I would like to know if there is a calculation that allows me to look up at how much current my PCB would draw in the worse-case scenario if all my MOSFETs would have to change at the same time. In the PCB I use 11 IRLL014 in a SOT223, and 4 IRF3704 in TO 220.
https://www.mouser.de/datasheet/2/427/sihll014-1768836.pdf http://www.irf.com/product-info/datasheets/data/irf3704.pdf It would be fine if you say what I have to look at in the datasheets.

Another thing is to recalculate the pull-down resistors, what I would need to look at as they are right now with 10K they don't work right.

enter image description here

on the GAIN there are SSR and 12V, SOURCE is gnd GATE comes from the ESP32(ttl)

On the Load Side there are either




As you can see there is already a 10k Ohm pull down but it seems that this one doesn't work as the fet doesn't turn off all the time or delayed. Should I just put in a 100k/1k and check if it works or is there a way to calculate it?

Thanks !

  • 1
    \$\begingroup\$ A schematic fragment isn't very helpful. We need a lot more. Voltages, loads, etc. \$\endgroup\$
    – Mattman944
    Apr 11, 2021 at 9:09

1 Answer 1


The 10k resistor is to ensure the mosfets are turned off during reset. In normal operation the gate should be actively driven high or low.

The IRll04 is a n-channel mosfet. The source goes to 0V, the drain goes to the load -. 12V goes to the other side of the load. To have it work as you've described requires the gate to be higher than 12V to turn it on - this would be a challenge with the ESP32.

  • \$\begingroup\$ Yes sorry, seems I misunderstood the scemantic a bit. There is 12vdc after the load (ssr) and the source will switch to the ground from the source if turned on over gate. They both work with ttl signal if I look at the datasheet. My question would be how to calculate the min and maximal current drawn at the gate if it switches or where can I find it. \$\endgroup\$
    – Daniel Do
    Apr 11, 2021 at 20:33
  • \$\begingroup\$ In the datasheet for the FET you should find the gate capacitance, which will likely be of the order of tens of nanofarads and so the 10k pull-down will give a switch-off time of the order of hundreds of microseconds. If you need a faster switch-off then you'll either need a lower value pull-down or better still drive the gate actively in both directions. \$\endgroup\$
    – Frog
    Apr 11, 2021 at 21:04

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