# LED Light Circuit

jtaekoJuly 20, 2013

Hi, I know this is a specialized post but was wondering if someone here could answer. It's pretty basic.

I want to make a DC (battery) powered circuit of 2 LEDs in parallel. The space for the battery is very limited. So I was thinking about using a 9v battery (a 1.5v won't last as long). But the LEDs I could find are usually around 1.x volts. So I know I have to use appropriate resistors before each LED. I was wondering if the difference between the battery and the LEDs is large, would this be advisable? Like if the 2 LEDs were 3v combined and the battery was 9v. Would the 6v difference that the resistors had to reduce voltage by drain the battery more than if I could find LEDs that combined were closer to the 9v battery? I don't know much about how this works. Thanks!

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Ron Natalie

Without getting into an explanation of how diodes work electrically, you need to limit the current flowing through the diode to not exceed what it is rated for lest it will burn out.

A typical (red lets say) LED has a 1.8v voltage drop and wants to be drive with 2ma of current (look up the specs on the diodes you wan to use). That means you need to size the resistor so that it flows no more than 2ma. Since the LED voltage drop is constant (for our purposes) the resistor must drop 7.2 (9v - 1.8v) volts.

Ohms law E=IR. 7.2 = .002 R
R = 3600 ohms

Now the only downside to using a battery bigger than 1.8v is that the resistor is going to dissapate some power (as heat). Twinkle, twinkle, little star, power equals i squared r.

P = .002*.002*3600 or 0.0144 watts. Not too much.

July 20, 2013 at 10:30AM
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yosemitebill

Ron's explanation is correct, however, a typical 5mm red LED current is actually 20ma - not 2ma. So, using the same math the current limiting resistor becomes 360 ohms. Standard 10% resistors in that range come in 330 and 390 ohm values - so you'd use a 390 ohm resistor in this case.

July 20, 2013 at 12:17PM
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