Question

A flyback converter is switched at 50 kHz with a duty cycle of 0.4. It is supplying 25 Watts load. The turn’s ratio of transformer is 5. The input voltage of this converter is derived from 12 V battery. The flyback inductor value is viewed from the primary side is 10 mH. Determine (i) output voltage (ii) voltage developed across the primary switch when it is off (iii) peak current through the primary switch (iv) peak inverse voltage of secondary diode

Answer 1

(i) output voltage

$V_o=12 \times 0.4 \sqrt{\frac{12^2}{\frac{25}{2\pi *10* 50}}}=645.7 V$

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

$V_{switch}=\frac{12}{5}+645.7=648.1 V$

(iii) the peak current through the primary switch

$I_{max}=2 I_{ave}=2(12 \times \frac{25}{648.1})=0.928 A$

(iv) peak inverse voltage of the secondary diode

$V=\frac{25}{645.7}=0.0387 V$