Buck Converter

FINAL REPORT BUCK CONVERTER MODULE ANALOG BASED

Group Members : Rifadi Ardiansyah Ahmad Syihan Ramadhan

NRP. 1310161015 NRP. 1310161029

Class : 3 D4 Industrial Electrical Engineering

Lecturer : Ir. Moh. Zaenal Efendi, M.T.

INDUSTRIAL ELECTRICAL ENGINEERING ELECTRO DEPARTEMENT ELECTRONICS ENGINEERING POLYTECHNIC INSTITUE OF SURABAYA

2019 BUCK CONVERTER MODULE ANALOG BASED I. 1.

Introduction Buck Coverter The Buck Converter is used in SMPS (Switched Mode Power Supplies) circuits where the DC output voltage needs to be lower than the DC input voltage. The DC input can be derived from rectified AC or from any DC supply. It is useful where electrical isolation is not needed between the switching circuit and the output, but where the input is from a rectified AC source, isolation between the AC source and the rectifier could be provided by a mains isolating transformer.The switching transistor between the input and output of the Buck Converter continually switches on and off at high frequency. To maintain a continuous output, the circuit uses the energy stored in the inductor L, during the on periods of the switching transistor, to continue supplying the load during the off periods. The circuit operation depends on what is sometimes also called a Flywheel Circuit. This is because the circuit acts rather like a mechanical flywheel that, given regularly spaced pulses of energy keeps spinning smoothly (outputting energy) at a steady rate. The output voltage can be changed by adjusting the width of the duty cycle (D). While the output of a buck converter according to the theoretically is : 𝑽𝑶 = 𝑽𝑺 . 𝑫 To reduce voltage ripple, filters made of capacitors (sometimes in combination with inductors) are normally added to such a converter's output (load-side filter) and input (supplyside filter).

Figure 1.1. The Basic Schematic of Buck Converter Switched mode supplies can be used for many purposes including DC to DC converters. Often, although a DC supply, such as a battery may be available, its available voltage is not suitable for the system being supplied. For example, problem with batteries, large or small, is that their output voltage varies as the available charge is used up, and at some point the battery voltage

becomes too high to power the circuit being supplied. However, if this high output level can be bucked to a useful level again. The DC input to a buck converter can be from many sources as well as batteries, such as rectified AC from the mains supply, or DC from solar panels, fuel cells, dynamos and DC generators. The buck converter is different to the boost Converter in that it’s output voltage is equal to, or lower than its input voltage. Buck Converter Operation 1. Operation at switch on In Fig. 1.2 therefore, when the switching transistor is switched on, it is supplying the load with current. Initially current flow to the load is restricted as energy is also being stored in L1, therefore the current in the load and the charge on C1 builds up gradually during the ‘on’ period. Notice that throughout the on period, there will be a large positive voltage on D1 cathode and so the diode will be reverse biased and therefore play no part in the action.

Figure 1.2. Buck Converter Operation at Switch On When the switch is closed, the diode is reverse-biased. Kirchhoff’s voltage law around the path containing the source, inductor, and closed switch is : 𝑉𝐿 = 𝑉𝑆 − 𝑉𝑂 = 𝐿

𝑑𝑖𝐿 𝑑𝑡

or

𝑑𝑖𝐿 𝑑𝑡

=

𝑉𝑠 − 𝑉𝑂 𝐿

The rate of change of current is constant, so the current increases linearly while the switch is closed, as shown in Fig 3.2 (b). The change in inductor current is computed from soving for ∆iL when the switch is closed gives : ∆𝑖𝐿 ∆𝑖𝐿 𝑉𝑆 − 𝑉𝑂 = = ∆𝑡 𝐷𝑇 𝐿 𝑉𝑆 − 𝑉𝑂 (∆𝑖𝐿 )𝑐𝑙𝑜𝑠𝑒𝑑 = ( ) . 𝐷𝑇 𝐿

Figure 1.3. Buck Converter Waveforms 2. Operation at switch off When the transistor switches off as shown in Fig 1.4 the energy stored in the magnetic field around L1 is released back into the circuit. The voltage across the inductor (the back e.m.f.) is now in reverse polarity to the voltage across L1 during the ‘on’ period, and sufficient stored energy is available in the collapsing magnetic field to keep current flowing for at least part of the time the transistor switch is open. The back e.m.f. from L1 now causes current to flow around the circuit via the load and D1, which is now forward biased. Once the inductor has returned a large part of its stored energy to the circuit and the load voltage begins to fall, the charge stored in C1 becomes the main source of current, keeping current flowing through the load until the next ‘on’ period begins.

