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This question already has an answer here:

I currently have 60 LEDs in a parallel circuit. I am using 4 1.5v batteries as the source. The LEDs have a forward voltage of 3v-3.4v (8000mcd, wavelength 460nm).

My questions are: Can I just use one resistor at the start of the circuit (as I'd rather not de-solder all 60 LEDs and add them onto each one)? and if so.. What kind (ohm) of resistor should I buy? How, in future, would I figure all this out myself?

schematic

simulate this circuit – Schematic created using CircuitLab

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marked as duplicate by JRE, Asmyldof, PeterJ, uint128_t, Nick Alexeev Apr 22 '16 at 4:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ That won't work if they are all in series. 3.4V*60=204V. So, either your description is wrong, or else you will need a LOT more than 4X1.5V battery. \$\endgroup\$ – JRE Apr 21 '16 at 15:21
  • \$\begingroup\$ Hm perhaps I am wrong with my description JRE. If I dont use a resistor they are really bright! Im afraid they will burn out though... \$\endgroup\$ – fire_head Apr 21 '16 at 15:23
  • \$\begingroup\$ Can you add a schematic to your question to show how things are connected? Edit the question and press CTRL M \$\endgroup\$ – JRE Apr 21 '16 at 15:24
  • \$\begingroup\$ You probably have the LEDs in parallel. \$\endgroup\$ – JRE Apr 21 '16 at 15:25
  • \$\begingroup\$ @JRE I have added the circuit. Yes, I may. Im a bit of a noob! Trying to learn \$\endgroup\$ – fire_head Apr 21 '16 at 15:30
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Now that you've posted a schematic, it is clear how things are wired.

You MUST have a resistor for each LED. If you try using just one, then the chances are very good that one fo the LEDs will draw more current than the others. It will then heat up and draw even more current - repeat until it burns out and another one starts getting more current.

You are probably getting by right now because the batteries can't deliver enough current to all of the LEDs. The voltage probably drops considerably when all of the LEDs are on.

You should have asked this question before building the circuit.

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  • \$\begingroup\$ Thanks for your answer. When you say "... getting by right now..." is that ok? ( I don't care too much if they burn out. Just worried about a fire :S Is that even possible?) Otherwise, my best bet is to de-solder the positive side of the LED and add the resistors there? I should probably note that the resistors I have are 470ohm but I believe those work best with a 12v source. \$\endgroup\$ – fire_head Apr 21 '16 at 15:39
  • \$\begingroup\$ You describe the LEDs as lighting up "very bright" which probably means they are running at the top end of their ability. At some point, one of them will start hogging the current and burn out. Then another and another. Each burned out LED leaves more current for the other LEDs, so the more burn out the more likely it becomes for the next one to burn out. Given that you are using batteries, you might be lucky in that the batteries give out before LEDs. \$\endgroup\$ – JRE Apr 21 '16 at 15:42
  • \$\begingroup\$ There are plenty of questions on stackexchange about calculating LED series resistors. Like this one \$\endgroup\$ – JRE Apr 21 '16 at 15:45

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