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L.E.D. voltage drop resistor

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Member - Posts: 1
Member spacespace
Joined: November 30, 2003
Posted: April 09, 2005 at 9:15 PM / IP Logged  
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
Platinum - Posts: 5,041
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Joined: December 23, 2002
Location: Arizona, United States
Posted: April 11, 2005 at 10:30 PM / IP Logged  
Yes 1/4 watt is fine. Anywhere from 400 to 820 will work.
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Member - Posts: 1
Member spacespace
Joined: May 19, 2003
Location: United States
Posted: June 04, 2005 at 5:56 PM / IP Logged  

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.


Member - Posts: 41
Member spacespace
Joined: September 15, 2004
Location: United States
Posted: June 06, 2005 at 2:48 PM / IP Logged  

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

a 1/4W would be more than enough, good luck L.E.D. voltage drop resistor -- posted image.

" will85iroc "
Silver - Posts: 586
Silver spacespace
Joined: May 18, 2003
Posted: June 06, 2005 at 3:55 PM / IP Logged  
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 L.E.D. voltage drop resistor -- posted image.

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...

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Member spacespace
Joined: June 12, 2005
Location: United States
Posted: June 12, 2005 at 12:33 AM / IP Logged  
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