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I have a moderately high power inverter board I've made. Standard 3 phase H-bridge with bootstrapped high side gate drivers. I've built a few, all have worked fine. They've all been pushing many kilowatts for many hours. This one works.

The issue/oddity is that the bootstrap diodes are the hottest thing on the board. My thermal camera says 43 degrees. This isn't obviously a problem, it's just odd given the current for the bootstraps is barely a few mA.

The diode is an SM4007, the board was built by jlcpcb and I used C64898 because it is a 1kV basic part. Essentially free.

I actually used this in a previous inverter board and never looked at it with the camera and have used it for hundreds of hours without issue.

Now I suspect the issue is the reverse recovery time is listed as about 1us for 4007 diodes, but what does this really mean? Clearly it's not dead shorting the bootstrap capacitors for 1us otherwise the circuit simply wouldn't work. The rise time is <50ns from 0 to 80V (20s battery,) essentially instant as far as 1us recover time is concerned.

Is this a problem? Is this going to bite me in some strange unforeseen way?

Now I'm aware of it, I'll spend the few extra pence for a fast recovery diode next time, but is it advisable to go and change the existing boards? I have about 10 with these diodes so 30 to change. Not a disaster but replacement isn't risk free so I'm not going to do it without good reason.

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Edit: I swapped one of the bootstraps for a BAT46Z and it runs much cooler. Definitely the reverse recovery. Absolutely no change in performance and only a minimal change to the bootstrap capacitor voltage.

Unfortunately I have to change it back because I need at least 120V capability, but next time.

JRE
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David Molony
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2 Answers2

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There's probably some cool scholarly articles on this, but I'll tell you what I know from experience and a very little bit of reading on the subject.

it's just odd given the current for the bootstraps is barely a few mA.

Well, apparently it's more than that or they wouldn't be getting hot!

Basically that diodes is built with a P-type, intrinsic, N-type construction, with the intrinsic layer serving as a barrier to increase the reverse voltage. In order to conduct, that intrinsic area has to fill with carriers. In order to shut off, those carriers have to be removed from the diode. At the speeds you're going, it's probably more accurate to say that there's a reverse-recovery charge than a reverse recovery time (note that reverse-recovery times are usually given with a specific voltage or reverse current).

I've actually kind of seen this, in driving transistors and LEDs fast. If you really want one to turn off quickly, you can pull the junction reverse-biased by a volt or three, and the shut-off happens much faster.

Of note (and ominously -- I'll get to that) this is the same construction -- and reason for construction -- as an RF switching PIN diode, which is designed specifically to have a really long \$t_{rr}\$.

So, you switch fast, you create hard fast changes in voltage on the diodes -- and current needs to flow to suck the carriers out of the junction. Because the voltage on the diode is close to a square wave, of necessity that somewhat fixed charge will flow out at high voltage -- and they'll get warm.

I suspect that in addition to getting warm, your diodes are also not doing a terrific job at bootstrapping -- especially when the high-side on-time is low (see the word "ominous", above). The diode's going to be taking charge away from the bootstrap capacitor; it may be taking enough that the top-side FET won't turn on all the way at extremes of duty cycle.

You may want to verify this, either through measurement or simulation.

TimWescott
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  • This may be relevant: https://electronics.stackexchange.com/questions/637374/behaviour-if-we-give-forward-voltage-during-the-reverse-recovery-time-of-a-diode/637478#637478 I measured such a (HV, probably PIN type, standard recovery) diode. With the fast dI/dt of a transistor switch, peak reverse current is many times forward bias. Here, what might be a few mA forward, could be a peak of 10s of mA reverse. Maybe even some A? Dunno. – Tim Williams Oct 15 '22 at 21:01
  • And then there are FREDs if you need really low recovery time but also don't want a schottky. They have an additional dopant (either gold, or more commonly platinum) to intentionally create traps and lower the carrier lifetime. This also increases the forward voltage, so it's a tradeoff. – Hearth Oct 15 '22 at 22:03
  • The bootstrapping is sufficient. I'll check tomorrow for additional evidence of excessive bootstrap losses but it isn't dropping out, the gates are getting driven to 12.5V (the nominally 12V line ended up 13.8 because of resistors and the DCDC used). I have no objection to Schottky, I vaguely remember debating it 2.5 years ago and thinking whatever, I'll get these essentially free ones and worst case i have to replace them... With Schottky... Them it worked and it wasn't until i got the thermal camera i thought anything else of it. – David Molony Oct 15 '22 at 23:28
  • I think the answer from Tim above, if correct, is kinda what I'm thinking through. If it's a charge rather than a time that truly defines the behaviour them it's probably all ok. It just charges... Really fast compared to the datasheet test value. If it is/might be time based there could be a nasty "shoot through"kind of behaviour. – David Molony Oct 15 '22 at 23:30
  • @DavidMolony It's more complicated than just a charge or just a time, as it depends on a number of circuit conditions. I believe it is closer to being a constant charge than a constant time, but it's still not perfectly constant. I would avoid PN diodes as a matter of course for fast switching applications, really. Schottky diodes have zero reverse recovery time/charge (as they're majority-carrier devices) other than charging the parasitic capacitance. (and if the lower breakdown voltage of Schottkys is a problem, there's always SiC Schottkys, with breakdown ratings in the kV.) – Hearth Oct 16 '22 at 14:33
  • @Hearth can you link to any articles or keywords? I just have a bit of practical experience, and the memory of an app note from a now-defunct semiconductor company (high performance BJTs in a world of MOSFETs; since bought by Diodes Inc). But I can't remember the name of the company. – TimWescott Oct 16 '22 at 16:12
  • @TimWescott I'm afraid the knowledge I have here comes from working for a company that makes FREDs, stuff I learned from more experienced coworkers and from practical experience running tests myself. Looking up details on double-pulse tests for diodes might get you something. The manufacturers usually just specify under specific conditions. The thing that matters the most is the diode current slew rate (\$\frac{dI_d}{dt}\$), but initial current and reverse voltage affect it too. – Hearth Oct 16 '22 at 18:48
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But is it a... Problem...

I would run a simulation with the MOSFET threshold voltage set to worst case (high), the duty cycle set to the worst case (max) and the diode recovery time set to worst case.

I would then check to make sure the MOSFET had sufficient gate voltage to turn on completely. If the loss of charge to the bootstrap capacitor by the diode's reverse recovery current is too large, there may be insufficient turn-on for the MOSFET, (at max duty cycle) and consequently there may be overheating of the MOSFET.

If the MOSFET gates are adequately driven, and the diodes don't overheat beyond their rating, you are probably OK.

Btw, what limits the duty cycle in your circuit?

Math Keeps Me Busy
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  • There's no issue with turning the MOSFET on. I've checked with a differential probe, the bootstrap cap remains 12V+and the gates are driven correctly. – David Molony Oct 15 '22 at 23:32
  • The duty cycle is limited in firmware, either written entirely by me or VESC (depending what I flash on any given day). It's 95% max. It's generating a sin wave in phase with the back EMF of the motor essentially. – David Molony Oct 15 '22 at 23:35