# Huge reverse current leakage in fast recovery diode for boost converter

I have a boost converter circuit that will boost 12 V to around 180 V, the circuit is works pretty good but there is huge reverse current in fast recovery diode right before the load. Here is my circuit and the simulation The green line is voltage in R3 (ignore the value because I know how to control that), the blue line is the current through D1, as you can see, in the very beginning, the current has opposite direction which is I assume as reverse current in D1. My questions are

1. How can the current go reverse the diode? It must be blocked since the maximum blocked voltage around 1 kV?

2. If it is how supposed to be, can I assume it as surge current of diode?

• Have you tried expanding the time scale to see what's going on there on a cycle-by-cycle basis? – Dave Tweed Apr 25 '18 at 2:23
• Create a circuit consisting of V2, D1, and R3 in series. Crank up V2's voltage. Run a time domain simulation of that circuit and observe the diode's current. Next, replace D1 with a fast recovery diode or a Schottky diode, and then re-run the simulation. (Hints: reverse bias capacitance; reverse recovery time; majority charge carrier recombination time.) – Jim Fischer Apr 25 '18 at 6:18
• Link the diode data sheet. – Andy aka Apr 25 '18 at 7:31
• Start using the SHIFT key. – winny Apr 25 '18 at 13:00

The 10ns spikes are given by parasitic oscillation with L or transistor pins parasitic inductance and diode capacitance when switching while L is still charged, you can notice that if you show the current through one inductance on the graphic.

Make some soft start like use variable frequency first then fixed to avoid that

Update

If you look at the drain current you can notice the same negative current glitch while turning on meaning of that is that it's parasitic inductance is causing that.

Same solution.

Zoomed graphic The positive glitch is the drain current , the negative is the diode current , down, in purple, the gate voltage. The zoom is in the region where M1 is switched on while L still charged

1) and 2) Nothing to do with the surge current , a real diode has a parasitic capacitance, the current is going through it no matter the breakdown voltage. Together with the drain parasitic inductance it makes a resonant circuit that makes the current even higher.