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Sorry if this question has been asked before, I've tried looking through google and Stack Exchange and I can't seem to find the answer.

Resistors on the gate (R3 on the picture) are usually in the K ohms with good reason. The part I don't understand is the voltage drop that to my knowledge should occur to that resistor due to it being in series. Considering the other side of the resistor after the mosfet is connected to the ground, wouldn't applying let's say 5V to the left side of R3, where the OUT pin on the picture is, make all the current go through the resistor and nothing left for the mosfet ?

I'm guessing my assumption is wrong and the other side of the gate is not connected to the ground, or I simply don't understand how a PMOS works well enough, but basically I don't get how Vgs(th) could ever get enabled considering there is a high resistance resistor in front of the gate.

Sorry for the stupid question, I don't exactly know what to ask on google and I've been at it for a couple hours looking and can't find a good answer.

Thank you !

PMOS with gate resistor

Image is taken from this Stack Exchange link

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  • \$\begingroup\$ replace the mosfet with a resistor (call it R4) of several meg ohms between R3 and ground ... what would be the voltage across R4? \$\endgroup\$ – jsotola Jun 14 at 23:00
  • \$\begingroup\$ Also, that is an N-channel MOSFET, not a P-Channel (P-MOS). \$\endgroup\$ – Daniel Jun 15 at 3:29
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A MOSFET gate looks like a capacitor. The resistor will slow down the MOSFET turn-on and turn-off, and it would protect a delicate driver pin (unlikely, but possible), but in the end the thing will still switch.

In that circuit the resistor is to slow down the MOSFET switching, to reduce the speed that the drain voltage changes, which in turn reduces EMI at the expense of a loss of efficiency.

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