I will use the gate driver configuration shown below:

Gate Driver datasheet

MOSFET Datasheet enter image description here

As you can see, the 5V and 12V are isolated. I chose this AC/DC converter for the 5V supply. It's 1 Output 5V/3A. It will supply a STM32 and a Raspberry Pi.

I decided to step up the 5V to 12V using this Isolated Module DC DC Converter.

That module is just 1W/ 84mA max. My question is: Can it effectively supply the peak currents to charge gate capacitance using decoupling capacitors?


Do the math. Find the total effective gate capacitance of the FET. Determine the maximum rate you will ever switch the FET. From that you can get the average current required on from the 12 V supply. If that's more than 84 mA, then you chose inappropriate parts.

If less than 84 mA, then it is doable. Find the total charge that is required to drive the get high once. Assume all that charge will come from the cap across the power input of the gate driver. Use a capacitance large enough so that the voltage drop is acceptable.

Once you do all that, you'll realize that you don't really need much. Then just plunk down a 1 µF 20 V cap and be done with it.


I just looked at the FET datasheet, and the total effective gate charge is 200 nC. That will cause a drop of 200 mV on a 1 µF cap. That shouldn't be a problem. If you don't like that for some reason, put a 10 µF cap in parallel with the 1 µF. In that case, put the 1 µF physicall closest to the gate driver IC.

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  • \$\begingroup\$ When you said: if that's less than 84 mA, then you chose inappropriate parts. Did you mean if that's "more than"? \$\endgroup\$ – Blue_Electronx Jul 6 '18 at 19:58
  • \$\begingroup\$ @Xav: Fixed. -- \$\endgroup\$ – Olin Lathrop Jul 7 '18 at 14:20
  • \$\begingroup\$ This is my approach: The 12V step-up module will supply an average current of I = Qxfs, where Q is the total charge of the mosfet and fs my pwm frequency. According to my mosfet datasheet the total charge is 160nC, my pwm frequency is 10 kHz. So I = 160nC * 10000 = 1.6 mA... am I right? \$\endgroup\$ – Blue_Electronx Jul 8 '18 at 1:20
  • \$\begingroup\$ Your units don't check. \$\endgroup\$ – Olin Lathrop Jul 8 '18 at 12:07
  • \$\begingroup\$ Do you mean writing the right expression? I = (160x10^-9)x10000 = 1.6 mA average. I suppose that the isolated module chosen above would work for this application. I don't know why some people have told me that the module provides too low current capability (84mA max). I guess that the 6A peak currents are provided by the decoupling capacitors. \$\endgroup\$ – Blue_Electronx Jul 9 '18 at 19:08

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