1
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ It depends on a lot of things about your circuit and your load. \$\endgroup\$ – Hearth Mar 11 at 19:44
  • \$\begingroup\$ Why four 1.5 Ohm switches with 10k different FETs to choose that may be better? criteria? \$\endgroup\$ – Sunnyskyguy EE75 Mar 11 at 19:53
  • \$\begingroup\$ What is your concern about current? Why? \$\endgroup\$ – Ale..chenski Mar 11 at 20:19
  • 1
    \$\begingroup\$ One of the main parameters to consider is the total gate charge \$Q_{G}\$ of the MOSFET, but without knowing the kind of driver you want to use nor the gate voltage \$V_{GS}\$ nor the drain voltage \$V_{GS}\$ it is difficult to you cannot estimate its value. \$\endgroup\$ – Daniele Tampieri Mar 11 at 20:29
  • 1
    \$\begingroup\$ If you mean what is the current necessary to drive the gate, could you say so? As the comments suggest, "output current" is ambiguous. A schematic of that portion of your circuit, showing driver, FET, load, and any catch diodes, would be very helpful. \$\endgroup\$ – TimWescott Mar 11 at 22:47
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.