collector to base biased transistor is_________ a) (VCC-VBE)/RB b) (VCC+VBE)/RB c) (VCE-VBE)/RB d) (VCE+VBE)/RB View Answer Answer: a Explanation: The collector current is analysed by the DC analysis of a transistor. It involves the

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DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, IC=IBβ. 2. The collector to emitter voltage (VCE) is obtained by_________ a) VCC – RC(IC-IB) b) VCC – RC(IC+IB) c) VCC + RC(IC+IB) d) VCC + RC(IC-IB) View Answer Answer: b Explanation: The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch. 3. What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode? a) IC = βIB b) IC > βIB c) IC >> βIB d) IC < βIB View Answer Answer: d Explanation: When in a transistor is driven into saturation, we use VCE(SAT) as another linear parameter. In, addition when a transistor is biased in saturation mode, we have IC < βIB. This characteristic used to prove that the transistor is indeed biased in saturation mode. 4. The thermal runway is avoided in a collector to base bias because_________ a) of its independence of β b) of the positive feedback produced by the base resistor c) of the negative feedback produced by the base resistor d) of its dependence of β View Answer Answer: c Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the base resistor in a collector to base bias. The IC which is responsible for the damage is reduced by decreased output signal. 5. When the temperature is increased, what happens to the collector current after a feedback is given? 288

a) it remains same b) it increases c) it cannot be predicted d) it decreases View Answer Answer: d Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the base current increases with decreasing collector current and the thermal runway too. 6. The demerit of a collector to base bias is_________ a) its need of high resistance values b) its dependence on β c) its independence on β d) the positive feedback produced by the base resistor View Answer Answer: a Explanation: When the stability factor S=1, the collector resistor value should be very large when compared to the base resistor. So, when RC is large we need to provide large power supply which increases the cost. At the same time, as the base resistor is small we need to provide small power supply. 7. The negative feedback does good for DC signal by_________ a) decreasing the gain b) increasing the gain c) stabilising the operating point d) increasing the stability factor View Answer Answer: c Explanation: The resistor RB can provide negative feedback for both AC and DC signals. The negative feedback for DC signal is done good as it can provide stable operating point. On the other side, the negative feedback is badly done for AC signal by decreasing the voltage gain.

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8. In the circuit, transistor has β =60, VBE=0.7V. Find the collector to emitter voltage drop VCE.

a) 5V b) 3V c) 8V d) 6V View Answer Answer: d Explanation: We know, IC=(VCC-VBE)/RB By putting the values, we have IC=5.9mA. IE=IC/α. So, IE=5.99mA. VCE= VCC-RC(IC+IB). We have VCE=6V. 9. In the circuit shown below, β =100 and VBE=0.7V. The Zener diode has a breakdown voltage of 6V. Find the operating point.

a) (6.7V, 5.3mA) 290

b) (5.7V, 5.3mA) c) (6.7V, 5mA) d) (6V, 5mA) View Answer Answer; a Explanation: We know, by KVL -12+(IC+IB)1K+6+VBE=0 We have IE=5.3. IC= αIE=5.24mA. From another loop, -12+IEIK+VBE=0 We have, VCE=12-5.3m*1000=6.7V. Hence the Q point is (6.7V, 5.3mA). 10. When the β value is large for a given transistor, the IC and VCE values are given by_________ a) (VCC-VBE)/RB, VCC-RCIC b) (VCC+VBE)/RB, VCC-RC(IC+IB) c) (VCC+VBE)/RB, VCC+RC(IC+IB) d) (VCC+VBE)/RB, VCC+RC(IC-IB) View Answer Answer: a Explanation: The base current IB is zero when β value is large. So, the VCE changes to VCC-RCIC. The collector current IC is changed to (VCC-VBE)/RB from β(VCC-VBE)/(1+ β)RE+ RB.

Questions & Answers (MCQs) focuses on “Self-Bias”. 1. The collector current (IC) that is obtained in a self biased transistor is_________ a) (VTH – VBE)/RE b) (VTH + VBE)/RE c) (VTH – VBE)/RE d) (VTH + VBE)/RE View Answer

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Answer: a Explanation: The collector current is analysed by the DC analysis of a transistor. It involves the DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, IC=IBβ. 2. The collector to emitter voltage (VCE) is obtained by_________ a) VCC – RCIC+RBIB b) VCC – RCIC-REIE c) VCC + RCIC d) VCC + RCIB View Answer Answer: b Explanation: The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch. 3. The thermal runway is avoided in a self bias because_________ a) of its independence of β b) of the positive feedback produced by the emitter resistor c) of the negative feedback produced by the emitter resistor d) of its dependence of β View Answer Answer: c Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the emitter resistor in a self bias. The IC which is responsible for the damage is reduced by decreased output signal. 4. When the temperature is increased, what happens to the collector current after a feedback is given? a) it remains same b) it increases c) it cannot be predicted d) it decreases View Answer Answer: d Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the drop across the emitter resistor increases with decreasing collector current and the thermal runway too. 292

5. What is the Thevenin’s voltage (VTH) in a self bias shown below?

a) VCCR2/R1+R2 b) VCCR1/R1+R2 c) VCCR2/R1-R2 d) VCCR2/R1-R2 View Answer Answer: a Explanation: The base current cannot be obtained directly from the KVL or KCL applications. The VCC and VBE cannot come under a single equation. So, the circuit is changed with a Thevenin’s voltage (VTH) and Thevenin’s resistance.

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6. What is the Thevenin’s resistance (RTH) in a self bias shown below?

a) R1R2/R1+R2 b) R2/R1+R2 c) R1R2/R1-R2 d) R1/R1-R2 View Answer Answer: a Explanation: The base current cannot be obtained directly from the KVL or KCL applications. A potential divider network is formed by R1 and R2.The VCC and VBE cannot come under a single equation. So, the circuit is changed with a Thevenin’s resistance. 7. The stability factor for a self biased transistor is_________ a) 1 – RTH/RE b) 1 + RTH/RE c) 1 + RE/RTH d) 1 – RE/RTH View Answer Answer: b Explanation: The stability of the circuit is inversely proportional to the stability factor. The emitter resistor is very large when compared to the Thevenin’s resistance. When β is not that large, then S=(1+ β)( RTH+ RE)/ (1+ β)RE+ RTH.

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8. In the circuit, the transistor has a large β value (VBE=0.7V). Find the current through RC.

a) 0.5mA b) 2mA c) 1mA d) 1.6mA View Answer Answer: c Explanation: We know, IC=VTH-VBE/RE =9*3/9=3V. IC=3-0.7/2.3=1mA. 9. A silicon NPN transistor is used and it has a large value of β. Find the required value of R2 when IC=1mA.

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a) 10kΩ b) 20kΩ c) 30kΩ d) 40kΩ View Answer Answer: d Explanation: For silicon, VBE=0.8V, VCE=0.2V. IC=VTH-VBE/RE. By pitting the values, we have VTH=1.3V. R2 can be found from, VCCR2/R1+R2. We get R2=40KΩ. 10. The value of αac for all practical purposes, for commercial transistors range from_________ a) 0.5 to 0.6 b) 0.7 to 0.77 c) 0.8 to 0.88 d) 0.9 to 0.99 View Answer Answer: d Explanation: For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

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Questions & Answers (MCQs) focuses on “Bias Compensation”. 1. The compensation techniques are used to_________ a) increase stability b) increase the voltage gain c) improve negative feedback d) decrease voltage gain View Answer Answer: b Explanation: Usually, the negative feedback is used to produce a stable operating point. But it reduces the voltage gain of the circuit. This sometimes is intolerable and should be avoided in some applications. So, the biasing techniques are used. 2. Compensation techniques refer to the use of_________ a) diodes b) capacitors c) resistors d) transformers View Answer 297

Answer: a Explanation: Compensation techniques refer to the use of temperature sensitive devices such as thermistors, diodes, transistors, sensistors etc to compensate variation in currents. Sometimes for excellent bias and thermal stabilization, both stabilization and compensation techniques are used. 3. In a silicon transistor, which of the following change significantly to the change in IC? a) VCE b) IB c) VBE b) IE View Answer Answer: c Explanation: For germanium transistor, changes in ICO with temperature contribute more serious problem than for silicon transistor. On the other hand, in a silicon transistor, the changes of VBE with temperature possesses significantly to the changes in IC. 4. What is the compensation element used for variation in VBE and ICO? a) diodes b) capacitors c) resistors d) transformers View Answer Answer: a Explanation: A diode is used as the compensation element used variation in VBE and ICO. The diode used is of the same material and type as that of transistor. Hence, the voltage across the diode has same temperature coefficient as VBE of the transistor. 5. The expression for IC in the compensation for instability due to ICO variation_________ a) βI+βIO+βICO b) βI+βIO c) βIO+βICO d) βI+βICO View Answer Answer: a Explanation: In this method, diode is used for the compensation in variation of ICO. The diode used is of the same material and type as that of transistor. Hence, the reverse saturation current IO of the diode will increase with temperature at the same rate as the transistor collector saturation current ICO.

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6. Which of the following has a negative temperature coefficient of resistance? a) sensistor b) diode c) thermistor d) capacitor View Answer Answer: c Explanation: The thermistor has a negative temperature coefficient of resistance. It means, its resistance decreases exponentially with increasing T. The thermistor RT is used to minimize the increase in collector current. 7. Which of the following has a negative temperature coefficient of resistance? a) capacitor b) diode c) thermistor d) sensistor View Answer Answer: d Explanation: The sensistor has a positive temperature coefficient of resistance. It is a temperature sensitive resistor. It is a heavily doped semiconductor. When voltage is decreased, the net forward emitter voltage decreases. As a result the collector current decreases. 8. Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance. a) 20kΩ b) 10kΩ c) 50kΩ d) 60kΩ View Answer Answer: b Explanation: ro=∆VCE/∆IC =3/0.3m=10kΩ. 9. The output resistance of CB transistor is given by _________ a) ∆VCB/∆IC b) ∆VBE/∆IB c) ∆VBE/∆IC d) ∆VEB/∆IE View Answer 299

Answer: a Explanation: The ratio of change in collector base voltage (∆VCB) to resulting change in collector current (∆IC) at constant emitter current (IE) is defined as output resistance. This is denoted by ro. 10. The negative sign in the formula of amplification factor indicates_________ a) that IE flows into transistor while IC flows out it b) that IC flows into transistor while IE flows out it c) that IB flows into transistor while IC flows out it d) that IC flows into transistor while IB flows out it View Answer Answer: a Explanation: When no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. αdc=-IC/IE. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

Questions & Answers (MCQs) focuses on “Thermal Runaway”. 1. Thermal runaway is_________ a) an uncontrolled positive feedback b) a controlled positive feedback c) an uncontrolled negative feedback

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d) a controlled negative feedback View Answer Answer: a Explanation: Thermal runaway is a self destruction process in which an increase in temperature creates such a condition which in turn increases the temperature again. This uncontrolled rise in temperature causes the component to get damaged. 2. The thermal runway is avoided in a self bias because_________ a) of its independence on β b) of the positive feedback produced by the emitter resistor c) of the negative feedback produced by the emitter resistor d) of its dependence on β View Answer Answer: c Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the emitter resistor in a self bias. The IC which is responsible for the damage is reduced by decreased output signal. 3. When the temperature is increased, what happens to the collector current after a feedback is given? a) it remains same b) it increases c) it cannot be predicted d) it decreases View Answer Answer: d Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the drop across the emitter resistor increases with decreasing collector current and the thermal runway too. 4. The thermal runway is avoided in a collector to base bias because_________ a) of its independence on β b) of the positive feedback produced by the base resistor c) of the negative feedback produced by the base resistor d) of its dependence on β View Answer

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Answer: c Explanation: The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the base resistor in a collector to base bias. The IC which is responsible for the damage is reduced by decreased output signal. 5. When the temperature is increased, what happens to the collector current after a feedback is given? a) it remains same b) it increases c) it cannot be predicted d) it decreases View Answer Answer: d Explanation: Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the base current increases with decreasing collector current and the thermal runway too. 6. Discrete transistors T1 and T2 having maximum collector current rating of 0.75A are connected in parallel as shown in the figure. This combination is treated as a single transistor to carry a single current of 1A, when biased with a self bias circuit. When the circuit is switched ON, T1 had draws 0.55A and T2 draws 0.45A. If the supply is kept ON continuously, it is very likely that_________

a) both T1 and T2 get damaged b) both T1 and T2 will be safe c) only T1 gets damaged d) only T2 gets damaged View Answer 302

Answer: c Explanation: The T1 transistor is having more power dissipation as it is drawing 0.55A. When power dissipation increases, the temperature increases and this leads to the ultimate further increase in the current drawn by T1. The current drawn by T2 will be reduced as the sum of currents drawn by T1 and T2 should be constant. 7. When the collector current is increased in a transistor_________ a) the reverse current is increased b) the temperature is increased c) collisions of electrons decrease d) the emitter does not emit electrons View Answer Answer: b Explanation: As the collector current is increased, the emitter releases more number of electrons. This causes more collisions of electrons at collector. This happens in a cycle and produces such a condition in which temperature is further more increased. 8. Which of the following are true? a) TJ – TA = θPd b) TJ – TA = θ/Pd c) TJ – TA = θ+Pd d) TJ – TA = θ-Pd View Answer Answer: a Explanation: The TJ is called as junction temperature which varies and TA is called as the ambient temperature which is fixed. The difference between these temperatures is directly proportional to the power dissipation. Here, θ is called as thermal resistance which is proportionality constant. 9. When the power dissipation increases in a transistor, the thermal resistance_________ a) increases b) cannot be predicted c) decreases d) remains same View Answer Answer: c Explanation: The power dissipation is directly proportional to thermal resistance. We have, TJ – TA = θPd in which we can observe θ ∝ 1/Pd. So, a device with low power dissipation has high thermal resistance. 303

10. Which of the following biasing techniques are prone to thermal runaway? a) self bias b) collector to base bias c) fixed bias d) the biasing technique is identified by temperature effect View Answer Answer: c Explanation: The collector current of a fixed bias transistor is IC= β(VCC-VBE)/RB. When the temperature is increased, the reverse saturation increases. The collector current also increases. This in turn increases the current again which leads to damage of transistor.

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Questions & Answers (MCQs) focuses on “Thermal Stability”. 1. For a given transistor, the thermal resistance is 8°C/W and for the ambient temperature TA is 27°C. If the transistor dissipates 3W of power, calculate the junction temperature (TJ). a) 51°C b) 27°C c) 67°C d) 77°C View Answer Answer: a Explanation: We know, TJ-TA=HPD TJ=TA+HPD=27+8*3=51°C. 2. Which of the following are true? a) TJ-TA=θpd b) TJ-TA=θ/pd c) TJ-TA=θ+pd d) TJ-TA=θ-pd View Answer Answer: a Explanation: The TJ¬ is called as junction temperature which varies and TA is called as the ambient temperature which is fixed. The difference between these temperatures is directly proportional to the power dissipation. Here, θ is called as thermal resistance which is proportionality constant. 3. A silicon power transistor is operated with a heat sink HS-A=1.5°C/W. The transistor rated at 150W (25°C) has HJ-C=0.5°C/W and the mounting insulation has HC-S=0.6°C/W. What maximum power can be dissipated if the ambient temperature is 40°C and (TJ)MAX=200°C? a) 70.6W b) 61.5W c) 37.8W

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d) 56.9W View Answer Answer: b Explanation: PD=(TJ-TA)/ HJ-C +HC-S +HS-A =200-40/0.5+0.6+1.5=61.5W. 4. The total thermal resistance of a power transistor and heat sink is 20°C/W. The ambient temperature is 25°C and (TJ)MAX=200°C. If VCE=4V, find the maximum collector current that the transistor can carry without destruction. a) 3.67A b) 7.56A c) 2.19A d) 4.16A View Answer Answer: c Explanation: PD =(TJ-TA)/ H =200-25/20=8.75W. Now, VCEIC = 8.75/4=2.19A. 5. The total thermal resistance of a power transistor and heat sink is 20°C/W. The ambient temperature is 25°C and (TJ)MAX=200°C. If VCE=4V, find the maximum collector current that the transistor can carry without destruction. What will be the allowed value of collector current if ambient temperature rises to 75°C? a) 3.67A b) 7.56A c) 2.19A d) 1.56A View Answer Answer: d Explanation: PD =(TJ-TA)/ H =200-75/20=6.25W. Now, IC = 6.25/4=1.56A. 6. Which of the following is true? a) HC-A = HJ-C – HJ-A b) HC-A = HJ-C + HJ-A c) HJ-A = HJ-C – HC-A d) HJ-A = HJ-C + HC-A View Answer 306

Answer: d Explanation: HJ-C is thermal resistance between junction and case and HC-A is thermal resistance between case and ambient. The circuit designer has no control over HJ-C. So, a proper approach to dissipate heat from case to ambient is through heat sink. 7. The condition to be satisfied to prevent thermal runaway? a) ∂PC/∂TJ > 1/Q b) ∂PC/∂TJ < 1/Q c) ∂PC/∂TJ > 1/Q d) ∂PC/∂TJ < 1/Q View Answer Answer: b Explanation: PC is the power dissipated at the collector junction. TJ is junction temperature which varies. The difference between these temperatures is directly proportional to the power dissipation. Here, Q is called as thermal resistance which is proportionality constant. 8. Thermal stability can be obtained by_________ a) shifting operating point b) increasing power supply c) heat sink d) decreasing current at collector View Answer Answer: c Explanation: As power transistors handle large currents, they always heat up during operation. Generally, power transistors are mounted in large metal case to provide a large area from which the heat generated by the device radiates. 9. Thermal stability is dependent on thermal runaway which is_________ a) an uncontrolled positive feedback b) a controlled positive feedback c) an uncontrolled negative feedback d) a controlled negative feedback View Answer Answer: a Explanation: Thermal runaway is a self destruction process in which an increase in temperature creates such a condition which in turn increases the temperature again. This uncontrolled rise in temperature causes the component to get damaged.

