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I am testing reverse voltage protection using Schottky diode and found small but not trivial level of voltage leakage when combined with boost module in open circuit.

Here, as you can see, the 12 V is connected to output not input, that is reverse direction voltage. To protect reverse voltage, a 1N5822 Schottky diode is inserted. The circuit is not closed intentionally.

I probed voltage of at input and output of boost converter.

When the power supply turned on, small voltage is probed and increases slowly up to about 0.29 V.

I have tested without boost module but no such behavior observed.

Is this behavior is normal? Why do I see 0.29 V in reverse bias of Schottky diode instead of zero here?

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The output of a boost converter usually has a resistive potential divider to to provide feedback to the regulator IC:

schematic

simulate this circuit – Schematic created using CircuitLab

There is a path between +12V and ground, via your "protection" diode D2, and R1 and R2, through which any diode leakage current will pass (shown in red). From the 1N5819 datasheet, we find that reverse leakage current can be many hundreds of microamps, even milliamps, and this current will develop a potential difference across R1 and R2, which is probably what you are measuring.

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  • \$\begingroup\$ Thank you! Is this behavior safe to ignore? I have tested with other boost converter and it even shows over 1V which makes me concern. Or, should I use different approach for reverse polarity protection of boost converter? \$\endgroup\$
    – slyx
    Jul 24, 2023 at 8:58
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    \$\begingroup\$ @slyx I can't imagine damage occurring from such a small current flow. It's not clear what you mean by "reverse polarity protection". Do you want to prevent damage in the event that someone connects the power source to the output instead of input, or are you worried that the source might be connected with positive/negative swapped? \$\endgroup\$ Jul 24, 2023 at 9:20
  • \$\begingroup\$ This is a part of my UPS circuit. UPS circuit have output from direct DC12V and battery + 12V boost converted. The purpose of schottky diode is to prevent unintentionally current flow from DC12V output to boost output for subtle difference of two output voltage. Of course, there exists another schottky diode in DC12V line too. \$\endgroup\$
    – slyx
    Jul 24, 2023 at 9:37
  • \$\begingroup\$ @slyx Normally we would avoid having a regulator like this sink current, but it's such a small amount, I doubt anything will break. \$\endgroup\$ Jul 24, 2023 at 9:44
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Is this behaviour is normal? Why do I see 0.29V in reverse bias of Schottky diode instead of zero here?

It's pretty normal for a lot of Schottky diodes to have poor reverse leakage current compared to their silicon counterparts and, it can get significantly worse with a rising temperature: -

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Image taken from 1N5819 data sheet.

As you can see, the reverse leakage current at 25 °C and 12 volts is around 3 to 4 μA and the voltage you see (0.29 volts) is due to that current flowing into the disabled circuit of your boost converter. Imagine if your boost converter was a perfect open-circuit when disabled; you'd see 12 volts instead of 0.29 volts.

I estimate that your disabled circuit is looking like a circa 83 kΩ resistor: -

enter image description here

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  • \$\begingroup\$ Thank you. I tried with other boost converter and I probed over 1V at the same point which does not look neglectable. Is this behavior safe to ignore? Better not to use this method for boost converter? \$\endgroup\$
    – slyx
    Jul 24, 2023 at 9:01

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