voltage reduction

i have an old car phone that i have fitted into a jaguar of the same era, i wanted it to light up when the ignition is on so i replaced all the leds in the habdset with 12 volt leds and connected it, it looks just right, but at night its just to bright, what would be the cheapest method to drop the voltage on the supply side to, say, 10volts.
i am a complete idiot when it comes to electical tech talk so please use plain english,
as i am pretty good at parrot fashion. :>)

U need a voltage divider to drop 10v from 12vdc

Cheapest method would likely be to place a resistance in series with the total LED load depending on the LED configuration. Do you have any data on the LEDs used other than them being 12 volt? Another method would be to do what is called PWM on the LEDs. Either of those methods would be inexpensive but the total current load of the LEDs needs known. The switching between light and dark would be manual unless you want to get more complex and add a sensor for light.

Ron

i think its sorted out i have fitted a Standard Potentiometers with Switch
Power rating: 0.4W linear, 0.2W log.
Max. voltage: 500Vdc.
Tolerance: ±20%.
Switch rating: 4A at 250Vac.
i dont know what these numbers mean but it is controling the light very well and not heating at all, could i have chose a better suited pot. for the job?

If it works it should be fine. Again, less the LED requirements as to current and exactly how you have how many configured it is hard to say. If it works and the pot runs cool then I would say run with it.

Ron

thanks guys, for your fast input and, appreciated, advice.
rex647
uk

I'm not sure 0.4 W will be enough to do what you need. The track in the pot may burn out. You should use a fixed resistor rated for more watts. I suspect 240 ohms to 1k will be enough. Get ones with a power rating of about 1W or more.

P.S. With LEDs, you are not looking to reduce the voltage, you are looking to limit the current. For example if you have 5 LEDs rated for 20mA each, then total current would be 0.1A to light them normally. Approx half brightness will need 0.05A (50mA). Using ohms law we can calculate the resistor needed...
R = V / I

12V / 0.05A = 240 ohms

You can then work out the power rating needed needed like this..

P = I x I x R

0.05 x 0.05 x 240 = 0.6 Watts