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10. Which of the following biasing techniques are affected by thermal runaway? a) self bias b) collector to base bias c) fixed bias d) the biasing technique is identified by temperature effect View Answer Answer: c Explanation: The collector current of a fixed bias transistor is IC= β(VCC-VBE)/RB. When the temperature is increased, the reverse saturation increases. The collector current also increases. This in turn increases the current again which leads to damage of transistor.

8. Questions on Signals and Amplifiers The section contains questions and answers on basics of signals, amplifiers, sinusoidal steady state analysis, amplifier circuit models and frequency response. Signals Amplifiers Circuit Models for Amplifier

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Frequency Response of Amplifier Sinusoidal Steady State Analysis

Questions & Answers (MCQs) focuses on “Signals”. 1. A cosine wave voltage signal has a 10V RMS value and 60Hz frequency. Also at time, t=0, the value of the voltage signal is equal to its RMS value. Which of the following is the correct mathematical representation of the voltage signal? a) 10 cos(60t) b) 10 cos (120πt) c) 14.14 cos(60t + π/4) d) 14.14 cos(120πt + π/4) View Answer Answer: d Explanation: Only equation 14.14 cos(120πt + π/4) satisfies all the given parameters. 2. Which of the following is a characteristic of digital signal? a) It takes quantized value 309

b) Its waveform is a continuous function c) The maximum number of signals that can be produced by N bits is 2N-1 d) There is no loss of value after converting an analog signal to digital signal View Answer Answer: a Explanation: Digital signal is non continuous and has discrete sets of possible value which it can take. 3. Consider an N-bits ADC (Analog to Digital Converter) whose analog input varies from 0 to Vmax, then which of the following is not true? a) The least significant bit correspond to a change of Vmax/2N -1 in the analog signal b) The resolution of the ADC is Vmax/2N -1 c) The maximum error in the conversion (or quantization error) is Vmax/2(2N -1) d) None of the mentioned View Answer Answer: d Explanation: None of the statements are true. 4. In compact disc (CD) audio technology, the signal is sampled at 44kHz. Each sample is represented by 16bits. What is the speed of the system in bits/second? a) 1.34 bits/second b) 2,750 bits/second c) 704,000 bits/second d) 1,441,792,000 bits/second View Answer Answer: c Explanation: 44000 X 16 = 704000 bits/s 5. An electrical signal can be represented in either Thevenin form or Norton From. However, Thevenin representation is preferred when a) The load resistance is very large as compared to the source resistance b) The load resistance is very low compared to the source resistance c) There is no preferred case. d) Both of the cases mentioned are the preferred case View Answer Answer: a Explanation: a is preferred in Thevenin’s case and b in Norton’s case

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6. An electrical signal can be represented in either Thevenin form or Norton From. However, Norton representation is preferred when a) The load resistance is very large as compared to the source resistance b) The load resistance is very low as compared to the source resistance c) There is no preferred case d) Both of the cases mentioned are the preferred case View Answer Answer: b Explanation: a is preferred in Thevenin’s case and b in Norton’s case

Consider the DAC (digital to Analog Converter) shown below. Where b1, b2….bn can be either 0 or 1. (Q.7 to Q.9) 7. What will be the value of the output current io, if the digital signal is in the form of b1b2b3…bn a) Vref/2R (b1/2+ b2/4 + b3/8….+bn/(2N)) b) 2Vref/R (b1/2+ b2/4 + b3/8….+bn/(2N)) c) Vref/R (b1/2+ b2/4 + b3/8….+bn/(2N)) d) none of the above View Answer Answer: c Explanation: It is the total current that will flow individual resistances. 8. Which is the Most Significant Bit and Least Significant Bit in this case? a) MSB: bN, LSB: bN b) MSB: bN, LSB: b1 c) MSB: b1, LSB: bN d) MSB: b1, LSB: b1 View Answer

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Answer: b Explanation: LSB will have maximum resistance (hence least contribution to the total current) and MSB will have least resistance (Hence maximum contribution to the total current). 9. For vref = 10V, n = 6, and R= 5000 ohm, which of the following is true? a) The maximum value of the output current is 1.9375 mA b) The change in the output current if the LSB is changed from 0 to 1 is 0.03125 mA c) The change in the output current is the MSB is change from 0 to 1 is 0.5 mA d) None of the mentioned View Answer Answer: b Explanation: According to the figure, the LSB is given by 10 / 50006 mA. 10. What is the binary representation of 57? a) 0011 1101 b) 0101 1001 c) 0111 1001 d) 0011 1001 View Answer Answer: d Explanation: The binary equivalent of 57 is 00111001.

Questions & Answers (MCQs) focuses on “Amplifiers”.

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1. An amplifier operating from ±3V provide a 2.2V peak sine wave across a 100 ohm load when provided with a 0.2V peak sine wave as an input from which 1.0mA current is drawn. The average current in each supply is measured to be 20mA. What is the amplifier efficiency? a) 20.2% b) 25.2% c) 30.2% d) 35.2% View Answer Answer: a Explanation:

2. In order to prevent distortion in the output signal after amplification, the input signal must be a) Higher than the positive saturation level of the amplifier b) Lower than the negative saturation level of the amplifier c) Must lie with the negative and the positive saturation level of the amplifier d) Both higher than the positive saturation level of the amplifier and lower than the negative saturation level of the amplifier View Answer Answer: c Explanation: Higher than the positive saturation and lower than the negative saturation level of the amplifier are the desired characteristics in order to prevent distortion. 3. The voltage gain of the amplifier is 8 and the current gain is 7. The power gain of the amplifier is a) 56 db b) 17.481 db c) 34.963 db d) 1 db View Answer 313

Answer: b Explanation: The power gain is given by 10 log (7 X 8) db. 4. Statement 1: Voltage gain of -5 means that the output voltage has been attenuated. Statement 2: Voltage gain of -5db means that the output voltage has been attenuated. a) Statement 1 and Statement 2 are true b) Statement 1 and Statement 2 are false c) Only Statement 1 is true d) Only Statement 2 is true View Answer Answer: d Explanation: A negative voltage gain means that a phase difference of 1800 has been introduced in the output waveform when compared to the input waveform. A voltage gain of -5db means that the signal has been attenuated. 5. Which of the following isn’t true? a) Both transformer and amplifier can provide voltage gain b) Both transformer and amplifier can provide current gain c) Both transformer and amplifier can provide power gain d) None of the mentioned View Answer Answer: c Explanation: For an ideal transformer the power input is always equal to the power output. In real conditions there is slight loss of power when transferring the power from an input source to an output source. Amplifiers only provide power gain. 6. Symmetrically saturated amplifiers operating in clipping mode can be used to convert a sine wave to a a) Square wave b) Pseudo Square wave c) Sawtooth wave d) Triangular wave View Answer Answer: b Explanation: Clipping circuits with low peak values of the output signals are used to generate pseudo square waves if the input signal is very large as compared to the output signal. 7. What is meant by stability of the an amplified signal? a) The amplified signal must have a finite amplitude 314

b) The amplified signal should not have self oscillation c) The input and the output signal must be proportional d) The ratio of the input and the output signal must be finite View Answer Answer: b Explanation: The ability of the amplifier to prevent self oscillation is a measure of its stability. 8. If Av, Ai and Ap represents the voltage gain, current gain and power gain ratio of an amplifier which of the below is not the correct expression for the corresponding values in decibel? a) Current gain: 20 log Ai db b) Voltage gain: 20 log Av db c) Power gain: 20 log Ap db d) Power gain: 10 log Ap View Answer Answer: c Explanation: Power gain is given by 10 log Ap db. 9. An amplifier has a voltage gain of 100 V/V and a current gain of 1000A/A. the value of the power gain decibel is a) 30 db b) 40 db c) 50 db d) 60 db View Answer Answer: c Explanation: Power gain in db is given by 10 log (100 X 1000) db. 10. The units of voltage gain is a) It has no units, it is a ratio b) Decibels (db) c) All of the mentioned d) None of the mentioned View Answer Answer: a Explanation: Voltage gain (Vo) = output voltage/input voltage (Vi). It is also expresses as 20 log (Vo/Vi) db.

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Questions & Answers (MCQs) focuses on “Circuit Models for Amplifier”. 1. Buffer amplifier needs to have a) Low input resistance and low output resistance b) High Input resistance and high output resistance c) Low input resistance and high output resistance d) High input resistance and low output resistance View Answer Answer: d Explanation: Buffer amplifiers are used to connect high input resistance source to a low output resistance load. 2. The ideal values for the input resistance (Ri) and the output resistance (Ro) of a transconductance amplifier are a) Ri = 0 and Ro = 0 b) Ri = ∞ and R0 = ∞ c) Ri = 0 and R0 = ∞ d) Ri = ∞ and Ri = 0 View Answer Answer: b Explanation: It is a desired characteristics of transconductance amplifier ideally. 3. An amplifier has a voltage gain of 40db. The value of AVO is a) 10 b) 100 c) 20

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d) 200 View Answer Answer: b Explanation: The expression is given by 10 log AVO = 40. Solving for Avo gives 100 as the answer. 4. The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 kΩ is connected. What is the value of the amplifier output resistance? a) 50Ω b) 200Ω c) 250Ω d) 350Ω View Answer Answer: c Explanation: 250 / (1000 + 250) X 100% = 20%. Hence the output resistance is 250 ohm. 5. Which of the following is a transresistance amplifier?

a)

b)

c)

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d) View Answer Answer: d Explanation: Figure d is the correct representation rest are voltage, current, transconductance amplifiers. 6. The signal to be amplified is current signal and the output desired is a voltage signal. Which of the following amplifier can perform this task? a) Voltage amplifier b) Current amplifier c) Transconductance amplifier d) Transresistance amplifier View Answer Answer: d Explanation: It is a characteristic of transconductance amplifier. 7. You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-kΩ source (S) and a 100-Ω load (L). The amplifiers have voltage gain, input resistance, and output resistance as follows: for A, 100 V/V, 10 kΩ, 10 kΩ, respectively; for B, 1 V/V, 100 kΩ, 100 Ω, respectively. Your problem is to decide how the amplifiers should be connected so that the voltage gain is maximum. a) SABL b) SBAL c) Both have the same voltage gain d) None as neither combination is able to amplify the input signal View Answer Answer: a Explanation: None 8. A transconductance amplifier with Ri = 2 kΩ, Gm = 40 mA/V, and Ro = 20 kΩ is fed with a voltage source having a source resistance of 2 kΩ and is loaded with a 1-kΩ resistance. Find the voltage gain realized. a) 18.05 V/V 318

b) 19.05 V/V c) 20.05 V/V d) 21.05V/V View Answer Answer: b Explanation:

9. The ratio of the short circuit current gain of a current amplifier (Ai) to the open circuit voltage gain of a voltage amplifier (AV), given that both amplifiers have the same value of the input resistance (Ri) and output resistance (R0), is a) Ri b) Ro c) Ri / R0 d) Ro / Ri View Answer Answer: c Explanation: It is a standard mathematical relation. 10. The ratio of the open circuit voltage of a voltage amplifier (AV) to the short circuit transconductance of a (Gm) of a transconductance amplifier, given that both have the same value of the internal resistance (Ri) and the output resistance (R0), is a) Ri b) R0 c) 1/Ri

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d) 1/R0 View Answer Answer: b Explanation: It is a standard mathematical relation.

Questions & Answers (MCQs) focuses on “Frequency Response of Amplifier”. 1. Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at a) f = 10 Hz is 55 db b) f = 10 kHz is 45 db c) f = 100 kHz is 25 db 320

d) f = 1Mhz is 0 db View Answer Answer: d Explanation: Use standard formulas for frequency response and voltage gain. 2. STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true? a) HP network passes dc and low frequencies and attenuate high frequency and opposite for LP network b) LP network passes dc and low frequencies and attenuate high frequency and opposite for HP network c) HP network passes dc and high frequencies and attenuate low frequency and opposite for LP network d) LP network passes low frequencies only and attenuate high frequency and opposite for HP network View Answer Answer: b Explanation: By definition a LP network allows dc current (or low frequency current) and an LP network does the opposite, that is, allows high frequency ac current. 3. Single-time-constant (STC) networks are those networks that are composed of, or can be reduced to a) One reactive component (L or C) and a resistance (R) b) Only capacitive component (C) and resistance (R) c) Only inductive component (L) and resistance (R) d) Reactive components (L, C or both L and C) and resistance (R) View Answer Answer: a Explanation: STC has only one reactive component and one resistive component. 4. The signal whose waveform is not effected by a linear circuit is a) Triangular Waveform signal b) Rectangular waveform signal c) Sine/Cosine wave signal d) Sawtooth waveform signal View Answer

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Answer: c Explanation: Only sine/cosine wave are not affected by a linear circuit while all other waveforms are affected by a linear circuit. 5. Which of the following is not a classification of amplifiers on the basis of their frequency response? a) Capacitively coupled amplifier b) Direct coupled amplifier c) Bandpass amplifier d) None of the mentioned View Answer Answer: d Explanation: None of the options provided are correct. 6. General representation of the frequency response curve is called a) Bode Plot b) Miller Plot c) Thevenin Plot d) Bandwidth Plot View Answer Answer: a Explanation: General representation of frequency response curves are called Bode plot. Bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes. 7. Under what condition can the circuit shown be called a compensated attenuator.

a) C1R1 = C2R2 b) C1R2 = C2R1 c) C1C2 = R1R2 322

d) R1 = 0 View Answer Answer: a Explanation: Standard condition of a compensated attenuator. Here is the derivation for the same.

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8. When a circuit is called compensated attenuator? a) Transfer function is directly proportional to the frequency b) Transfer function is inversely proportional to the frequency c) Transfer function is independent of the frequency d) Natural log of the transfer function is proportional to the frequency View Answer Answer: c Explanation: Transfer function does not has frequency in its mathematical formula. 9. Which of the following is true? a) Coupling capacitors causes the gain to fall off at high frequencies b) Internal capacitor of a device causes the gain to fall off at low frequencies c) All of the mentioned d) None of the mentioned View Answer Answer: d Explanation: Both the statements are false. 10. Which of the following is true? a) Monolithic IC amplifiers are directly coupled or dc amplifiers b) Televisions and radios use tuned amplifiers c) Audio amplifiers have coupling capacitor amplifier d) All of the mentioned View Answer Answer: d Explanation: These all are practical applications of different types of amplifiers.

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Questions & Answers (MCQs) focuses on “Sinusoidal Steady State Analysis”. 1. i(t) = ?

a) 20 cos (300t + 68.2) A b) 20 cos(300t – 68.2) A c) 2.48 cos(300t + 68.2) A d) 2.48 cos(300t – 68.2) A View Answer

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Answer: d

Explanation: 2. Vc(t) = ?

a) 0.89 cos (1000t – 63.43) V b) 0.89 cos (1000t + 63.43) V c) 0.45 cos (1000t + 26.57) V d) 0.45 cos (1000t – 26.57) V View Answer Answer: a

Explanation:

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3. Vc(t) = ?

a) 2.25 cos (5t + 150) V b) 2.25 cos (5t – 150) V c) 2.25 cos (5t + 140.71) V d) 2.25 cos (5t – 140.71) V View Answer Answer: d

Explanation: 4. i(t) = ?

a) 2 sin (2t 5.77) A b) cos (2t 84.23) A c) 2 sin (2t 5.77) A 327

d) cos (2t 84.23) A View Answer Answer: b

Explanation: 5. In the bridge shown, Z1 = 300 ohm, Z2 = 400 – j300 ohm, Z3 = 200 + j100 ohm. The Z4 at balance is

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a) 400 + j300 ohm b) 400 – j300 ohm c) j100 ohm d) -j900 ohm View Answer Answer: b Explanation: Use Z1 x Z4 = Z2 x Z3. Circuit for Q.6 and Q.7

6. i1(t) = ? a) 2.36 cos (4t 41.07) A b) 2.36 cos (4t 41.07) A c) 1.37 cos (4t 41.07) A d) 2.36 cos (4t 41.07) A View Answer

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Answer: c

Explanation: 7. i2(t) = ? a) 2.04 sin (4t 92.13) A b) 2.04 sin (4t 2.13) A c) 2.04 cos (4t 2.13) A d) 2.04 cos (4t 92.13) A View Answer

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Answer: b

Explanation: 8. In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be a) R C b) L C c) R L d) R R View Answer Answer: a Explanation: RC circuit causes a positive shift in the circuit. 9. P = 269 W, Q = 150 VAR (capacitive). The power in the complex form is a) 150 – j269 VA b) 150 + j269 VA c) 269 – j150 VA d) 269 + j150 VA View Answer Answer: c Explanation: S = P – jQ. 10. Q = 2000 VAR, pf = 0.9 (leading). The power in complex form is a) 4129.8 j2000 VA b) 2000 j4129.8 VA c) 2000 j41.29.8 VA d) 4129.8 j2000 VA View Answer Answer: d Explanation: Use cos T = 0.9 or T = 25.84 degrees. Q = S sin T or S = 4588.6 VA p = S cos T or P = 0.9 X 4588.6 4129.8 VA.

