I have a boost converter circuit that will boost 12 V to around 180 V. The circuit works pretty well but there is a huge reverse current in the fast-recovery diode right before the load. Here is my circuit and the simulation:

enter image description here

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 the opposite direction which, I assume, is a reverse current in D1. My questions are:

  1. How can there be a reverse current through the diode? It must be blocked since the maximum blocked voltage is around 1 kV.

  2. If it is, can I assume it is a surge current of the diode?

  • 9
    \$\begingroup\$ Have you tried expanding the time scale to see what's going on there on a cycle-by-cycle basis? \$\endgroup\$
    – Dave Tweed
    Apr 25, 2018 at 2:23
  • 3
    \$\begingroup\$ 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.) \$\endgroup\$ Apr 25, 2018 at 6:18
  • 3
    \$\begingroup\$ Link the diode data sheet. \$\endgroup\$
    – Andy aka
    Apr 25, 2018 at 7:31
  • 2
    \$\begingroup\$ Start using the SHIFT key. \$\endgroup\$
    – winny
    Apr 25, 2018 at 13:00
  • \$\begingroup\$ September 2nd Close voter - please explain what is unclear to you about this question. It has a good answer that explains the factors involved. \$\endgroup\$
    – Russell McMahon
    Sep 1, 2022 at 14:47

1 Answer 1


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


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

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.