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The inverter circuit I made from N-channel MOSFETs is given below. I used bootstrap for the high sides. Bootstrap capacitors are C3 and C1.

The gate voltage of the MOSFET is working properly, the output seems fine, but capacitors C3 and C1 always remain stable around 11 V.

In the picture below where the measurements are made, the red color belongs to the C3 capacitor, you can see that it is going straight. I can only see very minimal changes when I zoom in. The zoomed measurement is also available below.

I made changes in the switching times and capacitance values, but the result is still the same. Additionally, I simulated the bootstrap circuit (using only PWM, transistor, 12 V voltage source, diode and a capacitor) separately and I could not see much voltage change on the capacitor.

What could be the reason for this problem on the capacitor? What should I do?

enter image description here

Source: My own study

enter image description here Source: My own study

enter image description here Source: My own study

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2 Answers 2

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What could be the reason for this problem on the capacitor? what should I do ?

You are chasing shadows.

The capacitors are doing what they are meant to do i.e. when the relevant low-side MOSFET turns on, the capacitor is charged up to 12 volts minus 1 diode drop (call it 11 volts) then, when the lower MOSFET disengages and the upper MOSFET starts to come into play, the voltage across the capacitor remains about 11 volts so that it can provide adequate voltage to lift the upper MOSFET's gate about 11 volts higher than its source (the output).

I will add this; the presence of R4 and R6 is not conducive to fully deactivating the upper MOSFETs so, why don't you use proper drivers? Here's an article that gives more details about bootstrapping upper MOSFETs: -

enter image description here

Note that they use a driver circuit that does not try to drag down the upper MOSFET's gate to 0 volts but, instead, shorts the gate to its source. This protects the MOSFET from damage due to excessive gate-source voltages. This is why I recommend that you use a proper driver chip.

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  • \$\begingroup\$ I got the point of capacitor should stay same so that it can provide adequate voltage for upper mosfet but ı have another question. How does the voltage on the capacitor stays same although there is a path for it to discharge through gate and R3 (load) during M1(high side) is on? @Andy aka \$\endgroup\$
    – Mhan
    Commented Mar 25 at 12:21
  • \$\begingroup\$ @Mhan the gate-source region is a high impedance electrically. Sure, there is a transient charge taken each time the MOSFET switches and, that is why we choose a bootstrap capacitor that has a capacitance that is maybe 30 to 100 times the upper MOSFET's gate-source capacitance. Your value of 6.8 uF seems a tad high to me but, your simulator will allow you to see the effects of lowering it. \$\endgroup\$
    – Andy aka
    Commented Mar 25 at 13:26
  • \$\begingroup\$ Sir, thanks a lot for your explanations but I couldnt understand it. The capacitor has an easy path through the gate-source and load to discharge but it is almost stable, even if i change the capacitance it is slightly changing. I can understand it from the stable gate voltage point of view but cant grab how it doesnt discharge? the source of the M1 is high when it is open so no potential difference, can this be a reason ? @Andy aka \$\endgroup\$
    – Mhan
    Commented Mar 25 at 15:15
  • \$\begingroup\$ @Mhan no, the gate-source is not an easy path at all. When M5 is on, it's source is close to the positive supply voltage (as you would expect) so the low pin of capacitor C1 is connected to 240 volts. The high pin of capacitor C1 is about 11 volts higher and, will not discharge through the gate-source because it is mega ohms. There will be a slight loss of charge from the activation of M5 but that's all. \$\endgroup\$
    – Andy aka
    Commented Mar 25 at 15:33
  • \$\begingroup\$ Yes Sir exatly as you said. You said M5 and is this is still true in case of M1 and M6 is open where the conduction is happening ? @Andy aka \$\endgroup\$
    – Mhan
    Commented Mar 25 at 15:42
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Just for study, try to replace R3 with and inductor. It will pull a current through the capacitors because with 12V behind it, that is the path of low resistance. Lower than the body diode of the mosfet. But don't use it in a real schematic as the mains voltage, and the resulting current, is much to high to work without a gate driver. Also a 1k resistor in the gate connection will not switch fast enough. Have a look at the gate capacitance in the datasheet and calculate the RC time.

I have never seen a schematic with an H-bridge driving a resistive load, always a motor or something else with a coil.

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