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I'm trying to determine the proper power supply type and output current requirement to drive a three phase IGBT driver (this is a follow up to a previous question I asked, although focused on a different aspect)

I am driving these with PWM at 4 kHz, as that seems a common number (although, as a bonus question: what are the criteria for choosing the optimal frequency?). For the purposes of safety, I want to use an isolated DC/DC converter to supply the voltage that is driven into the gates. The input to that will be 12v (as this is for a electric vehicle motor) What are the current requirements for this supply?

My understanding is that the key component of this is the transistor switching time, and, more specifically, the rise time for the gate voltage. This is determined by the gate capacitance for the IGBT and the supply current (let's assume this IGBT module for the sake of clarity: Mitsubishi CM300DY-24T). The datasheet defines the input capacitance Cies as 61.5 nF. I can determine the rise time by looking at that and the current, but, how fast do I need it to turn on? Each of the six transistors will switch twice during the 250 microsecond PWM cycle, but that's not very useful by itself.

Second part of this question: for my first board spin, I used a PDSE1-S12-S15 chip. The output current for this is defined as 67 mA, which I'm guessing is way too small, although I've been hard pressed to find these modules that source much more. Although the internal details of this design are a bit vague, I'm guessing it is a switched capacitor design. As such, I'm wondering how relevant the 67 mA figure is for this use case, since it is an average. When the IGBTs switch on, the initial current flow will presumably be much larger and cause the output capacitors in the DC/DC converter to fully discharge. I don't think adding external capacitance to the output of the DC/DC solves the problem.

I'm considering using a transformer to generate the isolated gate driver voltage. How is this generally done for these sorts of applications? I've looked at a bunch of application notes on designing gate driver circuits, but they don't seem to talk about the input power supply.

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  • \$\begingroup\$ The gate supply you use depends on what power sources are available. Having 480V 3 phase available is different than if you have batteries. The PDSE1 is just a buck converter. If it was switched cap, it would tell you. Switched cap can't be isolated anyways. "When the IGBTs switch on, the initial current flow will presumably be much larger and cause the output capacitors in the DC/DC converter to fully discharge." Why do you think this? Your gate-source capacitance is magnitudes smaller than the output caps. \$\endgroup\$
    – DKNguyen
    Commented May 14, 2022 at 20:00
  • \$\begingroup\$ well, not *just a buck since a normal buck isn't isolated. You choose how fast you need to turn on based on heat and losses. I like at least 3/5ths of my shortest pulse to be spent not in transition but ultimately it comes down to heat \$\endgroup\$
    – DKNguyen
    Commented May 14, 2022 at 20:15
  • \$\begingroup\$ In this case, the input power supply would be 12v (for a car, a PMAC electric drive motor controller). I will add that to the description. And, good point, now that you say it, the input gate capacitance is obviously much smaller than the power supply output caps, I hadn't thought that through. :) \$\endgroup\$
    – JeffB
    Commented May 14, 2022 at 20:46
  • \$\begingroup\$ You could almost use that directly after conditioning it then but is this still running off 300V like in your previous question? Doesn't seem to match. \$\endgroup\$
    – DKNguyen
    Commented May 14, 2022 at 20:48
  • \$\begingroup\$ Do you mean use the 12v supply directly to power the drivers? My concern would be that, a component failure on the HV side could directly connect the high voltage supply to the 12v system. The isolated supply makes this less likely. \$\endgroup\$
    – JeffB
    Commented May 14, 2022 at 20:51

1 Answer 1

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For drive supply requirements I go by the highschool physics approach:

\$Q=CV\$

and

\$I=dQ/dt = f_{pwm}*Q\$

(note that dQ is the difference in charge to create the gate leading edge, the trailing edge is created without drawing current from the supply)

So a ~62nf gate charge @ 15V @ 4kHz gets me 3.72mA - not a lot of current, your gate driver and isolation solution will probably take more.

The bulk capacitance on the isolated side will provide the transient loads for charging the IGBT gate, if that capacitor is only 1uf = 1000nf it's 16 times larger than the gate capacitance, so a complete short between them would result in a 1/16 drop in voltage at the bulk cap, or 6.25% - probably within the tolerance of your driver/isolator, but then again a few more uf wouldn't hurt the budget.

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  • \$\begingroup\$ Yes, I need to switch faster than 4 kHz in order to have control over the duty cycle. For example, for a 10% duty cycle, the PWM would switch on at the beginning of the cycle, but then switch off 25 microseconds later. I guess one approach would be to specify the required resolution. For 1% resolution, I guess I would need 372 mA. \$\endgroup\$
    – JeffB
    Commented May 14, 2022 at 21:06
  • \$\begingroup\$ @JeffB Actually my equation is a bit misleading in that regard: only the leading edge of the gate drive will contribute to the load on the supply. The trailing edge is just the discharging of the gate through the driver and network to ground, so we only need to could the fundamental switching frequency, we get short pulsewidth for free. I will try to tweak the math or the answer to make that clearer \$\endgroup\$
    – Bryan
    Commented May 14, 2022 at 23:17

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