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I`m trying to create a basic delay circuit using a capacitor. I am using a momentary switch in series with a 220 Ohm resistor, a H332 100uf 25V rated capacitor and an LED, all connected to a PP3 8.4V 200mAh battery. I've tested all of my components, with the exception of the capacitor, which I tried substituting for another, which also didn't work.

I am sometimes observing small, short, dim bursts of light from the LED when I`m messing with the power supply, but otherwise the LED remains unlit, even after I have been holding the switch closed for several minutes. My understanding is that the LED should be unlit whilst the capacitor charges, then constantly lit once the capacitor has charged up and then should the power supply be interrupted by the opening of the switch the light should still shine for a short time before the capacitor runs out of charge. I could be completely flawed in my thinking - I'm a self-taught hobbyist and this is my first time using a capacitor to create delays.

Am I wiring the capacitor up wrong? Is that an unsuitable rating of capacitor for a short delay? Am I just expecting the wrong result? If the capacitor rating is the problem can you recommend a more suitable rating for a delay of roughly a second? I`m just using what I have on hand for the while, but I've got an order of assorted capacitors that should be arriving soon.

Thanks.

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  • \$\begingroup\$ Electronics works mostly with schematics, not breadboard photos of devices with unknown properties. This site has a built in circuit editor to provide quick and simple schematics. \$\endgroup\$ – PlasmaHH Sep 21 '15 at 12:20
  • \$\begingroup\$ Try putting the capacitor in parallel with the combined LED and resistor instead of in series. The capacitor will act like a reservoir, which is more like what you were expecting. \$\endgroup\$ – HandyHowie Sep 21 '15 at 12:23
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When you first push the button with the capacitor discharged, a current flows through the capacitor, the resistor and the LED, and the LED lights up. That current charges the capacitor, and after about 0.1 s it will be fully charged, no more current will flow, and the LED will be off.
The capacitor will have no way to discharge, except for its internal loss, which can require a lot of time.
The next time you push the button, the capacitor is still charged, and no current flows.
So the behaviour that you are experiencing is exactly what you might expect form the circuit that you realized.

If you put the capacitor in parallel of the LED, you will see the LED remain on for a brief period of time after you release the button, and turn on whth a little delay when you push it.

If you want delays of approximately 1 second, you need at least a 1000 uF capacitor.

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