EE 567 Mid Term

# EE 567 Mid Term - EE 567: INTEGRATED POWER ELECTRONICS...

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EE 567: INTEGRATED POWER ELECTRONICS MIDTERM

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Nutukurthi Krishna Chaitanya #36727542 "I believe in honesty and ethical code of my profession. My signature attests that I have neither given nor received any information relating to this exam _______________________________"
FLYBACK CONVERTER Problem : X = the number of day of month born = 1 => Average output current = 1A. Y = number of month born = 5 => Nominal input voltage = 5V Given the load is 30W. The input voltage can vary +/- X%, i.e. dv in the input voltage is +/- 1% The change in the output current is +/- X/3%, i.e 1/3 = +/- 0.33%. 1. TOPOLOGY AND REASON FOR SELECTION: From the given values we can calculate the output voltage and also the input current from the source. Average output voltage = 30/1 = 30V Average input current = 30/5 = 6A We can see that the input voltage is less than the output voltage. Therefore the converter to be designed may be a boost converter. 1.1 Boost Converter The simple boost converter circuit is shown in the figure

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Figure 1: Simple Boost converter To find the duty ratio in the Boost converter Vout = {1/(1-D1)} Vin => 30 = {1/(1-D1)} 5 => D1 = 5/6 = 0.83. Duty cycle of 83% implies that the inductor charges for the 83% of the total time and discharges for the remaining 17% when the diode is turned on. Hence very less time for the discharge of the inductor, this may put a voltage stress on the output. Also we know that V = L di/dt => L = V dt/di Since the input voltage and the ripple current is constant, L α di. This implies that large duty cycle results in large inductance which will increase the cost of the inductor. Also limiting the duty cycle increase the number of controller IC’s to choose from because many available today have a maximum duty cycle limitation of 50%. Hence now trying to limit the duty cycle we try using Buck boost Converter.
1.2 Buck Boost Converter: Figure 2: Buck Boost converter For Buck Boost Converter Vout = {D/(1-D)} Vin 30 = {D/(1-D)} 5 D = 6/7 = 0.85. Even in Buck Boost the duty ratio is very high same as the boost converter. Therefore we can reduce the duty ratio by using a coupled inductor instead of a single inductor in the Buck Boost converter. This is known as Flyback Converter.

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The main advantage in using flyback converter is isolation is achieved by using a coupled inductor in the circuit. The turns ratio in the coupled inductor acts as a trade off for the duty ratio. Hence the duty ratio can be decreased by using a Flyback converter. Flyback is also advantageous for power levels below 150W. In flyback the diode turn on, occurs when the inductor output coil voltage flies back as the input switch turns off. In buck boost a single inductor i.e. with one winding the flux and current are proportional, but in flyback with two windings the flux is proportional to the total of the ampere turns products of all windings. Figure 3: Flyback converter
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## This note was uploaded on 02/04/2011 for the course EE 587 taught by Professor Dr.mohammedsafiuddin during the Spring '11 term at SUNY Buffalo.

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EE 567 Mid Term - EE 567: INTEGRATED POWER ELECTRONICS...

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