015 v for a current change of 3 ma 0015v003a

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Unformatted text preview: 2.1 2.2 2.3 2.4 For a duty cycle of 25%, Vo is independent of Io and thus appears as an ideal voltage source of 10 V. For a duty cycle of 7.9%, the voltage changes by 10.06 - 9.91 = 0.015 V for a current change of 3 mA. (0.015V/(.003A) = 5 ohms. Converter appears as an ideal voltage source of 10V in series with a 5 ohm resistor. 3. Design a flyback converter operating in the discontinuous conduction mode (DCM) shown shown in figure below for the following specifications: 300V<Vin <400V (nominal 400V) 0W <Po <50W (nominal 50W) Vo =27V fs=50kHz ∆vOp_p ≤ 20mV 3.1, 3.2, 3.3: Solution: The solution below is generalised for the situation that output diodes are not ideal which means they have certain voltage drop UD. In the exam question the diodes are assumed to be ideal so you can replace UD with 0. The solution is generalised for the situation that output diodes are not ideal which means they have certain voltage drop UD. In the exam question the diodes are assumed to be ideal so you can replace UD with 0. 4. The problem with ripple in the output current from a single- phase full bridge inverter is to be studied. The first harmonic of the output voltage is given by Vo1=220V at f = 47 Hz. The load is given in the figure as L = 100 mH in series with an ideal voltage source eo(t). The converter works in square wave mode. The converter operates in sinusoidal PWM- mode, bipolar modulation mf =21 and ma = 0.8. 4.1 See the book, Chapter 8 on Inverters or Lecture presentation Chapter 8, slide 13 4.2 4.3...
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