1
\$\begingroup\$

So I am playing around with designing a UPS when another question hit me.

If I have a source feeding a small amount of current into a non-rechargeable battery is this OK?

By OK, I mean is this acceptable as a circuit to be put into products.

enter image description here

The battery will self discharge and begin to receive very small amounts of leakage current. After reading a bit I was surprised to find that you can actually recharge non-rechargeable batteries(disposable).

What scared me: Wikipedia quotes, "manufacturers do not support recharging, and warn that it may be dangerous". So this of course makes me worry about using this simple circuit. My intention is for the consumer to not use a rechargeable battery.

\$\endgroup\$
4
  • \$\begingroup\$ @Kaz while I do see the obvious connection here, I am more concerned with the validity of my circuit... Though in essence this is trickle charging the battery it would "never" actually recharge the battery due to the self discharge of the battery being higher then the reverse leakage current of the diode. \$\endgroup\$
    – tman
    May 28, 2014 at 2:49
  • \$\begingroup\$ You're asking two unrelated questions. One is whether your circuit makes a valid and safe charger. The other is whether you can trickle charge a disposable battery. The latter is a duplicate. I'd suggest trying to separate out your questions so you're only asking one at a time, instead of gluing two unrelated ones together. \$\endgroup\$ May 28, 2014 at 12:46
  • \$\begingroup\$ @StephenCollings I don't believe that to be true at all per the answer received below...I got exactly the answer I was asking. The question relates to disposable batteries and whether or not the reverse leakage of a diode is enough to warrant an unsafe design. They are entirely related. I don't believe this to be a duplicate because I am in no way interested in recharging the battery it is just a by product of the circuit. Good day to you! \$\endgroup\$
    – tman
    May 28, 2014 at 14:11
  • \$\begingroup\$ @tman And to you! \$\endgroup\$ May 28, 2014 at 14:17

1 Answer 1

4
\$\begingroup\$

It may not be acceptable, but with minor modifications it can be.

You must keep the current flowing into the battery within the limits specified by the battery manufacturer (not some random dude editing Wikipedia). For safety reasons, you should also ensure that no single point of failure can cause a high current (beyond the manufacturer-specified limit) to flow into the battery.

The only 3.6V primary cells that come to mind are those made by the Israeli company Tadiran, so just for an example, let's check their application information, page 22:

Maximum reverse (charging) currents are listed in the Tadiran Batteries Product Data Catalogue. To obtain full performance reverse currents should be kept below 10 μA.

So there's a "maximum current", which varies by battery type (and may, presumably, affect the battery performance), and a general maximum of 10\$\mu\$A that will not affect the battery performance. The "maximum current" is safety-related, not lifetime related. A PN silicon diode such as a 1N4148 will not leak more than 10\$\mu\$A unless you get it really hot (over 100°C). A 200mA Schottky might be okay. But, that's only one constraint.

The Batteries Product Data can be found here. Page 4 contains safety information including the maximum reverse current by battery type (15mA to 100mA), and a number of other cautions.

According to the UL recommendations you should use two series diodes (in case one fails) OR a diode and a current-limiting resistor that will ensure the maximum current is never exceeded, even if the diode fails. That resistor could be just a few hundred ohms (or less, depending on battery type) and meet the requirements of the Tadiran cells listed. For a memory backup or other very low current situation (low tens of uA), the resistor is probably the best because it will drop little voltage compared to a diode.

\$\endgroup\$
3
  • \$\begingroup\$ Wow this is an amazing response...the 3.6 came from using 3 AA batteries sorry for not suggesting that originally. THANKS FOR THIS EXCELLENT RESPONSE! \$\endgroup\$
    – tman
    May 28, 2014 at 4:49
  • \$\begingroup\$ Can you give me a source to the following, "According to the UL recommendations you should use two series diodes (in case one fails) OR a diode and a current-limiting resistor that will ensure the maximum current is never exceeded, even if the diode fails." \$\endgroup\$
    – tman
    May 29, 2014 at 15:59
  • \$\begingroup\$ Source is in the linked document above. \$\endgroup\$ May 29, 2014 at 16:07

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