In an ideal flyback diode selection, one would seek a diode which has very large peak forward current capacity (to handle voltage transients without burning out the diode), low forward voltage drop, and a reverse breakdown voltage suited to the inductor's power supply.
from Wikipedia

I'm confused. Shouldn't the forward voltage drop of the diode be higher so that energy can be dissipated in fewer cycles.

  • \$\begingroup\$ I think this quote is from a TV context, where flyback means the horizontal flyback of the beam, which is used to generate the anode voltage. In this context the energy must not be dissipated but transferred to the high-voltage capacitor. \$\endgroup\$ – Wouter van Ooijen Mar 31 '14 at 21:23

You are right to be confused.

What is the meaning of "flyback diode"?

For a "flyback" diode in parallel with an inductance, and meant to dissipate inductor energy when current through the inductor is interrupted, larger Vf for fixed current dissipates this energy more quickly. (By the way, you should not depend only on maximum current rating to judge if the diode is a good fit. You MAY ALSO have to calculate power dissipated in the junction, the junction temperature rise over ambient, know the max ambient temperature, and ensure the max junction temperature is less than the data sheet. If the frequency of operation gets high enough, the diode will burn out even if the current is less than the maximum peak current on the data sheet.)

Now, for a "flyback diode" used as a diode in a "flyback" circuit which boosts voltage, the efficiency of the circuit will be higher for a diode with low Vf at a fixed current. Reverse recovery figures prominently into the efficiency calculation so reverse recovery is important too.

So you can see there is some confusion because there are two different answers.

In general, it is much better to discuss circuit questions with a schematic. Otherwise, improper generalizations are inevitable.

  • \$\begingroup\$ In the case of a fast switching motor, like in a H-bridge, a diode with a larger forward voltage drop would dissipate energy quicker. So in this case a higher Vf makes sense, doesn't it ? \$\endgroup\$ – emperor_penguin Apr 1 '14 at 10:43
  1. The diode forward current capacity is wholly determined by the current flowing in the inductor/transformer at the time when the switch/contact/transistor opens. If the current is 10A then the diode can be rated at 10A - it won't see any more than 10A and even if there was a slight initial nsec transient, the peak current handling capabilities of the diode will easily do the job without breaking sweat.
  2. Low voltage drop serves some applications but in a lot of applications you want rid of the stored energy as soon as possible and the larger the forward voltage (and it aint gonna be much bigger than 1 V) the quicker the energy is dissipated. On this, some will some won't!
  3. Reverse breakdown voltage I agree with

In addition, I'd pick a diode with a low reverse recovery time because when the flyback switch starts conducting again the diode will draw reverse current for a time period that may be excessive. For instance a tiny BAS16 has 4ns RR time whereas the 1N4004 is 2 micro seconds.

  • \$\begingroup\$ Exactly ! So a higher forward voltage drop is desired to get rid of the energy quicker. So why does it say low forward voltage drop as a desirable quality ? \$\endgroup\$ – emperor_penguin Mar 31 '14 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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