1. Power Electronics Lecture-10 D.C to D.C Converters (Choppers)
Dr. Imtiaz Hussain
Assistant Professor
URL :

2. Power Electronic Interface
Power Electronics is an enabling technology providing the needed interface between the electrical source and electrical load.
The source and load often do differ in frequency, voltage amplitudes and number of phases.

3. Powering the information Technology
Most of the consumer electronics equipment supplied from the mains, internally needed very load dc voltages.
Figure shows the distributed architecture typically used in computers.
In which incoming voltage from the utility is converted into dc voltage (24V).
This semi regulated voltage is distributed within the computer where on-board power supplies convert this 24V into tightly regulated lower voltage.

4. Powering the information Technology
Many devices such as cell phones operates from low battery voltages.
However, the electronic circuitry requires higher voltages.
Thus necessitating a circuit to boost input dc to higher dc voltages

5. Introduction to D.C to D.C Converters (Choppers)
DC to DC converters are important in portable electronic devices such as cellular phones and laptop computers, which are supplied with power from batteries primarily.
Such electronic devices often contain several sub-circuits, each with its own voltage level requirement different from that supplied by the battery or an external supply.
They are also widely used in dc-motor drive applications.
Often input to these converters is an unregulated dc voltage, which is obtained by rectifying the line voltage.

6. D.C to D.C Converter System
Introduction
Uncontrolled
Diode Rectifier
Load
D.C to D.C Converter
Battery
Filter
AC Line Voltage
1-Phase or
3-Phase
DC Unregulated
DC Regulated vc
D.C to D.C Converter System

7. Efficiency & Power Losses
High efficiency is essential in any power processing application.
The efficiency of a converter is
The power lost in converter is

8. Efficiency & Power Losses
Efficiency is a good measure of the success of a given converter technology.
With very small amount of power lost, the converter elements can be packaged with high density, leading to a converter of small size and weight, and of low temperature rise.
How can we build a circuit that changes the voltage, yet dissipates negligible power?

9. Efficiency & Power Losses
The various conventional circuit elements are illustrated in Following figure.
The available circuit elements fall broadly into the classes of resistive elements, capacitive elements, magnetic devices including inductors and transformers, semiconductor devices operated in the linear mode and semiconductor devices operated in the switched mode.

10. Types of dc-dc Converters
Types of D.C to D.C converters
AC Link Choppers
Linear Converters
Switch Mode
Magnetic
E.t.c

11. AC Link Choppers
First dc is converted to ac with the help of an inverter.
After that, AC is stepped-up or stepped-down by a transformer, which is then converted back to dc by a diode rectifier.
Ac link chopper is costly, bulky and less efficient as the conversion is done in two stages.

12. Simple dc-dc Converters
Let us now construct a simple dc-dc converter. The input voltage vg is 100 V. It is desired to supply 50 V to an effective 5Ω load, such that the dc load current is 10 A.

13. Resistive dc-dc Converters
Using Voltage divided rule.

14. Linear dc-dc Converters
Linear Mode dc-dc converter

15. Switch Mode dc-dc Converters

16. Conclusion
Capacitors and magnetic devices are important elements of switching converters, because ideally they do not consume power.
It is the resistive element, as well as the linear-mode semiconductor device, that is avoided.
Semiconductor in switch mode however dissipate comparatively low power in either states (ON and OFF).
So capacitive and inductive elements, as well as switched-mode semiconductor devices, are available for synthesis of high-efficiency converters.

17. Switch Mode D.C to D.C Converters
Switch-mode DC to DC converters convert one DC voltage level to another, by storing the input energy temporarily and then releasing that energy to the output at a different voltage.
This conversion method is more power efficient (often 75% to 98%) than linear voltage regulation (which dissipates unwanted power as heat).
This efficiency is beneficial to increasing the running time of battery operated devices. 

18. Switch Mode D.C to D.C Converters
PWM or PFM regulates the output dc voltage.
The power flow through these converters is only in one direction thus their voltages and currents remain unipolar.

19. Switch Mode D.C to D.C Converters
A Chopper is a high speed on/off semiconductor switch.
It connects source to load and disconnects source from load at very high speed.
In this manner a chopped dc voltage is obtained from a constant dc supply Vs is obtained.

