I'm flummoxed. I have 33 LEDs (17 "lagoon blue" and 16 white) wired in parallel on 22 AWG stranded stereo wire. Blue LEDs have 20 ohm resistor in series, white have 27 ohm. The circuit has a slide switch in-line and the whole thing is powered by a 3.7V Li-Ion battery.

I did this once before, no slide switch and only 21 white LEDs w/ resistors. My on/off was to unhook the battery. Worked like a charm.

In my current project the big changes are 1) 12 LEDs added 2) two different kinds of LEDs used and 3) slide switch added.

Now, when I put a protected Li-Ion battery in the circuit the LEDs only light if I insert the battery while the switch is in the ON position. Then, if I switch the circuit off, everything turns off...and when I switch it back on, nothing. I have to pull out the battery and put the switch in its on position, then reassemble for the LEDs to light. In addition, if I leave it on, two of the cyan LEDs start to turn deep blue.

If I put in the unprotected battery it will turn back on but the LEDs fade on slowly to full brightness, and the battery gets hot.

What is going on?

UPDATE One battery is protected AW Li-Ion 14500 3.7V 750mAh Other is unprotected UltraFire Li-Ion 14500 3.7V 1200mAh

UPDATE 1/16/18 FYI this is an LED hoop. I disconnected the switch and ran straight from AW battery. LEDs still change color from cyan to blue, but sounds like that may be a separate issue with my resistor choice. Hard to fix or test that now. I wired another switch and the same issue occurred. Here is a sketch not to scale.

enter image description here

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    \$\begingroup\$ you added more LEDs. did you not think that adding LEDs would require more power? \$\endgroup\$
    – jsotola
    Jan 16 '18 at 2:19
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    \$\begingroup\$ You have passed the point of overloading the battery. Not a good thing to do with Li-Ion batteries. Paraphrasing @jsotola, did you think there was no limit to the number of LEDs you could drive? \$\endgroup\$
    – user105652
    Jan 16 '18 at 2:29
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    \$\begingroup\$ "when I put a protected Li-Ion battery in the circuit the LEDs only light if I insert the battery while the switch is in the ON position. Then, if I switch the circuit off... when I switch it back on, nothing." - switch shorting out as it is turned on and off? Protected battery will disconnect when it sees a short, unprotected battery will just get hot. Remove the switch and plug the battery in to turn the LEDs on. Still have the same problem? \$\endgroup\$ Jan 16 '18 at 3:05
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    \$\begingroup\$ So the unprotected battery sort of works but it gets hot. The protected battery doesn't work at all. Has it occurred to you that the protected battery is protected from those conditions in which it will get hot? In other words, the reason the protected battery is not working at all is that the protection circuit is actually protecting the battery. \$\endgroup\$ Jan 16 '18 at 18:17
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    \$\begingroup\$ @Sparky256 true I thought it was all caused by one issue, if not then it should be multiple questions \$\endgroup\$
    – sclarky
    Jan 17 '18 at 4:38

From this diagram:

apparently miswired hoop

it looks like your switch is wired incorrectly. When the switch is in one position, it will short out all the LEDs and the battery, essentially sucking all the power out of the battery and away from the LEDs. In the other position it will allow the LEDs to light.

This is not the correct way to wire a switch and you are lucky your battery hasn't exploded or something.

With this configuration I would expect the battery to get very hot whenever the switch is "off"; does this match your observations?

The switch should be wired in series, in between the battery and the LEDs.



simulate this circuit – Schematic created using CircuitLab

Estimated current (depends on LED specs)

BG 14mA*17=238mA (Blue Green or Cyan)
W 18.5mA*16=296mA (White)
Total = 534mA

If the LiPo is getting hot , I imagine some of the LED's are getting hotter due to missing series R's or wrong values.


Ahem... In theory you can add as many LED's as you want, but you MUST add or increase the resistor in series with each LED. 27 ohms sounds good for starters. Beg, borrow or buy a DVM so you can see the current drain from the battery vs. its amp/hour rating. The total current should not exceed 10% of the amp/hour rating, and that lets the LEDs run for ten hours.

LEDs are very efficient so do not be afraid to double the resistor value in series with the LED. They are non-linear devices so that should dim them only slightly, yet save much current drain on the battery.

Be sensible about the total number of LEDs used. The more you add the dimmer each of them will become as you are forced to increase the resistor values. Red LEDs have the lowest turn-on voltage so use a 33 ohm resistor as a minimum.

If the battery gets hot, measure your current and increase resistor values as needed. You may be close to the maximum number of LEDs you can use already, so double the resistor values to see if that cools the battery down, or limit the number of extra LEDs. There is a finite limit.

Thanks to @TonyStewart for doing the basic math, which shows you are already reaching the batteries limit. Double the resistor values, then check the battery to see if it cools off. I personally would remove some of the extra LEDs, or at least NOT add any more.

A very hot Li-Ion battery is like a small bomb, so go easy on it.


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