Tipuri De Tranzistori

The Bipolar Junction Transistor ( B.J.T.)

1 Electrical Science

Symbols

2 Electrical Science

How to remember the Symbols • • • •

The NPN transistor has an arrow that is: N not P pointing N in

3 Electrical Science

Transistor Biasing • The base emitter junction must be forward biased. • The base collector must be reverse biased.

Add the two batteries to the circuit on the right.Use the correct bias. What A type of transistor is this ? 4 d d

Electrical Science

Transistor Currents

This Rule must be memorised!

I E = I B + IC 5 Electrical Science

Question

A transistor has a collector current of 0.1 Amps. The base current is 100 micro Amps. Find the emitter current, in milli amps.

6 Electrical Science

Solution: First we convert all units to milli amps: 1)

Collector current = 0.1 Amps. This is 100 milli Amps. Base current is 100 micro Amps. This is 0.1 milli amps.

2) The formula:

IE = IB + IC Emitter current = 0 . 1 mA + 100 mA. Emitter current = 100 . 1 mA 7 Electrical Science

How transistors work

A small tap can control the flow of water from a pressure hose.

Tap = Base

A small signal on the base of a transistor can control the flow of current from collector to emitter. 8 Electrical Science

A Transistor amplifies signals. Unlike a resistor or capacitor. a transistor must be powered By a battery. We say it is an active device. V out V in

The transistor makes the signal louder.

A small signal is sent in.

A large signal comes out.

For example a radio receives a weak signal at the aerial Vin. It is amplified using a transistor and sent out to the speaker V out. The battery power supply is called Vcc, because the voltage is connected indirectly to the collector. 9 Electrical Science

Using an analogue meter to test Transistors

10 Electrical Science

Testing a Transistor in Circuit with a multimeter Vcc 12 V +

V

oV

Vce Vc

Vbb 5V

V

V

Vbe

The Base Emitter PN junction uses 0.7 volts to turn on the transistor, which joins the collector to the emitter. Question: a)State the voltage reading on each meter. b) Name these voltage readings. Vcc,Vce,Vbe, Vbb. Electrical Science

11

Parameters 1) Current Gain:

BDC

IC collector current = = base current IB this is also called h fe

2) CURRENT Transistor can only carry a limited current. If the maximum current is exceeded the transistor will overheat and be damaged. 3) POWER: Transistor can only dissipate a limited amount of power. If the maximum power is exceeded the transistor will overheat and be damaged. 12 Electrical Science

Question A transistor has a maximum power rating of half a watt. It’s maximum voltage supply VCE is 20V and maximum current IC is 50 mA. a) Fill in the following grid. b) Will any of these situations damage the transistor?

13 Electrical Science

Solution a)

b) Yes. When the current is100mA the limit is exceeded, and the transistor damaged. 14 Electrical Science

Heat Sinks • To cool down some transistors metal plates are attached to them. • Without these heat sinks they would overheat. • These remove heat from the transistor just like the metal fins on a motor bike, or a radiator in a car.

15 Electrical Science

Transistor Cases

base collector emitter

16 Electrical Science

D1 LED1

The Transistor as a Switch. The LED will light because the Base has more than 0.7V which turns the transistor on so the collector is connected to the emitter.

Find the voltage at A.

R3 1k R1 100 + V2 10V A

Resistor

Q1 NPN

R2 100 R4 50

500

17 Electrical Science

I r3 and r4

Vr3 and r4 = R3 + R4 10 1000 + 500 = 0 ⋅ 0066 A.

I r3 and r4 = I r3 and r4

-------------------Vr 3 = I r3 . R 3 Vr 3 = 0 ⋅ 0066.1000 Vr 3 = 6 ⋅ 6 volt -------------------Vr 4 = I r4 . R 4 Vr 4 = 0 ⋅ 0066. 500 Vr 4 = 3.3 volts 18 Electrical Science

The Transistor as a Switch. This LED will NOT light because the Base has less than 0.7V which turns off the transistor and so the collector is disconnected from the emitter.

1K

A Resistor

Find the voltage at point A.

50

19 Electrical Science

I r3 and r4

Vr3 and r4 = R3 + R4 10 1000 + 50 = 0 ⋅ 00952 A.

I r3 and r4 = I r3 and r4

-------------------Vr 3 = I r3 . R 3 Vr 3 = 0 ⋅ 00952.1000 Vr 3 = 9.52 volt -------------------Vr 4 = I r4 . R 4 Vr 4 = 0 ⋅ 00952. 50 Vr 4 = 0.48 volts 20 Electrical Science

Is the LED on ? Explain your answer

1k

Solution: The LED is off because the transistor is on and so short circuiting it.

