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Registered:
July 2002
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 Re: need help from u electronic ppl
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Wed, 20 November 2002 15:03
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the wattage would actually tell you what bulbs to use, but I'm sure the wirings would tolerate anything around 5W anyway, but if you want to use LEDs, you'll need a resistor as Roger mentioned,
Because the dash usually has 12v on it while most LEDs uses 1.2v, you'll need to work out / lookup the current (mA) your LED drains, then minus 12V from the voltage of the LED, eg. 1.2V, for example if your LED needs 1.2V, 50mA...
then the difference in voltage is 10.8V, the LED needs 50mA which is 0.05A, V=IR, which means R=V/I, which is 10.8 / 0.05, which is 216 ohms, that means you'll need a 216 ohm resistor or anything close to that..
or you can work it out the other way around, ie. LED 1.2V 50mA that means the LED's working resistance is 24 ohm, and you'll need 12V-1.2V of voltage drop which is 10.8, 10.8/1.2 means 9 times the voltage drop of the LED, so 9 x 24 ohm = 216 ohms
so work out your resistor using one of the above method (whichever method u understand more, you'll get the same result) and make sure you solder/join the resistor in series..
and even if you've got a regulated supply of say 1.2V, make sure you still get a resistor in line with the LED, as I think I remember, the LEDs will "decrease" in resistance as it gets hotter, so it will continue to drain more current and gets hotter until it destroys itself, so a resistor will regulate the current, and prevent the internal of the LEDs to get hot..
btw.. not all LEDs are "brighter" it's just their light output is more "concentrated", that would also mean, if you're going to be using them behind your dash, you'll need something to "diffuse" the light output of your LEDs (eg. a roughen piece of clear perspex)
almost forgot, the rating of the resistor to use can be worked out by.. 12v x 50mA = 0.6, slightly more than 1/2w, so you can use a 1w resistor etc. (using 12v assuming the resistor is taking the full load, you could use 10.8, but it's just a safety margin)..
P.S. As I work out the above, it seems like 1W resistor is too big, I think most LEDs drains 20mA, if that's the case, then 12V x 20mA = 0.24, which is less than 1/4W, if that's the case, then you could use the cheap $0.01c 1/4W from dick smith..
I hope the above clearly answers your question then I'm happy that I've helped someone..  I hope the post wasn't too long either! 
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