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### NEED HELP WITH LEDS!

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### NEED HELP WITH LEDS!#119159

By LEDquestions
#119159
I have never used LEDs before and am working on making a greeting card for an electrical engineer. I bought 2 Red LED Assemblies from RadioShack (product #276-0084) and I do not know what type of battery that I can use. I want to make something very simple, similar to a throwie, but I brought an LED with a built in resistor because I think resistors are supposed to be important? The information about the LEDs that I have:

"Red LED with holder and wire leads. Built in 1/4 W, 680-ohm reistor for 12VDC operation, 1/4" (6 mm) lens diameter, 2.78" overall length, including wire leads, 13/6" mounting hole.

Absolute maximum ratings:
Power dissipation: 180 mW
Forward Voltage: 12V
Forward Current: 15mA

Can I use one of those coin type batteries, like CR2032? Do I have to use a 12V battery? I want to use the smallest batter that I can, as I am making a greeting card.

Thank you to anyone who can help me.

### Re: NEED HELP WITH LEDS!#119168

By coyote20000
#119168
Try the coin cell, see what you get. It will either not work or be very dim.

First off you have two things going against you.
The resistor and the fact that the cell is probably 1.5v (Guessing).

If you get plain red leds, you could go with two coin cells in series (3.0v)
then limit the current to 15mA or whatever the normal forward current rating is.

(SupplyVoltage-LedForwardVoltage)/LedForwardCurrent=CurrentLimitingResistor
(3.0-2.0)/15mA=66 Ohms

Dave

### Re: NEED HELP WITH LEDS!#119221

By AndyC_772
#119221
If the LED is designed to draw 15mA @ 12V, then we can check the approximate value of the built-in resistor...

The LED alone will have a forward voltage of about 2V if it's a red one, more or less independently of the current. So that leaves the resistor to drop the other 10V at 15mA.

R=V/I = 10/0.015 = 667 Ohms

(I see they quote 680 Ohms, which is close enough... at least their specs match up!)

A CR2032 delivers about 3.0V, so that's 2.0V for the LED and just 1V across the resistor. So your current is:

I=V/R = 1/680 = 1.5mA

This will be very, very dim.

For what you're doing I'd suggest a thinner battery - say, a CR2016 - which still gives you 3V but just won't last as long. (I'm guessing you don't care too much about this). Then you want a standard LED without an integral resistor, so you can add your own to get the correct brightness.

I'd suggest a current of 5mA for reasonable brightness, so the resistor you need is given by:

R=V/I = (3-2)/0.005 = 200 Ohms

### Re: NEED HELP WITH LEDS!#119224

By angelsix
#119224
If any part of the LED legs are exposed at the LED head end (so above the resistor) you could try to get a wire connection there, and jumper it over the resistor back to the bottom of the leg, then you are bypassing the resistor all together, so a 3V cell without any resistor will be just fine although if you are concerned about blowing it instead of a wire jumper, use another resistor, about 80R or whatever is closest available, which will give you about 71R series resistance (1 / (1/680) + (1/80)).
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