# Wiring LED's to typical batteries? **URL:** \ **Category:** Factual Questions\ **Created:** [October 31, 2012, 10:40pm UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491 "2012-10-31T22:40:20Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![Tristan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tristan/32/18058_2.png) [@Tristan](https://boards.straightdope.com/u/Tristan)\ **Post date:** [October 31, 2012, 10:40pm UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/1 "2012-10-31T22:40:20Z")
So, after brainstorming about my son’s robot halloween costume, I thought it would be cool to put a few LED’s on it. I was under the impression that I could just wire one to a watch battery or somesuch, and all would be well. I only need it to last for a few hours tomorrow night (we are doing our ToT’ing on Thursday). The problem is that it wasn’t nearly that easy. I need a quick and dirty tip on how many AA or AAA batteries I am going to need to connect to these LED’s. The information on the back says: Supply Voltage: 5VDC Forward Current: 80mA Reverse Voltage: 0.4VDC 5mm blinking Red LED 5 volt \* 20mA \* 4.0mcd I’ve got an pack of 2 green LED’s with different values Forward (supply) voltage: 2.1 typ 2.8 max 2.1 Volt \* 30mA \* 630mcd I’m stymied. HALP!!! ---
**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:** [October 31, 2012, 11:25pm UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/2 "2012-10-31T23:25:46Z")
Use 9v batteries. Then, for #1, which takes 5v @ 80mA, use Ohm’s law to determine the dropping resistor value: V = IR V = 9v-5v= 4v I = .08A R = V/I = 4v/.08 = 50Ω Same for LED #2, but that’s 20mA, so V = IR R = V/I = 4/.02 = 200Ω Same for LED #3 R = V/I = (9-2.1)/.03 = 230Ω ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [November 1, 2012, 12:27am UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/3 "2012-11-01T00:27:20Z")
what **beowulff** said. And you need a current limiting resistor in the circuit if you want the LEDs to last more than a few minutes. some of those cheaper keychain flashlights use coin cells (typically a pair of CR2016s) and rely on the cells’ internal resistance to limit current. ---
**Author:** ![Michael63129](https://avatars.discourse-cdn.com/v4/letter/m/3da27b/32.png) [@Michael63129](https://boards.straightdope.com/u/Michael63129)\ **Post date:** [November 1, 2012, 3:21am UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/4 "2012-11-01T03:21:58Z")
Putting a resistor in series with a flashing LED isn’t the proper way to limit current, because they have a built-in chip which flashes the LED and when it is off very little current is drawn, so it sees the full supply voltage, possibly damaging it; in your case, it appears that it is intended for 5 volts only (some will specify a range); you want a 5.1 volt zener diode across the LED, or perhaps a 3.9 volt zener in series (no resistor needed). Not that it really matters at this point, I guess, given the time. Also, for the two green LEDs, it would have been better to put them in series, with a 160 ohm resistor, then they would draw only half the total current. In fact, if you had four green LEDs, you could have been fancy and put them in parallel with each other (91 or 100 ohm resistor in series with each LED, then each resistor+LED combination in parallel) and then put that in series with the 5 volt, 80 mA LED, so current draw would be 80 mA total, with only about 2 volts “lost” across the resistors (note that you wouldn’t want to do this with two flashing LEDs since they will interfere with each other, but you could put ordinary LEDs in series). ---
**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:** [November 1, 2012, 12:11pm UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/5 "2012-11-01T12:11:16Z")
> [@Michael63129](#): > > Putting a resistor in series with a flashing LED isn’t the proper way to limit current, because they have a built-in chip which flashes the LED and when it is off very little current is drawn, so it sees the full supply voltage, possibly damaging it; in your case, it appears that it is intended for 5 volts only (some will specify a range); you want a 5.1 volt zener diode across the LED, or perhaps a 3.9 volt zener in series (no resistor needed). Not that it really matters at this point, I guess, given the time. Or… just use a 7805 and be done with it. 🙂 $2 at RadioShack, or $0.35 from Jameco. ---
**Author:** ![Johnceniyo](https://avatars.discourse-cdn.com/v4/letter/j/839c29/32.png) [@Johnceniyo](https://boards.straightdope.com/u/Johnceniyo)\ **Post date:** [November 2, 2012, 6:10am UTC](https://boards.straightdope.com/t/wiring-leds-to-typical-batteries/639491/6 "2012-11-02T06:10:24Z")
you need a present restricting resistor in the routine if you want the LEDs to last more than a few minutes. some of those less expensive key chain lights use money tissues (typically a couple of CR2016s) and depend on the cells’ inner level of ability to resist restrict present.