Step 7: Circuit time!
The idea is the 555 timer generates a square wave which charges up the capacitors in the on phase. In the off phase, the capacitors light the LEDs, depleting over time and creating the fade out effect. Theoretically, during the on phase, the current used to charge the capacitors should detract from the LEDs, creating a fade in effect, but often the current is so large that this is nearly instantaneous. Alternatively, you can put in a switch that bypasses the pulse circuit (so that the ground from the LEDs goes straight to the mains ground) so you can go from pulse to steady on.
I more or less used capacitors and resistors that were lying around. You can customize the frequency using the 555 timer calculator. Alternatively, you could sub in a potentiometer for RB to get a variable rate.
The 12000 uF capacitor is just a bunch of large 12V capacitors wired together. You may need more or less depending on the number and rating of the LED strings to get the right effect.
Test out the circuit on a breadboard connected to the car's electronics before putting it onto a PCB. And then test it when you've soldered it before boxing it up.
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And that's my question, i have installed neons and not leds. can i somehow connects your "pulse generator" to my neons to have that effect?
thanks
thanks
'Points' closed, voltage will be close to zero. Period of time goes by and the points open. Voltage immediately spikes to 100 or more volts. This is due to the same reason the voltage on the secondary suddenly jumps to thousands of volts. The magnetic field in the coil collapses. A turns ratio difference causes the spikes to be lower on the negative side of the coil than the secondary. The spike drops back down as fast as it rose. There can be some steps in this waveform caused by the breakdown of air/fuel charge on the spark plug. The voltage finally levels out at about 12 to 14 volts until the points close again.
Perhaps you could attach a voltage regulator to the negative terminal to keep it kicked down to a lower voltage? http://www.allaboutcircuits.com/vol_6/chpt_5/7.html
Or even easier. use the negative terminal to to kick a relay that would deliver the 12+ volts to the lights.