I was going to use a 3.3V Zener diode to protect the analog inputs of an Arduino Due, but I found that the voltage drop across it was only 2V.

So I tested connecting it to my power source with a series resistor (see diagram) and I found that the voltage seems to vary somewhat widely:

R=15k   1.78V
1.5k    2.44V
150     3.3V

Incidentally, the datasheet for 1N4728A shows a test current of 76mA, which is close to the 58mA current that is obtained via the 150ohm resistor.


simulate this circuit – Schematic created using CircuitLab

I guess I shouldn't be too surprised by my findings, because it just means that the Zener diode has a positive effective resistance, but I was somewhat surprised that the variation was so high.

Also, it's surprising that even when the Zener diode is not "fully conducting" (or whatever the appropriate term is), it shorts out the lower resistor in my resistor network (see below). When I remove the diode, for an input voltage of 12V the voltage at the ADC is 2.89 V, but when I put the diode in, the voltage drops to 1.86V. Should I just choose smaller resistor values, or is the Zener diode useless for this purpose?


simulate this circuit

  • 3
    \$\begingroup\$ Zener diodes have V-I curves in their datasheets. If you can work with their limitations, they are fine to use. Otherwise take a look at something like a TL431. Note: there is nothing wrong with using a Zener that has a higher nominal voltage if it works for your circuit. You can use a 3.9V (or higher) Zener, if you have a lot of resistance in series. \$\endgroup\$
    – user57037
    Commented May 31, 2016 at 2:05
  • 2
    \$\begingroup\$ Watch out for the temperature coefficient, too. If that is important to you, then you better test your circuit over temperature. \$\endgroup\$
    – user57037
    Commented May 31, 2016 at 2:05

1 Answer 1


Putting the zener diode across the lower resistor in the divider is always going to disturb the voltage divider if the zener conducts any current at all. And apparently the zener that you have chosen conducts a fair amount of current and is far from an ideal unit.

If I was designing a protection for the A/D inputs I would use one of two approaches.

  1. Use the zener diode on the signal voltage source before the voltage divider.

  2. Use an actual diode clamp on the A/D input made of a dual diode component such as a BAT54S. The upper diode points into a clamp current sink made of a shunt supply that is designed such that the clamp works at approximately the 3.3V level. Something like this:

enter image description here

  • \$\begingroup\$ Thanks. I understand "approach 1" but I'm still parsing "approach 2". The symbols for D1 and D2 are Schottky diodes, and you are using them because of the low voltage drop? Why do they need to be matched? Thank you for the helpful advice. \$\endgroup\$ Commented May 31, 2016 at 7:41
  • \$\begingroup\$ The clamp diodes can be any diode you choose that has very low reverse bias leakage. I use the BAT54S because both diodes come conveniently interconnected in one small SOT-23 package. Low drop diodes are good because the lower diode clamps the signal to just below GND. The upper diode clamps the signal to just above the voltage of the current sink. The current sink can be a zener diode or other type of shunt regulator such as TL431. In some applications where the clamp level would be OK if it was above the normal signal level (3.3V in your case) it may be possible to eliminate the (continued) \$\endgroup\$ Commented May 31, 2016 at 12:18
  • \$\begingroup\$ (continued from above) current sink all together and then just connect the upper clamp diodes to a supply rail such as the 3.3V rail. If you do that make sure that the nominal load on the 3.3V supply is say 10x or more the current you would expect coming in from any one of the signal clamps. \$\endgroup\$ Commented May 31, 2016 at 12:19

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