I'm working on a project and i need to switch IRF830 N-Channel MOSFET at roughly 100KHz as low-side.so, how much output current I need to drive this MOSFET 4 in parallel?
First, gate drive current is not typically chosen to just meet the capacitance of the MOSFET. Instead the drive is selected to move the gate voltage as quickly as necessary to maximize the switching efficiency of the circuit. This is why even small gate drivers are specified to deliver a few Amps (in a brief pulse).
Now let me answer your question. The datasheet shows Qg = 38 nC with Vgs ending at 10 V and Vds starting at 400 V. Of this 5 nC is Qgs and 22 nC is Qgd.
Roughly speaking the gate driver has to drive 2 capacitors. One between Gate and Drain and one between Gate and Source. I tried a simple diagram below.
Vdrive --| Cgd |-- Drain
Vdrive --| Cgs |-- Source
The capacitance is junction and oxide capacitance so it is not constant but changes significantly -- Cgd especially -- with voltage. This is why both gate charge and gate capacitance are given.
Assume the drive is 10 V then 5 nC is put onto the gate 100 k times a second = 5 uA. For the gate charge only.
The gate to drain capacitance dominates but the drain is probably not a fixed voltage so this an calculation is an estimate. With the scant information provided it's a wild guess. But 38 nC * 100 kHz = 3.8 mA.
Of course scale all this up by 4 when you put 4 FETs in parallel.