enter image description here


I have the circuit shown above. Its made to measure analogue voltages up to 10V and converting currents up to 20mA to a voltage output. 10V or 20mA = ~3.3V output

Now I want to test the voltage measurement functionality: Vgate = 0V V_I_IN = 0V to 10V The Zener Diode has a nominal Vz=3.6V

The output voltage is linear up to 4V Input, if I go higher the output voltage rises slowly.

Voltage measured at Vgate=0V at J7 = 0V

There are no mistakes in the pcb layout, checked it very often.

Datasheet Zener

What could be the error? Here is the simulation result: enter image description here

  • \$\begingroup\$ What’s your simulation of the same circuit showing? \$\endgroup\$
    – winny
    Mar 1, 2018 at 9:18
  • \$\begingroup\$ @winny Hey, I edited the first post with a screenshot \$\endgroup\$
    – Fatih Y
    Mar 1, 2018 at 9:21

1 Answer 1


Your mistake is in understanding how a zener diode works. It just doesn't turn into a clamp at 3.6 volts and below this voltage looks like an open circuit.

If you read the data sheet you will see that this zener (MMBZ5227) with a reverse voltage of 1 volt applied will take a current of 15 uA. Now imagine what happens as you approach the zener voltage (where it is taking 20 mA). The 15 uA might double or quadruple at 2 volts and this current flows through the 100 k resistor and produces an error voltage. As you get higher in voltage this current rapidly increases towards the 20 mA current.

enter image description here

Picture source.

If you look at the knee point in the above picture you can see that current is conducting at lower voltages. This has the effect of lowering the 49.9k resistor value as applied voltage is increased.

  • \$\begingroup\$ So you mean that my current is too low and the Zener is clamping too early? \$\endgroup\$
    – Fatih Y
    Mar 1, 2018 at 9:41
  • \$\begingroup\$ The zener doesn't ever clamp it just alters its resistance non-linearly with applied voltage and can give the impression it clamps. With your resistor values you are hoping for a 3:1 reduction in voltage. If 3 volts is applied you would expect 1 volt on the output yes? 1 volt across 50k gives a current of 20 uA but, the zener might also be taking 15 uA too (according to the data sheet). As input voltage rises the zener current rises disproportionately and you get ever increasing errors. Zeners are by no means perfect for this type of job. \$\endgroup\$
    – Andy aka
    Mar 1, 2018 at 9:46
  • \$\begingroup\$ I get 1V output at 3V input but I think this is due tolerances in the zener diode leakage current \$\endgroup\$
    – Fatih Y
    Mar 1, 2018 at 9:52
  • \$\begingroup\$ what can you recommend for me to protect my µC Input instead a zener? \$\endgroup\$
    – Fatih Y
    Mar 1, 2018 at 9:53
  • \$\begingroup\$ A schottky diode from V_I_OUT to your positive supply is commonly used - it will only begin to conduct when the voltage rises above your Vcc level. But, you need to read the data sheet for your uC to see if they recommend other ideas. A lot of people will tend to regard the net series resistance of tens of kohm as sufficient to avoid a bad thing happening - the data sheet may specify an allowable input current that guarantees protection without having to use a schottky diode. \$\endgroup\$
    – Andy aka
    Mar 1, 2018 at 10:04

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