Answer to Question #185705 in Electrical Engineering for dv f

Question

Isolated dc-dc converter is switched at 30 kHz with a duty cycle of 0.65. It is supplying 48 Watts load. The turn’s ratio of the centertapped transformer is 6.5. The dc link of this converter is taken from 48 V battery. The output inductor value is 10 mH. The output inductor current ripple is 12 % of load current. Determine following parameters for half bridge, full bridge and push-pull dc-dc converters (i) output voltage (ii) average and rms current of secondary diode (iii) voltage developed across the primary switch when it is off (iv) peak inverse voltage of secondary diode.

Answer 1

(i) output voltage

$Z=X_L=2\pi fL=2 \pi \times 30 \times10^3 \times 10 \times 10{^-3}=1884.96 \Omega$

$V_{out}=\frac{1884.96}{1884.96 + 1884.96 \times \frac{12}{100}} \times 48=42.858 V$

(ii) average and RMS current of secondary diode

$I_{Avaerage}= \sqrt{\frac{48}{1884.96}+(\frac{12}{100} \times 0.02546)^2}0.02564 A$

$I_{rms}=0.5 I_{Avaerage}=0.5 \times 0.02564=0.01282 A$

(iii) the voltage developed across the primary switch when it is off

$\frac{V_{PRI}T_{ON}}{N_{PRI}}=\frac{V_{SEC}T_{OFF}}{N_{SEC}}$

$V_{SEC}=\frac{V_{PRI}T_{ON}}{N_{PRI}} \times \frac{V_{SEC}}{T_{OFF}}=48 \times \frac{1}{6.5} \times 0.65 =0.48 V$

(iv) peak inverse voltage of the secondary diode.

$V_{peak}=V_{dc} \pi =48 \pi = 150.796 V$