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I would like to use the TC4427 MOSFET driver chip, in order to switch the current through the IGBT AUIRG4PH50S, which will be placed in series with a resistive load. The power supply providing the main current through the load will be around 1V, and 100A. Another detail is that I would like to have up to x8 IGBTs (each with their own TC4427) - in parallel, to handle higher total currents through the main load. All x8 of these IGBTs will then be switched at the same time. The load will be being switched only around once per second.

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The TC4427 will drive the gate of the IGBT up to +15V when "on", and so I need to choose a +15V supply to power the VDD pin of the TC4427. To isolate the supply from other parts of the circuit, I would like this +15V supply to be an isolating DCDC converter (this DCDC converter power supply is seperate to the one providing the main current through the load, although they share the same ground.)

My question is how to choose the power capability of this DCDC converter in Watts? The datasheet for the TC4427 says peak output current of 1.5A - does this mean that during the switching process the +15V DCDC converter will need to provide P = (1.5A x 15V) = 22.5W for each IGBT? Or would I have to somehow calculate the actual current output by the driver based on the gate capacitance of this particular model of IGBT?

I am looking at something like the XP Power JSM1012S15 (10W, 670mA output), but I am not sure either whether this will be sufficient, or will be completely overkill for this application. Datasheet is here: https://www.xppower.com/fr/Product/JSM10-Series

Any guidance on how to go about calculating these things would be great.

edit: I later found a useful reference for a breakdown of the various power consumptions of a MOSFET driver: Avoiding MOSFET Driver Overstress http://ww1.microchip.com/downloads/en/AppNotes/01327A.pdf

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  • \$\begingroup\$ Will the IGBT even work at ONE VOLT? And if it will work, how FAST? Regarding the GateDrivers, place a bypass cap, value of 0.1uF, within 1milliMeter of the IC. And use a GroundPlane with no slits. Do not use sockets for the IC. \$\endgroup\$ – analogsystemsrf Dec 11 '18 at 15:04
  • \$\begingroup\$ You are right, datasheet says VCE(on) ~1.5V, so I will probably have a couple of volts here in the end. \$\endgroup\$ – teeeeee Dec 11 '18 at 15:11
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Driving FETs or IGBTs is effectively charging and discharging a capacitor at whatever your switching frequency is. The faster you charge and discharge the gate capacitance, the more efficient the switch is since there is a shorter period of partial conductance.

Any FET or IGBT will have a charge in coulombs listed for the gate at different drive voltages. Typically there is a plot like below. Given your driving ~15V into the gate of your AUIRG4PH50S IGBT, every charge and discharge you need to move that amount of charge. $$ q=130nC $$

                                        Gate Charge Chart

Since the Amp is defined as follows the math gets very easy. $$ 1 Amp = \dfrac{1C}{1 Sec} $$

The Average Current your driver will use is your switching frequency times the amount of charge moved every period. For your application with a 1Hz switching frequency you will use ~ 1hz * 130nC = 130nA per IGBT. $$ I_{avg} = f_{sw} * q_g $$

The Peak Current your driver will use is just ohms law. You have resistance in your drive, as well as a series resistor with the gate, and some dynamic gate resistance in the IGBT that is often given in the data sheet. There are some higher level complexities also depending on voltages across the transistor and miller capacitance etc. For a good first order approximation, you can just use whatever series resistor you have between the driver and the gate. Right now that looks like nothing, but you may want to consider adding some to help protect the driver, reduce ringing and allow slew rate control on the transistor (how fast it turns on and off to control EMC, ringing etc.) $$ I_{pk} \approx v_{driver} / r_g $$


Other Notes:

  1. IGBTs have a constant voltage drop ~1-2 Volts, so for low voltage applications like this they are not good. Use a FET instead.

  2. While your average drive current is very low, you need a lot of capacitance on the driver board to buffer the large (10+ Amps) it sounds like you will be pulling for the few nano or micro seconds it takes to charge and discharge your gates during switching.

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  • \$\begingroup\$ Since I will be switching once every second, or every 10 seconds or so, it is essentially only a single pulse of current to the gate? \$\endgroup\$ – teeeeee Dec 11 '18 at 15:17
  • \$\begingroup\$ Yes, although it will be a big pulse, it will be short,and buffered by capacitors on your supply... So you can have a fairly small supply. \$\endgroup\$ – MadHatter Dec 11 '18 at 16:25
  • \$\begingroup\$ Thanks @MadHatter. So I can add a gate resistor of say 10 Ohms each, plus the output impedance of the TC4427 (7 Ohms), giving a peak current of 15V / 17Ohm = ~900mA / IGBT. You're saying that this current should primarily be provided by reservoir caps on the VDD pin of each TC4427? If so, is there a rough calc I can do to estimate the size of each cap needed to provide this? Is it simply at least 150nC has to be stored on each cap, making it have to be at least C=150nC/15V=10nF? Secondly, I still have the problem of how to get the wattage of the main DCDC converter, which recharges the caps? \$\endgroup\$ – teeeeee Dec 12 '18 at 8:53
  • \$\begingroup\$ Your on the right track. Have some bulk capacitance like 10s of uF electrolytic etc. Along with 0.1uF ceramics. As for the wattage, you average current times the voltage, so 15v * 130nA, effectively nothing. To be safe I would just pick something like 1W if you have space for it, for a small design I would consider like a 500mA linear regular.... But bigger is always fine. \$\endgroup\$ – MadHatter Dec 12 '18 at 12:51
  • \$\begingroup\$ Also ensure your resistor can handle the peak current / power. I would not use anything smaller then ~1206 here. Just a guess though. \$\endgroup\$ – MadHatter Dec 12 '18 at 12:52

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