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I am trying to construct a simple MOSFET switch to control the input line of a linear regulator, but I am quite baffled as to how this works from the simulation I have done. The pulse input simulates a microcontroller drive, and I've simulated the input impedance of the regulator with a big resistor.

I was expecting the output to go to close to 0V when the drive pulse is high.

The images show my test circuit and the gate and output voltage waveforms, can anyone shed any light on what basic error or misunderstanding I have made in my circuit or simulation?

circuit

gate pulse

output voltage

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  • \$\begingroup\$ I realised that right after I hit send, and subsequently updated. Inverting the polarity of the pulse just changes the polarity of the current spikes on the output. Surely I want a NC switch, hence the p-channel FET? \$\endgroup\$ – droseman Sep 25 '17 at 16:35
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Googling "pmos high side switch" results in tons of wrong schematics, as they mostly come from questions on SE asked by people why have the same problem ;)

Here's a good one:

enter image description here

With VCC as the input voltage, you want the PMOS gate to be:

  • At VCC to turn it off
  • At VCC minus say 10-12V to turn it fully on, but never below VCC-20V as that would exceed the max Vgs and kill the MOSFET.

The NPN transistor works in saturation and pulls a current through R1 and R2. This sets the PMOS gate to the voltage you want. Adjust the R1/R2 divider to get the right voltage. The zener D1 is not really necessary, as the voltage divider will do the job of limiting Vgs...

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    \$\begingroup\$ +1 for working sample. D1 does add a faster gate charge dump path though while turning off. Does not need to be a zener though. Though it does add a recovery delay so it may be a wash in some cases. \$\endgroup\$ – Trevor_G Sep 25 '17 at 16:48
  • \$\begingroup\$ @peufeu thanks for the circuit, and the explanation, thats pretty spot on what I'm trying to do \$\endgroup\$ – droseman Sep 25 '17 at 17:19
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You are confused about how a P-Channel mosfet works.

In that configuration, to turn it on you need to apply a voltage less than your supply voltage (28V) but not more than 20V less. By spec, in this application 23V on the gate gets the quoted R values. To turn it off, the gate needs to go back up near the 28V level.

Since you are feeding it 0-5V absolute, that is really -28V and -23V Vgs. So a) it's always on in the simulator, and b) it's dead or dying on a breadboard.

The spikes you see is simply capacitive coupling from the gate.

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  • \$\begingroup\$ Great explanation, Trevor thanks. I have a follow on question - when the gate is switched off again, I get a slow decay in the output, which I assume is the gate capacitance discharging, is there a way to speed that up, aside from selecting a different device? \$\endgroup\$ – droseman Sep 25 '17 at 16:49
  • \$\begingroup\$ @droseman with 1M load it will take a while anyway, output of mosfet will just be a charged cap at that point. \$\endgroup\$ – Trevor_G Sep 25 '17 at 16:51

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