20. Switch Mode D.C to D.C Converters
During the period Ton, the chopper is on and load voltage Vo is equal to source voltage Vs.
During the period Toff, the chopper is off and load current io flows through the freewheeling diode.

21. Switch Mode D.C to D.C Converters
The load voltage is given by
Thus the load voltage can be varied by varying the switching duty ratio D.
𝑉 𝑜 = 𝑇 𝑜𝑛 𝑇 𝑜𝑛 + 𝑇 𝑜𝑓𝑓 𝑉 𝑠
𝑉 𝑜 = 𝑇 𝑜𝑛 𝑇 𝑉 𝑠
𝑉 𝑜 =𝐷 𝑉 𝑠
𝑤ℎ𝑒𝑟𝑒, 𝐷= 𝑇 𝑜𝑛 𝑇 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑠𝑤𝑖𝑡𝑐ℎ𝑖𝑛𝑔 𝑑𝑢𝑡𝑦 𝑟𝑎𝑡𝑖𝑜
𝑉 𝑜 =𝑓 𝑇 𝑜𝑛 𝑉 𝑠

22. Control of D.C to D.C Converters
Average value of output voltage Vo can be controlled by opening and closing the semiconductor switch periodically.
The control strategies for varying duty ratio D are
Constant frequency system
Variable frequency system

23. Constant frequency system
In this scheme Ton is varied but frequency is kept constant.
Variation of Ton means adjustment of pulse width. Therefore, this scheme is called PWM scheme.
𝑉 𝑜 = 1 4 𝑉 𝑠
𝑇 𝑜𝑛 = 1 4 𝑇
𝑉 𝑜 = 3 4 𝑉 𝑠
𝑇 𝑜𝑛 = 3 4 𝑇

24. Constant frequency system (PWM)
amplifier + -
Comparator
Vo (desired)
Vo (actual)
Sawtooth Wave
Switch Control Signal
Vcontrol
vcontrol t Ts
vst
Ton
Toff
Switch Control Signal

25. Variable frequency system
In this scheme Ton is kept constant but the frequency is varied.

26. Variable frequency system
Or Toff is kept constant but the frequency is varied.

27. Switch Mode D.C to D.C Converters
Types of Switch Mode D.C to D.C Converters
Step-Down (Buck) converter
Step-up (Boost) converter
Step Down/Up (Buck-Boost) converter

28. Step-Down Converter (Buck Converter)
As name implies a step-down converter produces a lower average output voltage than the dc input voltage Vd.
Its main application is in regulated dc power supplies and dc-motor speed control.

29. Step-Down Converter (Buck Converter)
Circuit operation when S is ON (closed)
Diode is reversed biased. Switch conducts inductor current 𝑖 𝐿 .
This results in positive inductor voltage, i.e:
It causes linear increase in inductor current
𝑣 𝐿 = 𝑉 𝑑 − 𝑉 𝑜
𝑖 𝐿 = 1 𝐿 𝑣 𝐿 𝑑𝑡

30. Step-Down Converter (Buck Converter)
Circuit operation when S is OFF (open)
Because of inductive energy storage, 𝑖 𝐿 continues to flow.
Diode is reversed biased. Therefore current flows through the diode.
𝑣 𝐿 =− 𝑉 𝑜

31. Step-Down Converter (Buck Converter)

32. Step-Down Converter (Buck Converter)
Continuous conduction mode
A buck converter operates in continuous mode if the current through the inductor (IL) never falls to zero during the commutation cycle.

33. Design procedure for Buck Converter
Calculate D to obtain required output voltage.
Select a particular switching frequency:
–preferably >20KHz for negligible acoustic noise
Higher fs results in smaller L, but higher device losses.
Thus lowering efficiency and larger heat sink.

