Re: Interior light timer/dimmer

On Sun, 11 Nov 2007 17:30:45 -0000, PCPaul <urd3@xxxxxxxxxxxx> wrote:

On Sun, 11 Nov 2007 17:24:27 +0000, Duncan Wood wrote:

On Sun, 11 Nov 2007 16:23:31 -0000, PCPaul <urd3@xxxxxxxxxxxx> wrote:

On Sun, 11 Nov 2007 15:17:47 +0000, SimonJ wrote:

Lots of these on the aftermarket which dim out the light after a set
period. But any which fade it *on* as well - like modern BMWs and
others? Or even a circuit of one? Lots of circuits for dimming out
ones found on Google, but not any for both on and off.


Just wire a capacitor in line with the +12V. That way you'll get both
fade in and out. You can set the time it takes using a variable
resistor. I think I'd be wanting at least a few thousand uF for the
capacitance and a 10k Ohm variable resistor.

Except of course that it wouldn't work.


Hardly rocket science.

Quite.

Let's see if I remember my electronics, it's a long time ago, but...

You'd want the capacitor to be wired across the bulb, e.g. from the
+12V to ground. That way the cap would initially act as a short
circuit, leaving 0V across the bulb, and as it charged the voltage
across it would increase until it was at +12 when fully charged. The
bulb would stay off at first and light as it charged up.

The delay then becomes the crucial bit. this would depend on the size
of the capacitor and on the resistance in the circuit - since you need
the bulb to be on full when the cap is fully charged I'd have the
resistor in series with the cap between the supply lines, and the pair
of them across the bulb.

You'd definitely have to play with the values to get a sensible delay.
And what it would do when the soft-off circuit kicked in I have no
idea...


But the resistor would be in series with the bulb & stop it coming on at
full. Or if the delay was more tham microseconds at all.


That would be the 'since you need the bulb to be on full when the cap is
fully charged I'd have the resistor in series with the cap between the
supply lines, and the pair of them across the bulb.' bit then...

I'd do ASCII art but it *always* goes wrong.

Basically imagine C+R in series, then take the pair and put them in
parallel with the bulb.

In which case there'll always be 12V acroos the bulb, wether the capacitors charged or not. You're going to need an active device.
.

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