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Old 6th February 2008, 10:19 PM   #1
qguy is offline qguy  Canada
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Default Reducing LED intensity

What value of resistor (or a range of from 1 to 100 so that i can play with it) should i put to reduce the intensity by half of those 3mm, 2 volt LED- thanks
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Old 6th February 2008, 10:23 PM   #2
jol50 is offline jol50  United States
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What is your supply voltage? Most just play around with some to get it the way they want. Also if your supply changes with load consider that. I don't know if 1v would give half light output, I doubt it.
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Old 6th February 2008, 10:35 PM   #3
qguy is offline qguy  Canada
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Have not checked the voltage, but should be under 4v I cant change the power supply as the Amp is not a DIY, its a commercial amplifier. Blocking part of the light is also not possible for aesthetics reasons, so a resistor in series is my only option
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Old 6th February 2008, 10:46 PM   #4
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Play with the current trough the LED.

10 mA is a good starting point...
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Old 7th February 2008, 05:14 AM   #5
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10ma is about 220 ohm. 5ma would be 430 according to a calculator on the net at 4v supply and 2v LED. So I guess I would get a 220 to 430+ ohm pot if possible, if the max current is 10ma. Or you know, put a 220 ohm on it and then a pot from 0 up to dim it. http://www.luxeonstar.com/resistor-calculator.php
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Old 7th February 2008, 11:45 PM   #6
qguy is offline qguy  Canada
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Cant I just put a 500 ohm pot w/o a resistor ?
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Old 7th February 2008, 11:53 PM   #7
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Put a 500 Ohms pot in series with a 100 Ohms resistor...
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Old 7th February 2008, 11:58 PM   #8
jol50 is offline jol50  United States
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Depending on max ma it can take, if you go lower ohms you could over power it.
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Old 8th February 2008, 12:04 AM   #9
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Quote:
Originally posted by jol50
Depending on max ma it can take, if you go lower ohms you could over power it.
Usually a LED can take 20 mA without any worries ...
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