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I plan on connecting 8 LEDs in parallel with 3 AA alkaline batteries (i.e. 4.5 volts) being the power source, and only one 230 Ohm resistor. The circuit is really simple/generic, so I assume no schematic is needed. The issue that I'm facing is that the more LEDs I add in parallel to the circuit, the dimmer the rest of the LEDs become. This is the kind of behavior one would expect if the LEDs we're connected in series, so I am a bit lost. It would also be worth mentioning that I also tried using a battery pack (the ones for phones) with an output current of 2A, but the results were the same.

Thanks for the help!

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marked as duplicate by The Photon, Harry Svensson, winny, DoxyLover, Voltage Spike Jan 6 '18 at 22:18

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.

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    \$\begingroup\$ Use separate resistors... electronics.stackexchange.com/questions/22291/… \$\endgroup\$ – Trevor_G Jan 5 '18 at 17:41
  • \$\begingroup\$ What made you think this plan would work? \$\endgroup\$ – Sunnyskyguy EE75 Jan 5 '18 at 18:40
  • \$\begingroup\$ all your LED share the same amount of current, limited by the resistor. this is why it's not the most flexible setup, in terms of dynamic LED counts. Use separate resistors, or calc your one resistor to feed enough current to light them all up. \$\endgroup\$ – dandavis Jan 5 '18 at 20:45
  • \$\begingroup\$ Schematics are always useful. \$\endgroup\$ – Peter Bennett Jan 6 '18 at 2:03
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This behavior is expected since the total current is divided over the 8 LED's. I suggest you use seperate resistors to insure current is flowing through all LED's.

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First look at the datasheet of the leds in order to associate brightness, current and voltage forward. One choosen these parameters You can calculate the final resistor required for that conditions. If You add one by one the behaviour is the expected because the current across the resistor is increased and the voltage forward the leds will be less.

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