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I recently came across a circuit with a drive signal which never went to ground, and instead always kept the MOSFET biased. What it the reason/benefit of doing this?

I've drawn an example like the circuit I saw, where the PWM signal would go from +5 to 0V meaning the gate voltage would either be +5 or +2.5V. This is shown in the below waveform.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ "I recently came across a circuit with a drive signal which never went to ground, and instead always kept the MOSFET biased. What it the reason/benefit of doing this?", the context matters 100% here. How does the rest of the circuit look like? Where did you find this? On its own as you've shown it, it makes no sense. \$\endgroup\$ – Harry Svensson Jul 10 at 17:17
  • \$\begingroup\$ Default the FET to on state during driver startup etc. \$\endgroup\$ – MadHatter Jul 11 at 4:38
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If you never wanted the load current to go to zero this would be a good way to do it, a little odd because generally mosfets should be in the on or off state, not in between as they dissipate a lot of power.

A circuit like this might be useful for a hit and hold on solenoid or something like that.

If you want to use this circuit, make sure you check the power dissipation in the mosfet. FYI, holding alt and selecting the mosfet (or any part) which will give you power in lt spice.

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  • \$\begingroup\$ Yes and I'd say to right-click the MOSFET and pick a specific one - the generic model is lackluster to say the least. \$\endgroup\$ – rdtsc Jul 10 at 17:18
  • \$\begingroup\$ Its a discharge circuit so I think they are using the big FET to share some of the power dissipation so they can use smaller SM resistors. \$\endgroup\$ – Fat Diode Jul 11 at 17:15

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