Boost converter high output impedance

I agree with Austin that a voltage doubler should be more than sufficient for your mosfet - the gate uses virtually no current at all.

From your diagram, it is unclear whether the gate driver and the level shifter are controlled from different MCU outputs or a common one.

Is the "Gate Driver IC" the core of your booster? From what is shown, I don't see why you'd need the "level shifter" - drive the driver from ONE pin, and when it is high, the H-bridge FET will switch.

Unless I'm misunderstanding something in your diagram, I'd think you could simply use an optocouper driven from an MCU pin to control current into a voltage doubler circuit to drive your H-bridge. Any number of OC's would work, since you're not driving much current. You're working with 12V, which your uC doesn't run off of, so the OC is necessary.

In fact, a Dickson Charge Pump clocked from a uC pin (i.e. program a pin for PWM) would provide an isolated voltage doubler: opto + 2 caps and 2 diodes. If you're not clocking, it's not doubling and your gate won't have enough voltage to conduct.

Basically, you'd be keeping the uC (MCU), H-bridge, and V1, and the Dickson would replace the rest of the components shown there, outputting to M2, running to ground, and sourcing from the 12V Vcc, with the clock as a solitary input from the uC. Opto is < US$0.50, caps and diodes are cents apiece.

Energy stored in a capacitor is \$ \frac{1}{2}CV^2 \$.

Energy stored in a inductor is \$ \frac{1}{2}LI^2 \$.

So the conservation of energy equation would be: $$ \frac{1}{2}LI_i^2 - \frac{1}{2}LI_f^2 = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 $$ (subscript i means initial, f means final)

Using an assumption that \$V_{out}\$ is near constant over one cycle, that is \$ V_{out} \gg V_f-V_i \$, then \$ 2 V_{out} \approx V_{f} + V_{i} \$. $$ \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 = \frac{1}{2}C(V_f^2-V_i^2) = \frac{1}{2}C(V_f+V_i)(V_f-V_i) = C V_{out} \Delta V$$ This is the answer given by jp314.