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I am using the IR2110 drivers and this is the schematic of a half bridge, to obtain the complete H bridge I repeat the same circuit 2 times

I measured the voltage between gate and source of each MOSFET, the low side MOSFETs work fine, the high side MOSFETs do not work, I realized that the Vgs is 12v for a short moment(1 ms) until a capacitor is charged, once the capacitor is shorted between gate and source and therefore the voltage between them is 0, I have no idea why

enter image description here

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    \$\begingroup\$ R3 is going to discharge your bootstrap capacitor(s) very quickly. Increase R3/R4 to 1M for example. Realize that a gate driver like this needs to be switching periodically to give the bootstrap capacitors a chance to recharge, you can't keep the high side on indefinitely. \$\endgroup\$ Commented Dec 23, 2020 at 20:41
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    \$\begingroup\$ Is C1 perhaps too large? Also, how are you probing Vgs of Q1? When Q1 is on, the gate is at 312V or so, and the source is at 300V, so you can't just connect oscilloscope ground to Q1 source. \$\endgroup\$
    – user57037
    Commented Dec 23, 2020 at 20:41
  • \$\begingroup\$ Oh, also, how long is a "short moment?" can you please edit your question and change that to a specific time? Like 1us, or 1ms or whatever the case may be? \$\endgroup\$
    – user57037
    Commented Dec 23, 2020 at 20:43
  • \$\begingroup\$ edited................. \$\endgroup\$ Commented Dec 23, 2020 at 20:45
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    \$\begingroup\$ You should do PWMing all the time, DT=100% isn't allowed, since the bootstrap capacitor won't never charge. \$\endgroup\$ Commented Dec 23, 2020 at 21:06

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Vgs=12 for 1 ms ... that's plenty long enough as using this circuit as designed, you'll be pulsing the high side driver much faster than that; typically every 50 us (20kHz) PWM frame rate.

You can't run this circuit at a 100% duty cycle (high side driver continually on) without coming up with alternate power arrangements instead of the bootstrap.

Removing R3 would improve matters, and allow a more reasonable value for C1, eliminating the leaky electrolytic capacitor.

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