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The datasheet of the IR2110 shows this as the typical connection:

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And in some cases even a gate-source resistor is used, as in this and this post

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But the app note shows this as the connection for a buck converter, no resistance whatsoever:

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Short answer, yes you want a gate resistor.

Longer answer, it reduces EMI by slowing down the MOSFET which in turn reduces ringing and high frequency components of the switching waveforms. On negative side this causes more power losses as the on and off times are increased. You can see in one of the schematics that there's a diode across the resistor, typically switching the mosfet off can be done quicker without ill effects depending on topology.

None of these circuits actually seem to be SMPS circuits so unless there is fast switching involved, it may not matter.

Gate source resistance is probably just to keep the mosfet OFF if circuit is not functioning so there's no 300V in output. In SMPS applications you may run into dV/dT induced spontaneous turn-on as mosfet parasitic capacitances conspire to raise gate voltage above threshold voltage. However in this instance a small cap is added from gate to source, not a resistor.

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  • \$\begingroup\$ How 'fast' is fast switching? I'm working at 160khz and I'm having a lot of troubles with noise. \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Apr 19 '16 at 21:26
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    \$\begingroup\$ @LuisRamonRamirezRodriguez "fast" means rise/fall time, not repetition rate. It could easily be in the few nanoseconds (hundreds MHz) range. \$\endgroup\$ – Oleksandr R. Apr 19 '16 at 21:31
  • \$\begingroup\$ 160kHz is plenty fast already. You have nice harmonics on top of that fundamental frequency and the driver can source and sink gate charge very quickly to accommodate large/small duty cycles. Also the interaction between inductive and capacitive elements in SMPS makes it worse, hence the need for snubber circuits. It's not just for EMI testing, if you have sensitive analogue circuitry it may pick up the noise. \$\endgroup\$ – Barleyman Apr 19 '16 at 23:50
  • \$\begingroup\$ Could a gate-source resistor also help to reduce noise? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Apr 20 '16 at 23:33
  • \$\begingroup\$ If your noise is coming through the gate, a resistor would add impedance to the capacitance forming a high pass filter with a resistor. \$\endgroup\$ – Voltage Spike Apr 20 '16 at 23:53
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Gate resistors should be used for input protection, the mosfet gate is a really thin layer of insulation sometimes nm's (tens to hundreds of atoms) thick. This creates a capacitor with some resistance. The problem is this layer can be blown away if a low enough impedance source is used, the ramp up time can damage the layer. ESD can also easily blow right through this layer (it doesn't take much joule heating in the material to create a hole because there isn't much material to start with and kV's of ESD create a lot of heat) The is really dependent on the mosfet, and on how much you care about protecting the input. There are many mosfets that have some kind of input protection. If they don't have any, you'll need it.

The IR2110 datasheet's recommendation in the example schematic has limiting resistors, I would put them in. With other drivers, I would check for some kind of limiting if they don't recommend them. DC to DC converters need to be very fast, and I don't know what the circuitry is but I'm willing to bet they limit the inrush current to cgs by design.

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  • \$\begingroup\$ Could a gate-source resistor also help to reduce noise? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Apr 20 '16 at 23:33
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The gate resistance influences the Mosfet turn on and turn off time. So the value or its absence depends on the Mosfet gate capacitance and the desired on/off time.

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No, don't use it! It's lamer's way to reduce EMI. The real way is to ensure short trace, close return path, etc. With resistor your rise and fall time will be longer and it will heat the MOSFET. Why would you do it? Even 10R may seriously harm your efficiency. Current limiting must be integrated in the gate driver.

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    \$\begingroup\$ So, how would you propose reducing the trace inductances below the single-digit nH range? At 10 nH/inch, it is not easy. Then you have the bond wire inductance of the MOSFET and driver to contribute about 5 nH each--so let's say 15 nH in total. This will ring with the gate capacitance and the MOSFET will not have a stable gate voltage, leading it to turn on and off rapidly and probably fail before long, unless instead this energy is dissipated in some way. Large gate resistors need not be used, but at least a few Ohm is highly advisable because one cannot completely eliminate parasitics. \$\endgroup\$ – Oleksandr R. Apr 19 '16 at 21:57
  • \$\begingroup\$ Just to be sure, i will check gate voltage tomorrow, on one of my drives. And transistor has some equivalent resistance... But anyway, in my layout the gate is really close to the output. \$\endgroup\$ – Gregory Kornblum Apr 19 '16 at 22:01
  • \$\begingroup\$ I have also attempted to absolutely minimize parasitics because I need very fast rise and fall times (<3ns) and short on-times (<50ns). But it seems to me that a gate resistor of 2-6 Ohm is still desirable for realistic values of the parasitics. Of course it depends on the value of the gate capacitance as well, so for some MOSFETs it will be a problem, while for others it won't. \$\endgroup\$ – Oleksandr R. Apr 19 '16 at 22:10
  • \$\begingroup\$ Judging by 3nsec i guess you have a really fast system. Mine is more common, up to 100kHz. So i definitely use different components. But i also MUST ensure efficiency, and sometimes current to the gate (so it will not just rise because voltage rises in drain). Bottom line- even 1R is a problem \$\endgroup\$ – Gregory Kornblum Apr 19 '16 at 22:14
  • \$\begingroup\$ Okay, but if you have ringing on the gate, then there may be significant inefficiency due to shoot-through, which will probably outweigh any losses in the gate resistors. My system is not really that fast (up to 2MHz or so) but I need the short pulses for a physics experiment. Anyway, comments are not the best way to have this discussion, so I'll leave it at that, I think. \$\endgroup\$ – Oleksandr R. Apr 19 '16 at 22:23

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