Re: <ALL> Power window switches

[Michiel van Wessem <jmvw@xxxxxxxxx>]

Hi Kwok,

It's possible and easy and you'll never have to replace bulbs again. As far
as I know, it's believed that LEDs will last at least 100 years. By that
time fossil fuels will be exhausted so you'll be driving a new car.

I know a simple way to do it. First find a LED that has a good color and
the same light output as the incandescent bulb you want to replace. Light
output is given in lumen. For the LED this will be given in its specs, but
you may have trouble finding out the output of the bulb. Some experimenting
may be necessary here. You'll need some kind of high power LED, possible
several in parallel.

Then you need to know the voltage drop over the LED and the current that
should go through the LED. These are given in LED specs.

You'll need to take up a resistor in series with the LED to limit the
current through it. The schematic is like this:

      _____      _____|\|____
  12 v     \/\/\/     |/|    |
         resistor    LED     |
ground_______________________|
                   

To calculate the value of the resistor you need use the following formula.
I'm assuming the car's voltage will be 14 rather then 12 Volts, because
sometimes it will.

R = (14 - Vled) / I

where R will be the value of the resistor that you need, Vled is the
voltage drop over the LED (given in its specs) in volts and I is the
current that should go through the LED (given in its specs) in amps.

Resistors come in certain values. Buy a resistor that is close to what you
got out of the formula. 

Resistors also come in certain wattages. They can be 1/8 Watt, 1/4 Watt,
1/2 Watt and so on. Use this formula to see what you need:

P = (14 - Vled) * I

Where P is the power in Watts. Get a resistor that can take more then that.

Example: LED's voltage drop is 1.2V, current is 20mA. R is (14 - 1.2) / .02
= 640 Ohm. I think the closest resistor to that that you can buy would be
620 Ohm. Power consumption is (14 - 1.2) * .02 = .256 Watts. A 1/4 Watt
resistor should be thus be good enough. 

Final thing about carbon-film resistors is tolerance. In this case, we
don't give a sh!t about tolerance. :)

It's important to connect the LED the right way. The cathode is the side of
the LED that should go to ground and has a longer lead in general. If you
switch polarity on the LED, you'll probably blow it up.

If you use two LED's in series, you get a schematic like this:

      _____      _____|\|______|\|____
  12 v     \/\/\/     |/|      |/|    |
         resistor    LED1     LED2    |
ground________________________________|
                   

The current through both LEDs will be the same (get LEDs that need the same
current, i.e. the same LEDs). For the Vled in the above formulas use:

Vled = Vled1 + Vled2

Where Vled1 obviously is the voltage drop over LED1 and Vled2 the drop over
Vled2. If you got the same LED's Vled1 will be the same as Vled2, in which
case Vled = 2 * Vled1.

Similar for 3 LEDs and so on.

Extra information: resistors are color coded. You can find the conversion
tables on the web. I saw one at http://www.ttk.com/TriaUtil/resistor.htm

Please let me know if this description is understandable or totally
incomprehensible. I'm pretty sure there are several people around here who
know much better what they're talking about then I am, but this works for me.

Good luck!

Michiel

At 05:48 PM 1/27/99 -0800, Kwok Seto wrote:
>To all digesters,
>
>Could anyone who knows electronics fairly well tell
>me if it is possible to convert a 12v incandescent bulb
>power source to "light up" a LED instead.  The reason I'm
>asking is because I have a older style power window
>switch (1993 318is) that uses a incandescent bulb 
>rather than a LED.  Any help on this topic is much
>appreciated...
>
>thanks
>Kwok 
>  
>
>

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