Figure 1.4. Buck Converter Operation at Switch Off

Assuming thet the output voltage Vo is a constant, the voltage across the inductor is opened switch is : 𝑉𝐿 = − 𝑉𝑜 = 𝐿

𝑑𝑖𝐿 𝑑𝑡

𝑑𝑖𝐿 − 𝑉𝑜 = 𝑑𝑡 𝐿 The Since the derivative of 𝑖𝐿 is a negative constant, the current 𝑖𝐿 decreases linierly. The change in current while switch opened is : ∆𝑖𝐿 ∆𝑖𝐿 𝑉𝑜 = =− ∆𝑡 (1 − 𝐷)𝑇 𝐿 − 𝑉𝑂 (∆𝑖𝐿 )𝑜𝑝𝑒𝑛 = ( ) . (1 − 𝐷 ) 𝑇 𝐿 For steady-state operation, the net change in inductor current must be zero over one period. Using euation above we get (∆𝑖𝐿 )𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖𝐿 )𝑜𝑝𝑒𝑛 = 0 𝑉𝑆 − 𝑉𝑂 −𝑉𝑂 ( ) . 𝐷𝑇 + ( ) . (1 − 𝐷 ) 𝑇 = 0 𝐿 𝐿 So, solving Vo is :

𝑉𝑜 = 𝐷𝑉𝑆

2.

Inductor An inductor is a passive electronic component that stores energy in the form of a magnetic field. In its simplest form, an inductor consist of a wire loop or coil. Inductance also depends on the radious of the coil and on the type of material around which the coil is wound. For a given coil radius and number of turn, air core result in the least inductance. Materials such as wood, glass, and plastic-known as dielectric materials are essentially the same as air for the purposes of inductor winding. Ferromagnetic subtances such as iron, laminated iron, and powdered iron increase the inductance obtainable with a coil having a give number of turns. In some cases, this incrase on the order of thousands of times. The shape of the core is also significant. Toroidal (donut-shaped) cores provide more inductance, for a given core material and number of turns, than selenoidal (rod-shaped) cores. In the international system of Units (SI), the unit of inductance is the henry (H). Inductors have values that typically range from 1 µH (10 -6 H) to 1 H. Many inductors have a magnetic core made of iron or ferrite inside the coil, which serves to increase the magnetics field and thus the inductane. It is difficult to fabricate inductors on to integrated circuit (IC) chips. Fortunately,

resistors can be substituted for inductors in most microcircuit applications. In some cases, inductance can be simulated by simple electronic circuits using transistors, resistors, and capacitors fabricated onto IC chips. Inductors are used with capacitor in various wireless communications applications. An inductor connected in series or parallel with a capacitor can provide discrimination against unwanted signals. Large inductors are used in the power supplies of electronic equipment of all types, including computers and their peripherals. In these systems, the inductors help to smooth out of rectifier utility AC, providing pure, battery like DC. 3.

Design of Buck Converter Module The buck converter has following parameters : 𝑉𝑆(𝑚𝑎𝑥) = 60 𝑉𝑜𝑙𝑡 𝑉𝑆(𝑚𝑖𝑛) = 50 𝑉𝑜𝑙𝑡 𝑉𝑜 = 20 𝑉𝑜𝑙𝑡 𝐼𝑂 = 3 𝐴 𝑆𝑤𝑖𝑡𝑐ℎ𝑖𝑛𝑔 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑠 ) = 40𝑘𝐻𝑧

Figure 3.1. Boost Converter Circuit 

Duty Cycle : 𝑉𝑜 20 2 𝐷= = = = 0.333 𝑉𝑠 60 6



The inductor value 1

𝑉0 + 𝑉𝐹

𝑓

𝑉𝑠(𝑚𝑎𝑥) + 𝑉𝐹

𝐿 = [ ] × (𝑉𝑠(𝑚𝑎𝑥) − 𝑉𝑜 ) × [

] ×

1 ∆𝑖𝐿

1 20 + 1.7 1 ] × (60 − 20) × [ ] × 40000 60 + 1.7 0.6 1 21.7 1 = [ ] × (40) × [ ] × 40000 61.7 0.6 1 = [ ] × 0,586169638 1000 = 0,000586169638 = [

= 586.169638 𝜇𝐻 (Calculated) = 586 𝜇𝐻 (In Use) ∆𝑖𝐿 = 20% × 𝐼𝐿(𝑎𝑣𝑔): 𝐼𝐿(𝑎𝑣𝑔) = 𝐼𝑜 = 3 𝐴 = 20% × 3 = 0.6 A 𝑉𝐹 = Diode Forward – Voltage (from data sheet of diode red: MUR 1560) = 1.7 Volt 