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9. Questions & Answers on Operational Amplifiers The section contains questions and answers on ideal operational amplifiers, inverting and noninverting configuration, differentiators and differential amplifiers, operational amplifiers, finite open loop gain effect, circuit performance bandwidth and large signal operations. The Ideal Operational Amplifiers The Inverting Configuration The Non Inverting Configuration Difference Amplifiers

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Integrators and Differentiators DC Imperfections in Operational Amplifiers Effect of Finite Open-Loop gain and Bandwidth on Circuit Perfomance Large Signal Operations on operational Amplifiers

Questions & Answers (MCQs) focuses on “The Ideal Operational Amplifiers”. 1. What is the minimum number of terminals required in an IC package containing four operational amplifiers (quad op amps)? a) 12 b) 13 c) 14 d) 15 View Answer Answer: c Explanation: The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals(4 pins) and one output terminal(2 pins). Another 2 pins are required for power. Similarly, The minimum no of pins required by dual-op-amp is 14: 4*2 + 4*1 + 2 = 14. 2. Which of the following is not a property of an ideal operational amplifier? a) Zero input impedance b) Infinite bandwidth c) Infinite open loop gain d) Zero common-mode gain or conversely infinite common mode-rejection. View Answer Answer: a Explanation: An ideal operational amplifier does not has a zero input impedance.

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3. In an ideal op amp the open-loop gain is 103. The op amp is used in a feedback circuit, and the voltages appearing at two of its three signal terminals are measured as v2 = 0V and v3 = 2V where it is assumed that v1 and v2 are input terminals and v3 is the output terminal. The value of the differential (vd) and common-mode (vcm)signal is a) Vd = 2 mV and vcm = 1 mv b) Vd = 2 mV and vcm = -1 mV c) Vd = 2 mV and vcm = 2mV d) Vd = 2 mV and vcm = -2mV View Answer Answer: b Explanation: Vc = 0.5(V1 + V2) and Vd = V2 – V1. 4. Consider the figure given below. Known that vo = 4V and vi = 2V, determine the gain for the op amp assuming that it is ideal except for the fact that it has finite gain

a) 1001 b) 2002 c) 3003 d) 4004 View Answer Answer: b Explanation: The Voltage at the positive input has to be -3.000v, vi = -3.020v A = vo / vi – vr = -2 / -3.020 -(-3) = 100. 5. Which of the following is not a terminal for the operational amplifier? a) Inverting terminal b) Non-inverting terminal c) Output terminal d) None of the mentioned View Answer

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Answer: d Explanation: There are three terminals for the operational amplifier. 6. Operational amplifiers are a) Differential input and single-ended output type amplifier b) Single-ended input and single-ended output type amplifier c) Single-ended input and differential output type amplifier d) Differential input and differential output type amplifier View Answer Answer: a Explanation: It is another way to refer to op amps based on its terminal characteristics. 7. Express the input voltages v1 and v2 in terms of differential input (vd) and common-mode input(vc). Given v2 > v2. a) Vd = V1 – V2, Vc = 0.5(V1 + V2) b) Vd = V2 – V1, Vc = V1 + V2 c) Vd = V1 – V2, Vc = V1 + V2 d) Vd = V2 – V1, Vc = 0.5(V1 + V2) View Answer Answer: d Explanation: This is the correct mathematical representation. 8. What is the minimum number of pins for a dual operational amplifier IC package? a) 4 b) 6 c) 8 d) 10 View Answer Answer: c Explanation: The minimum no of pins required by dual-op-amp is 8. Each op-amp has 2 input terminals(4 pins) and one output terminal(2 pins). Another 2 pins are required for power. 9. For an ideal operational amplifier (except for the fact that it has finite gain) one set of the value for the input voltages (v2 is the positive terminal v1 is the negative terminal) and output voltage (v0) as determined experimentally is v1= 2.01V, v2=2.00V and v0= -0.99V. Experiment was carried with different values of input and output voltages. Which of the following is not possible considering experimental error? a) v1= 1.99V, v2= 2.00V, v0 = 1.00V b) v1= 1.00V, v2= 1.00V, v0 = 0V 335

c) v1= 1.00V, v2= 1.10V, v0 = 10.1V d) v1= 0.99V, v2= 2.00V, v0 = 1.00V View Answer Answer: d Explanation: Only option d does not satisfies the mathematical relation between the given quantities. 10. What are the units of slew rate? a) Second/Volt b) Volt/second c) It is a ratio, no units d) Ohm/second View Answer Answer: b Explanation: These units are obtained from the definition of the term slew rate.

Questions & Answers (MCQs) focuses on “The Inverting Configuration”. 1. When does a resistance provide a negative feedback to an amplifier? a) Resistance is connected between the positive input terminal and the output terminal b) Resistance is connected between the negative input terminal and the output terminal c) Resistance is connected between the input terminals d) Resistance is connected between the negative input terminal and ground View Answer Answer: b Explanation: An op amp is said to have a negative feedback when a resistance is connected between the input and output terminals respectively. 2. The effect of the inverting configuration is a) The output signal and the input signal are out of phase by 180o b) The output signal and the input signal are in phase 336

c) The output phase is leading the input phase by 90o d) The output phase is lagging behind the input phase by 90o View Answer Answer: a Explanation: Inverting introduces a phase shift of 180o or it ‘inverts’ a peak. 3. For an ideal negative feedback configuration which of the following is true? a) There is a virtual open circuit between the input terminals b) The closed loop gain for a negative feedback does not depend only on the external parameters c) There is a virtual short circuit between the input terminals d) There is a virtual ground at the negative input terminal View Answer Answer: c Explanation: There is always a virtual short circuit in this type of case. There will be a virtual ground if and only if one of the terminals is grounded. 4. The negative feedback causes a) The voltage between the two input terminals to the very small, ideally zero b) The voltage between the two input resistance very high, ideally infinite c) Current flow through the positive input terminal and no current flows through the negative input terminal d) Both a and c View Answer Answer: a Explanation: Ideally the input terminals are at the same potential but in real practice there is a very small potential between the two terminals. 5. The non-inverting closed loop configuration features a high resistance. Therefore in many cases unity gain follower called buffer amplifier is often used to a) Connect a high resistance source to high resistance load b) Connect low resistance source to low resistance load c) Connect low resistance source to a high resistance source d) Connect high resistance source to a low resistance load View Answer Answer: d Explanation: Buffer amplifiers are required to connect a high resistance load to a low input resistance output.

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6. The advantage of a weighted summer operational amplifier is a) It is capable of summing various input voltages together b) Each input signal may be independently adjusted by adjusting the corresponding input resistance c) If one needs both sign of a voltage signal then two operational amplifiers are needed d) All of the mentioned View Answer Answer: d Explanation: All of the mentioned are characteristics of a weighted summer operational amplifier over the traditional amplifier. 7. The following is a circuit of weighted summer capable of summing coefficients of both sign. The expressions for the output voltage v0 is

a) v0 = v1 (R1/Ra ) (Rc/Rb ) + v1 (Ra/R2 ) (Rc/Rb ) – v1 (Rc/R3 ) – v1 (Rc/R4 ) b) v0 = – v1 (R1/Ra ) (Rc/Rb ) – v1 (Ra/R2 ) (Rc/Rb ) + v1 (Rc/R3 ) + v1 (Rc/R4 ) c) v0 = v1 (Ra/R1 ) (Rc/Rb ) + v1 (Ra/R2 ) (Rc/Rb ) – v1 (Rc/R3 ) – v1 (Rc/R4 ) d) v0 = – v1 (Ra/R1 ) (Rc/Rb ) – v1 (Ra/R2 ) (Rc/Rb ) + v1 (Rc/R3 ) + v1 (Rc/R4 ) View Answer Answer: c Explanation: The voltages are increased first by the left side of the portion and then are also magnified by the right side of the circuit. There are four inputs given out of which two are magnified twice and the other are magnified only once. 8. You are provided with an ideal op amp and three 10kΩ resistors. Using series and parallel resistor combinations, how many different inverting-amplifier circuit topologies are possible? a) 2 b) 3 c) 4 d) 5 View Answer Answer: c Explanation: Consider series and parallel combination of the resistances provided and arrange then in the feedback region and as output resistance. 338

9. The loop gain for an ideal operational amplifier with R1 = 10kΩ and R2(negative feedback) = 1MΩ is a) 20 db b) 40 db c) 60 db d) 80 db View Answer Answer: b Explanation: Loop gain in this case is given by 20 log (1000000/10000). 10. In an inverting op-amp circuit for which the gain is −4 V/V and the total resistance used is 100 kΩ. Then the value of R1 and R2 (negative feedback) a) R1 = 20KΩ and R1 = 80KΩ b) R1 = 80KΩ and R1 = 20KΩ c) R1 = 40KΩ and R1 = 60KΩ d) R1 = 50KΩ and R1 = 50KΩ View Answer Answer = a Explanation: Solve R1 + R2 = 100 R2/R1 = 4 for R1 and R2 respectively.

Questions & Answers (MCQs) focuses on “The Non Inverting Configuration”.

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1. In the non-inverting configuration of operational amplifier a) The positive terminal is connected to the ground directly b) The negative terminal is connected to the ground directly c) The positive terminal is connected to the power source d) The negative terminal is connected to the power source View Answer Answer: c Explanation: Non inverting configuration requires a power source connected to the power source. 2. For ideal non-inverting operational amplifier a) Input and output resistances are infinite b) Input resistance is infinite and output resistance is zero c) Input resistance is zero and output resistance is infinite d) Input and output resistances are zero View Answer Answer: b Explanation: It is an ideal characteristic of the non-inverting op amp. 3. For an ideal non-inverting operational amplifier having finite gain (A), the ratio of output voltage (v0) to input voltage (vi) is (given R2 is the feedback resistance) a) (1+R2/R1)/(1+((1+R2/R1)/A)) b) (R2/R1)/(((1+R2/R1)/A)) c) (1+R2/R1)/(((1+R2/R1)/A)) d) (R2/R1)/(1+((1+R2/R1)/A)) View Answer Answer: a Explanation: It is a standard mathematical expression. 4. The gain for an ideal non-inverting operational amplifier is (given R2 is the feedback resistance) a) R2/R1 – 1 b) R2/R1 c) -R2/R1 d) R2/R1 + 1 View Answer Answer: d Explanation: It is a standard mathematical expression. 340

5. While performing an experiment to determine the gain for an ideal operational amplifier having finite gain, a student mistakenly used the equation 1 + R2/R1 where R2 is the feedback resistance. What is the percentage error in his result? Given A is the finite voltage gain of the ideal amplifier used. a) (R2/R1)/(A+ R2/R1) X 100% b) (1+R2/R1)/(A+R2/R1) X 100% c) (1+R2/R1)/(A+1+R2/R1) X 100% d) (R2/R1)/(A+1+R2/R1) X 100% View Answer Answer: c Explanation: The correct formula is (1+R2/R1)/(1+((1+R2/R1)/A)). 6. The finite voltage gain of a non-inverting operational amplifier is A and the resistance used is R1 and R2 in which R2 is the feedback resistance. Under what conditions it can one use the expression 1 + R2/R1 to determine the gain of the amplifier? a) A ~ R2/R1 b) A >> R2/R1 c) A << R2/R1 d) None of the mentioned View Answer Answer: b Explanation: The formula is valid for the ideal case in which the value of A is infinite, practically it should be very large when compared to R2/R1 . 7. Which of the following is not true for a voltage follower amplifier? a) Input voltage is equal to output voltage b) Input resistance is infinite and output resistance is zero c) It has 100% negative feedback d) None of the mentioned View Answer Answer: d Explanation: All the statements are false. 8. For designing a non-inverting amplifier with a gain of 2 at the maximum output voltage of 10 V and the current in the voltage divider is to be 10 μA the resistance required are R1 and R2 where R2 is used to provide negative feedback. Then a) R1 = 0.5 MΩ and R2 = 0.5 MΩ b) R1 = 0.5 kΩ and R2 = 0.5 kΩ c) R1 = 5 MΩ and R2 = 5 MΩ 341

d) R1 = 5 kΩ and R2 = 5 kΩ View Answer Answer: a Explanation: 1 + R2/R1 = 2 and 10/(R1+R2) = 10 μA. Solve for R1 and R2. 9. It is required to connect a transducer having an open-circuit voltage of 1 V and a source resistance of 1 MΩ to a load of 1-kΩ resistance. Find the load voltage if the connection is done (a) directly and (b) through a unity-gain voltage follower. a) 1 μV and 1 mV respectively b) 1 mV and 1 V respectively c) 0.1 μV and 0.1 mV respectively d) 0.1 mV and 0.1 V respectively View Answer Answer: b Explanation: When a unity gain follower is uses then input signal is equal to output signal. When connected directly, output signal is given by 1 X 1kΩ/1MΩ or 1mV. 10. Consider the figure given below. If the resistance R1 is disconnected from the ground and connected to a third power source v3, then expression for the value of v0 is

a) 2v1 + 4v2 − 3v3 b) 6v1 + 8v2 − 3v3 c) 6v1 + 4v2 − 9v3 d) 3v1 + 4v2 − 3v3 View Answer Answer: c Explanation: When a third power source is connected to the resistance of 1kΩ, then also the potential between the two input terminals of op amps remains the same. Using this fact the expression c is obtained. 342

Questions & Answers (MCQs) focuses on “Difference Amplifiers”. 1. For the difference amplifier which of the following is true? a) It responds to the difference between the two signals and rejects the signal that are common to both the signal b) It responds to the signal that are common to the two inputs only c) It has a low value of input resistance d) The efficacy of the amplifier is measured by the degree of its differential signal to the preference of the common mode signal View Answer Answer: a Explanation: All the statements are not true except for the fact that it responds only when there is difference between two signals only. 2. If for an amplifier the common mode input signal is vc, the differential signal id vd and Ac and Ad represent common mode gain and differential gain respectively, then the output voltage v0 is given by a) v0 = Ad vd – Ac vc b) v0 = – Ad vd + Ac vc c) v0 = Ad vd + Ac vc d) v0 = – Ad vd – Ac vc View Answer Answer: c Explanation: It is a standard mathematical expression. 3. If for an amplifier v1 and v2 are the input signals, vc and vd represent the common mode and differential signals respectively, then the expression for CMRR (Common Mode Rejection Ratio) is a) 20 log (|Ad| / |Ac|) b) -10 log (|Ac| / |Ad|)2 c) 20 log (v2 – v1 / 0.5(v2 + v1))

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d) All of the mentioned View Answer Answer: d Explanation: Note that all the expressions are identical. 4. The problem with the single operational difference amplifier is its a) High input resistance b) Low input resistance c) Low output resistance d) None of the mentioned View Answer Answer: b Explanation: Due to low input resistance a large part of the signal is lost to the source’s internal resistance. 5. For the difference amplifier as shown in the figure show that if each resistor has a tolerance of ±100 ε % (i.e., for, say, a 5% resistor, ε = 0.05) then the worst-case CMRR is given approximately by (given K = R2/R1 = R4/R3)

a) 20 log [K+1/4ε]. b) 20 log [K+1/2ε]. c) 20 log [K+1/ε]. d) 20 log [2K+2/ε]. View Answer Answer: a Explanation: None.

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6. For the circuit given below determine the input common mode resistance.

a) (R1 + R3) || (R2) || + (R4) b) (R1 + R4) || (R2 + R3) c) (R1 + R2) || (R3 + R4) d) (R1 + R3) || (R2 + R4) View Answer Answer: c Explanation: Parallel combination of series combination of R1 & R3 with the series combination of R3 and R4 is the required answer as is visible by the circuit. 7. For the circuit shown below express v0 as a function of v1 and v2.

a) v0 = v1 + v2 b) v0 = v2 – v1 c) v0 = v1 – v2 d) v0 = -v1 – v2 View Answer Answer: b Explanation: Considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result. 345

8. For the difference amplifier shown below, let all the resistors be 10kΩ ± x%. The expression for the worst-case common-mode gain is

a) x / 50 b) x / 100 c) 2x / (100 – x) d) 2x / (100 + x) View Answer Answer: d Explanation: None. 9. Determine Ad and Ac for the given circuit.

a) Ac = 0 and Ad = 1 b) Ac ≠ 0 and Ad = 1 c) Ac = 0 and Ad ≠ 1 d) Ac ≠ 0 and Ad ≠ 1 View Answer Answer: a Explanation: Consider the fact that the potential at the input terminals are identical and obtain the values of V1 and V2. Thus obtain the value of Vd and Vc.

