2
\$\begingroup\$

I recently bought a 3.7 V lithium-ion cell to replace an old 3.6 V NiCd one in a handheld vacuum cleaner. I am aware that Li-ion cells don't handle current surges nearly as well as NiCds do, and for that reason, I bought a protected cell.

The problem is, when I load the cell through a 2.2 kΩ resistor, the protection trips open, so at currents of around 1-2 mA, when I would expect the PCB not to trip before 3.3 A as I've read online somewhere.

Could the PCB be faulty? I don't think I'm doing anything wrong here.

EDIT: I am able to untrip the PCB by applying a voltage to the battery, but of course it just trips the moment I load it again, even though the battery reads at 3.8 V open-circuit.

\$\endgroup\$
3
  • 3
    \$\begingroup\$ Where's the datasheet link? "No datasheet? No sale!" Also you're "loading" the cell - not "shorting" it. Shorting it implies the current is taking a short-cut around the intended path. \$\endgroup\$
    – Transistor
    Commented Feb 12, 2022 at 12:15
  • \$\begingroup\$ @Transistor Good question, I bought it online from a retailer near me and there was no datasheet available. It's a generic 18650 cell though. I've tried looking around for it but can't find much. This is the link to the retailer's site I might contact them to ask them for a data sheet if they have one. \$\endgroup\$ Commented Feb 12, 2022 at 12:18
  • \$\begingroup\$ @Transistor Also, changed "Short" to "Load", thank you. \$\endgroup\$ Commented Feb 12, 2022 at 12:29

1 Answer 1

0
\$\begingroup\$

Your battery probably has over-discharge protection, which will shut down the output if the terminal voltage is too low (somewhere around 3V).

If the battery is sufficiently discharged to trigger this with a 2.2K resistor, then it will probably also trigger the soft-start in any charging circuitry, so if you briefly connect the battery to a charger, it won't acquire any meaningful charge - you have to leave it on for a while.

\$\endgroup\$
1
  • \$\begingroup\$ I am aware of over-discharge protection, but the open-circuit voltage of the battery was 3.8V, and on loading it, the voltage momentarily dropped, but nowhere near 3 V. I've ended up simply removing the protection circuit, though I'm running some tests to see if it'll get overcharged (I've already taking measurements that ensure the current draw isn't too much in the actual device). \$\endgroup\$ Commented Feb 12, 2022 at 18:00

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.