The maximum inductor current ∆𝑖

𝑉

iL(𝑚𝑎𝑥) = 𝑖𝐿(𝑎𝑣𝑔) + 2𝐿 ∶ 𝑖𝐿(𝑎𝑣𝑔) = 𝑅𝑜 0,6 𝑉𝑜 = 3+ ∶ 𝑖𝐿(𝑎𝑣𝑔) = = 𝐼𝑜 = 3 𝐴 2 𝑅 = 3 + 0.3 = 3.3 𝐴 

Winding number of inductor 𝑛= = = = = =



𝐿 ∙ 𝑖𝐿(𝑚𝑎𝑥) 𝐵𝑚𝑎𝑥 ∙ 𝐴𝑐

104 ∶ 𝐵𝑚𝑎𝑥 = 0.25 𝑇𝑒𝑠𝑙𝑎: 𝐴𝑐 = 1.96 𝑐𝑚2

586.169638 × 10−6 ∙ 3.3 104 0.25 ∙ 1.96 1934.35981 × 10−6 104 0.49 0.00394767308 × 104 39.4767308 (Calculated) 40 (In Use)

Wire size is based on RMS Current of inductor 2

2

∆𝑖 /2 𝑖𝐿(𝑟𝑚𝑠)𝑡 = √(𝑖𝐿(𝑎𝑣𝑔) ) + ( 𝐿 ) √3

= √ (3)2 + (

0.6/2 √3

2

)

= √9 + (0.173205081)2 = √9 + 0.0300000001 = √9.0300000001 = 3.00499584 𝐴



Calculation of wire size  Cross Sectional Area of Wire (𝑞𝑤 ) 𝑞𝑤(𝑡) =

𝑖𝐿(𝑟𝑚𝑠)𝑡 𝐽

𝐽 = 4.5 𝐴/𝑚𝑚2 (current density)

:

3.00499584 4.5 = 0.667776853 𝑚𝑚2 =



Diameter of Wire (𝑑𝑤 ) 4

𝑑𝑤(𝑡) = √ × 0.667776853 𝜋



= √0.850239896 = 0.922084538 𝑚𝑚 Recalculate by assuming number of split wire (∑ 𝑠𝑝𝑙𝑖𝑡) = 10 𝑖𝐿(𝑟𝑚𝑠)𝑡 3.00499584 𝑖𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = = = 0.3180862562 ∑ 𝑠𝑝𝑙𝑖𝑡 10 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 =

𝑖𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 0.3180862562 = = 0.0706858347 𝐽 4.5

4 4 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √ × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √ × 0.0706858347 = 0.3 𝜋 𝜋 (∑ 𝑠𝑝𝑙𝑖𝑡) 𝐶ℎ𝑜𝑜𝑠𝑒𝑛 = (∑ 𝑠𝑝𝑙𝑖𝑡) + (∑ 𝑠𝑝𝑙𝑖𝑡) 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 = 10 + 3 = 13  Wire Size Diameter of bobbin PQ3535 (𝐷𝑏𝑜𝑏 ) = 17 mm = 1.7 cm Circumference of Bobin (𝐾𝑏𝑜𝑏 ) = 𝜋 × 1.7 = 5.34070751 Total

Wire

Length

=

(𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑ 𝑠𝑝𝑙𝑖𝑡) + 40%∗ (𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 ×

∑ 𝑠𝑝𝑙𝑖𝑡) Total Wire Length = (40 × 5.340707511 × 13) + 40% × (40 × 5.340707511 × 13) Total Wire Length = 2777.16791 + 1110.86716= 3888.03507 = 3900 cm = 39 m

1. Output Capacitance 𝐶𝑜 =

∆𝑄 ∆𝑉𝑜

=

∆𝑖𝐿 ∙𝑇 8∆𝑉𝑜

=

∆𝑖𝐿 8×𝑓×∆𝑉𝑜

0.6 8 × 40000 × 0.02 0.6 = 6400 =

= 100 𝜇𝐹. 50 Volt ∆𝑉𝑜 =

∆𝑄 ∆𝑖𝐿 ∙ 𝑇 = 𝐶𝑜 8𝐶𝑜

𝑤ℎ𝑒𝑟𝑒 ∆𝑉𝑜 = ±0,1% × 𝑉𝑜 = 0,001 × 𝑉𝑜 ∆𝑉𝑜 = 0,001 × 20 ∆𝑉𝑜 = 0.02 2. Snubber Circuit 𝐶𝑠 ≈

