I'm trying to design a circuit that could be powered by an USB host OR by a wide voltage range power input. These are my requirements:

  1. the USB rail (VUSB) should be obviously protected from the auxiliary power input (VIN)
  2. VIN should have priority over VUSB
  3. THRU could be used to power a sibling device by connecting THRU(unit1)->VIN(unit2)
  4. voltage drops should be negligible (in particular when using VUSB)
  5. VIN should offer polarity inversion protection
  6. downstream voltage: 5VDC


I'm currently using two ideal diode controllers LM74700 (D1, D2) and a DC/DC stepdown converter LMZM23601V5SILR (U1), the circuit above synthesizes the power input stage. D3 is a clamp contained in a ST USB6B1.

So far, all the requirements are satisfied, with the exception of (5): by applying a reverse voltage to VIN/GND, I end up with a short.

This happens no matter whether I use the LM74700 or a diode, so I suspect I'm missing something really basic here. I added D3 because so far it's the only potential shunt on the upstream side of the circuit but I can't explain why is being engaged.

Does anybody have an explanation and a solution to fulfill all the reqs?

Any help is wildly appreciated!

  • \$\begingroup\$ Looking more carefully, when GND's potential is higher than VIN there's a forward conduction path along D3 and D2, that's the reason of the short for requirements (5) \$\endgroup\$
    – oxullo
    Aug 27, 2019 at 12:07

1 Answer 1


Just add one diode (any you have handy) on VIN, before the THRU cronnection. Then raise VIN by .7v (or use schotky and raise .3v). If you can't raise VIN, you can try to lower VUSB a bit, so that VIN keeps having priority; just check that output is working despite having 0.1-0.3v less than nominal voltage. From what I can see, the IC works down to 4v, so you have room to play with...

  • \$\begingroup\$ By adding another LM74700, req number (5) is granted and with no significant voltage drop, but could you explain me why D3 gets forward-biased without this additional diode? I expected D2 to be impeding this to happen. Thanks! \$\endgroup\$
    – oxullo
    Aug 27, 2019 at 7:33
  • \$\begingroup\$ In order to properly answer that question we should look at the whole circuit composed of the different parts (VIN source, VUSB source and THRU circuit). But if we consider those being black boxes, the point is that you placed that diode on VUSB only; to have the protection you imagined, you should put the same diode between VIN and GND as well. That's actually another way to solve your issue... maybe. \$\endgroup\$
    – Anichang
    Aug 27, 2019 at 7:47
  • \$\begingroup\$ Since in reality the THRU line is VIN itself (in order to prevent a confusing IN/OUT daisy-chain), adding the diode to VIN/GND would prevent requirement 3 to be fulfilled. Now that is clear how D3 contributes in creating a conduction path when reverse polarity is applied between VIN and GND, I guess one practical solution would be to remove the USB protection (D3). \$\endgroup\$
    – oxullo
    Aug 27, 2019 at 13:22
  • \$\begingroup\$ Well, I was not speaking of a zener diode (D3). I was thinking at normal diodes between gnd and v+ on both inputs; in other words, replace D3 with a normal diode and add one in the same direction on the VIN rail. In any case, if the solution works... no need to keep changing things: Just Do It (TM Shia LaBeouf). \$\endgroup\$
    – Anichang
    Aug 27, 2019 at 13:59
  • \$\begingroup\$ All diodes depicted are shottky, not zeners, so supposedly there's no backward bias in the given conditions. \$\endgroup\$
    – oxullo
    Aug 28, 2019 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.