Hi,
square wave....
I assumed something like this.
now let´s do some example:
let´s say 100Hz, 50% duty cycle (5ms OFF, 5ms ON),rise rate: in 10us from -1V to 3.3V and the same time back
* Gate voltage is -1V (OFF) continously for 4.09ms --> drive current is 0mA
* Gate voltage is -1V to 3.3V (OFF to ON) = 4.3V in 10us --> this means an average current of 5uF * 4.3V / 10us = 2.15A during this 10us
* Gate voltage is 3.3 (ON) continously for 4.09ms --> drive current is 0mA
* Gate voltage is 3.3V to -1V (ON to OFF) = -4.3V in 10us --> this means an average current of 5uF * (-4.3V) / 10us = -2.15A during this 10us
..repeating
The average current is 0mA (at the capacitor, independent of waveform)
The peak current is 2150mA (at capacitor, gate driver, power supply)
The RMS current is 96mA (at the capacitor, gate driver)
The average power supply current is: 2.15A x 10us / 10ms = 2.15mA (assuming a single 4.3V supply, independent of waveform)
You will get totally different peak and RMS results with modifed rise_times / modified_waveform.
Btw: Don´t be too lazy to draw a waveform chart. It doesn´t only give us the information what you expect, it will in future help you to understand what´s happening.
Klaus