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Date:   Sat, 3 Dec 2005 07:41:33 -0800
Reply-To:   John Bange <jbange@GMAIL.COM>
Sender:   Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:   John Bange <jbange@GMAIL.COM>
Subject:   Re: Bright LED Cheap source & usage
In-Reply-To:   <6.0.3.0.0.20051203093215.04ca0ec0@buncombe.main.nc.us>
Content-Type:   text/plain; charset=ISO-8859-1

> > Here's a link to the same item on a > website: http://snipurl.com/kgds

Heh. I got one of these on ebay once. Man, what junk. Good source of parts, I reckon, but a god-awful flashlight.

Now my question. What size resistor do I need to run one of these LED > bulbs on the 12V system of the vanagon? I want to put one in the back > light of an aftermarket tach I have, which is very dim at night... >

Well, assuming .4A and 4.5V, it's a fairly simple application of Ohm's Law. First, you figure out how much excess voltage you need to "get rid of" (we're actually limiting current flow, but it helps to think in reverse): 12V - 4.5V = 7.5V then you plug the numbers into the formula R=E/I (where E is volts and I is amps). 7.5v / .4a = 18.75ohms So a 20ohm resistor ought to do the trick.

One of the problems, of course, is that we only have measured and obvserved numbers to work with rather than the actual specs for the LEDs. Close enough estimate though, I think. Someone correct me if I've made a gross error. It's early and I'm not awake!

-- John Bange '90 Vanagon - "Geldsauger"


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