I have a problem with the MOSFET commutation. The fall time is very slow when the load resistor has a low value. It seems like there's a capacitor. However, this doesn't happen when I use a BJT transistor instead of a MOSFET.

I've tried with lots of MOSFETs but the result is the same. I read this may occur because of the scope probe although I'm not sure how. Can anyone help me?

  • 6
    \$\begingroup\$ circuit schematic, part values and links to description will all help someone analyze the situation. \$\endgroup\$ May 28, 2013 at 1:48
  • 2
    \$\begingroup\$ Yes, mosfets have a gate capacitance; you need either active drive in both directions or a resistor to ground. \$\endgroup\$
    – pjc50
    May 28, 2013 at 7:44
  • \$\begingroup\$ Mosfets can hold a charge on the gate if there is no 'escape' path. Even though the gate capacitance is small and the amount of charge is small the voltage can be significant enough to hold the device on (V = q/C). What you seem to be experiencing is a slow leakage of this charge through a very high resistance path in the circuit. As pjc50 (one point) has pointed out you need to ground this charge through a resistor. \$\endgroup\$ May 28, 2013 at 7:54
  • \$\begingroup\$ have a look at brunningsoftware.co.uk/FET.htm \$\endgroup\$ May 28, 2013 at 10:35
  • \$\begingroup\$ Your statement makes no sense. A FET will be driven slower with a larger series resistor from the gate. It is impossible to tell what you are talking about without a diagram and more details. \$\endgroup\$ May 28, 2013 at 12:58

1 Answer 1


Some general points, since the question is somewhat vague:

In order to get good switching speed with a MOSFET, you need to actively drive the gate, and actively discharge the gate. This can be accomplished with a discrete two transistor totem-pole gate driver, or (more commonly these days) with monolithic gate driver ICs that have amps of instantaneous sink/source capability.

One thing that some people don't consider is the internal gate resistance of the MOSFET, which manifests itself in series with the interal gate-to-source and gate-to-drain capacitances. Consider this simplified model (courtesy Vishay):

enter image description here

\$R_g\$ is one of the main reasons you need a stiff driver, since it limits the current available to charge and discharge \$C_{gs}\$. (\$C_{gs}\$ itself is another reason.)

\$C_{gd}\$ is also known as the 'Miller' capacitance, is quite nonlinear and also plays a role in gate turn-on and turn-off, since the gate needs to overcome both of these capacitances to get the device solidly 'on'.

enter image description here

Another area that can cause slower than necessary turn-off (with N-channel MOSFETs) is if the applied gate voltage is much higher than the voltage required to turn the device on. If you have a logic-level MOSFET and need only 5V to fully enhance the channel, and apply 12V to the gate, that extra 7 volts doesn't get you much except extra \$C_{gs}\$ charge to remove when you're trying to turn the device off. If you're using a gate driver that can sink or source amps of current, this extra voltage isn't really a concern. If you're passively discharging the gate, it can significantly affect the commutation time.


Not the answer you're looking for? Browse other questions tagged or ask your own question.