1
\$\begingroup\$

I'm a bit lost in the datasheet of the IRF3205 N-channel MOSFET: where should I look to find the maximum acceptable gate current?

Is it I = 10/4.5 as determined by the conditions V_GS=10V, G=4.5Ohm in the typical values for the turn on times array?

So I_max=2.22A? Or something else?

I wanna make it switch as fast as possible,I'm trying to get a better understanding of datasheets.

\$\endgroup\$
2
  • \$\begingroup\$ If there's any significant gate current for any significant length of time, your MOSFET's dead. There's no reason to spec a maximum gate current; not accounting for leakage, the total amount of energy delivered into the gate for any given switching event is approximately constant regardless of the peak gate current. \$\endgroup\$
    – Hearth
    Feb 20 at 1:50
  • \$\begingroup\$ Sometimes there is an R<sub>G</sub> parameter, which tells you roughly how fast it can go with an ideal driver. Since the gate circuit impedance can be no less than the internal resistance. (This one does not specify; at best we can assume it's somewhat near the test condition i.e. 4.5Ω.) There is no pulsed gate current limit to worry about. \$\endgroup\$
    – Tim Williams
    Feb 20 at 2:03

2 Answers 2

Reset to default
0
\$\begingroup\$

As long as you don't violate the maximum gate-source voltage, you can push whatever current you want without having any realistic chance of destroying the part.

That said, the maximum instantaneous current is your gate driver voltage divided by the gate impedance, so will be 2-3 A for the numbers you mentioned.

This is only the initial peak current and will decay exponentially as the gate charges.

\$\endgroup\$
3
  • \$\begingroup\$ Hmm, then why do almost all tutorials on mosfets tell you you need to protect the gate of your mosfet with a resistor? \$\endgroup\$
    – djfm
    Feb 20 at 11:11
  • \$\begingroup\$ @djfm Please show a tutorial that says that the resistor is there to "protect the gate". A resistor is sometimes placed to reduce the charging current, so one has more control over the switching speed. This makes the switching slower than it could be, but helps control EMI and gate ringing. \$\endgroup\$
    – tobalt
    Feb 20 at 11:37
  • \$\begingroup\$ you appear to be right, apologies. I'd swear I had read this everywhere when in fact, I cannot find anything. Maybe I misunderstood something or just remembered a wrong fact, I'll comment back if I find the source of my misconception. Thanks for clarifying this! It's actually making my life much easier in my beginner project I'm doing. \$\endgroup\$
    – djfm
    Feb 20 at 13:27
-2
\$\begingroup\$

You are unable to find gate current in the specs simply because MOSFETS, unlike BJT, do not work on current into the controlling element. They work based on voltage at the Gate.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ There is no DC current, but as FET gates are capacitive, the faster you can push the charges in and out means the faster the FET turns on/off and there will be less losses. If the FET is turned on/off too slowly, it can burn from the power dissipated. So you do want to use large currents to drive the gate to get to the target voltage fast. Now, there must be a sane limit if this current can be 1A or 100A, if the target is to switch it as fast as possible. \$\endgroup\$
    – Justme
    Feb 20 at 5:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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