34. Design procedure for Buck Converter
Inductor requirement
C Calculation
Possible switching devices: MOSFET, IGBT and BJT. Low power MOSFET can reach MHz range.
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−𝐷 2𝑓 𝑅
𝑟= 1−𝐷 8𝐿𝐶 𝑓 2

35. Example-1
A buck converter is supplied from a 50V battery source. Given L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4. Calculate: (a) output voltage (b) output voltage ripple.
Solution
(a) output voltage
𝑉 𝑜 =𝐷 𝑉 𝑠
𝑉 𝑜 =0.4×50
𝑉 𝑜 =20𝑉

36. Example-1 𝑟= 1−𝐷 8𝐿𝐶 𝑓 2 𝑟= 1−0.4 8×400𝜇×100𝜇× (20𝐾) 2 𝑟=4.6%
(b) Output Ripple
𝑟= 1−𝐷 8𝐿𝐶 𝑓 2
𝑟= 1−0.4 8×400𝜇×100𝜇× (20𝐾) 2
𝑟=4.6%

37. Example-2 𝑉 𝑜 =𝐷 𝑉 𝑠 𝐷= 𝑉 𝑜 𝑉 𝑠 𝐷= 25 50 =0.5
A buck converter has an input voltage of 50V and output of 25V. The switching frequency is 10KHz. The power output is 125W. (a) Determine the duty ratio, (b) value of L to ensure continuous current, (c) value of capacitance to limit the output voltage ripple factor to 0.5%.
Solution
(b) Value of L
(a) output voltage
𝑉 𝑜 =𝐷 𝑉 𝑠
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−𝐷 2𝑓 𝑅
𝐷= 𝑉 𝑜 𝑉 𝑠
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−0.5 2×10𝐾 𝑅
𝐷= =0.5
𝑃 𝑜 = 𝑉 2 𝑅

38. Example-2 Resistance is calculated as
L must at least be 10 times greater than Lmin.
𝑅= 𝑉 2 𝑃 𝑜 = =5Ω
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−0.5 2×10𝐾 ×5
𝐿≥ 𝐿 𝑚𝑖𝑛 = 𝐾 =125𝜇𝐻
𝐿=1.25𝑚𝐻

39. Example-2 𝑟= 1−𝐷 8𝐿𝐶 𝑓 2 0.005= 1−0.5 8×1.25𝑚×𝐶× (10𝐾) 2
(c) Value of C
Value of capacitance to limit the output voltage ripple factor to 0.5% can be calculated using following equation.
𝑟= 1−𝐷 8𝐿𝐶 𝑓 2
0.005= 1−0.5 8×1.25𝑚×𝐶× (10𝐾) 2
𝐶= 1−0.5 8×1.25𝑚×0.005× (10𝐾) 2
𝐶= =100𝜇𝐹

40. Example-3
Design a buck converter such that the output voltage is 28V when the input is 48V. The load is 8Ohm. Design the converter such that it will be in continuous current mode. The output voltage ripple must not be more than 0.5%. Specify the frequency and the values of each component. Suggest the power switch also.
Solution:
First of all determine the switching frequency.
𝑓=25𝐾𝐻𝑧
Then calculate the switching duty ratio
𝑉 𝑜 =𝐷 𝑉 𝑠
𝐷= =0.58

41. Example-3 𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−𝐷 2𝑓 𝑅 𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−0.58 2×25𝐾 ×80
Now value of inductor can be chosen to ensure continuous conduction
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−𝐷 2𝑓 𝑅
𝐿≥ 𝐿 𝑚𝑖𝑛 = 1−0.58 2×25𝐾 ×80
𝐿≥ 𝐿 𝑚𝑖𝑛 =0.67𝑚𝐻
𝐿=6.7𝑚𝐻

42. Example-3 𝑟= 1−𝐷 8𝐿𝐶 𝑓 2 𝐶= 1−0.58 8×6.7𝑚×0.005× (25𝐾) 2
Value of capacitance to limit the output voltage ripple factor to 0.5% can be calculated using following equation.
𝑟= 1−𝐷 8𝐿𝐶 𝑓 2
𝐶= 1−0.58 8×6.7𝑚×0.005× (25𝐾) 2
𝐶= =2.5𝜇𝐹

43. Example-3
Selection of Power semiconductor switch
𝑓=25𝐾𝐻𝑧
𝑇=40𝜇𝑠

44. Step-Up Converter (Boost Converter)
A boost converter (step-up converter) is a DC-to-DC power converter with an output voltage greater than its input voltage. 

