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I'm trying to regulate 12V to 4.7V with a zener diode. (I'm doing this to protect an AVR's gpio pin, so the current is very low.) I tried the simplest circuit with the zener, but it does not work as expected.

schematic

simulate this circuit – Schematic created using CircuitLab

I expect the multimeter to measure 4.7 Volts on the zener diode, but it only shows about 3.2-3.3 Volts. I replaced the diode with other ones with the same model, and the results are the same.

Are my diodes faulty, or am I doing something (very) wrong?

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  • \$\begingroup\$ Is the 12V a signal that you want to step down to 4.7 volts or so for the AVR's I/O? If not, what is it? \$\endgroup\$ – EM Fields Jun 3 '15 at 10:52
  • \$\begingroup\$ Yes it is. Technically, 2.5 volts should be enough for the avr to detect as high. The real problem here was that I didn't understand why the above circuit wasn't working properly, but I do now. \$\endgroup\$ – lszabi Jun 3 '15 at 13:56
  • \$\begingroup\$ I understand, I was just wondering - since the current into the AVR's I/O is so small - why you'd want to use a Zener (unless, of course, there were huge spikes on the 12 volt signal) instead of just going with a simple voltage divider. \$\endgroup\$ – EM Fields Jun 3 '15 at 15:38
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That is a fairly husky Zener diode. The test current shown in the datasheet is 45 mA. You are only allowing 2.5 mA, which is apparently not enough to get over the diode's "knee".

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  • \$\begingroup\$ I tried with 150R resistor, and it showed 4.4 Volts. Close enough, i think... Thanks for the answer. \$\endgroup\$ – lszabi May 26 '15 at 17:57
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You are using a 1W max zener diode at far below its rated current: the datasheet shows the zener voltage as 4.4 .. 5V at 45 mA. You are using it at ~ 7V / 4.7kOhm = 1.5 mA. The same datasheet shows the Zz as 13 Ohm, so the 'missing' 43.5 mA would account for a missing ~ 0.6V drop.

The effect you see is larger, probably because you are operating the zener outside the current where the stated Zz is a valid approximation.

Solutions:

  • use a smaller (400mW) zener
  • use a smaller resistor
  • use a diode clamp to Vcc instead of a zener (but make sure the input can handle the extra 0.6v, and that Vcc can absorb the extra current!)
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If you're just trying to drop 12 volt signal to around 4.7 volts, you ought to be able to use a simple voltage divider, like this:

enter image description here

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  • \$\begingroup\$ Why the downvote? \$\endgroup\$ – EM Fields May 28 '15 at 18:10
  • \$\begingroup\$ Not from me. But this doesn't explain why the zener didn't work. \$\endgroup\$ – Oskar Skog Feb 13 '17 at 16:44
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The zener voltage is guaranteed to be within 4.4V and 5.0V at 45mA.

You are putting about 1.8mA through it. It will have a lower voltage at such a low current.

If you want a more predictable result you can use a zener diode that has a lower test current.

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