My kids have a pair of IKEA night lights ("spoka") where the internal battery packs have died. I'm debating replacing the batteries (3x AAA NIMH, marked 3.6V so in series) and have looked at this answer but since we've pretty much only ever used them plugged in, that seems like it might be better to just bypass the battery pack.

The lights worked with the the battery leads shorted, but there was an inline fuse on the old packs, and the circuit board gets uncomfortably hot.

One guess is that if I put a resistor there rather than shorting it, it will limit the current and keep things from burning up. At the same time, I've got very little sense what size resistor to try to convince whatever very simple battery management circuit is in there that there's a full pack.

I do have a basic multimeter if it makes sense to measure the current or voltage across the battery charge leads.

The PSU is a cheap 5V, 500ma. (edited: incorrectly wrote 200ma)

  • 1
    \$\begingroup\$ You should determine the schematic of the device in question before trying to decide your next action. \$\endgroup\$ Jun 16, 2018 at 2:47
  • 2
    \$\begingroup\$ Someone did it for you. Cutting the batteries loose should suffice. \$\endgroup\$ Jun 16, 2018 at 3:07
  • 1
    \$\begingroup\$ Just remove the batteries and see what happens. If it doesn't work like that, add the resistor. If it still doesn't work, add a capacitor in parallel with the resistor. You will have to work out the power dissipation in the resistor and make sure it is reasonable. \$\endgroup\$
    – user57037
    Jun 16, 2018 at 5:34
  • \$\begingroup\$ @IgnacioVazquez-Abrams If the reverse-engineered schematic is correct, the wall-wart is one of those evil devices having a center-negative barrel-jack. They were a bit more common about ~30 yrs ago (at least here in Italy). Nothing conceptually wrong, until you forget about it and try to reuse one for other projects. When I was a boy I fried a circuit I worked on for days with one of those evil things! :-) \$\endgroup\$ Jun 16, 2018 at 7:26
  • \$\begingroup\$ @LorenzoDonati that is a 200 mA, 5V AC adapter. \$\endgroup\$ Jun 17, 2018 at 23:36

2 Answers 2


In the spirit of fooling the battery charging circuit into thinking that it is attached to a charged battery and also getting at least a little light in the case of a power failure, you can replace the battery pack with an electrolytic capacitor instead of a resistor.

The charging circuit will charge the capacitor to the normal battery charging voltage and then the current will stop flowing. This minimizes the heat generated by the charging circuit. Just make sure to connect the capacitor with the same polarity as the battery. For even more run time, you could use a supercapacitor.

The capacitor voltage rating must be larger than the battery charging voltage in order to avoid damage. Note that the charging voltage is often higher than the rated voltage of the battery pack, so a 5V or higher rated capacitor should be used.

Standard electrolytics are about $1 and a small supercapacitor is only about $2, so I would go with the supercapacitor and the much longer run time.

supercap pic

(Picture of Illinois Capacitor DGH504Q5R5, which has a particularly good shape to fit in place of the battery pack)

  • 1
    \$\begingroup\$ Please, don't add generic links to your posts. They may give unexpected and variable results in the future and they don't improve the post at all (people browsing this site are expected to know how to perform a google search for generic terms). It is much better to link directly to (for example) a Wikipedia article or a datasheet (possibly from the manufacturer site). And if you really want to make your posts shine, it would be better yet to quote some relevant excerpts (or give a summary by yourself) from the documents you are linking to. \$\endgroup\$ Jun 16, 2018 at 7:18
  • \$\begingroup\$ Good point, links changed to Wikipedia. \$\endgroup\$
    – crj11
    Jun 16, 2018 at 11:19
  • \$\begingroup\$ How would a super cap give more run time when running off an AC adapter? \$\endgroup\$ Jun 17, 2018 at 23:47
  • \$\begingroup\$ @Misunderstood The super cap can store more charge than a standard electrolytic cap, so it can power the lamp for more time after a power failure. For about $2, you can buy a 10,000uF standard electrolytic cap or you can buy a DGH504Q5R5 super cap with about 50 times the capacitance and thus 50 times the charge storage. \$\endgroup\$
    – crj11
    Jun 18, 2018 at 2:57
  • 1
    \$\begingroup\$ Worked perfectly, even with a very small electrolyic (I had a 470uF 10V around.) Thanks! \$\endgroup\$
    – nke
    Jun 18, 2018 at 3:54

If you have a multimeter that can measure resistance, then why not measure the resistance across the battery pack you ripped out of the appliance? And if you don't have it anymore (or if it's completely discharged), google says the internal resistance of a NiMh battery is 320 mOhm, so a 1 ohm resistor (if you have one) sounds like a good approximation to start out with.

However, that was all for the purpose of fooling the battery management circuit into thinking you had a full battery pack. 1 ohm doesn't sound like enough to mitigate the heat problem you were having. For the purpose of current-limiting, I think you should try a current-limiting diode - http://www.newark.com/solid-state/1n5283/diode-current-regulator-220a-do/dp/79T5000?pf=810461032&anyFilterApplied=true&regulated-current=220ma&ddkey=http%3Aen-US%2FElement14_US%2Fc%2Fsemiconductors-discretes%2Fdiodes%2Fcurrent-regulator-diodes

  • 4
    \$\begingroup\$ You can't measure the resistance of a battery the way you suggest. \$\endgroup\$ Jun 16, 2018 at 4:52
  • \$\begingroup\$ A circuit sensing the presence of a battery doesn't really care about that tiny resistance, which is negligible most of the time in most applications when the battery is fully charged. It is the battery voltage appearing across the sense terminals that makes the difference. \$\endgroup\$ Jun 16, 2018 at 7:10
  • \$\begingroup\$ Yep, you'd measure infinite resistance across a battery if there was no current flowing across it, right? I have no idea why I didn't think of that \$\endgroup\$
    – Tri
    Jun 16, 2018 at 22:08
  • \$\begingroup\$ No. An ohm meter sources a small constant current through the leads and measures the voltage across the leads. Putting the leads across the terminals of a battery kind of screws with the voltage measurement. \$\endgroup\$ Jun 17, 2018 at 23:31
  • \$\begingroup\$ I hae two of the battery packs, but they are completely discharged and have resisted any attempt to charge them in the original circuit. \$\endgroup\$
    – nke
    Jun 18, 2018 at 3:41

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.