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I found this reverse polarity circuit that uses a matched PNP pair and a MOSFET: https://projects-raspberry.com/raspberry-pi-b-power-protection-circuit/

I tested this circuit and with no load i get around 270uA of quiescent current when connected in direct polarity. That current is to high since i want to protect a battery operated circuit.

So i changed the 47k resistors for 1M resistors and i got around 15uA of current consumption. I reverse the polarity and the circuit blocked it, so it seems it still works.

My questions are:

  1. Is it ok to use 1M resistors instead of 47K? any caveat?
  2. any other ideas to lower current consumption?

Thanks

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  • \$\begingroup\$ First time I saw this "ideal diode" circuit, it had 100k on the left and 65k on the right. So it can tolerate a lot of variation. Probably has to do with performance behaviors that the designer wants. ie. how fast you want it to switch. \$\endgroup\$ Commented Oct 13, 2022 at 18:39
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    \$\begingroup\$ Also note, that the circuit as shown can only be used up to about 5V in reverse bias direction before Vbe of the BJT breaks down. \$\endgroup\$ Commented Oct 13, 2022 at 18:41
  • \$\begingroup\$ Thanks @Aaron for your response. So max Vbe will limit the reverse voltage protection, right? I was looking something up to 12V \$\endgroup\$ Commented Oct 17, 2022 at 13:43
  • \$\begingroup\$ This circuit fixes the Vbe breakdown issue: electro-tech-online.com/articles/… \$\endgroup\$ Commented Oct 17, 2022 at 14:12
  • \$\begingroup\$ Great, thanks, i'll try that. Just another question: in direct polarity, just Q1 is ON, so to reduce quiescent current i just need to have a high R1 value and keep R2 as original value right? I want to reduce quiescent current but keep circuit fast enough if reverse polarity occurs \$\endgroup\$ Commented Oct 17, 2022 at 15:19

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