1. Power Electronics Professor Mohamed A. El-Sharkawi
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2. Ideal Switch v v i - + v vs i sw t R V s sw
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3. Power Control  (toff) Power P Ps Time On-time (ton) Off-time
Period (t)
Off-time
(toff)
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4. Load Switching toff ton Time (t) Power P (toff) On-time (ton)
Period (t)
On-time
(ton)
Off-time
(toff)
Time (t)
Power P
toff
ton
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5. Energy Consumption (E)
Period (t)
On-time
(ton)
Off-time
(toff)
Time (t)
Power P Ps
Duty Ratio (K)
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6. El-Sharkawi@University of Washington

7. El-Sharkawi@University of Washington

8. El-Sharkawi@University of Washington

9. El-Sharkawi@University of Washington

10. Bi-polar Transistor (C) (C) (C) Collector N I Base (B) V (B) P (B)CE CB BE
(C)
(B)
(E) N P
(C)
(B)
(E)
Collector
Emitter
Base
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11. Characteristics of Bi-polar TransistorV CE CB BE
Characteristics of Bi-polar Transistor I B1
Saturation Region I C V CE I B V BE
0.6 I B2
< B1
Linear Region
I = 0 B
Cut Off Region
Collector Characteristics
Base Characteristics
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12. Closed switch Open switchR L V CC CE B I
B max B
= 0 V CE I C V CC R L
(1)
(2) V CC
Closed
switch
At point (1)
VCE is very small
At point (2)
IC is very small
Open
switch
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13. Example
Estimate the losses of the transistor at point 1 and 2. Also calculate the losses at a mid point in the linear region where IB=0.1A. The current gain in the saturation region is 4.9 and in the linear region is 50. I I
B max B
= 0 V CE I C C
10 V CC R L
(1)
IB max=2A V CE
100V
(2) V CC
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14. Solution At point 1 Total losses = base loses + collector losses I I V
B max I C V CE I C
10
Vcc 1
IB max=2A RL 3 V CE
100V 2 I B
= 0 V CC
At point 1
Total losses = base loses + collector losses
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15. Solution At point 2 Total losses = collector losses
B max I C V CE I C
10
Vcc 1
IB=0 RL 3 V CE
100V 2 I B
= 0 V CC
At point 2
Total losses = collector losses
Assume VCE=0.99 VCC
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16. Solution At point 3 Total losses = base loses + collector losses
B max I C V CE I C
10
Vcc 1
IB max=0.1A RL 3 V CE
100V 2 I B
= 0 V CC
At point 3
Total losses = base loses + collector losses
Power transistors cannot operate in the linear region
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17. Thyristors [Silicon Controlled Rectifier (SCR)]
Anode (A)
Cathode (K)
Gate (G) I A V RB
Ig = 0
Ig = max
Ig > 0 Ih V TO V BO AK
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18. Closing Conditions of SCR
Positive anode to cathode voltage (VAK)
Maximum triggering pulse is applied (Ig)
Anode (A)
Cathode (K)
Gate (G)
Closing angle is a
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19. Opening Conditions of SCRA V RB
Anode current is below the holding value (Ih)
Ig = 0 Ih AK
Opening angle is b
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20. Power Converters
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21. Power Converters
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22. AC/DC Converters
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23. Single-Phase, Half-Wave
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24. w t i vt v s a b
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25. Average Voltage Across the Loadw t i vt v s a b
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26. w t
Load voltage i vt v s a b
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27. V
ave p a
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28. El-Sharkawi@University of Washington

29. Root-Mean-Squares (RMS)
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30. Root Mean Squares of f Step 2: Step 1: Step 3:
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31. Concept of RMS v2 w t v Average of v2 Square root of the average of v2
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32. Root-Mean-Squares (RMS) of a sinusoidal voltage
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33. RMS of load voltage RMS of Supply Voltage
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34. El-Sharkawi@University of Washington

35. This looks like the negative of the average voltage across
the load. Why?
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36. Electric Power
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37. Power Factor
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38. Power Factor Real Power Complex Power
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39. Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes
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40. Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes
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41. Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes
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42. El-Sharkawi@University of Washington

43. El-Sharkawi@University of Washington

44. El-Sharkawi@University of Washington

45. Half Wave Versus Full Wave
Average Voltage
RMS Voltage
Power
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46. Example
A full-wave, ac/dc converter is connected to a resistive load of 5 . The voltage of the ac source is 110 V(rms). It is required that the rms voltage across the load to be 55 V. Calculate the triggering angle, and the load power.
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47. Solution
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48. DC/DC Converters
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49. DC-to-DC Conversion
Step-down (Buck) converter: the output voltage of the converter is lower than the input voltage
Step-up (Boost) converter: the output voltage is higher than the input voltage.
3. Step-down/step-up (Buck-Boost) converter.
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50. Step Down (Buck converter)I V l t V on
Time CE V t S I + V l - t on
Time t
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51. Example
Solution
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52. Step up (Boost converter)
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53. Inductor current is unidirectional
Keep in mind
Inductor current is unidirectional
Inductor cannot permanently store energy
Voltage across inductor reverses
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54. v i i ioff ton toff s t ion on off R L C Time
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55. i ton toff Inductor current Inductor voltage von voff
Time
von
Inductor voltage
voff
Time
Energy is acquired by inductor
Energy is released by inductor
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56. i on L VS i
off L vt R C VS
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57. Example
A Boost converter is used to step up 20V into 50V. The switching frequency of the transistor is 5kHz, and the load resistance is 10. Compute the following:
The value of the inductance that would limit the current ripple at the source side to 100mA
The average current of the load
The power delivered by the source
The average current of the source
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58. Solution
Part 1
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59. Part 2
Part 3
Part 4
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60. Buck-Boost converter it is vt vs L C R
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61. v t L R C i o ff + - i on L v s
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62. DC/AC Converters
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63. DC/AC Conversion VAB Q I A B 1 2 3 4 Q and Q are on Time Load voltage
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64. El-Sharkawi@University of Washington

65. El-Sharkawi@University of Washington

66. First Time Interval
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67. Second Time Interval
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68. Voltage Waveforms Across Load
Waveforms are symmetrical and equal in magnitude
Waveforms are shifted by 120 degrees
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69. AC/AC Converters
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70. 1. Single-Phase, Bidirectional
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71. El-Sharkawi@University of Washington

72. El-Sharkawi@University of Washington

73. 2. DC Link
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74. 3. Uninterruptible Power Supply (UPS) Wind Turbine Controller
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