45. Step-Up Converter (Boost Converter)
When the switch is closed, current flows through the inductor in clockwise direction and the inductor stores the energy.
Polarity of the left side of the inductor is positive. 
𝑣 𝐿 = 𝑉 𝑑
𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑑 𝐿

46. Step-Up Converter (Boost Converter)
When switch is opened, the output receives energy from the input as well as from the inductor.
Hence output is large.
𝑣 𝐿 = 𝑉 𝑑 − 𝑉 𝑜
𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑑 − 𝑉 𝑜 𝐿

47. Step-Up Converter (Boost Converter)
𝑣 𝐿 = 𝑉 𝑑
𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑑 𝐿
S Closed
𝑣 𝐿 = 𝑉 𝑑 − 𝑉 𝑜
S Open
𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑑 − 𝑉 𝑜 𝐿

48. Step-Up Converter (Boost Converter)
𝑣 𝐿 = 0 𝐷𝑇 𝑉 𝑑 𝑑𝑡+ 𝐷𝑇 𝑇 𝑉 𝑑 − 𝑉 0
0= 𝑉 𝑑 𝑡 0 𝐷𝑇 + (𝑉 𝑑 − 𝑉 𝑜 ) 𝑡 𝐷𝑇 𝑇
0= 𝐷𝑇𝑉 𝑑 + (𝑉 𝑑 − 𝑉 𝑜 )[𝑇−𝐷𝑇]
0= 𝑇𝑉 𝑑 −𝑇 𝑉 𝑜 + 𝐷𝑇𝑉 𝑜
𝑉 𝑜 = 𝑉 𝑑 1−𝐷

49. Step-Up Converter (Boost Converter)
Boost Converter Design
Minimum inductor value
Capacitor Value
𝐿 𝑚𝑖𝑛 = 𝐷 (1−𝐷) 2 𝑅 2𝑓
𝑟= 𝐷 𝑅𝐶𝑓

50. Example-4
The boost converter has the following parameters: Vd=20V, D=0.6, R=12.5ohm, L=65uH, C=200uF, fs=40KHz. Determine (a) output voltage, (b) output voltage ripple.
Solution
(b) output voltage ripple
(a) output voltage
𝑟= 𝐷 𝑅𝐶𝑓
𝑉 𝑜 = 𝑉 𝑑 1−𝐷
𝑟= ×200𝜇×40𝐾
𝑉 𝑜 = 20 1−0.6 =50𝑉
𝑟=0.6%

51. Example-5
Design a boost converter to provide an output voltage of 36V from a 24V source. The load is 50W. The voltage ripple factor must be less than 0.5%. Specify the duty cycle ratio, switching frequency, inductor and capacitor size, and power device.
Solution
First of all determine the switching frequency.
𝑓=30𝐾𝐻𝑧
Then calculate the switching duty ratio
𝑉 𝑜 = 𝑉 𝑑 1−𝐷
36= 24 1−𝐷
⇒36(1−𝐷)=24
⇒𝐷=0.33

52. Example-5 𝑃 𝑜 = 𝑉 2 𝑅 𝑅= 36 2 50 =25.92Ω 𝐿 𝑚𝑖𝑛 = 𝐷 (1−𝐷) 2 𝑅 2𝑓
Calculate the load resistance
𝑃 𝑜 = 𝑉 2 𝑅
𝑅= =25.92Ω
Calculate inductor value to ensure continuous current
𝐿 𝑚𝑖𝑛 = 𝐷 (1−𝐷) 2 𝑅 2𝑓
𝐿 𝑚𝑖𝑛 = 0.33 (1−0.33) 2 × ×30𝐾 =63.9𝜇𝐻
Inductance must be greeter than or equal to Lmin
𝐿≥𝐿 𝑚𝑖𝑛 =63.9𝜇𝐻
𝐿=100𝜇𝐻

53. Example-5
Then calculate the Capacitor Value for ripple factor less than 0.5%
𝑟= 𝐷 𝑅𝐶𝑓
0.005= ×𝐶×30𝐾
𝐶= ×0.005×30𝐾 =84.8𝜇𝐹
While selecting the power device we must take into account the switching frequency
𝑓=30𝐾𝐻𝑧
𝑇= 1 𝑓 = 1 30𝐾 =33.3𝜇𝑠

54. Example-5
Selection of Power semiconductor switch
𝑓=30𝐾𝐻𝑧
𝑇=33.3𝜇𝑠

55. Buck-Boost Converter If D>0.5, output is higher
The buck–boost converter is a type of DC-to-DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude.
If D>0.5, output is higher
If D<0.5, output is lower
Output voltage is always negative

56. Buck-Boost Converter
while in the On-state, the input voltage source is directly connected to the inductor (L). This results in accumulating energy in L. In this stage, the capacitor supplies energy to the output load.

