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What is this diode-resistor pair on a N-ch mosfet gate doing?

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Active Member
I'm looking at example layouts for the IRF2136 gate driver. Most of them are pretty similar. Here is the typical connection from the data sheet.


Here's one of the better complete layouts I've found


But some people's designs do this on the connection to the gates:


LO and HO go straight to the Gate driver pins. I recognize that Diode and 100 ohm resistor as a kind of snubber circuit. If it were to protect the driver chip I would think the diode would face the other way. So my best guess is that this is to allow the FET to turn off faster by letting more current flow when the driver starts pulling it low.

Any insights on this?



Well-Known Member
Most Helpful Member
Yes. It's just to have a different gate resistances for turn-on and turn-off, rather than the same resistor for both.


Active Member
there another use of such https://www.infineon.com/dgdl/an-1077.pdf?fileId=5546d462533600a40153559563801007 though not much coverage in text

... if i consider where i have used bi-di difference is at emitter collector or even base circuit of the bjt where the diode has following effects
  • @ reverse polarity
    • high ohmic resistor
    • small to medium boost capacitor
    • junction/electrode discharger , to "balance" another opposite PN junction (e.g an ac/charge termination/"routing")
    • reaction input current path to a capacitor (an ac termination)
  • @ forward polarity
    • log resistor / voltage dependent resistor
    • a "current/charge router" , to prevent reverse conduction where might occur but is not desired (trivial, default)
    • zener/stabistor
  • ?? - might be more
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