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I have tied the gate to a 10 kilo-ohm pull-up resistor, as shown below. The FET I am using is the NTD2955T4G. The control signal is a control PCB. The output of the control PCB has a low side switch internally when the output is activated (pulls the output low). This control PCB is a commercial unit I purchased.

I would like to know how long the FET will take to go from ON to OFF and also from OFF to ON.

I am not PWMing the relay.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Try to measure R3. If it is too small, M1 will stay stuck on forever. \$\endgroup\$ Jul 28, 2022 at 22:25
  • \$\begingroup\$ @TimWilliams trying to get the value from the manufacturer. The entire PCB is potted :( \$\endgroup\$
    – JoeyB
    Jul 28, 2022 at 22:27
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    \$\begingroup\$ You can measure it using Thevenin's theorem e.g. using my website seventransistorlabs.com/Calc/Thevenin.html [Edited by a moderator.] \$\endgroup\$ Jul 28, 2022 at 22:28

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That FET has an input capacitance of about 500 pF. Very roughly you can calculate the delays from the R.C time constant. So with 10k and 500 pF, you get tau=5 us. The FET actually will take somewhat longer to turn off as you need to discharge the gate to less than a threshold voltage. So perhaps 10-20 us might be the delay.

The turn-on delay will likely be limited by your driver, but the 100 ohm will cause a delay of about R.C also, or 50 ns.

Note that if you want slower turn-on, and increase the 100 ohm to (say) 10k, the turn on won't be as effective because R1 will cause the VGS to be dvided by about 2. Best to connect R1 between the control signal and 12 V, not gate and 12V.

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  • \$\begingroup\$ I managed to find the internal circuitry of the control PCB. It has a 2.8V that is pulled up. I will update the schematic above. \$\endgroup\$
    – JoeyB
    Jul 28, 2022 at 22:08
  • \$\begingroup\$ "Best to connect R1 between the control signal and 12 V, not gate and 12V."If he does the gate of the voltage will be 12V and the PCB will be destroyed \$\endgroup\$
    – Miss Mulan
    Jul 28, 2022 at 22:27
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    \$\begingroup\$ Now it will take an infinite time to turn off -- you can't turn off a p-channel FET powered from 12 V with a 2.8 V signal. Best to flip the whole thing (n-FFET to ground with the relay in the drain to 12 V. Drive the gate via 100 ohm and 10k to GND from the control signal. \$\endgroup\$
    – jp314
    Jul 29, 2022 at 0:41

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