Hi, I need to make a circuit which will make 3 LEDs (I'm using those 3mm blue LEDs) slowly dim off when I open the car door. Any idea or there is any schematic where I can see?
If you know the voltage drop of the LEDs and the series resistor then it's easy to calculate the current decay.
You'll need a pretty large capacitor though, assuming the LEDs are in series and the resistor is 150R for a current of 10mA, a 6800:mu:F capacitor will give an RC constant of just over a second meaning it'll take about 5 seconds to turn completely off.
If you know the voltage drop of the LEDs and the series resistor then it's easy to calculate the current decay.
You'll need a pretty large capacitor though, assuming the LEDs are in series and the resistor is 150R for a current of 10mA, a 6800µF capacitor will give an RC constant of just over a second meaning it'll take about 5 seconds to turn completely off.
You may use only one or two leds in each series (and a higher resistor), that will give yoiu more time with a smaller capacitor. Of course, you will have to connect those series strings in parallel
The capacitor will discharge from the supply's voltage to the led's foward voltage, with fewer leds you have more voltage difference and more discharge time.
[The extra power and current that will be used putting the 3 leds (and resistors) in parallel will be no problem - a few mA is little current for a car's barttery]
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Also, it would be a good idea to add a small resistor (like 10Ω) between the supply and the capacitor to limit the current surge when you turn on the leds
The problem with putting the LEDs in series is that they willl draw more current so you'll need a larger capacitor to provide the current for the same length of time.