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There are dedicated "MOSFET driver" IC's 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?

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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, two issues occur when driving larger MOSFETs:

  1. Higher gate capacitance - Digital signals are meant to drive small loads (on the order of 10-100pF). This is much less than the many MOSFETs, which can be in the thousands of pF.
  2. Higher gate voltage - A 3.3V or 5V signal is often not enough. Usually 8-12V is required to fully turn on the MOSFET.

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

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    \$\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\$ – Wouter van Ooijen Sep 24 '12 at 21:04
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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.

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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.

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    \$\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\$ – Chris Stratton Aug 1 '12 at 15:04
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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.

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  • \$\begingroup\$ And how do you do the voltage translation for the totem? \$\endgroup\$ – jpc Apr 16 '11 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 '11 at 22:08

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