57. Buck-Boost Converter
In Off-state, the inductor is connected to the output load and capacitor, so energy is transferred from L to C and R.

58. Buck-Boost Converter In ON-state (Switch Closed) 𝑣 𝐿 = 𝑉 𝑑 =𝐿 𝑑 𝑖 𝐿 𝑑𝑡
𝑣 𝐿 = 𝑉 𝑑 =𝐿 𝑑 𝑖 𝐿 𝑑𝑡
⇒ 𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑑 𝐿

59. Buck-Boost Converter In OFF-state (Switch Opened)
𝑣 𝐿 = 𝑉 𝑜 =𝐿 𝑑 𝑖 𝐿 𝑑𝑡
⇒ 𝑑 𝑖 𝐿 𝑑𝑡 = 𝑉 𝑜 𝐿

60. Buck-Boost Converter 𝑉 𝑑 𝐷𝑇 𝐿 + 𝑉 𝑜 1−𝐷 𝑇 𝐿 =0 𝑉 𝑜 =− 𝑉 𝑑 𝐷 1−𝐷
Steady state operation
𝑉 𝑑 𝐷𝑇 𝐿 + 𝑉 𝑜 1−𝐷 𝑇 𝐿 =0
𝑉 𝑜 =− 𝑉 𝑑 𝐷 1−𝐷

61. Buck-Boost Converter
𝑟= 𝐷 𝑅𝐶𝑓
𝐿 𝑚𝑖𝑛 = (1−𝐷) 2 𝑅 2𝑓

62. Example-6
Determine the switching duty ratio of a buck-boost converter such that the output voltage is -28V when the input is 100V. The load is 1Ohm. Design the converter such that it will be in continuous current mode. The output voltage ripple must not be more than 0.5%. Specify the frequency and the values of each component. Suggest the power switch also.
Solution:
First of all determine the switching frequency.
𝑓=50 𝐾𝐻𝑧
Then calculate the switching duty ratio
𝑉 𝑜 =− 𝑉 𝑑 𝐷 1−𝐷
⇒ 𝑉 𝑜 − 𝑉 𝑑 = 𝐷 1−𝐷
⇒0.28= 𝐷 1−𝐷
⇒𝐷= =0.2

63. Example-6 𝐿 𝑚𝑖𝑛 = (1−𝐷) 2 𝑅 2𝑓 𝐿>𝐿 𝑚𝑖𝑛 = (1−0.2) 2 ×10 2×50𝐾
Then calculate the Capacitor Value for ripple factor less than 0.5%
𝑟= 𝐷 𝑅𝐶𝑓
0.005= ×𝐶×50𝐾
𝐶= ×0.005×50𝐾 =80𝜇𝐹
Value of inductor can be calculated as
𝐿 𝑚𝑖𝑛 = (1−𝐷) 2 𝑅 2𝑓
𝐿>𝐿 𝑚𝑖𝑛 = (1−0.2) 2 ×10 2×50𝐾
𝐿> 𝐾
𝐿>64𝜇𝐻

64. Example-6 𝑓=50𝐾𝐻𝑧 𝑇= 1 𝑓 = 1 50𝐾 =20𝜇𝑠
While selecting the power device we must take into account the switching frequency
𝑇= 1 𝑓 = 1 50𝐾 =20𝜇𝑠
𝑓=50𝐾𝐻𝑧

65. Switch mode vs Linear Power Supplies
One of the major applications of switch mode dc-dc converters is in switch mode power supplies (SMPs).
SMPs offer several advantages over linear mode power supplies.
Efficient (70-95%)
Weight and size reduction
They also have some disadvantages
Complex design
EMI problems

66. Linear Mode Power Supply

67. Switch Mode Power Supply

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