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10. Determine the voltage gain for the given circuit known that R1 = R3 = 10kΩ abd R2 = R4 = 100kΩ.

a) 1 b) 10 c) 100 d) 1000 View Answer Answer: b Explanation: Voltage gain is 100/10.

Questions & Answers (MCQs) focuses on “Integrators and Differentiators”. 1. The other name for Miller Circuit is a) Non-Inverting Integrator b) Inverting Integrator c) Non-Inverting Differentiator d) Inverting Differentiator View Answer Answer: b Explanation: Miller Circuit is also called Inverting integrator. 2. The slope of the frequency response of an integrator is a) Linear with negative slope b) Linear with positive slope c) Exponential increase 347

d) Exponential decrease View Answer Answer: a Explanation: The slope is linear and negative. 3. The integrating transfer function has the value of a) jωCR b) –jωCR c) 1 / jωCR d) -1 / jωCR View Answer Answer: d Explanation: Standard mathematical expression for the transfer function. 4. The expression for the integration frequency is a) CR b) 1/CR c) R/C d) C/R View Answer Answer: b Explanation: Standard mathematical expression for the integrator frequency. 5. Determine the expression for the transfer function for the circuit shown below.

a) (Rf/R)/(1+jωCRfC) 348

b) (Rf/R)/(1-jωCRfC) c) – (Rf/R)/(1+jωCRfC) d) – (Rf/R)/(1-jωCRfC) View Answer Answer: c Explanation: It is a standard expression. 6. The frequency transfer function of a differentiator is given by a) jωCR b) 1/jωCR c) – jωCR d) – 1/jωCR View Answer Answer: a Explanation: Standard mathematical expression for the transfer function of a differentiator. 7. The slope of the frequency response of a differentiator is a) Linear with negative slope b) Linear with positive slope c) Exponential increase d) Exponential decrease View Answer Answer: b Explanation: The slope is linear with a positive slope. 8. The phase in the integrator and differentiator circuit respectively are a) +90 degrees and +90 degrees b) -90 degrees and -90 degrees c) -90 degrees and +90 degrees d) +90 degrees and -90 degrees View Answer Answer: d Explanation: These are the characteristics of the integrators and differentiators circuits respectively. 9. Consider a symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator. Find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude. 349

a) 0.25ms b) 0.50ms c) 2.5ms d) 5.0ms View Answer Answer: b Explanation: According to the question 1/CR = 2. 10. The expression for the differentiator time constant is a) CR b) 1/CR c) R/C d) C/R View Answer Answer: a Explanation: Standard mathematical expression for the time constant for the differentiators.

“DC Imperfections in Operational Amplifiers”. 1. Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. What is (approximately) the peak sine-wave input signal that can be applied without output clipping? a) 7 mV b) 10 mV 350

c) 13 mV d) 9mV View Answer Answer: a Explanation: The maximum that can be sent without clipping is 10V – 1000 X 3mV or 7V. (Q2 & Q.3) Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. If the effect of VOs(input offset voltage) is nulled at room temperature (250C), how large an input can one now apply if: 2. The circuit is to operate at a constant temperature? a) 8.5 mV b) 9 mV c) 9.5 mV d) 10 mV View Answer Answer: d Explanation: Explanation: Maximum signal that will not be clipped is 10mV because 10mV X 1000 = 10V. 3. The circuit is to operate at a temperature in the range 0°C to 75°C and the temperature coefficient of VOS is 10 μV/°C? a) 8.5 mV b) 9 mV c) 9.5 mV d) 10 mV View Answer Answer: c Explanation: Since the effect is nullified at 25oC, the peak that can be sent now is given by 10 – (75-25) X 0.1 mV. 4. One of the DC imperfections of the amplifiers are dc offset voltage which is a) Existence of output signal even when the common mode signal is zero b) Existence of common mode signal causing zero output signal c) Existence of output signal even when the differential signal is zero d) Existence of differential signal causing zero output signal View Answer

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Answer: c Explanation: DC offset voltage is existence of output signal even when the differential signal is zero. 5. For the amplifier shown determine the value of the bias current (Ib) and input offset current (Io) respectively.

a) Ib = IB1 + IB2 Io = IB1 – IB2 b) Ib = IB1 + IB2 Io = | IB1 – IB2 | c) Ib = 0.5(IB1 + IB2) Io = | IB1 – IB2 | d) Ib = 0.5(IB1 + IB2) Io = IB1 – IB2 View Answer Answer: c Explanation: Standard mathematical expressions are used with the given variables. 6. Consider the circuit shown below which reduces the impact of the input bias current. If IB1 = IB2 = Input bias current, then determine the value of R3 so that the output voltage (v0) is not impacted by the input bias current.

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a) (R1 R2)/(R1+R2) b) (R1 R2)/(R1-R2) c) R1-(R1 R2)/(R1+R2) d) R2- (R1 R2)/(R1+R2) View Answer Answer: a Explanation: This will be possible when R3 has the same value as the net effect of R1 and R2. 7. Consider an inverting amplifier circuit designed using an op amp and two resistors, R1 = 10 kΩ and R2 = 1 MΩ. If the op amp is specified to have an input bias current of 100 nA and an input offset current of 10 nA, find the output dc offset voltage resulting. a) 0.1 mV b) 1 mV c) 10 mV d) 100 mV View Answer Answer: d Explanation: Use the mathematical definition of bias current and offset current. (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS (offset voltage) = 2 mV and output saturation voltages of ±12 V. 8. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate? a) 3s b) 6s c) 9s d) 12s View Answer 353

Answer: b Explanation: Use vO = VOS (VOS/CR)t. 9. Select the largest possible value for a feedback resistor RF so that at least ±10 V of output signal swing remains available. a) 10 kΩ b) 100 kΩ c) 1 MΩ d) 10 MΩ View Answer Answer: d Explanation: Use vO = VOS (VOS/CR)t. 10. What is the corner frequency of the resulting STC network? a) 1 Hz b) 0.16 Hz c) 0.33 Hz d) 0.5 Hz View Answer Answer: b Explanation: The required answer is given by 1/6 Hz.

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Questions & Answers focuses on “Effect of Finite Open-Loop gain and Bandwidth on Circuit Performance”. 1. An internally compensated op amp is specified to have an open-loop dc gain of 106 dB and a unity gain bandwidth of 3 MHz. Find fb and the open-loop gain at fb. a) 15Hz and 103 db b) 30Hz and 103 db c) 15 Hz and 51.5 db d) 30 Hz and 51.5 db View Answer

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Answer: a Explanation: Use the equations below.

2. A single-pole model has __________ db/decade roll-off of the gain. a) -3 db/decade b) -6 db/decade c) -10 db/decade d) -20 db/decade View Answer Answer: d Explanation: It is a standard characteristic of a single-pole model. 3. Single-pole model is also known as a) Frequent pole b) Stable pole c) Dominant pole d) Responsive pole View Answer Answer: c Explanation: Single-pole model is also called dominant pole. 4. An op amp having a 106-dB gain at dc and a single-pole frequency response with ft = 2 MHz is used to design a non-inverting amplifier with nominal dc gain of 100. The 3-dB frequency of the closed-loop gain is a) 10 kHz b) 20 kHz c) 30 kHz d) 40 kHz View Answer Answer: b Explanation: 356

Use the equation below to obtain a frequency response curve and proceed further.

5. An internally compensated op amp has a dc open-loop gain of 106 V/V and an AC open-loop gain of 40 dB at 10 kHz. Estimate its gain–bandwidth product and its expected gain at 1 kHz. a) 0.1 MHz and -60 db b) 10 MHz and -60 db c) 10 MHz and 60 db d) 1 MHz and 60 db View Answer Answer: d Explanations: Use the following results.

6. An inverting amplifier with nominal gain of −20 V/V employs an op amp having a dc gain of 104 and a unity-gain frequency of 106 Hz. What is the 3-dB frequency f3dB of the closed-loop amplifier? a) 2π 23.8 kHz b) 2π 47.6 kHz c) 2π 71.4 kHz d) 2π 95.2 kHz View Answer

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Answer: b Explanation:

7. cascading two identical amplifier stages, each having a low-pass STC frequency response with a 3dB frequency f1, results in an overall amplifier with a 3dB frequency given by a) √(√2+1) f1 b) √(√3-1) f1 c) √(√2-1) f1 d) √(√3+1) f1 View Answer Answer: c Explanation:

8. Find the ft required for internally compensated op amps to be used in the implementation of the closed loop amplifiers with dc gain of +100 V/V and 3db bandwidth of 100kHz? a) 1 kHz b) 10 kHz c) 100 kHz d) 1 MHz View Answer 358

Answer: b Explanation: None 9. A particular op amp, characterized by a gain–bandwidth product of 20 MHz, is operated with a closed-loop gain of +100 V/V. What 3-dB bandwidth results? At what frequency does the closed-loop amplifier exhibit a −6° phase shift? a) 21 kHz b) 31.5 kHz c) 42 kHz d) 52.5 kHz View Answer Answer: a Explanation:

10. Find the ft required for internally compensated op amps to be used in the implementation of the closed loop amplifiers with dc gain of -2 V/V and 3db bandwidth of 10 MHz? a) 7.5 MHz b) 15 MHz c) 22.5 MHz d) 30 MHz View Answer Answer: d Explanation: -R2 ⁄ R1 = -2V⁄V f3db = 10 MHZ ft = 10 MHZ(2 + 1) = 30 MHZ.

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“Large Signal Operations on operational Amplifiers”. 1. Slew rate of an amplifier is defined as a) Minimum rate of change of the output possible in a real operational amplifier b) Maximum rate of change of the output possible in a real operational amplifier c) Average rate of change of the output possible in a real operational amplifier d) Ratio of the maximum and the average rate of change of the output in a real amplifier View Answer Answer: b Explanation: By definition slew rate is the maximum rate of change of the output possible in a real operational amplifier. 2. Determine the slew rate of the amplifier having full power bandwidth f0 and the rated output voltage as V0. Given that the input signal is of sinusoidal nature. a) 2πf0 V0 b) V0 / 2πf0 c) V0 / f0 d) f0 V0 View Answer Answer: a Explanation: v = V0sin wt dv/dt = wV0 sin wt max value of dv/dt = wV0 max value of w = w0 = 2πf0 w0 V0 = Slew Rate = 2πf0 V0. 3. The units of the full power bandwidth is a) Watt b) Joule c) Seconds d) Hertz View Answer 360

Answer: d Explanation: It has the units of frequency. 4. The full-power bandwidth, fM, is the maximum frequency at which a) an output sinusoid with an amplitude equal to the op-amp rated output voltage (Vo max) can be produced without distortion b) it is the range of the frequencies in which the amplitude of output signal is equal to or greater than half of the op-amp rated output voltage c) it is the range of the frequencies in which the amplitude of output signal is equal to or less than half of the op-amp rated output voltage d) It is the range of the frequencies in which the power gain is half or more than half of the maximum rated power gain of the op-amp View Answer Answer: a Explanation: This is the only statement that satisfies the definition of the full-power bandwidth. 5. Which of the following is not limitation of the operational amplifier a) Output voltage saturation b) Output current limits c) Slew rate d) None of the mentioned View Answer Answer: d Explanation: None of the mentioned are the limitations of the operational amplifier. 6. A particular op amp using ±15-V supplies operates linearly for outputs in the range −12 V to +12 V. If used in an inverting amplifier configuration of gain –100, what is the rms value of the largest possible sine wave that can be applied at the input without output clipping? a) 120 mV b) 60 mV c) 84.85 mV d) 42.42 mV View Answer Answer: c Explanation: Peak value of input wave = 12/100 or 120 mV. Hence the rms value is 120/√2 or 84.85 mV. 7. For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate 361

requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.) a) 10 V/µs b) 20 V/µs c) 40 V/µs d) 80 V/µs View Answer Answer: c Explanation: None. 8. An op amp having a slew rate of 20 V/µs is to be used in the unity-gain follower configuration, with input pulses that rise from 0 to 3 V. What is the shortest pulse that can be used while ensuring full-amplitude output? a) 0.10 µs b) 0.15 µs c) 0.20 µs d) 0.30 µs View Answer Answer: b Explanation: Time taken to reach 3V from 0V with slew rate of 20V/µs is 3/20 µs or 0.15 µs. (Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt = 2 MHz, SR = 1 V/µs, and V0 max = 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi. 9. If Vi = 0.5 V, what is the maximum frequency before the output distorts? a) 31.8 kHz b) 318 kHz c) 3.18 kHz d) 3.18 MHz View Answer Answer: a Explanation: Vi = 0.5v, V0 = 0.5 X 10 = 5V 2πf V0 = SR or f = 31.8 kHz. 10. If f = 20 kHz, what is the maximum value of Vi before the output distorts? a) 0.397 V 362

b) 0.795 V c) 1.192 V d) 1.590 V View Answer Explanation: V0 = 10Vi 2πf V0 = SR = 20πf Vi, here f is 20 kHz, SR is 1 V/µs. Hence the value of Vi is 0.795 V.

10. Questions on MOS Field Effect Transistors (MOSFETs) The section contains questions and answers on basics of MOSFET, device structure, physical and small signal operation of MOSFET, basics of MOSFET configurations and circuit biasing, body effect and discrete MOSFET circuits. MOSFETs Device Strucuture and Physical Operation MOSFETs Current-Voltage Characterisitcs MOSFETs Circuits at DC MOSFET in Amplfier Design MOSFET in Small Signal Operation

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Basic MOSFET Amplifier Configurations Biasing in MOS Amplifier Circuit Discrete-Circuit MOS Amplifiers The Body Effect

Questions and Answers for Campus interviews focuses on “MOSFETs Device Strucuture and Physical Operation”. 1. For NMOS transistor which of the following is not true? a) The substrate is of p-type semiconductor b) Inversion layer or induced channel is of n type c) Threshold voltage is negative d) None of the mentioned View Answer Answer: c Explanation: The threshold voltage is positive for NMOS. 2. Process transconductance parameter is directly proportional to a) Electron mobility only b) (Electron mobility)-1 only 364

c) Oxide capacitance only d) Product of oxide capacitance and electron mobility View Answer Answer: d Explanation: It is the product of the electronic mobility with the oxide capacitance (F/m2). 3. The SI Units of the Process transconductance Parameter (k’) is a) V2/A b) A/V2 c) V/A d) A/V View Answer Answer: b Explanation: k’ = μn Cox where μn is electronic mobility (m2/Vs) and Cox is oxide capacitance is (F/m2). 4. Aspect ratio of the MOSFET has the units of a) No units b) m c) m2 d) m-1 View Answer Answer: a Explanation: It is the ratio of the induced channel width (w) to the induced channel length (l). 5. The MOSFET transconductance parameter is the product of a) Process transconductance and inverse of aspect ratio b) Inverse of Process transconductance and aspect ratio c) Inverse of Process transconductance and inverse of aspect ratio d) Process transconductance and aspect ratio View Answer Answer: d Explanation: This statement only satisfies the mathematical expression. 6. With the potential difference between the source and the drain kept small (VDS is small), the MOSFET behaves as a resistance whose value varies __________ with the overdrive voltage a) Linearly b) Inversely 365

c) Exponentially d) Logarithmically View Answer Answer: b Explanation: For small VDS, resistance r is given by R = 1 / ((μn Cox)(w/l)(VOV)). 7. For a p channel MOSFET which of the following is not true? a) The source and drain are a p type semiconductor b) The induced channel is p type region which is induced by applying a positive potential to the gate c) The substrate is a n type semiconductor d) None of the mentioned View Answer Answer: b Explanation: The induced channel is p type region which is induced by applying a negative potential to the gate. 8. When the voltage across the drain and the source (VDS) is increased from a small amount (assuming that the gate voltage, VG with respect to the source is higher than the threshold voltage, Vt), then the width of the induced channel in NMOS (assume that VDS is always small when compared to the Vov) a) Will remain as was before b) Will become non uniform and will take a tapered shape with deepest width at the drain c) Will become non uniform and will take a tapered shape with deepest width at the source d) Will remain uniform but the width of the channel will increase View Answer Answer: c Explanation: The voltage across the source will be VOV and the voltage will decrease linearly to VOV – VDS as we reach the drain end. The width of the induced channel is proportional to the voltage. 9. The saturation current of the MOSFET is the value of the current when a) The voltage between the drain and drain becomes equal to the overdrive voltage b) The voltage between the drain and drain becomes equal to the threshold voltage c) The voltage between the drain and drain becomes equal to the voltage applied to the gate d) The voltage between the drain and drain becomes equal to difference the overdrive voltage and the threshold voltage View Answer 366

Answer: a Explanation: By definition of the MOSFET saturation current. 10. At channel pinch off a) The width of the induced channel becomes non linear b) The width of the induced channel becomes very large (resulting in very large resistance and very low, practically zero, current) c) width becomes 1/e times the maximum possible width d) The width of the induced channel becomes zero and the current saturates View Answer Answer: d Explanation: It is a characteristics of a channel pinch off.

