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Old 05-16-2003, 05:42 PM   #1
chrisdeaner
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Question Quick LED+resisitor question

I replaced the bulbs in my autometer gauges with blue LEDs (3.7volts typical, more info here if needed). And 440ohm (i believe) resisitors, to match the 12V running in fairly close to the 3.7V the LEDs like to see. However, they aren't bright enough...so I want to add another LED to each gauge. What resistor should I use for this (2 3.7V LEDs = 7.4V (+/-) per gauge) to work on the 12V coming in? Also, what formula gives me the right resistance?
Thanks!
Chris
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Old 05-16-2003, 06:52 PM   #2
DrD
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you use the current draw of the LED, coupled with ohms law (V=IR) - you can calculate everything you need with the info you already have. (i.e., LED voltage + resistance required for 1 LED)
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Old 05-16-2003, 08:49 PM   #3
chrisdeaner
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Ok, so I have two LEDs running at 3.7V, total of 7.4V, and each draw 20 mA each, total of 40 mA. Incoming voltage is 12V, so V=IR of 12V=(40mA)(R) for an R of 300ohms? Does that seem right?
-Chris
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Old 05-16-2003, 09:12 PM   #4
PARANOID56
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Old 05-17-2003, 08:41 PM   #5
dtstyle
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depends on how they are wired. One method takes more volts, the other takes more amps.
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Old 05-17-2003, 10:52 PM   #6
chrisdeaner
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I was just going to do them in parallel.
-Chris
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Old 05-17-2003, 11:23 PM   #7
DrD
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Quote:
Originally posted by chrisdeaner
I was just going to do them in parallel.
-Chris
that's the more amps way - I think you want around a 207 ohm resistor if you have one resistor in series with the two LED's which are in parallel - or you could treat them as individual circuits and do each resistor-LED pair seperately (so you would have the two resistor-LED series combinations in parallel) in which case you would use the same resistor size you had started with for each one
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