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LED Delay Circuit

Started by alex192, August 02, 2010, 06:36:44 PM

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alex192

OK, I have just completed the Arbiter Fuzz Face (Ge) clone which I started a few weeks back, and it all sounds very good with the AC128s. However, I am having a slight issue with the LED that I have added.

To avoid the pop when the LED turns on, I have grounded the input to the circuit board in the off position, and also added a second resistor and a capacitor, according to these instructions:
http://www.muzique.com/lab/led.htm

I am using a bright blue LED, so I can afford to get away with a 33k resistor for the brightness, and I am using a 47uF cap. However, whenever I hit the switch for it to turn on, the LED suddenly lights up very bright, then decays to the correct brightness -- this is the reverse of what the delay circuit should do; the LED should light up slowly as the cap is charged, thus reducing the pop.

I think the problem with the method listed above (in the link), is that the ground on the other end of the cap is always grounded, and as such the cap is always fully charged, so that when I come to flick the switch to ground the LED, the cap dumps its charge through the LED, as opposed to being filled up. I have noticed that when I unplug the input jack (thus unearthing the battery and the cap), then plug it back in with the switch in the on position, the LED does slowly light up.

Would I be correct in thinking that the ground from the capacitor should be attached to the top right lug of the switch, so that it is not grounded when in the off position, but as soon as the LED is grounded, so will the capacitor and as such it will slowly light up?

Cheers for reading.

Also, is a pulldown resistor on the circuit needed if the circuit input is grounded at all times?

pjwhite

I think you have figured out the correct solution.  As drawn, the circuit will charge the capacitor to 9 volts regardless of the switch position.  When you do turn on the switch, the LED lights with 9 volts (minus the LED voltage drop) across the 390 ohm resistor, slowly diminishing until the capacitor has discharged enough that the 4.7K resistor (or 33k in your case) acts to reduce the current.
To solve this problem, the negative side of the capacitor should connect to the cathode of the LED, not ground.

alex192

Yeah, I just tried rewiring the capacitor to the switch and it works very well now. The LED lights up gradually (still quite fast though), and elimintates any pop. I don't really see the need to add the pull down resistor now.