Questions & Answers (MCQs) focuses on “MOSFETs Current-Voltage Characterisitcs”.

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1. If a MOSFET is to be used in the making of an amplifier then it must work in a) Cut-off region b) Triode region c) Saturation region d) Both cut-off and triode region can be used View Answer Answer: c Explanation: Only in the saturation region a MOSFET can operate as an amplifier. 2. For MOSFET is to be used as a switch then it must operate in a) Cut-off region b) Triode region c) Saturation region d) Both cut-off and triode region can be used View Answer Answer: d Explanation: In both regions it can perform the task of a switch. (Q.3 & Q.4) Using the circuit shown below,

3. Determine the conditions in which the MOSFET is operating in the triode region. i. VGD > Vt (Threshold voltage) ii. VDS > VOV iii. ID ∝ (VOV – 0.5VDS)VDS a) i, ii, and iii are correct b) i and iii are correct c) i and ii are correct d) ii and iii are correct View Answer 368

Answer: b Explanation: Only the points I and iii are correct and ii is false. 4. Determine the conditions in which the MOSFET is operating in the saturation region i. VGD > Vt (Threshold voltage) ii. VDS > VOV iii. ID ∝ (VOV)2 a) i, ii, and iii are correct b) i and iii are correct c) i and ii are correct d) ii and iii are correct View Answer Answer: d Explanation: i is false and ii and iii are true. 5. In the saturation region of the MOSFET the saturation current is a) Independent of the voltage difference between the source and the drain b) Depends directly on the voltage difference between the source and the drain c) Depends directly on the overdriving voltage d) Depends directly on the voltage supplied to the gate terminal View Answer Answer: a Explanation: Saturation current does not depends on the voltage difference between the source and the drain in the saturation region of a MOSFET. 6. An n-channel MOSFET operating with VOV=0.5V exhibits a linear resistance = 1 kΩ when VDS is very small. What is the value of the device transconductance parameter kn? a) 2 mA/V2 b) 20 mA/V2 c) 0.2 A/V2 d) 2 A/V2 View Answer Answer: a Explanation: Use the standard mathematical expression to determine the value of kn. 7. An NMOS transistor is operating at the edge of saturation with an overdrive voltage VOV and a drain current ID. If is VOV is doubled, and we must maintain operation at the edge of saturation, what value of drain current results? 369

a) 0.25ID b) 0.5ID c) 2ID d) 4ID View Answer Answer: c Explanation: I0 is directly proportional to VOS. (Q.8-Q.10) Using the circuit below answer the question

8. Which of the following is true for the triode region? a. VDG > Vtp b. VSD < VOV c. ID ∝ VOV d. None of the mentioned View Answer Answer: d Explanation: VDG > |Vtp| and VSD < |VOV|. 9. Which of the following is true for the saturation region? a) VDG ≤ |Vtp| b) VSD ≤| VOV| c) VDG < |Vtp| d) VSD <| VOV| View Answer Answer: a Explanation: It is a characteristic for the saturation region. 10. The current iD a) Depends linearly on VOV in the saturation region b) Depends on the square of VOV in the saturation region 370

c) Depends inversely on VOV in the triode region d) None of the mentioned View Answer Answer: b Explanation: Use the standard mathematical expressions for i0 in different regions. Questions and Answers focuses on “MOSFETs Circuits at DC”. 1. The transistor in the circuit shown below has kn = 0.4 mA/V2, Vt = 0.5 V and λ = 0. Operation at the edge of saturation is obtained when

a) (W/L)RD = 0.5 kΩ b) (W/L)RD = 1.0 kΩ c) (W/L)RD = 1.5 kΩ d) (W/L)RD = 2.0 kΩ View Answer Answer: c Explanation: Use the standard formula for edge saturation. 2. The PMOS transistor in the circuit shown has Vt = −0.7 V, μpCox = 60 μA/V2, L = 0.8 μm, and λ = 0. Find the value of R in order to establish a drain current of 0.115 mA and a voltage VD of 3.5 V.

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a) 12.5 KΩ b) 25 kΩ c) 37.5 kΩ d) 50 kΩ View Answer Answer: a Explanation:

3. The NMOS transistors in the circuit shown have Vt = 1 V, μnCOX = 120 μA/V2, λ = 0, and L1 = L2 = L3 = 1μm. Then which of the following is not the value of the width of these MOSFETs shown a) 2 µm b) 8µm c) All of the mentioned d) None of the mentioned View Answer Answer: d Explanation:

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4. The MOSFET shown has Vt = 1V, kn = 100µA/V2 and λ = 0. Find the required values of W/L and of R so that when vI = VDD = +5 V, rDS = 50 Ω, and VO = 50 mV.

a) W/L = 25 and R = 4.95 kΩ b) W/L = 25 and R = 9.90 kΩ c) W/L = 50 and R = 4.95 kΩ d) W/L = 50 and R = 9.90 kΩ View Answer Answer: c Explanation:

(Q.5-Q.7) For each of the circuits shown find the labeled voltages. For all transistors, kn(W/L) = 1 mA/V2, Vt = 2V, and λ = 0 5. Find V3

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a) 2.41V b) 3.41V c) 4.41V d) 1.41V View Answer Answer: b Explanation: V3 = 10- 4 * 2 + 1.4 = 3.4v. 6. Find V4 and V5

a) 4V and -5V respectively b) -4V and 5V respectively c) 4V and 5V respectively d) -4V and -5V respectively View Answer 374

Answer: a

Explanation: 7. Find V1 and V2 a) 2V and -4V b) -2V and 4V c) 2V and 4V d) -2V and -4V View Answer Answer: c Explanation: ID = 1 = 1⁄2 * 1 * (VGS – 2)2 => VGS = 3.41v. V3 = 3.41v. (Q.8 & Q.9) For each of the circuits shown find the labeled node voltages. The NMOS transistors have Vt = 1 V and kn( W/L ) = 2 mA/V2 and λ = 0 8. Find V1 and V2

a) 2.44 and -1.28 V b) 2.44 and -2.56 V c) 1.22 and -2.56 V d) 1.22 and -1.28 V View Answer 375

Answer: b Explanation:

9. Find V3 and V4

a) 3.775V and 5V b) 3.775V and 2.55V c) 7.55V and 2.55V 376

d) 7.555V and 5V View Answer Answer: d Explanation:

10. For the PMOS transistor in the circuit shown kn= 8 µA/V2, W/L = 25,|Vtp| = 1V and I = 100μA. For what value of R is VSD = VSG?

a) 0 Ω b) 12.45 kΩ c) 25.9 kΩ d) 38.35 kΩ View Answer

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Answer: a Explanation: VSG will be equal to VSD only when the resistance shown is zero or in other words there should not be any resistance.

Questions & Answers (MCQs) focuses on “MOSFET in Amplfier Design”. (Q.1 & Q.2) Various measurements are made on an NMOS amplifier for which the drain resistor RD is 20 kΩ. First, DC measurements show the voltage across the drain resistor, VRD, to be 2 V and the gate-to-source bias voltage to be 1.2 V. Then, ac measurements with small signals show the voltage gain to be −10 V/V. 1. What is the value of Vt for this transistor? a) 0.7V b) 0.8V c) 0.9V d) 1.0V View Answer

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Answer: b Explanation:

2. If the process transconductance parameter is 50μA/V2, what is the MOSFET’s W/L? a) 25 b) 50 c) 75 d) 100 View Answer Answer: a

Explanation:

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(Q.3-Q.5) Consider the amplifier below for the case VDD = 5 V, RD = 24 kΩ, (W/L) = 1 mA/V2, and Vt = 1 V.

3. If the amplifier is biased to operate with an overdrive voltage VOV of 0.5 V, find the incremental gain at the bias point. a) -3 V/V b) -6 V/V c) -9 V/V d) -12 V/V View Answer Answer: d Explanation:

4. For amplifier biased to operate with an overdrive voltage of 0.5V, and disregarding the distortion caused by the MOSFET’s square-law characteristic, what is the largest amplitude of a 380

sine-wave voltage signal that can be applied at the input while the transistor remains in saturation? a) 1.61 V b) 1.5 V c) 0.11 V d) 3.11 V View Answer Answer: b Explanation: Use the standard mathematical formula to obtain the result. 5. For the input signal of 1.5V what is the value of the gain value obtained? a) -12.24 V/V b) -12.44 V/V c) -12.64 V/V d) -12.84 V/V View Answer Answer: c Explanation: The amplitude of the output voltage signal that results is approximately equal to Voq – VOB = 2 – 0.61 = 1.39v. The gain implied by amplitude is Gain = -1.39/0.11 = -12.64 V/V. 6. Which of the following is the fastest switching device? a) JEFT b) Triode c) MOSFET d) BJT View Answer Answer: c Explanation: MOSFET is the fastest switching device among the given four options. 7. Bias point is also referred by the name a) DC Operating Point b) Quiescent Point c) None of the mentioned d) All of the mentioned View Answer

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Answer: d Explanation: Bias point is called dc operating point as the MOSFET functions best at this point. Also since at the bias point no signal component is present it is called quiescent point (he reason why it is represented by the symbol ‘Q’) (Q.8 –Q.10) Consider the amplifier circuit shown below. The transistor is specified to have Vt = 0.4 V, kn = 0.4 mA/V2, W/L = 10 and λ = 0. Also, let VDD = 1.8V, RD = 17.5kΩ, VGS = 0.6V and vgs = 0V.

8. Find ID. a) 0.08 mA b) 0.16 mA c) 0.4 mA d) 0.8 mA View Answer

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Answer: a

Explanation: 9. Find VDS. a) 0.1V b) 0.2 V c) 0.4 V d) 0.8 V View Answer Answer: c

Explanation: 10. Find Av. a) -12 V/V b) -14 V/V c) -16 V/V d) -18 V/V View Answer Solution: b Explanation: Av = – kn Vov RD = -0.4 * 10 * 0.2 * 17.5 = – 14.4v

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Questions and Answers focuses on “MOSFET in Small Signal Operation”. 1. An NMOS technology has μnCox = 50 μA/V2 and Vt = 0.7 V. For a transistor with L = 1μm, find the value of W that results in gm 1mA/V at ID = 0.5 mA. a) 10 μm b) 20 μm c) 30 μm d) 40 μm View Answer Answer: b Explanation:

2. Consider an NMOS transistor having kn= 2 mA/V2. Let the transistor be biased at VOV = 1V. For operation in saturation, what dc bias current ID results? If a +0.1-V signal is superimposed on VGS, find the corresponding increment in collector current by evaluating the total collector 384

current ID and subtracting the dc bias current ID. a) ID = 1mA and Increment = 0.21 mA b) ID = 1mA and Increment = 0.42 mA c) ID = 2mA and Increment = 0.21 mA d) ID = 2mA and Increment = 0.42 mA View Answer Answer: a Explanation:

3. We know ID =1/2 kn (VGS + vgs – Vt)2. Let the signal vgs be a sine wave with amplitude Vgs, and substitute vgs = Vgs sin ω t in Eq.(5.43). Using the trigonometric identity show that the ratio of the signal at frequency 2ω to that at frequency ω , expressed as a percentage (known as the second-harmonic distortion) is a) Vgs/Vov x 100% b) 1/2Vgs/Vov x 100% c) 1/4Vgs/Vov x 100% d) 1/8Vgs/Vov x 100% View Answer Answer: c Explanation:

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4. If in a particular application Vgs is 10 mV, find the minimum overdrive voltage at which the transistor should be operated so that the second-harmonic distortion is kept to less than 1%. a) 1V b) 0.75V c) 0.5V d) 0.25V View Answer Answer: d Explanation:

(Q.5-Q.7) An NMOS amplifier is to be designed to provide a 0.50-V peak output signal across a 50-kΩ load that can be used as a drain resistor. 5. If a gain of at least 5 V/V is needed, what value of gm is required? a) 0.1 mA/V b) 0.2 mA/V c) 0.4 mA/V d) 0.8 mA/V View Answer Answer: a Explanation: gmRd = 5 or gm= 5/50 mA/V. 6. Using a dc supply of 3 V, what values of ID and VOV would you choose? a) 0.34 mA and 0.35 V respectively b) 0.34 mA and 0.69 V respectively c) 0.034 mA and 0.35 V respectively d) 0.034 mA and 0.69 V respectively View Answer

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Answer: d Explanation:

7. What W/L ratio is required if μnCox = 200 μA/V2? a) 1.23 b) 1.23 c) 1.43 d) 1.53 View Answer Answer: c Explanation:

(Q.8-Q.9) For a 0.8-μm CMOS fabrication process: Vtn= 0.8 V, Vtp = −0.9 V, μnCox = 90 μA/V2, μpCox = 30 μA/V2, Cox = 1.9 fF/μm2, VA (n-channel devices) = 8L (μm), and |VA| (p-channel devices) = 12L (μm). 8. Find the small-signal model parameters (gm, ro and gmb) for an NMOS transistor having W/L = 20 μm/2 μm and operating at ID = 100 μA and |VSB| = 1V. a) gm= 0.42mA/V, ro= 160 kΩ, gmb = 0.084 mA/V b) gm= 0.21mA/V, ro= 160 kΩ, gmb= 0.042 mA/V 387

c) gm= 0.42mA/V, ro= 80 kΩ, gmb = 0.042 mA/V d) gm= 0.24mA/V, ro= 80 kΩ, gmb = 0.084 mA/V View Answer Answer: a Explanation:

9. Find the small-signal model parameters (gm, ro and gmb) for a PMOS transistor having W/L = 20 μm/2 μm and operating at ID = 100 μA and |VSB| = 1V. a) gm= 0.24mA/V, ro= 240 kΩ, gmb = 0.024 mA/V b) gm= 0.24mA/V, ro= 120 kΩ, gmb = 0.048 mA/V c) gm= 0.24mA/V, ro=240 kΩ, gmb = 0.048 mA/V d) gm= 0.12mA/V, ro= 240 kΩ, gmb = 0.048 mA/V View Answer Answer: c Explanation:

10. The overdrive voltage at which each device must be operating is a) NMOS = 0.83V and PMOS = 0.48V b) NMOS = 0.48V and PMOS = 0.83V 388

c) NMOS = 0.24V and PMOS = 0.41V d) NMOS = 0.41V and PMOS = 0.24V View Answer Answer: b Explanation: NMOS case

PMOS case

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Questions & Answers (MCQs) focuses on “Basic MOSFET Amplifier Configurations”. 1. In which of the following configuration does a MOSFET works as an amplifier? a) Common Source (CS) b) Common Gate (CG) c) Common drain (CD) d) All of the mentioned View Answer Answer: d Explanation: There are three basic configurations for connecting the MOSFET as an amplifier. Each of these configurations is obtained by connecting one of the three MOSFET terminals to ground, thus creating a two-port network with the grounded terminal being common to the input and output ports. 2. The MOSFET in the following circuit is in which configuration?

a) Common Source (CS) b) Common Gate (CG) c) Common Drain (CD) d) None of the mentioned View Answer Answer: b Explanation: It is the circuit for Common gate configuration.

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3. The MOSFET in the following circuit is in which configuration?

a) Common Source (CS) b) Common Gate (CG) c) Common Drain (CD) d) None of the mentioned View Answer Answer: c Explanation: It is the circuit for Common drain configuration. 4. The MOSFET in the following circuit is in which configuration?[/expand]

a) Common Source (CS) b) Common Gate (CG) c) Common Drain (CD) d) None of the mentioned View Answer Answer: a Explanation: It is the circuit for Common source configuration.

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(Q.5-Q.10) Reference circuit for Q.5-Q.10 The circuit below is the characterization for the amplifier as a functional block.