𝐼𝑂𝑁 ×𝑡𝑓𝑎𝑙𝑙 2×𝑉𝑜𝑓𝑓

≈

3 × 58 × 10−9 2 × 60

≈

174 × 10−9 120

≈ 1.45 𝑛𝐹 (Calculated) ≈ 2 𝑛𝐹. 1 𝐾𝑣𝑜𝑙𝑡 (In Use)

𝑅𝑠 =

𝐷𝑇 2 × 𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟

0.333 × 25 × 10−6 = 2 × 1.45 × 10−9 8.325 × 10−6 = 2.9 × 10−9 = 2.87068966 × 103 Ω = 2870.68966 Ω (Calculated) = 1 𝑘Ω. 10 Watt (In Use)

Winding Step 1. The first step that should determine the parameter – pareameters that will be used as reference calculations like Vs, Vo, Io, switching freuency, Bmax, etc. 2. Then calculate the value of inductane that we need based on all parameters. 3. Determine how bobbin diameter to be used, so that the diameter is known. For this time we used P3535 wire diameter to the size of 0,3 mm. 4. Then calculate n (number of turns) of high frequency inductor. 5. Then measure , where this parameter is used to determine how large a cross-sectional area if the wire used. 6. After the meet, see the data sheet. 4.

Component  Modul buck converter : 1. Terminal 2 PTR 2. Terminal 3 PTR 3. MOSFET IRFP460 4. Diode MUR 1560 5. Ferrite Core PQ35/35 6. Bobbin 7. Copper wire 0,3mm 8. Capacitor 100µF, 50V 9. Snubber Resistor 1kΩ, 10W 10. Snubber Capacitor 2nF, 1kV 11. Snubber Diode FR 3017 12. Heatsink 13. Spacer PCB  Power supply  Avometer  Amperemeter DC  Oscilloscope  Load (lamp)  Load (variabel resistor)  Probe wire  Jumper wire  Power wire

5.

Design Procedure a. Acrylic Template Design

b. Schematic Circuit of PWM

c. Board Circuit of PWM

d. Schematic Circuit of Buck Converter

e. Board Circuit of Buck Converter

f. Implementation Boost Converter

6.

Experiment Result 1. Functional Test

Duty 𝑽𝑺 (V) 50%

60%

10,02 25,01 50 10,03 25,01 50,1

𝑰𝑺 (A) 0,11 0,15 0,2 0,14 0,2 0,26

𝑽𝑶 Prak (V) 3,85 11,7 23,7 5,22 13,85 24,23

𝑽𝑶 Teori (V) 5,01 12,501 25 6,018 15,006 30,06

𝑰𝑶 (A) 0,22 0,31 0,41 0,25 0,35 0,45

Functional test data image attachment : Frequency = 40kHz Duty Cycle = 50% 1. Vin = 10 Volt a. Vpulse

b. VGS

𝑷𝑺 (W) 1,102 3,751 10 1,404 5,002 13,153

𝑷𝑶 Prak (W) 0,847 3,627 9,717 1,305 4,847 13,026

𝑷𝑶 Teori (W) 1,1022 4,6255 10,25 1,5045 5,2521 13,527

η (%) 76,84 96,68 97,17 92,93 96,91 99,03

Error (%) 23,15 21,58 5,2 13,26 7,7 3,7

c. VDS

d. Vout

2. Vin = 25 Volt a. Vpulse

b. VGS

c. VDS

d. Vout

3. Vin = 50 Volt a. Vpulse

b. VGS

c. VDS

d. Vout

Duty Cycle = 60% 1. Vin = 10 Volt a. Vpulse

b. VGS

c. VDS

d. Vout

2. Vin = 25 Volt a. Vpulse

b. VGS

c. VDS

d. Vout

3. Vin = 50 Volt a. Vpulse

b. VGS

c. VDS

d. Vout

2.

Reliability Test

Duty 𝑽𝑺 (V)

50%

50,01 50 50,2 50,3 50,2

𝑰𝑺 (A)

𝑽𝑶 (V)

0,22 0,46 0,72 0,91 1,21

23,68 23,32 23,2 23,11 23,07

RELIABILITY TEST A. I = 0,5 A

B. I = 1 A

𝑰𝑶 (A) 0,5 1 1,5 2 2,5

η (%) 93,09 98,62 96,28 94,95 99.03

C. I = 1,5 A

D. I = 2 A

E. I = 2,5 A

3. Partial Test Duty 𝑽𝑺 (V) 50%

60%

10 25 50 10 25 50

𝑰𝑺 (A) 0,08 0,12 0,17 0,11 0,16 0,24

𝑽𝑶 Prak (V) 3,9 11,5 23,54 5 13 24,45

𝑽𝑶 Teori (V) 5 12,5 25 6 15 30

𝑰𝑶 (A) 0,2 0,25 0,35 0,2 0,3 0,4

𝑷𝑺 (W) 0,8 3 8,5 1,1 4 12

𝑷𝑶 Prak (W) 0,78 2,875 8,75 1 3,9 11,78

𝑷𝑶 Teori (W) 1 3,125 8,75 1,2 4,5 12

η (%) 97,5 95,83 96,92 90,9 97,5 98,1

Error (%) 22 8 5,84 16,66 13,3 1,83

7.

