There are dedicated "MOSFET driver" ICs available (ICL7667, Max622/626, TD340, IXD*404.)

Some also control IGBTs.

What is the practical purpose of these? Is it all about maximizing the switching speed (driving gate capacitance) or are there other motives?


4 Answers 4


A MOSFET driver IC (like the ICL7667 you mentioned) translates TTL or CMOS logical signals to a higher voltage and higher current, with the goal of rapidly and completely switching the gate of a MOSFET.

An output pin of a microcontroller is usually adequate to drive a small-signal logic level MOSFET, like a 2N7000. However, three issues occur when driving larger MOSFETs:

  1. Higher gate capacitance - Digital signals are meant to drive small loads (on the order of 10–100 pF). This is much less than the gate capacitance of many MOSFETs, which can be in the thousands of pF.
  2. Higher gate voltage - A 3.3 V or 5 V signal is often not enough. Usually 8–12 V is required to fully turn on the MOSFET.
  3. A switching MOSFET can cause a back-current from the gate back to the driving cicruit. MOSFET drivers are designed to handle this back current. (Ref: Laszlo Balogh Design And Application Guide For High Speed MOSFET Gate Drive Circuits.)

Finally, many MOSFET drivers are designed explicitly for the purpose of controlling a motor with an H-bridge.

  • 11
    \$\begingroup\$ There is a third issue: a switching MOSFET can cause a back-current from the gate back to the driving cicruit. MOSFET drivers are designed to handle this back current. ([ref](www.ti.com/lit/ml/slup169/slup169.pdf) p12) \$\endgroup\$ Sep 24, 2012 at 21:04
  • \$\begingroup\$ For PMV19XNEA a logic level MOSFET, with a combined turn-on and rise time delay of approx. 25 ns, can we conclude that a MOSFET driver would not be needed for switching 2 amps at 19V? Typical delay times for the ICL7667 indicate that the driver may even exacerbate the overall delay. \$\endgroup\$
    – Mike
    Feb 20, 2022 at 16:04

Yes, it's about maximizing the switching speed by dumping lots of current into the gate, so that the power MOSFET spends the least amount of time possible in the transition state, and therefore wastes less energy and doesn't get as hot.

It says as much in the datasheets of the parts you listed :)

The ICL7667 is a dual monolithic high-speed driver designed to convert TTL level signals into high current outputs ... Its high speed and current output enable it to drive large capacitive loads with high slew rates and low propagation delays ... The ICL7667’s high current outputs minimize power losses in the power MOSFETs by rapidly charging and discharging the gate capacitance.

  • \$\begingroup\$ Why not just use a transistor instead? there's ones that can easily handle 1.5A. why get the IC instead of a single transistor? \$\endgroup\$ Mar 30, 2021 at 9:37
  • \$\begingroup\$ @user3033693 because now you need to charge that transistor again. If it can handle 1.5A, it will have a high input capacitance too. To mitigate this, you'll need several stages, and at this point, you know why you use an IC. There also are the other problems mentionned int the other answers. \$\endgroup\$ Nov 13, 2021 at 11:49

If you want to calculate the gate current during switching you can use this formula:

Ig = Q/t

where Q is the gate charge in Coulomb (nC from the data sheet) and t is the switching time (in ns if you use nC).

If you need to switch in 20 ns, a typical FET with a total gate charge of 50 nC will need 2.5A. You can find nimbler parts with gate charge below 10 nC. I prefer to use 2 BJTs in a totem configuration for driving MOSFETs instead of the expensive driver ICs.

  • \$\begingroup\$ And how do you do the voltage translation for the totem? \$\endgroup\$
    – jpc
    Apr 16, 2011 at 14:19
  • \$\begingroup\$ Lately I've had good results using logic level MOSFETs and running the totem on the 3V3 rail. You can also use a BJT for voltage translation if you are ok with the signal being inverted. \$\endgroup\$
    – morten
    May 2, 2011 at 22:08

Yes. And another reason is to drive "high side" of the bridge. For this those ICs have an external capacitor and internal oscillator with diode voltage multiplier, so the gate driving output is providing voltage few volts higher than bridge and/or bus voltage.

  • 2
    \$\begingroup\$ Yes - special high side drivers exist so that the better-performing N-channel devices can be used on the high side of the bridge as well as the low side. Otherwise - without a gate voltage above the positive supply rail - a P-channel device must be used there. There's a point where the superiority of the N-channel devices justifies the additional circuit complexity of this technique. \$\endgroup\$ Aug 1, 2012 at 15:04

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