# Is it possible to repeatedly use a schottky diode at its surge current without damage

I've been working on a circuit with an inductive load and a MOSFET, but I'm having trouble finding a suitable freewheeling diode. If I choose a diode with a maximum surge current higher than the maximum current through the load, would it be dangerous (for the component and/or for me) to frequently switch the MOSFET on and off? (Minimum of a few seconds between switchings)

I believe the biggest issue is that I don't know how long the inductive voltage spike will last, so I can't properly estimate how long the diode will be powered at max surge current.

As I understand it, it's a temperature thing. As long as I keep the temperature under the maximum rated temperature everything should be fine. (When the temperature starts getting worrisome, give it some time to cool off before using it again.)

• If you provided the circuit with component listings, it is possible someone could provide better advice. – Eric Urban Jun 25 '16 at 3:46
• @EricUrban The load is an electromagnet, so to maximize pulling force I have the circuit powered with a maximum current of ~200A. (the circuit is only powered for a few moments at a time, so the overall temperature rise is only a few °C.) It's very difficult to find Schottky diodes rated for 200+A, but I can find diodes with a surge rating of 250+A (adding a bit of a margin there.) – CoilKid Jun 25 '16 at 3:53
• For example, I was just looking at this datasheet. It shows that the diode is only rated for 5A of continuous current, but it can handle 220A if it's a ~8ms surge. *Note: I'd probably keep looking for a diode that is rated for an average current of more than 5A. (Maybe 100-150A, if I can find it?) – CoilKid Jun 25 '16 at 3:55
• What voltage is powering the electromagnet? – Bruce Abbott Jun 25 '16 at 4:09
• This answer of mine could be relevant. – Lorenzo Donati supports Monica Jun 25 '16 at 7:11

The single pulse surge rating is usually a 'maximum stress level' parameter that should only occur under adverse conditions and not in normal operation. If you expect it to occur regularly then the diode's reliability may suffer.

I would use a diode rated for at least double the expected surge current. Heating shouldn't be a problem if it only has to operate once every few seconds. The inductive current will probably decay within a few hundred milliseconds of the solenoid being turned off.

A couple of 60A diodes in parallel should do the job. These are readily available from most suppliers. If you can't find any suitable diodes locally then perhaps you could try scavenging some from old PC power supplies.

• So you're suggesting using two 60A diodes, instead of one? If the surge rating is ~triple the normal expected would a single diode work, rather than two? Also, I found a diode that I think might work but I've had trouble finding the breakdown voltage on the datasheet. (I've had this problem while looking at other diodes.) Thanks! – CoilKid Jun 25 '16 at 5:42
• ...Unless the breakdown voltage is the Vrrm, which would explain a lot. – CoilKid Jun 25 '16 at 5:46
• If each diode in parallel receives only $$\frac{1}{\text{n}}*\text{amperage}$$ then I should be able to use several diodes such that the maximum repetitive surge currents on each sum to greater than the maximum current in my circuit, right? *Note: n is the number of diodes. – CoilKid Jun 25 '16 at 21:55
• Vrrm is maximum permitted reverse voltage. DSS 60-0045B is rated at 45V, which should be plenty enough for a 14.8V power supply. Fig. 5 in the datasheet shows ~34W dissipation at 20A average with 8% duty cycle, = 250A for 240ms every 3 seconds. – Bruce Abbott Jun 25 '16 at 22:38
• If I understand you correctly, you're saying a DSS 60-0045B could work as long as the inductive voltage spike lasts for less than 240ms, and I let it "rest" for at least 3 seconds before switching it again? That sounds great, but again I'm still not sure how to find the duration of said voltage spike. Should I ask that as a question? – CoilKid Jun 25 '16 at 22:49

This addresses a subset of the question which is needed to understand what the diode "sees":

Vfd - diode forward voltage drop
Pd - diode dissipation etc

The length of the Current duration and waveform is based on energy dissipation.
At turn off you have energy of 0.5Li^2 and this dissipates in the circuit it flows in.
The diode dissipates Vfd.I and if this was the only element and ideal then diode voltage would remain essentially at Vfd until energy was dissipated so
Vf.i.t = 0.5Li^2
and t = 0.5Li/Vfd. (In fact Vfd will decrease with decreasing i but is liable to be close enough to constant for this purpose.)(And the diode is liable to NOT be the main initial loss source - see below).

If you have series resistance (and you do, even if small) then you have P=I^2.R dissipation and as energy dissipates I will fall so P in R will fall and Vr will fall.
Initially at 200A Pd_peak = Vfd x 200.
If Vfd = say 1V then Pd = 200W.
For R to have equal losses to the diode then
Pd = Pr or Vfd.I = I^2.R
so R = Vfd.I/I^2 = Vfd/I = 1/200 Ohm = 5 milliOhm.
If your coil + diode lead resistance exceeds 5 millOhm then resistive dissipation will initially dominate.

You can easily model the exponential energy and voltage decreases using eg SPICE.