I am having difficulties in trying to switch the high-side IGBT in the circuit below, that is connected to a capacitor. The input is from a power supply of 50V. When S1 is ON the capacitor charges from 0V to 50V. (There is a reason to why I need the capacitor to charge from 0V - 50V).

I give a switching signal of 3.3V from a micro-controller and 12V from a regulator.

I use V0D3120AD, which to the best of knowledge acts as an opto-coupler. I connected the opto-coupler as shown below to drive the high-side IGBT but it does not work as intended.

enter image description here

I also thought of using a gate driver shown below (where the bottom MOSFET would be a capacitor in my case and grounding LO pin) but the problem is that my bootstrap capacitor needs to be charged from Vcc via a ground path for it to provide the required voltage to switch on the IGBT, and therefore I suppose I cannot use this gate drive circuitry, since the bootstrap capacitor doesn't have a path to ground to charge up and I don't know if it can charge up through the 10mF capacitor.

enter image description here

My other solution is shown below, where I use 12V regulator and the ground of the regulator is connected to the emitter of the IGBT thus always giving 12V from gate to emitter. The 10k resistor is to discharge the parasitic capacitance of the IGBT.

enter image description here

I would appreciate some advice whether my last solution of using just a regulator is a safe way of achieving high side driving or is there some modification I can make to the other circuits.

  • \$\begingroup\$ Are all the GNDs the same? 12V and 3.3V? If so, your first diagram has no ground on the output side of U9. \$\endgroup\$
    – Aaron
    Jun 30, 2022 at 17:14
  • 1
    \$\begingroup\$ You need to drive the gate with 65 V if your emitter is at 50 V. You need a floating supply for your high side drive. \$\endgroup\$
    – winny
    Jun 30, 2022 at 18:05
  • \$\begingroup\$ @Aaron Yes all GNDs are the same. Vee pin of U9 would have been connected to the GND but then I wouldnt get 12V across Gate and Emitter so I connected Vee pin to the Emitter of IGBT to get 12V across Gate and Emitter \$\endgroup\$
    – Mayur
    Jul 1, 2022 at 8:03

2 Answers 2


If you want to drive your IGBT high-side, you need a floating/isolated gate drive voltage referenced to its source, which none of your schematics have.

Note, if you manage to turn it on hard, it will short the +50V power supply to the discharged 10.000µF capacitor. If the power supply has large output capacitance, the IGBT will most likely explode. If that doesn't happen, the power supply will most likely trip an overcurrent limit and either shut down, foldback or hiccup (none of which will charge your capacitor quickly) or limit current to the maximum allowed value (which will charge your cap). What it does depends on the model.

I would recommend using a PMOS, which does not require a boosted supply. You can use it as a switch with a resistor to limit current, or use the PMOS as constant current.

Or a constant current buck converter for efficiency if you plan to charge/discharge the cap frequently.


You need a gate drive DC voltage, that is relative to the emitter of the IGBT, not relative to GND. The "safety off" resistor at the gate should be between gate and emitter or you risc excessive negative gate voltages.

As I understand, this is a simple, low frequency, on/off application, no PWM involved. In this context you can use an optocoupler with photovoltaic output. The input side is a LED and the output is a DC voltage of around 7-10 V as long as the LED is driven.

The voltage of such a coupler is rising slowly, the IGBT will turn on slow as well. An internal pull down resistor is built in. You need a coupler >= 8 V here to be above the plateau voltage (e.g. APV1122AX). The thermal stress in the IGBT is relevant but looks OK for the chosen type.

An isolated DC/DC converter will work as well.

The IGBT has a saturation voltage of around 2 V, so the capacitor cannot reach the 50V. A standard N-MOSFET (with lower threshold voltage) would perform better here.

  • \$\begingroup\$ Thank you. Just need to clarify couple of things. Why would my 3rd circuit diagram not be a good idea? If I use isolated DC-DC converter, lets assume I have chosen one that can take an input of 3.3V and output 12V. The input 3.3V is referenced to GND, and because of isolation the output 12V would be reference to GND1 (different ground). pin1 - 3.3V (Vin) pin2 - GND pin3 - 12V (Vout) pin4 - GND1 Should I then connect pin4 to the emitter of the IGBT and pin3 to the gate? \$\endgroup\$
    – Mayur
    Jul 1, 2022 at 8:12
  • \$\begingroup\$ @Mayur From your 3rd schematic I could not see that it is an isolated converter because there are only 3 pins shown. I assumed again a wrong GND path. R6 should be connected to the emitter, not to GND. You need a slow turn on, R/C delay at the gate or something else to limit the current in the IGBT. \$\endgroup\$
    – Jens
    Jul 1, 2022 at 13:31
  • \$\begingroup\$ Yes you right the 3rd schematic is not an isolated regulator. Let us assume in place of that regulator I replace it with an isolated DC-DC converter that has the following pin configuration. Pin1 - 3.3V (Vin) Pin2 - GND Pin3 - 12V (Vout) Pin4 - GND1 (different ground). Am I correct in saying that I can now connect pin3(Vout) to the Gate and pin4 to the Emitter, and either have a gate resistor (R6) to limit the current to the gate for slow turn on, or as you suggested have R6 connected between Gate and Emitter? Would this work out? I prefer your suggestion of the isolated converter \$\endgroup\$
    – Mayur
    Jul 2, 2022 at 7:57
  • \$\begingroup\$ @Mayur I prefer the solution from bobflux with the P-FET, but you can use the isolated converter the way you said. It is important to slow down the turn on, use a resistor (1k) between +12 and the gate and a capacitor (>=10uF) between gate an emitter. \$\endgroup\$
    – Jens
    Jul 4, 2022 at 21:47

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