5. If the value of Rin for the common source configuration is R1 and that for common source with a source resistance configuration is R2 ideally. The ratio of R1/R2 will be a) R1/R2 = 1 b) 0 < R1/R2 < 1 c) R1/R2 > 1 d) R1/R2 = 0 View Answer Answer: a Explanation: Ideally both must have infinite resistance. 6. Which is true for the value of Avo for common source (Represented by A1) and common source with a source resistance (represented by A2). a) A1 = A2 b) A1 > 2 c) A1 < A2 d) |A1| < |A2| View Answer Answer: c Explanation: A1 = -gmRD and A2 = -gmRD/1+gmRS Reference circuit for Common source configuration

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Reference circuit for common source with source resistance RS

7. Which of the following is true for the voltage gain (AV) for the common source configuration (represented by A1) and the common gate configuration (represented by A2)? a) A1 = A2 b) |A1| = |A2| and A1 ≠ A2 c) |A1| > |A2| d) |A1| < |A2| View Answer Answer: b Explanation: A1 = -gm(RL||RD) and A1 = gm(RL||RD) Reference figure for common source configuration

Reference figure for common gate configuration

8. The value of the voltage gain (Av) for the common source with source resistance (represented by A1) and common gate configuration (represented by A2) are related to each other by a) A1 > A2 393

b) |A1| > |A2| c) A1 < A2 d) A1 > A2 and |A1| > |A2| View Answer Answer: c Explanation: A1 = – gm(RL||RD)/ 1 + gmRS and A2 = gm(RL||RD) Reference figure for common source with source resistance configuration

Reference figure for common gate configuration

9. In which of the following configuration is the input resistance (Ri) not equal to zero ideally? a) Common source configuration b) Common source configuration with source resistance c) Common gate configuration d) Source follower configuration View Answer

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Answer: c Explanation: Refer to the circuit for the common gate configuration

10. Which of the following has AVO independent of the circuit elements? a) Common source configuration b) Common gate configuration c) Source follower configuration d) None of the mentioned View Answer Answer: c Explanation: AVO = 1 source follower. “Biasing in MOS Amplifier Circuit”. (Q.1-Q.2) A discrete MOSFET common source amplifier has Rin = 2 MΩ, gmm = 4mA/V, ro = 100 kΩ, RD = 10 kΩ, Cgs = 2pF and Cgd = 0.5pF. The amplifier is fed from a voltage source with an internal resistance of 500 kΩ and Is connected to the a load of 10 kΩ. 1. The value of the overall mixed gain AM is a) -15.2 V/V b) -16.2 V/V c) -17.2 V/V d) -18.2 V/V View Answer

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Answer: a Explanation:

2. The upper 3-db frequency fH is a) 11.1 kHz b) 22.1 kHz c) 33.1 kHz d) 44.1 kHz View Answer Answer: c Explanation:

(Q.3-Q.4) The amplifier in the figure shown below is biased to operate at ID = 1mA and gm = 1mA/V.

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3. Find the midband gain. a) 0.43 V/V b) 1.43 V/V c) 2.43 V/V d) 3.43 V/V View Answer Answer: b Explanation:

4. Find the value of CS that places FL at 10Hz a) 6.57 µF b) 12.57 µF c) 18.57 µF d) 24.57 µF View Answer

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Answer: c Explanation:

(Q.5-Q.6) )In the NMOS transistor of the circuit shown below is biased to have gm = 1mA/V and r0 = 100 kΩ.

5. Find AM a) 1.02 V/V b) 2.04 V/V c) 3.06 V/V d) 4.08 V/V View Answer Answer: c Explanation:

6. Find fH if Cgs = 1 pF and Cgd = 0.2 pF. a) 875 kHz 398

b) 875 kHz c) 875 kHz d) 875 kHz View Answer Answer: c Explanation:

(Q.7-Q.8) For a particular depletion mode NMOS device, Vt = -2V, kn W/L = 200 µA/V2 and λ = 0.02V-1. For VDS = 2V 7. What is the drain current that flows a) 104 µA b) 208 µA c) 312µA d) 416 µA View Answer Answer: d Explanation:

8. What is the value of the drain current if both L and W are doubled? a) 408 µA b) 412 µA c) 416 µA d) 420 µA View Answer

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Answer: a Explanation:

(Q.9-Q.10) A depletion type N channel MOSFEt with knW/L = 2 mA/V2 and Vt = 3V has its source and gate grounded. For Vd = 0.1V and neglecting channel length modulating effect 9. Find drain current. a) 1.18 mA b) 0.89 mA c) 0.59 mA d) 0.3 mA View Answer Answer: b Explanation:

10. In which region is the triode operating? a) Triode region b) End of saturation region c) Saturation region d) None of the mentioned View Answer

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Answer: a Explanation:

Questions and Answers focuses on “Discrete-Circuit MOS Amplifiers”. 1. Calculate the overall voltage gain Gv of a common source amplifier for which gm = 2mA/V, RD = 10 kΩ, R0 = 10 kΩ and RG = 10 MΩ. The amplifier is fed from a signal source of Thevenin resistance of 0.5MΩ and the amplifier is coupled with a load of 10 kΩ. a) -11.2 V/V b) -22.4 V/V c) -33.6 V/V d) -44.8 V/V View Answer Answer: a Explanation:

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(Q.2-Q.6) The MOSFEt circuit below has Vt = 1V, knW/L = 0.8 mA/V2 and VA = 40V

2. Find the value of RG so that iD = 0.1 mA, the largest possible value of RD is used while the maximum signal swing at the drain is of 1V and the input resistance at the gate is 10 MΩ. a) 1 MΩ b) 10 MΩ c) 0.1 MΩ d) 0.01 MΩ View Answer Answer: b Explanation:

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3. Find the value of gm at the bias point a) 0.1 mA/V b) 0.2 mA/V c) .0.3 mA/V d) 0.4 mA/V View Answer Answer: d Explanation:

4. If terminal Z is grounded, X is connected to a signal source having a resistance of 1 MΩ and terminal Y is connected to a load resistance of 40 kΩ, find the voltage gain from the signal source to the load. a) 3.25 v/V b) 6.5 V/V c) 9.75 V/V d) 13 V/V View Answer Answer: b Explanation:

5. If terminal Y is grounded find the voltage gain from X to Z with Z open-circuit. a) 0.33 V/V b) 0.66 V/V c) 0.99 V/V 403

d) None of the mentioned View Answer Answer: c Explantion:

6. If terminal X is grounded and terminal Z is connected to a current source delivering a current of 10 µA and having a resistance of 100 kΩ, find the voltage signal that can be measured at Y neglecting the effect of V0 . a) 0.34V b) 0.68 V c) 3.4 V d) 6.8 V View Answer Answer: a Explanation:

(Q.7-Q.8) The NMOS transistor in source follower circuit shown has gm = 5mA/V and a large r0 .

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7. Find the output resistance. a) 0.1 kΩ b) 0.2 kΩ c) 0.3 kΩ d) 0.4 kΩ View Answer Answer: b Explanation:

8. Find the open-Circuit voltage gain. a) 1 V/V b) 2 V/V c) 3 V/V d) 4 V/V View Answer

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Answer: a Explanation (Q.7-Q.8):

9. The NMOS transistor in the common gate amplifier as shown in the circuit below has gm = 5 mA/V. Find the input resistance and the voltage gain.

a) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V b) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V c) Input resistance: 0.2 kΩ and Voltage gain: 3.05 V/V d) Input resistance: 0.2 kΩ and Voltage gain: 7.1 V/V View Answer

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Answer: d Explanation:

10. If the output of the source follower in (I) is connected to the input of the common gate amplifier of (II). Determine the overall voltage gain (V0 /Vi ).

a) 1.55 V/V b) 3.55 V/V c) 5.55 V/V d) 7.55 V/V View Answer 407

Answer: b Explanation:

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Questions & Answers (MCQs) focuses on “The Body Effect”. 1. The _____________ of a MOSFET is affected by the voltage which is applied to the back contact. a) Threshold Voltage b) Output Voltage c) Both threshold and output voltage d) Neither of the threshold nor the output voltage View Answer Answer: a Explanation: The voltage difference between the source and the bulk, VBS changes the width of the depletion layer and therefore also the voltage across the oxide due to the change of the charge in the depletion region. This results in a difference in threshold voltage which equals the difference in charge in the depletion region divided by the oxide capacitance. 2. The variation of the threshold voltage with the applied bulk-to-source voltage is typically observed by plotting the _________________ as a function of the source-to-drain voltage. a) drain current b) square root of the drain current c) square of the drain current d) natural logarithm of the drain current View Answer Answer: b Explanation: The change in threshold current is directly proportional to the square root of the drain current. For further assistance check the mathematical expression for the same. 3. The SI units of the body effect parameter is a) Volt b) Volt X Volt c) √Volt d) It has no units View Answer

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Answer: c Explanation: Vt = Vt0 + k[√(Vsb + 2φf) – √2φf]. In this expression k is the body effect parameter hence its units can be determined. 4. An NMOS transistor has Vt0 = 0.8 V, 2 φf = 0.7 V, and γ = 0.4 V1/2. Find Vt when VSB = 3 V. a) 0.12 V b) 1.23 V c) 2.34 V d) 3.45 V View Answer Answer: b Explanation: Vt = Vt0 + k[√(Vsb + 2φf) – √2φf]. use this expression to obtain the desired result. 5. The threshold voltage is a) Increases on increasing temperature b) May increase or decrease on increasing temperature depending upon other factors c) Temperature independent d) Decreases on increasing temperature View Answer Answer: d Explanation: The threshold voltage depends only on the temperature and it decreases by roughly 2 mV for every degree Celsius increase in the temperature. 6. As the voltage on the drain is increased, a value is reached at which the pn junction between the drain region and substrate suffers avalanche breakdown known as a) Weak avalanche b) Strong avalanche c) Weak storm d) Punch-through View Answer Answer: a Explanation: As the voltage on the drain is increased, a value is reached at which the pn junction between the drain region and substrate suffers avalanche breakdown. This breakdown usually occurs at voltages of 20 V to 150 V and results in a somewhat rapid increase in current (known as a weak avalanche). 7. A breakdown effect that occurs in modern devices at low voltages (of around 20 V) is a) Weak avalanche 410

b) Strong avalanche c) Weak storm d) Punch-through View Answer Answer: d Explanation: Punch-through occurs in devices with relatively short channels when the drain voltage is increased to the point that the depletion region surrounding the drain region extends through the channel to the source. The drain current then increases rapidly. Normally, punchthrough does not result in permanent damage to the device. 8. At ______________ the drain current is no longer related to the Vgs by square law relationship. a) When the temperature is high (around 700 Celsius) b) When temperature is very low (around -50 Celsius) c) Velocity saturation d) None of the mentioned View Answer Answer: c Explanation: At velocity saturation the current depends linearly on the Vgs. 9. In MOSFETs a breakdown may occur at around 30 V. This is due to a) Velocity saturation b) Breakdown of the gate diode c) Sudden decrease in the depletion region d) Fall of the threshold voltage due to impurities View Answer Answer: b Explanation: The breakdown of the oxide at the gate may occur when the voltage is around 30 V. This may also permanently damage the device. 10. Which of the below issues may not be experienced when using MOSFETs? a) Weak avalanche b) Velocity saturation c) Punch-through d) All of the mentioned View Answer

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Answer: d Explanation: All of the mentioned are some of the common issues that one may face while dealing with MOSFETs.

11. Questions & Answers on Bipolar Junction Triodes (BJTs) The section contains questions on basics of BJT, device structures and physical operations, circuits, current-voltage properties, amplifier design, signal operations and circuit configuration and biasing. BJTs Device Strucutres and Physical Operations BJTs Current-Voltage Characteristics BJT Circuits at DC BJT in Amplifier Design

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Small Signal Operations and Model Basic BJT Amplifier Configuration Biasing in BJT Amplifier Circuits Spread Spectrum

Questions & Answers (MCQs) focuses on “BJTs Device Strucutres and Physical Operations”. 1. Which of the following is not a part of a BJT? a) Base b) Collector c) Emitter d) None of the mentioned View Answer Answer: d Explanation: BJT consists of three semiconductor regions, base region, emitter region and collector region.

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2. The number of pn junctions in a BJT is/are a) 1 b) 2 c) 3 d) 4 View Answer Answer: b Explanation: There are two pn junctions, base-emitter junction and collector-emitter junction respectively. 3. In which of the following modes can a BJT be used? a) Cut-off mode b) Active mode c) Saturation mode d) All of the mentioned View Answer Answer: d Explanation: These three are the defined regions in which a BJT operates. 4. If a BJT is to be used as an amplifier, then it must operate in___________ a) Cut-off mode b) Active mode c) Saturation mode d) All of the mentioned View Answer Answer: b Explanation: A BJT operates as an amplifiers in active mode and as a switch in cut-off or saturation mode. 5. If a BJT is to be used as a switch, it must operate in____________ a) Cut-off mode or active mode b) Active Mode or saturation mode c) Cut-off mode or saturation mode d) Cut-off mode or saturation mode or active mode View Answer Answer: c Explanation: A BJT operates as an amplifiers in active mode and as a switch in cut-off or saturation mode. 414

6. In cut off mode a) The base-emitter junction is forward biased and emitter-collector junction is reversed biased b) The base-emitter junction is forward biased and emitter-collector junction is forward biased c) The base-emitter junction is reversed biased and emitter-collector junction is reversed biased d) The base-emitter junction is reversed biased and emitter-collector junction is forward biased View Answer Answer: c Explanation: In cut-off mode there is no current flowing through the BJT hence both junctions must be reversed biased else if either of them is forward biased then the current will flow. 7. On which of the following does the scale current not depends upon? a) Effective width of the base b) Charge of an electron c) Electron diffusivity d) Volume of the base-emitter junction View Answer Answer: d Explanation: The saturation current does not depends upon the volume of the base-emitter junction. Instead it depends upon the area of the cross section of the base-emitter junction in a direction perpendicular to the flow of current. 8. On which of the following does the collector current not depends upon? a) Saturation current b) Thermal voltage c) Voltage difference between the base and emitter d) None of the mentioned View Answer Answer: d Explanation: Collector current depends linearly of the saturation current and exponentially to the ratio of the voltage difference between the base and collector and thermal voltage. 9. The range for the transistor parameter also referred as common-emitter current gain has a value of__________ for common devices. a) 50-200 b) 400-600 c) 750-1000 d) > 1000 View Answer

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Answer: a Explanation: Most commonly used transistors have a voltage gain of in the range of 50-200. Only some specially designed transistors have a transistor parameter in the range of 1000. 10. The collector current Ic is related to the emitter current Ie by a factor k. If b is the transistor parameter then the value of k in terms of b is a) k = b/(b + 1) b) k = (b + 1)/b c) b = (k + 1)/k d) None of the mentioned View Answer Answer: a Explanation: Ic = k Ie (given) and also Ie = (b + 1)/b Ic (standard result). Equating these two results we get k = b/(b + 1).

Questions & Answers (MCQs) focuses on “BJTs Current-Voltage Characteristics”.

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1. The curve between the collector current versus the potential difference between the base and emitter is a) A straight line inclined to the axes b) A straight line parallel to the x-axis c) An exponentially varying curve d) A parabolic curve View Answer Answer: c Explanation: The natural logarithm of the collector current depends directly on the the potential difference between the base and the emitter. 2. The curve between the collector current and the saturation is a) A straight line inclined to the axes b) A straight line parallel to the x-axis c) A straight line parallel to the y-axis d) An exponential curve View Answer Answer: a Explanation: The collector current depends directly on the saturation current. 3. The magnitude of the thermal voltage is given by a) k/Tq b) kT/q c) q/Kt d) Tk/q View Answer Answer: b Explanation: kT/q is the correct mathematical expression for the thermal voltage. 4. The correct relation between the transistor parameters α and ß are related by a) ß = 1 – α/α b) ß = 1 + α/α c) α = ß + 1/ß d) α = ß/ß + 1 View Answer Answer: d Explanation: Only expression α = ß/ß + 1 is the correct expression that relates α and ß.

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5. The correct expression relating the emitter current Ie to the collector current Ic is a) Ie = α Ic b) Ic = α Ic c) Ie = ß Ic d) Ic = ß Ic View Answer Answer: b Explanation: Ie = Ic/α or Ic = α Ie 6. The value of the thermal voltage at room temperature can be approximated as a) 25 mV b) 30 mV c) 35 mV d) 40 mV View Answer Answer: a Explanation: Thermal voltage is given by kT/q which at T = 25 degrees Celsius is approximately 25 mV. 7. The correct relation between the emitter current Ie and the base current Ib is given by a) Ib = (1 + α) Ie b) Ib = (α – 1) Ie c) Ie = (1 – ß) Ib d) Ie = (1 + ß) Ib View Answer Answer: d Explanation: The correct mathematical expression are Ie = (1 – ß) Ib and Ib = (1 – α) Ie respectively. 8. The Early Effect is also called as a) Base-width modulation effect b) Base-width amplification effect c) Both of the mentioned d) None of the mentioned View Answer Answer: a Explanation: At a given value of vBE, increasing vCE increases the reverse-bias voltage on the collector–base junction, and thus increases the width of the depletion region of this junction. This 418

in turn results in a decrease in the effective base width W. Also the saturation current is inversely proportional to the width, the saturation current will increase and also makes collector current increases proportionally. This is the Early Effect. For the reasons mentioned above, it is also known as the base-width modulation effect. 9. For the BJT to operate in active mode Collector-Base junction must be a) Heavily doped b) Must reversed bias c) Must be forward bias d) Lightly doped View Answer Answer: b Explanation: The BJT operates in active mode when the collector-Base junction is reversed bias. Also doping cannot prevent saturation of the transistor. 10. Collector current (Ic) reaches zero when a) Vce = Vt ln (Isc/I) b) Vt = Vce ln (Isc/I) c) Vce = Vt ln (I/Isc) d) Vce = Vt ln (Isc + I/I) View Answer Answer: a Explanation: Ic = Is exp (Vbe/Vt) – Isc exp(Vbc/Vt). In this expression put ic = 0 and simplify.

