I need to protect the gate driving circuit by putting the resistor between the gate and the control circuit:

enter image description here

During the testing of the mockup MOSFETs can go be destroyed which leaded to high voltage on the gate circuit.

I'm going to place 0805 SMD resistor (R19) which will be destroyed in milliseconds as well (as it will be exposed to 50 W continuous overheating). However it will be much simplier to replace just this resistor insted of looking up for the whole circuit for destroyed elements which was a real pain.

My concerns are:

  1. If R19 will be overheated and destroyed - will it break the circuit? Or it is possible that it will go to short circuit (as FETs usually do)?
  2. If the resistor destroyed - how likely that the PCB will be destroyed too? I'm using 35 um foil.
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    \$\begingroup\$ Resistors fail open, so that's fine. Heat can cause the copper foil to come off of the board, so that's bad. I think the real solution lies in finding and fixing whatever is killing the FETs, though. That shouldn't happen often enough that you need to consider it. \$\endgroup\$ – JRE Aug 10 '18 at 9:26
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    \$\begingroup\$ Maybe if you post the rest of the circuit someone could tell you why the FETs keep burning out. \$\endgroup\$ – JRE Aug 10 '18 at 9:28

It is likely that R19 will fail open circuit. It is rare but not impossible for a film resistor failure to result in a short. I've never seen it happen.

Consider using an appropriately rated resettable fuse in place of R19.

Predicting the damage to the PCB foil is hard to do. I've seen 35um (1oz) foil pads apparently survive thermal destruction of a resistor, but then fall apart during (very careful) rework.

  • \$\begingroup\$ +1 there are resettable fuses that are rated to break 230VAC mains voltage but most are not. 1210 is the smallest SMT one I see. \$\endgroup\$ – Spehro Pefhany Aug 10 '18 at 12:16

I agree with @amb with using a resetable fuse, or even a normal fuse. I would also try to find a fuse that trips as quick as possible.

A resistor that should break exactly what a fuse does, but a fuse is designed to do this well. So why not use a fuse?

Also, while the resistor heats up (before it burns up), all that current is still running through the rest of the circuit. So other components can still be damaged. I'm sure a fuse will burn out much quicker and more consistently.

  • \$\begingroup\$ Resistor will protect the rest of the circuit as it will take most of the voltage (which is not true for a fuse). So it should still protect the circuit. \$\endgroup\$ – Roman Matveev Aug 10 '18 at 10:27
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    \$\begingroup\$ PTC fuses become resistors very quickly in a fault condition. \$\endgroup\$ – amb Aug 10 '18 at 10:31

The Zener will also see some substantial dissipation, especially if the voltage is positive. If it was +240VDC, the Zener would dissipate more than 5W and the resistor ~50W. A TVS may be better for this purpose than a Zener.

As far as your question goes, carbon film through-hole (eg. 1K) resistors can arc and drop in resistance with 220V on them (becoming a glowing ceramic rod), but usually thin film SMD resistors, even cheap ones, use something like ruthenium oxide, which probably has less tendency to arc. In my experience a metal film resistor (through hole, such as 1K 1%) will probably work.

You can certainly test a dozen resistors or so and see what happens (just solder them onto a test board and try it- a scope with slow sweep will let you measure the duration of conduction). Maybe try something like an 0805 size. I expect it will work.

As an aside, this circuit looks like it would switch very, very slowly (especially in the 'on' direction - several milliseconds most likely), which can be very hard on the MOSFETs if they are carrying substantial current. This could be what is causing your trouble.

I assume you are trying to switch AC mains voltage and are driving the gates with an isolated voltage source.

  • \$\begingroup\$ Why milliseconds? You said that because of 1K resistor? LTSpice similation much much more optimistic. \$\endgroup\$ – Roman Matveev Aug 10 '18 at 11:41
  • \$\begingroup\$ No, because of the 330K resistor. \$\endgroup\$ – Spehro Pefhany Aug 10 '18 at 11:46
  • \$\begingroup\$ The current to the gates goes through the NPN transistor. 330k resistor just feeds its base. \$\endgroup\$ – Roman Matveev Aug 10 '18 at 11:48
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    \$\begingroup\$ Yes, I know, but the voltage will rise very slowly at the base of the transistor, and thus at the gates of the MOSFETs. See [page 7](file:///C:/Users/Spehro/Downloads/TLP291(SE_datasheet_en_20130529.pdf) toff vs. load resistance. It's roughly proportional to load resistance and exceeds 1ms at 100K, so maybe 3.5ms typical with 330K. There are various ways to clean that up. \$\endgroup\$ – Spehro Pefhany Aug 10 '18 at 11:51
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    \$\begingroup\$ PS in the 'off' direction 10K+1K= 11K is what discharges the gate charge, which is bad but not nearly as bad as the opto. An added PNP would help with that and the opto could be swapped for a digital one or cleaned up with a ST. \$\endgroup\$ – Spehro Pefhany Aug 10 '18 at 12:24

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