I have just answered a question here, and it has spurred me to ask a question:

How fast will this type of protection work? Would it work with, say, a 500kHz signal? And is there an easy way to work this out fro the zener datasheet? (Using zener and resistive divider to protect from overvoltage)

I would expect, at some point, that oscillations would occur where the zener doesn't break down fast enough?

The previous question: 12V input on microcontroller pin

Schematic example given by Michael on that thread:


  • \$\begingroup\$ This zener (onsemi.com/pub/Collateral/BZG03C15-D.PDF) has a response time less than 1 ns. \$\endgroup\$
    – Andy aka
    Jul 24, 2017 at 14:52
  • \$\begingroup\$ It is not so much the signal frequency, but the edge rate of the event. \$\endgroup\$ Jul 24, 2017 at 14:54
  • \$\begingroup\$ Zeners have capacitance, so at some point that'll have an effect too. \$\endgroup\$
    – brhans
    Jul 24, 2017 at 15:18
  • \$\begingroup\$ Andy - Yes, initially I used to work with the response rate but since working a lot with AC and inductors / RF etc, I have realised the answer may be more complex. Peter - would you be able to elaborate? I am assuming a square wave signal. brhans - This is one of the reasons I asked the Q :) ta \$\endgroup\$ Jul 24, 2017 at 15:43

1 Answer 1


Diode capacitance likely limits the time response of this circuit. Avalanche diode capacitance depends on its reverse-bias voltage, and is maximum for low voltage, while minimum as you approach the reverse-breakdown voltage. It acts very much like a varactor diode (and can be used as a voltage-variable capacitor).
Since the 12V_INPUT is some kind of a logical signal, the diode is swept from a low voltage up to near its breakdown limit, so its capacitance changes over a wide range, perhaps from 100 pf to 10 pf.
Taking worst case (100 pf), a 500 kHz input signal would be very close to the upper limit, and would cause some edge distortion. This might be a good thing, offering some minor noise immunity.

  • \$\begingroup\$ Thanks Glen, very useful thank you - builds my understanding a bit as to whats happening. The 12V Vcc is just an example. I have had some success taking pulses at a couple of hundred kHz and have been wondering if the diode will catch the spike or there will be a delay where some delay in response will potentially damage something on the protected side if the signal is too fast? . For example - if this was feeding the uC a frequency pulse from a switched inductor with reverse emf - would this be enough to stop spikes damaging anything? (Assuming the 500kHz case) \$\endgroup\$ Jul 24, 2017 at 15:48
  • \$\begingroup\$ @Rendeverance Avalanche clamping can be very fast. The zener also protects against transients trying to swing below ground, acting as a normal diode. Spec sheets tend to focus on the avalanche region. That 6.8K series resistor (R1) also helps to protect the MCU_INPUT quite well, when combined with the zener. \$\endgroup\$
    – glen_geek
    Jul 24, 2017 at 16:15
  • \$\begingroup\$ thanks, so as long as I can read the signal, it should not be an issue with respect to protection to operate at 500kHz (for sake of the example)? \$\endgroup\$ Jul 24, 2017 at 16:37

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