Power Electronics Professor Mohamed A. El-Sharkawi El-Sharkawi@University of Washington

Ideal Switch v v i - + v vs i sw t R V s sw El-Sharkawi@University of Washington

Power Control  (toff) Power P Ps Time On-time (ton) Off-time Period (t) Off-time (toff) El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

Energy Consumption (E) Period (t) On-time (ton) Off-time (toff) Time (t) Power P Ps Duty Ratio (K) El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

Characteristics of Bi-polar Transistor V 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 El-Sharkawi@University of Washington

Closed switch Open switch R 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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

Opening Conditions of SCR A V RB Anode current is below the holding value (Ih) Ig = 0 Ih AK Opening angle is b El-Sharkawi@University of Washington

Power Converters El-Sharkawi@University of Washington

Power Converters El-Sharkawi@University of Washington

AC/DC Converters El-Sharkawi@University of Washington

Single-Phase, Half-Wave El-Sharkawi@University of Washington

w t i vt v s a b El-Sharkawi@University of Washington

Average Voltage Across the Load w t i vt v s a b El-Sharkawi@University of Washington

w t Load voltage i vt v s a b El-Sharkawi@University of Washington

V ave p a El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

Root-Mean-Squares (RMS) El-Sharkawi@University of Washington

Root Mean Squares of f Step 2: Step 1: Step 3: El-Sharkawi@University of Washington

Concept of RMS v2 w t v Average of v2 Square root of the average of v2 El-Sharkawi@University of Washington

Root-Mean-Squares (RMS) of a sinusoidal voltage El-Sharkawi@University of Washington

RMS of load voltage RMS of Supply Voltage El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

This looks like the negative of the average voltage across the load. Why? El-Sharkawi@University of Washington

Electric Power El-Sharkawi@University of Washington

Power Factor El-Sharkawi@University of Washington

Power Factor Real Power Complex Power El-Sharkawi@University of Washington

Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes El-Sharkawi@University of Washington

Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes El-Sharkawi@University of Washington

Single-Phase, Full-Wave, AC-to-DC 2-SCRs and 2 Diodes El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

Half Wave Versus Full Wave Average Voltage RMS Voltage Power El-Sharkawi@University of Washington

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. El-Sharkawi@University of Washington

Solution El-Sharkawi@University of Washington

DC/DC Converters El-Sharkawi@University of Washington

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. El-Sharkawi@University of Washington

Step Down (Buck converter) I V l t V on Time CE V t S I + V l - t on Time t El-Sharkawi@University of Washington

Example Solution El-Sharkawi@University of Washington

Step up (Boost converter) El-Sharkawi@University of Washington

Inductor current is unidirectional Keep in mind Inductor current is unidirectional Inductor cannot permanently store energy Voltage across inductor reverses El-Sharkawi@University of Washington

v i i ioff ton toff s t ion on off R L C Time El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

i on L VS i off L vt R C VS El-Sharkawi@University of Washington

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 El-Sharkawi@University of Washington

Solution Part 1 El-Sharkawi@University of Washington

Part 2 Part 3 Part 4 El-Sharkawi@University of Washington

Buck-Boost converter it is vt vs L C R El-Sharkawi@University of Washington

v t L R C i o ff + - i on L v s El-Sharkawi@University of Washington

DC/AC Converters El-Sharkawi@University of Washington

DC/AC Conversion VAB Q I A B 1 2 3 4 Q and Q are on Time Load voltage El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

First Time Interval El-Sharkawi@University of Washington

Second Time Interval El-Sharkawi@University of Washington

Voltage Waveforms Across Load Waveforms are symmetrical and equal in magnitude Waveforms are shifted by 120 degrees El-Sharkawi@University of Washington

AC/AC Converters El-Sharkawi@University of Washington

1. Single-Phase, Bidirectional El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

El-Sharkawi@University of Washington

2. DC Link El-Sharkawi@University of Washington

3. Uninterruptible Power Supply (UPS) Wind Turbine Controller El-Sharkawi@University of Washington