L.E.D. voltage drop resistor

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Forum Name: Lights, Neon, LEDs, HIDs
Forum Discription: Under Car Lighting, Strobe Lights, Fog Lights, Headlights, HIDs, DRL, Tail Lights, Brake Lights, Dashboard Lights, WigWag, etc.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=53615
Printed Date: May 05, 2024 at 6:03 PM

Topic: L.E.D. voltage drop resistor

Posted By: DavenJo
Subject: L.E.D. voltage drop resistor
Date Posted: April 09, 2005 at 9:15 PM

This may have been discussed here and if it was I appologize, I would like to use a led for a indicator light and need the value of a resistor for my truck and how it would be wired in...Led volt...1.85v @ 20 ma. Would a 1/4 watt be big enough? Thanks

Replies:

Posted By: auex
Date Posted: April 11, 2005 at 10:30 PM

Yes 1/4 watt is fine. Anywhere from 400 to 820 will work.

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Posted By: icuppu
Date Posted: June 04, 2005 at 5:56 PM

Your dropping resistor value can be calculated with a simple calculator.

The formulae:

Resistor drop = ( System voltage - LED voltage) / LED current.

Spelled out:

The value of the resistor drop (current limiting resistor) is equal to the charging alternator voltage and not the battery voltage minus the L.E.D. voltage and then divided by the L.E.D. current will give you the required resistor value in ohms.

You can check for the charging voltage at the battery with a volt meter while the vehicle is running.

In your case you would input into your calculator in the following format; assuming you have 14 volt charging.

(14 volt - 1.85 volt) / 0.020 amp = 607.5 ohm resistor value

Hope this is of any use to you and good luck.

Francisco

Posted By: WILL85IROC
Date Posted: June 06, 2005 at 2:48 PM

I X E = P SO.... 1.8V X .020 A = 0.036W or 36mW

a 1/4W would be more than enough, good luck posted_image

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" WILL85IROC "

Posted By: xetmes
Date Posted: June 06, 2005 at 3:55 PM

WILL85IROC wrote:

I X E = P SO.... 1.8V X .020 A = 0.036W or 36mW
a 1/4W would be more than enough, good luck

Thats the power dissipated by the LED

The power burned off on the resistor is (using 14V like previous) is (14-1.85)*0.02=0.243 W, a little under a 1/4 W so it might get warm, you may want to use more than one LED or a 1/2 W resistor...

Posted By: Scuba_04
Date Posted: June 12, 2005 at 12:33 AM

Ohm's Law Man!

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