Energy stored in a cap = QV/2.
Power needed to repeatedly charge a cap from a certain voltage:
P = freq * QV/2 * 2.
Average current needed: I = P/V = freq*Q.
Using the total gate charge Qg = 180nC from the datasheet (even though Qg depends on the actual gate and drain voltage), and 10KHz:
I = 180nC * 10kHz = 1.8mA
which is not much.
With the amount of current being switched at the load, it is easy to guess that minimizing the switching loss of the MOSFET is important. Just guessing a number to get some ideas, say 90ns for the gate charge/discharge time.
The gate drive charge/discharge current needed would be Qg/t (at 10V Vgs, same as the condition given in the datasheet): Igs = Qg/t = 180nC/90ns = 2A.
There is a big difference between the average and the gate-charge current because the duty cycle of gate-charge to total time is 90ns/100us = 0.0009. 100us comes from the inverse of 10kHz.
- A DC/DC converter of 83mA average current should be more than enough.
- Enough bypass capacitance would be needed to supply the gate-charge
switching current of thousand times that of the average.
- Driving the MOSFET directly with a optocoupler may provide only a
fraction of the drive current wanted. The drive current probably should be in the amps range. But the actual required depends on the optimization of the MOSFET switching which takes much more to work out.
Edit to include a way to estimate bypass capacitance on floating supply:
Take the Input Capacitance from the datasheet, e.g. Ciss = 10880pF. If you want the voltage ripple caused by charging the base to be less than 1:100 on the supply, then use a bypass cap that is 100*Ciss*2 (the times 2 is for extra margin for output capacitance of MOSFET), that is 2.2uF in this example. It is a lot less than my impression when I wrote the above. Use ceramic capacitor for low ESR and high current capability.