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Questions & Answers (MCQs) focuses on “BJT Circuits at DC”. 1. Which of the following condition is true for cut-off mode? a) The collector current Is zero b) The collector current is proportional to the base current c) The base current is non zero d) All of the mentioned View Answer Answer: a Explanation: The base current as well as the collector current are zero in cut-off mode. 2. Which of the following is true for the cut-off region in an npn transistor? a) Potential difference between the emitter and the base is smaller than 0.5V b) Potential difference between the emitter and the base is smaller than 0.4V c) The collector current increases with the increase in the base current d) The collector current is always zero and the base current is always non zero View Answer Answer: b Explanation: Both collector and emitter current are zero in cut-off region. 3. Which of the following is true for a typical active region of an npn transistor? a) The potential difference between the emitter and the collector is less than 0.5 V b) The potential difference between the emitter and the collector is less than 0.4 V c) The potential difference between the emitter and the collector is less than 0.3 V d) The potential difference between the emitter and the collector is less than 0.2 V View Answer Answer: c Explanation: Most commonly used transistors have Vce less than 0.4 V for the active region. 420

4. Which of the following is true for the active region of an npn transistor? a) The collector current is directly proportional to the base current b) The potential difference between the emitter and the collector is less than 0.4 V c) All of the mentioned d) None of the mentioned View Answer Answer: c Explanation: The base current and the collector current are directly proportional to each other and the potential difference between the collector and the base is always less than 0.4 V. 5. Which of the following is true for the saturation region of BJT transistor? a) The collector current is inversely proportional to the base current b) The collector current is proportional to the square root of the collector current c) The natural logarithm of the collector current is directly proportional to the base current d) None of the mentioned View Answer Answer: b Explanation: The collector current is directly proportional to the base current in the saturation region of the BJT. 6. Which of the following is true for a npn transistor in the saturation region? a) The potential difference between the collector and the base is approximately 0.2V b) The potential difference between the collector and the base is approximately 0.3V c) The potential difference between the collector and the base is approximately 0.4V d) The potential difference between the collector and the base is approximately 0.5V View Answer Answer: d Explanation: The commonly used npn transistors have a potential difference of around 0.5V between he collector and the base. 7. The potential difference between the base and the collector Vcb in a pnp transistor in saturation region is ________ a) -0.2 V b) -0.5V c) 0.2 V d) 0.5 V View Answer

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Answer: b Explanation: The value of Vcb is -0.5V for a pnp transistor and 0.5V for an npn transistor. 8. For a pnp transistor in the active region the value of Vce (potential difference between the collector and the base) is a) Less than 0.3V b) Less than 3V c) Greater than 0.3V d) Greater than 3V View Answer Answer: a Explanation: For a pnp transistor Vce is less than 0.3V, for an npn transistor it is greater than 0.3V. 9. Which of the following is true for a pnp transistor in active region? a) CB junction is reversed bias and the EB junction is forward bias b) CB junction is forward bias and the EB junction is forward bias c) CB junction is forward bias and the EB junction is reverse bias d) CB junction is reversed bias and the EB junction is reverse bias View Answer Answer: a Explanation: Whether the transistor in npn or pnp, for it be in active region the EB junction must be reversed bias the CB junction must be forward bias. 10. Which of the following is true for a pnp transistor in saturation region? a) CB junction is reversed bias and the EB junction is forward bias b) CB junction is forward bias and the EB junction is forward bias c) CB junction is forward bias and the EB junction is reverse bias d) CB junction is reversed bias and the EB junction is reverse bias View Answer Answer: b Explanation: Whether the transistor in npn or pnp, for it be in saturation region the EB junction must be forward bias the CB junction must be forward bias.

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Questions & Answers (MCQs) focuses on “BJT in Amplifier Design”. 1. Find the maximum allowed output negative swing without the transistor entering saturation, and a) 1.27 mV b) 1.47 mV c) 1.67 mV d) 1.87 mV View Answer

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Answer: d Explanation:

2. The corresponding maximum input signal permitted is a) 1.64 mV b) 1.74 mV c) 1.84 mV d) 1.94 mV View Answer Answer: d Explanation: If we assume linear operation right to saturation we can use the gain Av to calculate the maximum input signal. Thus for an output swing ∆ Vo = 0.8 we have ∆ Vi = ∆ Vo / Av = -0.7 / -360 = 1.94 mV. (Q.3- Q.5) For the amplifier circuit in Fig. 6.33(a) with Vcc = +10 V, Rc = 1 kΩ and the DC collector bias current equal to Ic, 3. Find the voltage gain. a) 100 Ic b) 200 Ic C) 400 Ic

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d) 800 Ic View Answer Answer: c Explanation:

4. The maximum possible positive output signal swing as determined by the need to keep the transistor in the active region. a) 9.7 + Ic b) 9.7 – Ic c) 10.3 + Ic d) 10.3 – Ic View Answer Answer: a Explanation: Assuming the output voltage Vo = 0.3v is the lowest Vce to stay out of saturation. Vo = 0.3 = 10 – IcRc = 10 – IcRc + ∆Vo ∆ Vo = -10 + 0.3 + Ic*1. 5. The maximum possible negative output signal swing as determined by the need to keep the transistor in the active region. a) 0.1 Ic b) Ic c) 10 Ic d) 100 Ic View Answer

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Answer: b Explanation: Maximum output voltage before the Transistor is cutoff. Vce + ∆Vo = Vcc ∆Vo = Vcc – Vce = 10 – 10 + 10 Ic = 10 Ic. 6. The transistor in the circuit below is biased at a dc collector current of 0.5 mA. What is the voltage gain?

a) -1 V/V b) -10 V/V c) -100 V/V d) -1000 V/V View Answer

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Answer: c Explanation:

7. For a BJt Vt is 5 V, Rc = 1000 ohm and bias current Ic is 12 mA. The value of the voltage gain is __________ a) -1.2 V/V b) -2.4 V/V c) -3.6 V/V d) -4.8 V/V View Answer Answer: b Explanation: Voltage gain is (Ic X Rc ) / Vt. (Q.8–Q.10) (Q.3- Q.5) For the BJT amplifier circuit with Vcc = +10 V, Rc = 1 kΩ and the DC collector bias current equal to 5 mA, 8. The value of the voltage gain is _______________ a) -2 V/V b) -4 V/V c) -10 V/V d) -20 V/V View Answer Answer: a Explanation: The voltage is 400 X Ic where Ic is 5 mA.

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9. The maximum possible positive output signal swing as determined by the need to keep the transistor in the active region. a) -1.7 V b) -2.7 V c) -3.7 V d) -4.7 V View Answer Answer: d Explanation: The maximum voltage swing is given by -10 + 0.3 + (Ic X Rc). Putting Ic as 5 mA, we get -4.7 mV. 10. The maximum possible negative output signal swing as determined by the need to keep the transistor in the active region. a) 0.5 V b) 1 V c) 5 V d) 10 V View Answer Answer: c Explanation: It is given by -10 + 10 + (Ic X Rc). Putting Ic as 5 mA we get 5V. Questions & Answers (MCQs) focuses on “Small Signal Operations and Model”. 1. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of gm. a) 12mA/V b) 24 mA/V c) 36 mA/V d) 48 mA/V View Answer

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Answer: d Explanation:

2. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of R?p. a) 625 ohm b) 1250 ohm c) 2500 ohm d) 5000 ohm View Answer Answer: c Explanation:

3. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of Re. a) 2.5 ohm b) 20.6 ohm c) 25.2 ohm d) 30.4 ohm View Answer

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Answer: b Explanation:

4. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the smallest value found of the resistance looking into the base? a) 347 ohm b) 694 ohm c) 1041 ohm d) 1388 ohm View Answer Answer: b Explanation:

5. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the largest value 430

found of the resistance looking into the base? a) 1050 ohm b) 21000 ohm c) 3150 ohm d) 4200 ohm View Answer Answer: d Explanation:

6. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What is the minimum ß he can tolerate for the transistor used? a) 100 b) 150 c) 200 d) 250 View Answer

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Answer: a Explanation:

7. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What emitter bis current should he choose? a) 1.06 mA b) 1.16 mA c) 1.26 mA d) 1.36 mA View Answer Answer: c Explanation:

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8. Which of the following is true? a) Ib = ß Ic b) Ib = ß + 1/ Ic c) Ib = Ic/ß d) Ib = Ic/ ß – 1 View Answer Answer: c Explanation: The correct relationship between Ic and Ie is Ib = Ic/ß. 9. The SI units of transconductance is a) Ampere/ volt b) Volt/ ampere c) Ohm d) Siemens View Answer Answer: a Explanation: Transcoductance is given by Ic/Vt. 10. Which of the following represents the correct mathematical form of the term denoted by the symbol Rp? a) ß/gm b) Vt/Ib c) All of the mentioned d) None of the mentioned View Answer Answer: c Explanation: Both of the expressions are identical.

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Questions & Answers (MCQs) focuses on “Basic BJT Amplifier Configuration”. (Q1-Q.6) An amplifier is measured to have an internal resistance of 10 kΩ, voltage gain of 100V/V and output resistance of 100 Ω. Also, when a load resistance of 1 kΩ is connected between the output resistance if found to decrease to 8 kΩ. If the amplifier is fed with the signal source having an internal resistance of 2 kΩ, then 1. Find Gm. a) 1 A/V b) 10 A/V c) 100 A/V d) 1000 A/V View Answer Answer: a Explanation: Gm = (voltage gain in V/V) / (output resistance) or 100/100 A/V. 2. Find Av. a) 9.09 V/V b) 10 V/V c) 90.9 V/V d) 100 V/V View Answer Answer: c Explanation: Av = Avo X Rl/Ro+Rl or 100 X 1000/1000+100 or 90.9 V/V. 3. Find Gvo. a) 53.3 V/V b) 63.3 V/V c) 73.3 V/V 434

d) 83.3 V/V View Answer Answer: d Explanation: Gvo = (Avo X input resistance) / (input resistance + signal resistance). 4. Find Gv. a) 53.4 V/V b) 72.7 V/V c) 83.3 V/V d) 90.9 V/V View Answer Answer: b Explanation: Gv = (Gvo X Av) / Avo or 83.3 X 90.9 / 100 V/V. 5. Find R out. a) 146 Ω b) 292 Ω c) 584 Ω d) 1168 Ω View Answer Answer: a Explanation: Rout = Rl (1 – Gvo/Gv). Put in the respective values and solve. 6. Find Ai. a) 182 A/A b) 364 A/A c) 546 A/A d) 728 A/A View Answer Answer: d Explanation: (Vo X R in) / (Vi X Rl) gives the required value of Ai.

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(Q.7-Q.10) The circuit shown below is a small sine wave signal with average zero and transistor ß=

7. Find the value of R(E) to establish a dc emitter current of about 0.5 mA. a) 28.57 kΩ b) 57.04 kΩ c) 114.08 kΩ d) 228.16 kΩ View Answer

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Answer: a Explanation:

8. Find R(C) to establish a dc collector voltage of about +5V. a) 5 kΩ b) 10 kΩ c) 15 kΩ d) 20 kΩ View Answer Answer: d Explanation: Vc = 15 – Rc.Ic 5 = 15 – Rc * 0.99 * 0.5m Rc = 20.2kΩ = 20kΩ. 9. For R (L) = 10 kΩ and transistor Ro = 200 kΩ, determine the overall voltage gain. a) -21 V/V b) -42 V/V c) -86 V/V d) -123 V/V View Answer

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Answer: c Explanation:

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Questions & Answers (MCQs) focuses on “Biasing in BJT Amplifier Circuits”. (Q.1-Q.6) consider the figure shown below and answer the questions that proceed.

1. For Vcc = 15V, R1 = 100 kΩ, R(E) = 3.9 kΩ, R(C) = 6.8 kΩ and ß = 100, determine the dc collector current for each transistor. a) 0.29 mA b) 0.48 mA c) 0.96 mA d) 1.92 mA View Answer

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Answer: c Explanation:

2. For Vcc = 15V, R1 = 100 kΩ, R(E) = 3.9 kΩ, R(C) = 6.8 kΩ and ß = 100, determine the dc collector voltage for each transistor. a) 4.25 V b) 8.5 V c) 12.75 V d) 17 V View Answer Answer: b Explanation:

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3. Find R (in 1) for R (sig) = 5 kΩ. a) 2.4 kΩ b) 4.8 kΩ c) 17.3 kΩ d) 34.6 kΩ View Answer Answer: a Explanation: It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively. 4. Find Vb1/Vsig for R(sig) = 5 kΩ. a) 0.08 V/V b) 0.16 V/V c) 0.24 V/V d) 0.32 V/V View Answer Answer; d Explanation:

5. Find R (in 2). a) 2.4 kΩ b) 4.8 kΩ c) 17.3 kΩ d) 34.6 kΩ View Answer Answer: a Explanation: It is the parallel combination of the 32 kΩ resistor and 2.6 kΩ resistor respectively. 6. Find Vb2/Vb1. a) -34 V/V b) -51 V/V c) – 68.1 V/V 441

d) -100 V/V View Answer Answer: c Explanation:

7. For Rl = 2 kΩ find Vo/Vb2. a) -59.3 V/V b) -29.7 V/V c) -89.1 V/V d) None of the mentioned View Answer Answer: a Explanation:

8. Find the overall voltage gain. a) 323 V/V b) 646 V/V c) 969 V/V d) 1292 V/V View Answer

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Answer: d Explanation:

Questions & Answers (MCQs) focuses on “Spread Spectrum”. (Q.1-Q.3) A pseudo-noise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second. 1. The PN sequence length is a) 10 b) 12 c) 15 d) 18 View Answer 443

Answer: c Explanation: The PN sequence length is N = 2m – 1 = 16 – 1 = 15. 2. The chip duration is a) 1µs b) 0.1 µs c) 0.1 ms d) 1 ms View Answer Answer: b Explanation: Tc = 1/(107) or 0.1 µs. 3. The period of PN sequence is a) 1.5 µs b) 15 µs c) 6.67 ns d) 0.67 ns View Answer Answer: b Explanation: The period of the PN sequence is T = NTc = 15 x 0.1 = 1.5 s 4. A slow FH/MFSK system has the following parameters. Number of bits per MFSK symbol = 4 Number of MFSK symbol per hop = 5 The processing gain of the system is a) 13.4 dB b) 37.8 dB c) 6 dB d) 26 dB View Answer Answer: d Explanation: PG = Wc/Rs = 5 x 4 = 20 or 26 db. 5. A fast FH/MFSK system has the following parameters. Number of bits per MFSK symbol = 4 Number of pops per MFSK symbol = 4 The processing gain of the system is a) 0 dB b) 7 dB 444

c) 9 dB d) 12 dB View Answer Answer: d Explanation: PG = 4 x 4 = 16 or 12 db. (Q.6-Q.7) A rate 1/2 convolution code with dfrec = 10 is used to encode a data sequence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz. 6. The coding gain is a) 7 dB b) 12 dB c) 14 dB d) 24 dB View Answer Answer: a Explanation: 0.5 x 10 = 5 or 7 db is the coding gain. 7. The processing gain is a) 14 dB b) 37 dB c) 58 dB d) 104 dB View Answer Answer: b Explanation: PG = (107)/(2 x 1000) = 5000 or 37 db. (Q.8-Q.9) An FH binary orthogonal FSK system employs an m 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection.[/expand] 8. If the hop rate is one per bit, the hopping bandwidth for this channel is a) 6.5534 MHz b) 9.4369 MHz c) 2.6943 MHz

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d) None of the mentioned View Answer Answer: a Explanation: The length of the shift-register sequence is L = 2m – 1215 = 32767 bits For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate. The hop rate is 100 hops/sec. Since the shift register has L 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz. 9. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is a) 3.2767 MHz b) 13.1068 MHz c) 26.2136 MHz d) 1.6384 MHz View Answer Answer: b Explanation: If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum frequency separation for orthogonality 2/T 400 Hz. Since there are N 32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 MHz. 10. In a fast FH spread spectrum system, the information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density 0.5No and an SNR 20 ~13 dB (total SNR over the three hops). The probability of error for this system is a) 0.013 b) 0.0013 c) 0.049 d) 0.0049 View Answer Answer: b Explanation: The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The

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probability of a chip error with non-coherent detection is

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12. Questions on Small-Signal Low-Frequency AC models of Transistors The section contains questions on ac models and analysis, transistor amplifier, biasing parameters, two port devices and hybrid model, transistor hybrid model, h-parameters and its measurement, cb transistor physical model, hybrid model in ce, cb and cc, ac and dc analysis problems, transistor circuit analysis and millers theorem. The AC Analysis of a Small-Signal Low-Frequency Common Emitter Transistor Biasing Parameters Problems on AC and DC Analysis Two-Port Devices and Hybrid Model Transistor Hybrid Model Determination of the h-parameters Measurement of h-parameters

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Conversion Formulas for the Parameters of the Three-Transistor Configurations Analysis of Transistor Amplifier Circuit using hparameters Comparision of Transistor Amplifier Linear Analysis of a Transistor Circuit Physical Model of a CB Transistor Approximate Hybrid model and its use in CE,CB,CC Millers Theorem

“The AC Analysis of a Small-Signal Low-Frequency Common Emitter Transistor”. 1. The feature of an approximate model of a transistor is a) it helps in quicker analysis b) it provides individual analysis for different configurations c) it helps in dc analysis d) ac analysis is not possible View Answer Answer: a Explanation: The small signal model helps in quicker ac analysis of a transistor. The approximate model is applicable for all the configurations. The dc analysis is not obtained by using a small signal model of transistor. 2 A transistor has hfe=100, hie=2kΩ, hoe=0.005mmhos, hre=0. Find the output impedance if the lad resistance is 5kΩ. a) 5kΩ b) 4kΩ c) 20kΩ d) 15kΩ View Answer Answer: b Explanation: RO=I/hoe=1/0.005m =20kΩ.ROI= RO || RLI=20||5 =4kΩ. 3. A CE amplifier when bypassed with a capacitor at the emitter resistance has a) increased input resistance and increased voltage gain b) increased input resistance and decreased voltage gain c) decreased input resistance and increased voltage gain d) decreased input resistance and decreased voltage gain View Answer 449

Answer: c Explanation: When a transistor is bypassed with a capacitor, it short circuits in the small signal analysis of transistor and the resistor too shorts. The input resistance becomes RI=hie. The value of the input resistance is decreased and the gain now will be increasing. 4. A transistor has hie =2kΩ, hoe=25µmhos and hfe=60 with an unbypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance? a) 90kΩ and 50kΩ respectively b) 33kΩ and 45kΩ respectively c) 6kΩ and 40kΩ respectively d) 63kΩ and 40kΩ respectively View Answer Answer: d Explanation: As the emitter is unbypassed, the input resistance Ri=hie+(1+hfe)Re =2+61=63kΩ. The output resistance RO=1/hoe=1/25MΩ=40kΩ. 5. A transistor has hie =1KΩ and hfe=60 with an bypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance? a) 90kΩ and 50kΩ respectively b) 33kΩ and 45kΩ respectively c) 6kΩ and 40kΩ respectively d) 63kΩ and 40kΩ respectively View Answer Answer: d Explanation: As the emitter is bypassed, the input resistance Ri=hie =1kΩ. The output resistance RO=1/hoe but the value is not given. So, hoe=0 and RO=1/0=∞.

