Electrical – MOSFET drivers blowing out

Yes, the 1K gate-to-source resistor is creating your problem. The good news is, you don't need it. A gate-to-source resistor is needed if your MOSFET cannot be actively driven off when it is needed to be, but that is not the case with the IR2101. You will often see a gate-to-source resistor used when a MOSFET gate is connected to a microcontroller to force the MOSFET off until the microcontroller can initialize the pin as an output.

Here is the datasheet for IR2101. And here is the Functional Block Diagram (from the datasheet):

As you can see from the functional block diagram, the gate drive outputs HO and LO are actively driven both high and low so the 1K ohm resistor is not needed.

The problem that you are seeing is that the bootstrap capacitor is being discharged through the 1K ohm resistor (in series with the 100 ohm resistor). From looking at the decay rate on your waveform, it looks as though that waveform was captured with only the 330nF capacitor in place (using the formula T=RC to get the time-constant).

A bit of housekeeping: The charge pump controller you found (LM2767) is only good up to a VCC of 12V. I did some quick searching, and I wasn't able to find one that works. But that's a minor detail, I'll move on to your main question, assuming that you can generate that 24VDC.

If you have 24V available as drawn, the gate driver and top power MOSFET will be in a race to see which blows up first. Your gate is rated for 20V, and your gate driver is rated for 16V, meaning the first time the high-side device turns off, the gate driver will blow. If the gate driver doesn't blow, then the gate will as soon as the gate driver puts the full 24V on to the gate.

My first question back to you is "Are you really sure you need a boost to your bootstrap capacitor?" Looking at the MOSFETs, you don't see a significant decrease in Rds(on) above 7V, so lets see how long it takes to go from your fully charged bootstrap voltage down to 7V (and you can calculate for lower as necessary). From your gate driver datasheet, the quiescent current is 35µA at 12V nominal, resulting in \$35\mu A=C\dfrac{5V}{t_{on}} \$ or \$t_{on}=C \dfrac{5V}{35\mu A}\$ which comes out to about 142 ms per 1 µF of capacitance. If you're holding the voltage high for a long period of time (multiple seconds), then a boost is necessary.

Since you only need a bit of current, you might be able to put a resistor (e.g. 100kΩ) in series with your 24V source, and a 12V zener diode across your high-side FET.

Another option is to have a high-side PNP set to switch on when your output is high.

Another option may be to have a small boost converter (charge-pump or inductor) that is enabled when the output is high.

The easy way to "just handle" this problem is to get an isolated DC-DC converter to provide a dedicated high-side source, though that is also the most expensive option.