I´ll be clear and fast in what I´m asking once my last question did not have many views.

So, a couple days ago I was at the bench switching a IRLZ34N with a 555 timer, powered by a 15V, @ 50kHz, \$33\Omega\$ gate resistor. As I started changing the drain resistor values, I could the see the mosfet, when switched off, would increase its drain voltage in a speed which relied on the drain resistor. To bring some numbers, @ \$R_{drain} = 47k\Omega\$ it was really slow to build up that 15V again, @ \$R_{drain} = 3k\Omega\$ it was able to reach 15V within 2us. @ @ \$R_{drain} = 330\Omega\$ it reached 15V in about 400ns.

Although I´ve not uploaded any image to clarify, I searched for what could be the reason why it has such behavior and realized it has to do with drain-gate capacitance (\$C_{DS}\$) and gate-source capacitance (\$C_{GS}\$) though I really don´t get how it affects the switching like that.

I know \$C_{gs}\$ affects the circuit in increasing \$V_{GS}\$ and switching the transistor on and off but I don´t understand how to relate that to the drain voltage as I just mentioned.

Why does the mosfet behave like that after all?

  • \$\begingroup\$ You have some formatting problems, here. Your math hasn't rendered.. \$\endgroup\$
    – Reinderien
    Dec 8, 2018 at 1:22
  • \$\begingroup\$ Fixed. You need to escape your dollar signs on this board (which is opposite the mathematics Stackexchange). So to get \$\Omega\$ you need to write \$\Omega\$, not $\Omega$ \$\endgroup\$
    – TimWescott
    Dec 8, 2018 at 2:09
  • \$\begingroup\$ Read up on Miller effect and Plateau. Also, if you switch 10amps in 10 nanoseconds, or 1 amp/1nS, thru 5 mm of wire (which is about 5 nanoHenries) you will see volts across that wire. If you grounds (the gate driver must share a GND with the FET, right?), then depending on how your GND is wired, you may experience negative feedback and the FET switching will be delayed; or you may get positive feedback, and the entire system becomes an oscillator. \$\endgroup\$ Dec 8, 2018 at 3:12

1 Answer 1


What you are seeing has little to do with the capacitances you mention. Instead, you are seeing the effect of large drain-source capacitance. This has little effect on turn-on performance, but determines the turn-off speed.

When the gate voltage drops to zero, the drain voltage will rise (approximately) as a simple RC circuit using the Coss and the drain resistance. For the IRLZ34, Coss is nominally 660 pf. So a 330 ohm drain resistor will produce a time constant of about 200 nsec, and 400 nsec is two time constants. 3k produces a time constant of about 2 usec, and 47k gives 31 usec. With no detailed responses provided by you, it seems pretty clear that these numbers are an OK match for what you seem to be seeing. The actual capacitances would seem to be a bit smaller than the data sheet numbers, but that's not unusual.

The effective capacitances involved are actually voltage-dependent, so the results are not as simple as might be wished, but this approximation to the real results seems to give reasonably close answers.

  • \$\begingroup\$ Interesting fact. I thought of Miller effect in the beginning but didn´t move ahead looking for it. Thank you for the clear answer, it makes more sense for me now. \$\endgroup\$ Dec 8, 2018 at 3:24
  • \$\begingroup\$ @IronMaiden - The output capacitance is not primarily a Miller capacitance. It's an artifact of the large FET structure. \$\endgroup\$ Dec 8, 2018 at 4:04

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