# How can I slow a FET's turn ON without slowing its turn OFF?

If as a designer you have run in to a situation where it is desirable to slow down the turn ON of a FET using a series gate resistor, such as in why put a resistor in series with the gate of a FET, but where a corresponding slow down in turn OFF is undesirable, what can you do?

Take the generic example below where a 5V gate driver (U1) is being used to turn ON/OFF a low-side n-channel FET that’s switching a mystery load.

R1 is used to slow down the turn ON of the FET. The FET’s gate capacitance is filled with a current that is peak limited to 500mA (5V/10Ohms) by the 10 Ohm resistor, and the gate voltage will rise according to the time constant set by R1 and the gate capacitance of Q1 (approximately anyway). The time it takes to rise to the FET’s threshold voltage will be the turn ON time, but what about the turn OFF?

Well, without any other intervention, the turn OFF time would be approximately the same as the turn ON time, since the FET’s gate capacitance would have to be discharged through R1 as well. However, adding the diode D1 with the polarity shown can speed up the turn OFF time without speeding up the intentionally slowed turn ON time.

Note that now the gate voltage will have an additional diode drop of voltage on it in the OFF state, but as long as that voltage is sufficiently below the FET threshold voltage it will be fully turned OFF and this is fine. Also note that if the opposite behavior is desirable, a fast turn ON with a slowed turn OFF, the diode simply needs to be reversed in polarity for the same concept (and similarly a diode drop off the turn on voltage must still be sufficiently above the FET threshold voltage for full turn ON).

One application where this may be necessary is with bridge drive circuits with high-side and low-side FETs where one FET must be fully turned OFF before turning ON the second FET to avoid shoot through (a temporary short of the power supply to GND through both FETs while they are both briefly ON).