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I was looking to protect my 5V device from 20V, since the charging port is Type-C, but used for USB 2.0 communication without any power delivery or bc. Expected to be charged with 1A-1.5A from the wall or USB 3.0 900ma (btw would I need any special IC to request 1A? I think not, but can be wrong)

The protection IC solutions always require power provided to that protection IC (3.3V or 5V). I wanted 0 consumption protection circuit, when charger is not plugged in.

So I quickly conjured up this little circuit on a napkin. Quickly threw it into circuitlab simulation thing, and it works (simulates) as intended. The load is disconnected as soon as Vin reaches 5.5-6V. So the thing works the way I want it to work.

Alternative solution would be to throw in some small linear regulator, tolerant to 20V+, which would power the protection IC (but I kinda feel proud that I invented this circuit in like 5 minutes).

I would like to have some feedback whether my circuit doesn't miss anything critical or makes sense at all. Of course I know I'll have to pick parts carefully in terms of leakage currents and threshold voltages etc. Or am I simply overcomplicating stuff and a tiny USB-powered LDO with protection IC is a no-brainer.

P.S. can't provide circuitlab link, the thing wants paid membership, if anyone wants to reproduce the circuit, I tweaked MOSFET thresholds down a little (default ones don't work at 5V I think). Caps optional.

enter image description here

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  • \$\begingroup\$ But why is this necessary? If there is nothing to negotiate and request more than 5V, why would any charger give out more than 5V? \$\endgroup\$
    – Justme
    Feb 16, 2021 at 15:07
  • \$\begingroup\$ I've heard enough stories about Type-C chargers occasionally sending 20V (usually cheaper ones). There is a handful of chips that exists solely to protect 5V from up to 22V, I think they exist for a reason \$\endgroup\$
    – Ilya
    Feb 16, 2021 at 15:15
  • \$\begingroup\$ You can treat it as a standalone problem of course. Just a 5V OVP circuit) \$\endgroup\$
    – Ilya
    Feb 16, 2021 at 15:16
  • \$\begingroup\$ Why not use a buck-boost converter (3 volt to 40 volts in and 5 volts out)? BTW the R2C2 delay could cause a 10 ms+ overload conditions. \$\endgroup\$
    – Andy aka
    Feb 16, 2021 at 15:18
  • \$\begingroup\$ Yeah I should think about the caps. Probably remove that one. Why would I need a buck-boost? Minimum input is 5V, and the protection ICs are either 5V or 3.3V and consume only a tiny bit of a current. Again, expected input is 5V, 20V protection is just in rare case it ever sees 20V (probably never, but you never know). Buck-boost just for that seems like a huge waste of space and cost \$\endgroup\$
    – Ilya
    Feb 16, 2021 at 15:24

1 Answer 1

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If you just try to protect against some "should never happen" case, a simpler solution may be to put a 5.6 - 6 V varistor like this across the power terminals.

The varistor can handle 250 A. I suggest to add also a fuse that would quickly disconnect the rogue power source. There are self-resetting fuses available on the market, say hold 2 A, trip at 3.5.

The only problem I see with this solution is the quite wide range of the trigger voltages of the varistor. If the lower boundary is 6 V or about, the upper boundary may go as high as 11 V. It will trigger at 20 V anyway, but under some 8 V it may not.

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  • \$\begingroup\$ Interesting idea. I have never used varistors, will do some research, read some datasheets, see if I can make anything fancy out of it. Even if I don't use it in the end (and maybe I will), still useful to know. \$\endgroup\$
    – Ilya
    Feb 16, 2021 at 15:29
  • \$\begingroup\$ If you're going to use a fuse, so that the overvoltage condition can't last very long, then why not a 5.6 V zener, say a 5 watt part, in place of the varistor? Wouldn't that have a more predictable threshold voltage? \$\endgroup\$
    – nekomatic
    Feb 18, 2021 at 14:08

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