# Vrrm on a secondary rectifying diode for a full bridge forward converter

Discussion in 'General Electronics Chat' started by reggiereggie, Jun 1, 2008.

1. ### reggiereggie Thread Starter New Member

Jun 1, 2008
2
0
Hi all,

I have found a formula that I would like to understand the derivation of.

I am trying to work out the peak reverse voltage on a secondary side full bridge forward converter rectifier diode. The secondary is centre tapped with two diodes, top diode D1 and bottom diode D2, an output choke and capacitor.

I understand the operation of the circuit and would assume the reverse voltage is the primary voltage referred to the secondary during the time that say D1 turns off. But I have found this formula that states

Vrrm >= [((Vout +Vf).Vinmax)/(duty max . Vinmax)] + leakage inductance spike

Vrrm = diode repeat reverse voltage
Vout= dc output voltage of converter
Vf = it doesnt state which diode voltage drop this is, maybe its the other secondary diode (when d1=off d2=on)
Vinmax = max dc bus voltage = input voltage to converter

I have found a link that has the waveforms which I understand but not how to work out the magnitude of the Vd1s reverse voltage. its the waveform titled: These are typical waveforms in a full-bridge PWM topology

http://www2.electronicproducts.com/Optimal_output_rectifier_selection-article-MAYINT1-MAY2002.aspx

Please can someone explain this formula or explain how they would work out the Vrrm on a secondary rectifying diode for a full bridge forward converter.

reggie