I am designing a device that can be powered either through two AAA alkaline batteries, or through its USB port, with priority with the USB port. I thought of a discrete circuit that fulfills my needs (simple, almost no power consumption and very low dropout when on batteries, ...).

So everything seems fine, except that, when the USB power is removed, there is a quite high reverse current transient that goes back to the batteries, because the main capacitors that were charged to 3.3V then have to go down to the batteries level. It can be as high as 200mA during about 10µS, then quickly decreasing but still being at about 1mA during as long as 15 seconds. (There is a 10µ ceramic cap and a 0.035 supercap in parallel).

Is this acceptable ? Or will it make the alkaline batteries release gas / leak / loose all their capacity / burn in fire / and if I repetitively plug-replug the USB cable, kill me in my sleep ?

And if some other people decide to use rechargeable batteries (any kind) in this device, will it cause the same / other problems ?


Ok, here is the schematics of said circuit, for reference. It tries to combine several constraints, which are not fully explained here, but anyway. Just keep in mind that the question is not about this schematic or how to switch between USB and batteries.

enter image description here

  • \$\begingroup\$ Please post a schematic of the discrete circuit which thought of. \$\endgroup\$ Commented Apr 26, 2016 at 19:41
  • \$\begingroup\$ I was afraid the answers would then derive in things that are unrelated to the original question, but... ok. \$\endgroup\$
    – dim
    Commented Apr 26, 2016 at 19:49
  • 1
    \$\begingroup\$ Not to make the circuit more complicated, but could that MosFET switch, Q201, help out a bit? Could you make it not turn on as long as pin 2 is greater than pin 3? That way, your device would run off of the capacitors until they were drained to the level of the batteries. \$\endgroup\$
    – Mark
    Commented Apr 26, 2016 at 21:34
  • \$\begingroup\$ @Mark: Indeed, I could. But I'd need some comparator that would draw power and the circuit would become more complicated. I wanted to avoid that if there is no need to. Hence the question. \$\endgroup\$
    – dim
    Commented Apr 27, 2016 at 6:45
  • \$\begingroup\$ I wasn't thinking about adding a comparator. I was thinking more along the lines of modifying your USB detection circuit. Rather than sensing the USB voltage on the left side, detect it on the right side, where the cap voltage also sits. Then, instead of the 3.5 volt reference from the divider, use the battery for the reference. Then when you plug in the USB, the higher cap voltage turns off Q201. It then turns Q201 back on again when the cap voltage drops back down, as a result of the USB disconnecting. That should be a minimal difference in parts, but a whole lot more thinking. \$\endgroup\$
    – Mark
    Commented Apr 27, 2016 at 8:37

1 Answer 1


Even an amp for a few milliseconds isn't going to harm them at all and a mA for a few seconds isn't going to much either, you'd need to be pushing several watts for quite some time before you would experience any real problems (you can always swap out the alkalines for rechargeables if you are still worried, there are NiZn cells that have the same voltage as alkalines but they're hard to find)


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