# Powering LEDs **URL:** \ **Category:** Factual Questions\ **Created:** [December 2, 2015, 9:42pm UTC](https://boards.straightdope.com/t/powering-leds/739255 "2015-12-02T21:42:56Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![KCB615](https://avatars.discourse-cdn.com/v4/letter/k/ea666f/32.png) [@KCB615](https://boards.straightdope.com/u/KCB615)\ **Post date:** [December 2, 2015, 9:42pm UTC](https://boards.straightdope.com/t/powering-leds/739255/1 "2015-12-02T21:42:56Z")
I am working on a project that will use four 1/4" LEDs to illuminate a bunch (around 100 or so) of fiber optic filaments. I need four LEDs, since they will be switched in four different groups. The LED kits that the supplier gives use a 9-volt battery to drive a single LED with an in-line switch. I am not enamored by the idea of mounting and maintaining four 9-volt batteries, I would prefer to use a single 120vAC to 9vDC transformer. Can I drive the four (presumably 9-volt) LEDs by simply rewiring the 9-volt battery leads to a common power supply, or do I have to start considering resistors and other electronic things that confuse and complicate what was supposed to be a simple project? ---
**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\ **Post date:** [December 2, 2015, 9:52pm UTC](https://boards.straightdope.com/t/powering-leds/739255/2 "2015-12-02T21:52:20Z")
Presumably, the LEDs you are using already have dropping resistors, since 9v is not a normal multiple of LED forward voltage. So, just wire them in parallel to your 9v source. ---
**Author:** ![KCB615](https://avatars.discourse-cdn.com/v4/letter/k/ea666f/32.png) [@KCB615](https://boards.straightdope.com/u/KCB615)\ **Post date:** [December 3, 2015, 1:51am UTC](https://boards.straightdope.com/t/powering-leds/739255/3 "2015-12-03T01:51:51Z")
Exactly what I was hoping to hear. Thanks. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [December 3, 2015, 11:57am UTC](https://boards.straightdope.com/t/powering-leds/739255/4 "2015-12-03T11:57:54Z")
> [@beowulff](#): > > Presumably, the LEDs you are using already have dropping resistors, since 9v is not a normal multiple of LED forward voltage. > So, just wire them in parallel to your 9v source. Agree, but you need to _damn sure_ the LEDs have built in resistors. If they don’t, - Connect the LEDs in series. - Use a 12 VDC power supply. - Use an LM317 (and appropriate resistor) configured as a [constant current source](http://users.telenet.be/davshomepage/current-source.htm). ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [December 5, 2015, 1:01pm UTC](https://boards.straightdope.com/t/powering-leds/739255/5 "2015-12-05T13:01:16Z")
> [@Crafter\_Man](#): > > Agree, but you need to _damn sure_ the LEDs have built in resistors. If they don’t, I’d say they are packages sold as “power on” LED’s… with inbuilt current control… as he said, just attach the 9V and the switch, and the LED can be switched on and off… They don’t cost much ,so if the first one blows when wired into the 9V, then add the 10K to 1K or perhaps down to 90 Ohm resistor in series… Limitting the current to approx 1mA, 10 mA, 100 mA…