Analysis The Buck Converter is used in SMPS (Switched Mode Power Supplies) circuits where the DC output voltage needs to be lower than the DC input voltage. The DC input can be derived from rectified AC or from any DC supply. The switching transistor between the input and output of the Buck Converter continually switches on and off at high frequency. To maintain a continuous output, the circuit uses the energy stored in the inductor L, during the on periods of the switching transistor, to continue supplying the load during the off periods. The inductor current has to be same at the start and end at one cycle. This means the overall change in the current is zero: (∆𝑖𝐿 )𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖𝐿 )𝑜𝑝𝑒𝑛 = 0 Subtituting and by their expressions yields: 𝑉𝑆 − 𝑉𝑂 −𝑉𝑂 ( ) . 𝐷𝑇 + ( ) . (1 − 𝐷 ) 𝑇 = 0 𝐿 𝐿 So, solving Vo is : 𝑉𝑜 = 𝐷𝑉𝑆 The above equation shows that the output voltage can be changed by adjusting the width of the duty cycle (D) (as duty cycle goes from 0 to 1 or 0% to 100%). According to the equation, when the duty cycle increase or higher with the same voltage input, than the voltage output will decrease. The design of buck converter using a analog PWM circuit to control voltage pulse width (Pulse Width Modulation) and frequency. IC NE555 is use to control PWM as a voltage regulation and IC LM393 to control the frequency as a voltage divider. Potensiometer is use to change to value of PWM and frequency. The module buck converter is supplyed with voltage input 50 Volt (minimum input voltage) and 60 Volt (maximum input voltage). There are 3 experiment to test the module buck converter, such as partial test, functional test and reliability test. For partial test and functional test it takes three data in different duty cycles. The duty cycle is 50% and 60% with input voltage is 10 Volt, 25 Volt and 50 Volt. The different of partial test and functional test is in the circuit. The partial test just use PWM circuit and Buck Converter circuit, otherwise in functional test all circuit is combined with the acrylyc (including transformator, power switch, fuse, etc). The result of this experiment is that the effiency of both test is good, it can near 100% but in functional test at 50%

duty cycle and 10 Volt input voltage the effiency is lower than the other. For errors, everytime input voltage increase with the same duty cycle, the errors increase too. When testing the reliability of buck converter module, we tested to produce a output current until it limit. The limit output current is 3 A theoritically, but it can be 2,5 A practically. From the data we used resistor to change the output current. The data start from 0,5 A with 0,5 step. The modul can still running at the 2,5 A, but the inductor became a little noisy and the MOSFET fumes and smells like a burning smell just for a while. From the experiment we know that the module can work as the design want. The minimum error voltage is 3.7% and the efficiency is up to 99.03% when the duty cylce is set 60%. The output voltage of every DC-DC Converter depend on range and the quality factor of the inductor. The standard minimum of the inductor quality factor is 1000. When the inductor quality factor is less than 1000, it can result losses of resistif in inductor very high and can lead the inductor became a load and give a high error of output voltage. In our partial experiment, the quality factor of our inductor is 1361. 8. Conclusion From the experiment we can get a conclusion: 1. The range of error is a. 3.7-13.6% on 50% duty cycle b. 5.2-23.15% on 60% duty cycle 2. The efficiency are good that can up to 99.03% 3. The modul is capable to operate under 2.5 A 4. The quality factor of inductor is 1361. 9. Suggestion 1. Design the buck converter with mathematic calculation and know your component design and cost based on the general market. 2. Use a good isolation for inductor winding and winding the inductor same as your design to get the same value and best quality factor of inductor. 3. Always to read datasheet for every component correctly. 4. To do the experiment, do as Standard Operational Procedure. 10. References 1. http://www.learnabout-electronics.org 2. https://www.theengineeringprojects.com/2018/01/introduction-to-buck-converter.html 3. https://www.allaboutcircuits.com/technical-articles/utilization-of-simple-converterscircuits/ 4. Hart, Daniel W, Power Electronics, New York: The McGraw-Hill Companies, Inc, 2011. 5. Ir. Zaenal Efeni, MT, Power Electronics Module.