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I have (had, actually) an El Cheapo pocket LED flashlight powered by 3 AAA cells. It was shipped with very low quality "salt" cells (regular, low weight - someone please suggest a proper English name for them). I tried to beef-up this flashlight by installing powerful Varta alkaline cells but flashlight blinked couple of times and gone dead. Now I have slight memories about what I read somewhere what LED drivers does not allow alkaline (higher current) feeding. Are my memories correct? Can I actually install alkaline cells to LED lighting devices? (Yes, No or Sometimes depending on driver)

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by installing powerful Varta alkaline cells but flashlight blinked couple of times and gone dead.

Yep. You definitely fried it there. I once had a cheap keychain flashlight that I didn't want so I hooked it up to 5V and it had exactly those symptoms. When I hooked it up again, it didn't work. No big deal, I just wanted to see what would happen, and I didn't really have a use for it.

However, if the person who designed that flashlight has greater than 0 electronic knowledge, they would add a resistor to limit the current. The only reason a resistor wouldn't limit the current correctly is if you had more voltage, which both more voltage and current allowed is a huge problem. Current is supplied on demand, but the LED sucks up as much as it can, but the resistor should limit it no matter the amount of current that can be supplied, unless the voltage is increased.


Edit: One thing I forgot: Quote from jippie's comment:

Alkaline batteries probably have much lower internal impedance than the original ones. The designer used the battery's internal impedance as series resistor. Done quite a lot in el Cheapo stuff.

Batteries have a internal resistance, and that's why the designer used the original ones to save on money. It's a smart trick, but you should have put a resistor in. How you would know this before, I don't know. But, if you have another one, you would just need to add a quick one from the store. At my RadioShack, I can get a five pack of 1/8W, 1/4W, or 1/2W for $1.49. (Yes they are all the same price... I have been wondering why anybody would want the more flimsy wires and a smaller piece that has less capacity but is the same price as the others.) But... you don't know how much current the LED can handle, unless you measured the current with the old batteries and used Ohm's Law.

If you have another and a multimeter, you can fix it on the other, but this one is toast. Gone. Bad. Broken. Fried.

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    \$\begingroup\$ Alkaline batteries probably have much lower internal impedance than the original ones. The designer used the battery's internal impedance as series resistor. Done quite a lot in el Cheapo stuff. \$\endgroup\$
    – jippie
    Commented Aug 18, 2013 at 16:07
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Your LED flashlight depended on the internal resistance of the batteries to limit current through the LEDs. With the original batteries, the internal impedance was high enough to limit LED current to safe levels.

This is common practice in many low price, non critical electronic devices. The saving is not just in the cost of the resistor, but also in incremental cost of assembly, however small. Anyone familiar with the so-called design of cheap consumer electronics will know how common this practice is.

Alkaline batteries typically have lower internal resistance i.e. can support much higher current than the cheaper zinc-carbon batteries. Without any other form of current limiting, the ”improved” capability of alkaline batteries has caused the death of the LED.

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  • \$\begingroup\$ I suspect what manufacturer was saving space too, battery "compartment" (second from the left) is in direct contact with LED assembly. Almost the same flashlight of German origin have spring there. \$\endgroup\$
    – user539484
    Commented Aug 18, 2013 at 18:23
  • \$\begingroup\$ Do you mean resistor, not LED?: The saving is not just in the cost of the ***LED***... That would be a pretty pointless LED flashlight if they didn't add a LED for cost. \$\endgroup\$ Commented Aug 18, 2013 at 18:46
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    \$\begingroup\$ @AnnonomusPerson Not that they wouldn't do it if they could get away with it. \$\endgroup\$ Commented Aug 18, 2013 at 21:04
  • \$\begingroup\$ @Annonomus Yes, obviously resistor :-) Fixed. \$\endgroup\$ Commented Aug 18, 2013 at 22:17

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