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6. In the given circuit, find the equivalent resistance between A and B nodes.

a) 100kΩ b) 50kΩ c) 40kΩ d) 60kΩ View Answer Answer: b Explanation: RAB=RO||100Ω = (RSI+hie/1+hfe)||100 =9+1/100||100=100||100=50Ω. 7. Which of the following acts as a buffer? a) CC amplifier b) CE amplifier c) CB amplifier d) cascaded amplifier View Answer Answer: a Explanation: The voltage gain of a common collector amplifier is unity. It is then used as a buffer. The CC amplifier is also called as an emitter follower. Though there is no amplification done, the output will be stabilised. 8. Which of the following is true? a) CC amplifier has a large current gain b) CE amplifier has a large current gain c) CB amplifier has low voltage gain d) CC amplifier has low current gain View Answer 451

Answer: b Explanation: The CE amplifier has high current and voltage gains. The CC amplifier has unity voltage gain which cannot be regarded as high. The common base amplifier has a unity current gain and high voltage gain. 9. In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO. a) 10µA b) 100µA c) 90µA d) 500µA View Answer Answer: b Explanation: IC=α IE +ICBO =0.995*10mA+0.5µA=9.9505mA. IB=IE-IC=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199 ICEO=9.9505-199*0.0495=0.1mA==100µA. 10. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current. a) 0.01mA b) 0.07mA c) 0.02mA d) 0.05mA View Answer Answer: c Explanation: Here, IC=4.9/5K=0.98mA α = IC/IE .So, IE=IC/α=0.98/0.98=1mA. IB=IE-IC=1-0.98=0.02mA.

Questions & Answers (MCQs) focuses on “Biasing Parameters”. 1. The current gain of BJT is_________ a) gmro b) gm/ro c) gmri

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d) gm/ri View Answer Answer: c Explanation: We know, current gain AV=hfe. In π model, hfe is referred to β. We know, ri= β/gm. From this, β=rigm. 2. For the amplifier circuit of fig. The transistor has β of 800. The mid band voltage gain VO/VI of the circuit will be_________

a) 0 b) <1 c) =1 d) 800 View Answer Answer: c Explanation: The circuit is PNP transistor, collector coupled amplifier. The voltage gain is unity for a CC amplifier. Hence on observation, the CC amplifier gives a unity gain. 3. In a bipolar transistor at room temperature, the emitter current is doubled the voltage across its base emitter junction_________ a) doubles b) halves c) increase by about 20mV d) decreases by about 20mV View Answer 453

Answer: c Explanation: The change in voltage with temperature can be found by, V(T) = 2.3m(∆T)VO . In a bipolar transistor at room temperature if the emitter current is doubled the voltage across its base emitter junction thereby doubles. 4. A common emitter transistor amplifier has a collector current of 10mA, when its base current is 25µA at the room, temperature. What is input resistance? a) 3kΩ b) 5kΩ c) 1kΩ d) 7kΩ View Answer Answer: c Explanation: We know, β/gm=ri = (IC/IB)/(IC/VT)=VT/IB=25m/25µ=1k. 5. For an NPN transistor connected as shown in below, VBE=0.7V. Give that reverse saturation current of junction at room temperature is 10-13A, the emitter current is_________

a) 30mA b) 39mA c) 29mA d) 49mA View Answer Answer: d Explanation: When the collector and base are shorted, the transistor behaves as a normal diode. So, the diode equations imply. IE=IO(eV/V0-1). We get, IE=49mA. 454

6. The voltage gain of given circuit below is_________

a) 100 b) 20 c) 10 d) 30 View Answer Answer: c Explanation: The gain for the given circuit can be found by, AV=RF/RS =100K/10K=10. 7. A small signal source V(t)=Acos20t+Bsin10000t, is applied to a transistor amplifier as shown. The transistor has β=150 and hie=3KΩ. What will be the VO?

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a) 1500(Acos20t+Bsin10000t) b) -150(Acos20t+Bsin10000t) c) -1500Bsin10000t d) -150Bsin10000t View Answer Answer: d Explanation: AV=-hfe RLI/hie=3*150/3=-150. So, VO=-150V(t) But cos20t has low frequency so capacitors are open circuited. Only, the sine component is allowed. So, Vo =-150Bsin10000t. 8. Which of the following statements are correct for basic transistor configurations? a) CB Amplifiers has low input impedance and low current gain b) CC Amplifiers has low input impedance and high current gain c) CE Amplifiers has very poor voltage gain but very high input impedance d) The current gain of CB Amplifier is higher than the current gain of CC Amplifiers View Answer Answer: a Explanation: The CE amplifier has moderate input and output impedances. The CC amplifier has unity voltage gain. The common ba se amplifier has a unity current gain and high voltage gain. 9. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current? a) 3mA and 55µA 456

b) 2.945mA and 55µA c) 3.64mA and 33µA d) 5.89mA and 65µA View Answer Answer: a Explanation: (IC – ICBO)/α=IE = (2.945-0.002)/0.98=3mA. IE=IC+IB . So, IB=3-2.495=0.055mA=55µA. 10. The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance. a) 20kΩ b) 10kΩ c) 50kΩ d) 60kΩ View Answer Answer: b Explanation: ro=∆VCE/∆IC =3/0.3m=10kΩ. 11. A transistor is connected in CB configuration. The emitter voltage is changed by 200mV, the emitter by 5mA. During this transition the collector base voltage is kept constant. What is the input dynamic resistance? a) 30Ω b) 60Ω c) 40Ω d) 50Ω View Answer Answer: c Explanation: The ratio of change in emitter base voltage (∆VEB) to resulting change in emitter current (∆IE) at constant collector base voltage (VCB) is defined as input resistance. This is denoted by ri. We know, ∆VEB/∆IE=ri =200/5=40Ω.

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Questions and Answers focuses on “Problems on AC and DC Analysis”. 1. In the circuit, transistor has β =60, VBE=0.7V. Find the collector to emitter voltage drop hoe.

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a) 5V b) 3V c) 8V d) 6V View Answer Answer: d Explanation: We know, IC=(VCC-VBE)/RB By putting the values, we have IC=5.9mA. IE=IC/α. So, IE=5.99mA. hoe= VCC-RC(IC+IB). We have hoe=6V. 2. For the circuit shown, find the quiescent point.

a) (6V, 1mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, hoe=12V (IC)SAT =VCC/RL=12/6K=2mA. IB=10V/0.5M=20µA. IC= βIB=1mA. I hoe=VCC-ICRL=12-1*6=6V. So, quiescent point is (6V, 1mA).

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3. For the circuit shown, find the quiescent point.

a) (10V, 4mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, IE=VEE/RE=10/5kΩ=2mA IC=α IE =IE =2mA VCB=VCC-ICRL=20-10=10V. So, quiescent point is (10V, 2mA). 4. In the circuit shown below, β =100 and VBE=0.7V. The Zener diode has a breakdown voltage of 6V. Find the operating point.

a) (6.7V, 5.3mA) b) (5.7V, 5.3mA) c) (6.7V, 5mA) d) (6V, 5mA) View Answer 460

Answer; a Explanation: We know, by KVL -12+(IC+IB)1K+6+VBE=0 We have IE=5.3. IC= αIE=5.24mA. From another loop, -12+IEIK+VBE=0 We have, hoe=12-5.3m*1000=6.7V. Hence the Q point is (6.7V, 5.3mA). 5. 10. When the β value is large for a given transistor, the IC and hoe values are given by_________ a) (VCC-VBE)/RB, VCC-RCIC b) (VCC+VBE)/RB, VCC-RC(IC+IB) c) (VCC+VBE)/RB, VCC+RC(IC+IB) d) (VCC+VBE)/RB,VCC+RC(IC-IB) View Answer Answer: a Explanation: The base current IB is zero when β value is large. So, the hoe changes to VCC-RCIC. The collector current IC is changed to (VCC-VBE)/RB from β(VCC-VBE)/(1+ β)RE+ RB. 6. For the circuit shown, find the quiescent point.

a) (10V, 4mA) b) (4V, 10mA) c) (10V, 3mA) d) (3mA, 10V) View Answer Answer: c Explanation: We know, IE=VEE/RE=30/10kΩ=3mA IC=α IE =IE =3mA VCB=VCC-ICRL=25-15=10V. So, quiescent point is (10V, 3mA). 7. The PNP transistor when used for switching the power, then it is called_________ a) sourcing current b) sinking current 461

c) forward sourcing d) reverse sinking View Answer Answer: a Explanation: Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it. 8. In which of the regions, the small capacitors are open circuited? a) high frequency b) medium frequency c) low frequency d) the region does not affect he capacitors View Answer Answer: c Explanation: In low frequency region, the small capacitors are open circuited and large capacitors are in active state. In high frequency region, the large capacitors are short circuited and small capacitors are active. 9. In mid frequency region, the large capacitors are short and small capacitors are open circuited. What happens to the RC coupled circuit? a) the circuit is now frequency blind b) it is DC isolated c) the circuit turns reactive d) the circuit is AC dependent View Answer Answer: a Explanation: In the mid frequency region, the whole circuit is resistive because the large capacitors are short and small capacitors are open circuited. The gain is constant in this region. So, the circuit is frequency blind as the gain is constant in this region. 10. In low frequency region, the gain_________ a) increases b) decreases c) remains same d) depends on value of capacitors View Answer

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Answer: b Explanation: At very low frequency, the gain decreases due to the coupling capacitor. When the frequency is decreased, the reactance of the circuit increases and the drop across the coupling capacitor increases. The gain is therefore decreased as the output decreased.

13. Questions & Answers on Field-Effect Transistors The section contains questions and answers on junction field effect transistor, pinch off voltage, insulated gate, fet small signal model, common source and drain amplifier, fet biasing, fet amplifier and unijunction transistors. The Junction Field-Effect Transistor The Pinch off Voltage Vp The JFET Volt-Ampere Characteristics The FET Small-Signal Model The Insulated-Gate FET(MOSFET)

The Common-Source Amplifier The Common-Drain Amplifier Biasing the FET A Generalized FET Amplifier The Unijunction Transistors

14. Questions on Large Signal Amplifiers The section contains questions on class a large signal amplifiers, second and higher order harmonic distortion. Class A Large Signal Amplifiers Second Harmonic Distortion

Higher-Order Harmonic Distortion

15. Questions & Answers on Low Frequency Transistor Amplifier Circuit The section contains questions and answers on cascading amplifiers, decibel, cc and cb configurations, ce amplifier, emitter follower, high input resistance transistor circuit, cascode transistor and amplifiers. Cascading Transistor Amplifiers N-Stage Cascading Amplifiers The Decibel Simplified CE Hybrid Model Simplified Calculations for the CC Configuration Simplified Calculations for the CB 463

CE Amplifier with an Emitter Resistance The Emitter Follower High Input Resistance Transistor Circuit The Cascode Transistor Configuration Difference Amplifiers

Configuration 16. Questions on High Frequency Transistor The section contains questions on high frequency t model, hybrid pi models, alpha cutoff frequency, ce short circuit current frequency response, resistive load and transistor amplifier response. The High Frequency T Model The CB Short Circuit Current Frequency Response The Hybrid PI CE Transistor Model Hybrid PI Conductances In Low Frequency H Parameters The CE Short Circuit Gain Obtained With The Hybrid PI Model

The Alpha Cutoff Frequency The CE Short Circuit Current Frequency Response Current Gain With Resistive Load Transistor Amplifier Response, Taking Source Resistance Into Account

17. Questions & Answers on Frequency Response The section contains questions on low and high frequency responses on cs and ce amplifiers, high frequency models of bjt and mosfet, millers theorem, high frequency response of source and emitter followers, differential amplifiers, mos and bipolar cascode amplifiers, cd-cs, cc-ce, cd-ce, cc-cb and cd-cg configurations, pole splitting, frequency and miller compensation.

Low-Frequency Response of the CS Amplifiers Low-Frequency Response of the CE Amplifiers High-Frequency Model of the BJT High-Frequency Model of the MOSFET High-Frequency Response of the CS Amplifiers High-Frequency Response of the CE Amplifiers The High Frequency Gain Function Determining the 3-dB Frequency fH 9 Approximate Determination of fH Miller’s Theorm High-Frequency Response of the CG

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High-Frequency Response of the MOS Cascode Amplifier High-Frequency Response of the Bipolar Cascode Amplifier High Frequency Response of the Source Followers High-Frequency Response of the Emitter Followers High Frequency Response of the Differential Amplifiers The CD-CS, CC-CE and CD-CE Configurations The CC-CB and CD-CG Configurations High Frequency Response of Multistage Amplfiers Frequency Compensation Miller Compensation and Pole Splitting

18. Questions on Differential and Multistage Amplifiers The section contains questions and answers on input bias, voltage, current, multistage and differential amplifiers, feedback structure and negative feedback properties, transconductance and transresistance amplifiers, a, b and ab output stages, feedback effects, power bjts, heat sinks and variations in ab classification. Input Bias and Offset Currents of the Bipolar Differential Amplifier Differential Amplifier with Active Load Multistage Amplifier The General Feedback Structure Properties of Negative Feedback Voltage Amplifiers Current Amplifiers Tranconductance Amplifiers Transresistance Amplifiers Feedback Voltage Amplifier (Series-Shunt) Determining the Loop Gain The Transfer Function of the Feedback Amplifier

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Effect of Feedback on the Amplifier Poles (The Nyquist Plot) Effect of Feedback on the Amplifier Poles Stability Study using Bode Plots Classification of Output Stages Class A output Stage Class B Output Stage Calss AB Output Stage Biasing the Class AB Circuit Power BJTs Transistor case and Heat Sinks Variations on Class AB Configuration

Best Reference Books – Electronic Devices and Circuits We have compiled a list of Best Reference Books on Electronic Devices and Circuits Subject. These books are used by students of top universities, institutes and colleges.

Here is the full list of best reference books on Electronic Devices and Circuits. 1. “Semiconductor Physics and Devices” by D A Neamen 2. “Microelectronic Devices” by E S Yang 3. “Solid State Electronic Devices” by B G Streetman. 4. “Microelectronics” by J Millman and A Grabel 5. “Microelectronic Circuits” by A S Sedra and K C Smith 6. “Microelectronics: An Integrated Approach” by R T Howe and C G Sodini 7 . “Electronic Devices and Circuits” by Theodore F Bogart

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8. “Introductory Electronic Devices and Circuits : Conventional Flow Version” by Paynter 9. “Electronics Devices And Circuits” by J B Gupta 10. “Basic Electronic Devices and Circuits” by Mahesh B Patil

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