9k

On means collector is connected to emitter.

Electrical Science

21

I r3 and r4 I r3 and r4 I r3 and r4

Vr3 and r4 = R3 + R4 10 = 10000 = 1 mA.

-------------------Vr 3 = I r3 . R 3 1 Vr 3 = .1000 1000 Vr 3 = 1volt -------------------Vr 4 = I r4 . R 4 1 .9000 1000 Vr 4 = 9 volts

Vr 4 =

22 Electrical Science

State the voltage bias on the base. VCC +30V

R1 22k

RC 1.0k

A Resistor

R2 10k

RE 1.0k

Solution : The voltage at the base is 9.38 Volts. 23 Electrical Science

Question

RC 1.0k

RB 10k

+ VBB 3V

+ VCC 20V

If the current gain BDC is 50 find a) IB b) IC c) IE d) VB e) VC Electrical Science

24

VC

RC 1.0k RB 10k

Ic

IB + VBB 3V

IE

+ VCC 20V

VB

25 Electrical Science

Solution a) • • • •

First we find IB IB is the current in the base. Current is V/R V is 3 - 0.7 dropped by the base emitter PN junction. • The base current = 2.3/10,000

• IB = 0.000230 Amps. • IB = 0.230 milli Amps. • IB = 230 micro Amps. 26 Electrical Science

Solution b) • To find IC • We know: Gain means output compared to input. • Gain = IC / IB • Multiply both sides by IB • Gain IB = IC IC = 50 . 230 u Amps IC = 11500 u Amps • IC = 11. 5 mA. 27 Electrical Science

Solution c) • To find IE. • IE = IC + IB • =11.7 mA • Solution d) • VB is always 0.7 volts higher than the voltage at the emitter. • In this case the emitter is grounded so VB is 0.7 volts

• • • •

Solution e) Vc is Vcc – the voltage dropped across Rc. Vc = 20 – (11.5 mA)(1 k) Vc = 8.5 volts.

28 Electrical Science

Question: The transistor is not fully on. We say it is not saturated.

D1 LED1 R3 1k

a) You measure the voltage across R1 as 4.5 volts. Find Vce.

R1 100 + V2 10V

b) R5 uses 0.1 m Volts. Find the current through it.

R5 = 120 ohms

Q1 NPN

c) Find Ic. d) Find the current Gain of the circuit. e) Find Ie.

R2 100 R4 50

29 Electrical Science

Answer • A) The supply is 10 volts – R1 uses 4.5 volts. – R2 uses 4.5 volts. Same resistance and current. – The diode used 0.7 volts. – These add up to 9.7 volts. – So the transistor uses 0.3 volts between collector and emitter and this is called Vce.

• B) Using Ohms Law : The current throuth a resistor = I R5 I R5 I R5

The voltage across THAT resistor THAT resistor

0 ⋅1 m Volts = 120 0 ⋅ 0001 Volts = 120 = 0.00000083A

I R5 = 0.83 uA

30 Electrical Science

c) Solution( Use Ohm’s Law)

The voltage across THAT resistor The current throuth a resistor = THAT resistor 4.5 volts I R1 = 100 I R1 = 0.045 amps I R1 = 45 m Amps

31 Electrical Science

Solution part d Ic Gain = Ib 45 ma Gain = 0 ⋅ 00083 ma Gain = 54216 32 Electrical Science

d) Ie = Ic + Ib Ie = 45 ma + .00083 mA Ie = 45.00083 ma 33 Electrical Science

Q15 March 2005

Redraw the transistors and show the direction of conventional current 34 Electrical Science

Q15 March 2005 -

+ N

P

+

P

N N

P

-

+

Solution: 1) Fill in the P and N terminals. 2) Fill in the battery + and -. The Base emitter must be forward biased. The base collector reverse biased. 35 Electrical Science

Q15 March 2005 -

+

N

P

+

N

P N

P +

Solution: Conventional current flows from the + of the battery to the – of the battery. 36 Electrical Science

Q15 May 2005

The hfe of an NPN transistor is 250. If the transistor is operating under non-saturated conditions in a circuit and the collector current found to be 25mA determine the base current.

37 Electrical Science

Solution 100 micro amps.

Saturation: means the transistor is turned on fully. Increasing the voltage on the base will have no effect.

38 Electrical Science

Question A transistor circuit has the follow measurements. Vcc 30 V IB = 0.25 mA Ic = 50 mA

Find IE.

39 Electrical Science