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I have a PCB as shown in the schematic.

Looking at only one of the dual MOSFET ICs, I apply 3.3V to MOSFET2 and MOSFET1, which should turn off MOSFET1 (because P-channel) and turn on MOSFET2 (because N-channel.) What happens is that both MOSFETs are on, even though the P-channel one should be off, so there is a short circuit. My power supply starts limiting the current and the IC gets hot. This shouldn't happen in my opinion. The gate signal is from STM32 GPIO pins.

I don't really understand the Vgs voltage in the P-channel part of the datasheet. If I had -3.0V Vgs and the source is at 38V, I would need 35V at the gate, which defeats the purpose of driving high currents & voltages with low voltages at the gate. Please help me understand why the P-channel MOSFET is not going off.

schematic

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  • \$\begingroup\$ Please use normal schematic symbols, not IC outlines. \$\endgroup\$
    – winny
    Jan 19 at 16:36

2 Answers 2

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The SI4559ADY is a dual MOSFET, right?

If you have 38V at the source of the P-MOSFET and 3.3V at the gate of the P-MOSFET, then it conducts and you are far above the allowed Vgs of +/- 20V, right? The gate should have about 28V, if you want to conduct and 38V (or up to 58V,) if you want it to block.

The N-MOSFET is conducting a little bit. but 10V at its gate would be better.

I think you need a proper MOSFET driver, like the MIC4604.

Or something like this: example circuit (u need to take care, that LOWsideOFF goes high clearly before HighSideON goes high... and vice versa... or you will have something called "shoot through"... once I wondered, which idiot bought candles... then I realized, that my FETs were burning the flame "retardant" PCB... lol)

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  • \$\begingroup\$ oops... the MIC4604 would be for two N-channel MOSFETs... \$\endgroup\$
    – RRIDDICC
    Jan 19 at 16:32
  • \$\begingroup\$ there was a similar question before: electronics.stackexchange.com/questions/407303/… \$\endgroup\$
    – RRIDDICC
    Jan 19 at 16:34
  • \$\begingroup\$ Why would I need a MOSFET driver? \$\endgroup\$
    – epicMan123
    Jan 19 at 19:47
  • \$\begingroup\$ @epicMan123 because u cannot turn those MOSFETs on with 3.3V... the gate needs a voltage that is 8V..18V higher(N-channel) or lower (P-channel)... so the P-channel FET needs 38V (off) and 28V (on)... and the N-channel FET needs 0V (off) and 10V (on)... u can do that with a discrete circuit (pull-up resistors and voltage dividers and zener diodes and small FETs, that are happy with 3.3V at their gate...) or u use an integrated driver chip... \$\endgroup\$
    – RRIDDICC
    Jan 19 at 20:23
  • \$\begingroup\$ Couldn't I just amplify the gate signal with a BJT? \$\endgroup\$
    – epicMan123
    Jan 19 at 20:30
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Your schematic shows a 0.7 V signal (across D5) amplified by 100 in a linear amplifier.

A 3.3 V opamp cannot produce a 70 V output signal. What is it you are trying to achieve? If the goal is a 3.3 V square wave, that can be done by replacing D5 with a simple 3.3 V zener diode, eliminating the opamp, all of its surrounding components, and D3.

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  • \$\begingroup\$ The op-amp is not what this question is about. Please ignore it. It is there for a completely separate reason. \$\endgroup\$
    – epicMan123
    Jan 19 at 20:32

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