In the datasheet of the IRF540 power MOSFET there are two different statements about the input gate resistance.

At the end of page one, it says that with VDD = 25 V and TJ = 25 °C, the input gate resistance equals Rg = 25 Ω.

Then, on page two, it says: Gate input resistance Rg = 0.5 - 3.6 Ω at 1 MHz.

Now, I have some questions about this:

  1. Does the first statement refers to DC analysis? If it does, the second measurement tells us the gate resistance strongly decreases with frequency. Which are the possible causes of this phenomenon? Parasitic capacitance (if yes, it should be called gate impedance, not resistance)?

I utilized this MOSFET with a gate driver (TC426) which has 6 Ω output resistance. At 12 V output voltage without load, I have measured 8 V with this MOSFET with a rectangular pulse of 100 ns at 10 kHz. Measuring 8 V means the input gate resistance would be like 10 Ω, which may be realistic as it is between 25 Ω (DC value) and 3.6 Ω (1 MHz).

  1. Do you know a possible alternative MOSFET for my application or a possible solution for this problem? I need to drive three of these or similar MOSFETs though my gate driver. With 3 of them, I've measured about 4 V. It appear to be in line with my previous computation (three input equivalent resistors of 10 Ω means 3.33 Ω which absorbs 4.29 V from 12 V with 6 Ω output resistance).
  • \$\begingroup\$ nS = nanosiemens. ns = nanosecond. ocrdu has already corrected it for you but please remember it. \$\endgroup\$
    – winny
    Jan 27 at 18:52

2 Answers 2


The first statement is from footnote (b), which applies to the avalanche parameter. The circuit is described in Fig.12. Rg is the external resistor shown in the schematic. It bears no relation to other tests or parameters.

The second statement is a measurement of the part itself, its internal resistance.

Without a schematic and waveforms, and preferably layout or photos of the setup, it's not clear what you were doing with the gate driver, or if your conclusions are correct. Note that short pulse durations are critically dependent on wiring length, circuit layout, component placement, probing technique, and probe and oscilloscope bandwidth. All of which must be ensured adequate for the measurement.

You also request alternative parts, but without any information at all regarding application, I'm afraid we are at a loss.

  • 1
    \$\begingroup\$ It may be worth noting that the internal series resistance of the gate is in series with the gate capacitance. At DC the gate will pass at most Gate-source leakage = 100 nA current. \$\endgroup\$
    – jpa
    Jan 28 at 6:54
  • \$\begingroup\$ Yes, a simple equivalent circuit can be drawn, for the gate-source loop, as a Rg + (Rlkg || Ceq) network, where Rg is small (ohms), Ceq is due to gate capacitance/charge, and Rlkg is the leakage. Such a model is valid from DC to switching harmonics, where stray inductance takes over. \$\endgroup\$ Jan 28 at 7:30

The gate driver is driving the paralleled gate capacitances. A gate driver is essentially a current source. You can achieve the same results by removing the mosfets and putting a SMD capacitor between gate-source pads of each mosfet. As you'll notice, the waveforms will be essentially the same.

So mosfet parameters other than gate capacitance don't really matter for what you're dealing with.

Gate drivers have a hard job enough driving just one mosfet. Typically, you want one gate driver per mosfet. And make sure the gate driver can push enough current to charge the gate capacitance fast enough.

At 12 V output voltage without load

What does that mean? Load where? If the gate driver is connected to the mosfet, it's loaded by the gate capacitance. It doesn't matter whether the mosfet itself is loaded. The drain can be left open. Adding a load to the drain increases the capacitance slightly if the load has little parasitic inductance. If it is inductive or connected with long wires, the gate-drain capacitance won't have much influence on the gate waveform I'd think.

I have measured 8 V with this MOSFET with a rectangular pulse of 100 ns at 10 kHz

I have no idea how you'd measure that, and what 8V means. It's a waveform, not a DC level. The gate waveform will look more-or-less like constant-current charge/discharge ramp with a fixed voltage level between the ramps. A mosfet gate is not a resistive load at all.


Your Answer

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

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