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Author: Subject: Resistor to Drop from ~12v to ~9v for LED Cluster
ko_racer

posted on 23/8/07 at 03:10 PM Reply With Quote
Resistor to Drop from ~12v to ~9v for LED Cluster

I know I should be able to do this myself (ohms law) but haven't done it for ages.

I have an LED cluster (well 2 actually) which has the following spec:

Typical Voltage 9.6v (11.2 Max)
Power Dsipation 2.5w Max
Forward Current 250mA

I want to use this from my car battery which can range from 12v to 14v.

Can anyone suggest how I work out what resistors (and what power rating) to use or a suitable voltage regulator?

Thanks

Keith

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smart51

posted on 23/8/07 at 03:21 PM Reply With Quote
It isn't quite as simple as ohms law as LEDs are not ohmic.

if you want to lose 4V at 250 mA (13.6V alternator voltage to 9.6V nomnam) then 16 ohms would do it. There is a danger though, 250mA is with all the LEDs on. With most of them off, the current would be less and so would be the volt drop. Your LEDs might be damaged. Best to fit a regulator and heat sink.

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02GF74

posted on 23/8/07 at 03:34 PM Reply With Quote
wot he ^^^^ said except for the truning the LEDS off.

I donlt know how the LEDs are wried up but am guesing it is some kind of lamp with two terminals. Then all the LEDS will be in parrallel* with their own current limiting resistors.

* you can connect one or more LEDs in series as long as thay have a reisistor but you do not connect LEDS in parrel and have one lresitor for both.

anyways, I digress. The only way you would be able to turn on any of the LEDS is by supplying a low voltage but the reistor will limit the curretn, if it is too low then no current flows (see voltage current graph for diode).

So I reckon all you need to do is to drop the extra voltage via a resistor - as ^^^ says, 16 ohm shold do it but a rough calc shows it is be quite a big one physcially to dissipate the heat. (1 W)

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smart51

posted on 23/8/07 at 08:54 PM Reply With Quote
What is the LED cluster for? Is it just a lamp or is it some sort of information display? 9.6 and 11.2V are very odd choices. I'd expect 9V or 12V if they were just lamps with series resistors.
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RazMan

posted on 23/8/07 at 09:10 PM Reply With Quote
Probably no help at all but I used a 2k5 resistor in series with my 6 Savage switches to bring down the brightness at night (they are suprisingly bright) but I think 3K would be better.





Cheers,
Raz

When thinking outside the box doesn't work any more, it's time to build a new box

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Stott

posted on 23/8/07 at 09:21 PM Reply With Quote
Here ya go:

This may help, enter your LED/Potential Divider parameters in the dark orange bits and out comes the answer in the dark green bits, easy!

Stott

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