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I have a DAC output from my STM32 microcontroller. The DAC output gives 0-3.3V and I want to protect it from high voltage backwards, e.g if I apply e.g 30V onto the DAC output by mistake.

Is this the correct way to do it? I have one Schottky diode that blocks up to 30V and one 3.3V Zener that makes sure that the voltage can only be at maximum 3.3V due to the leakage from the Schottky diode. The Schottky diode has a 0.2A limit but that's OK. DAC outputs cannot spit out more than 30mA.

What do you think of this solution?

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Nov 8 '20 at 19:26
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A 3v3 zener will have an awful slope resistance, i.e. it will load the DAC when the output is close to 3v3 but will not protect very effectively above 3v3. I’d suggest using a series resistor (not ideal if you need a low output impedance but essential if you want to protect the DAC) and then diodes to clamp the DAC from being pulled far above the supply or below ground. Typically ICs will tolerate about a diode drop beyond the supply rails.

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    \$\begingroup\$ Sorry. I cannot use clamping due to the 3.3v is connected to the linear voltage regulator. It will close down the voltage regulator's current. \$\endgroup\$ – MrYui Nov 7 '20 at 11:33
  • \$\begingroup\$ If you know the minimum load on the 3.3 rail then you can choose a resistor value that would feed less than this current through the clamp resistor if 30v is applied to the input. If that isn’t feasible then perhaps buffer the DAC output with an operational amplifier that runs from the 30v supply, if that’s available. \$\endgroup\$ – Frog Nov 7 '20 at 20:00
  • \$\begingroup\$ In any case, exposing your micro’s DAC (or any pin) directly to the outside world may be unwise, depending on your application. \$\endgroup\$ – Frog Nov 7 '20 at 20:03
  • \$\begingroup\$ At the PAx pins, there flows no current. The current only goes to the ground. So the load is realy realy large. \$\endgroup\$ – MrYui Nov 7 '20 at 20:28
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Assuming that your output load is a reasonably high impedance, a series resistor straight on the DAC output will really help iron out the finite Zener slope resistance. But don't make it too big or leakage will cause some drop in output voltage across the resistor. Just swap out the Schottky for it - that thing is dropping output voltage regardless and needs to go.

Also, go for the next zener voltage up, say 3.6V. You don't want say a 5% tolerance on its voltage rating switching the zener on at 3.15V. The small overvoltage will not be a problem.

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  • \$\begingroup\$ So replace the 3.3v zener to a 3.6v zener? Not 3.9v. Because the output can only handle 4.0v \$\endgroup\$ – MrYui Nov 7 '20 at 11:38
  • \$\begingroup\$ OK that is new information, you should add that to your question. I have made the change to my answer. Do you know the load input circuit/resistance as well? Also, any noise-decoupling capacitors change things. \$\endgroup\$ – Guy Inchbald Nov 7 '20 at 11:41
  • \$\begingroup\$ There is no noise at all here. That only can happen is that I by accident applying 30V to the DACx outputs. \$\endgroup\$ – MrYui Nov 7 '20 at 11:43
  • \$\begingroup\$ Putting a resistor here is not a smart idé. Because it will create a voltage drop. \$\endgroup\$ – MrYui Nov 7 '20 at 11:44
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    \$\begingroup\$ @DanielMårtensson You propose a diode in series, which will drop a stonking 100mV or more. Say it's into a 100M ohm load and you insert a 100 ohm resistor instead, that's a mere one-millionth of the DAC output dropped, a tiny fraction of 1 mV, one digit in a 20-bit resolution. Surely it's a trade-off worth considering? \$\endgroup\$ – Guy Inchbald Nov 7 '20 at 12:40

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