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What happens if I give DC to X2 type capacitor.

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Advanced Member level 4
Feb 5, 2013
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I have made AC indicating lamp, which can operate with min 63.5VAC to max 440VAC. But I want to make AC/DC,What happens if I give DC to X2 type capacitor....will it work?...PLEASE HELP:sad:

Okay.....My circuit is shown below....
will it work for DC??

Your circuit uses the reactance of the capacitor Xc to drop the AC voltage. It appears to the AC to be another resistor but doesn't heat up like a real resistor would. If you apply DC it will simply charge up, giving one brief pulse of current to your LED as it does so but after that, the only current to flow will be through the two resistors as Frank stated.


You can use an inductor in place of the capacitor because if has a reactance to AC that will limit the current.

And to second-guess your next question ................. No it cannot be used for DC because the resistance of the inductor will be too small.

No, not constant current but you can use it as a voltage dropper in the same way as the capacitor and it will then pass DC. The problem is the size of the inductor which could be physically large and it could be quite heavy. The DC current will only be limited by the resistance of the wire which will almost certainly be too low and you have another problem - when you switch off, the magnetic field in the inductor will collapse and could produce a VERY high voltage, possibly several KV which could damage the LED and surrounding circuitry.

You can work out the values needed with these formaulae:

For a capacitor, the effective resistance is Xc = 1/(2 * pi * f *C)
For an inductor, the effective resistance is XL = 2 * pi * f * L

In both cases 'f' in in Hz, C is in Farads and L is in Henries. You can work out what value for Xc or XL you need using Ohms law, X = V/I where V is the voltage you want to drop (input - LED voltage) and I is the LED current.

They have two wires and look like a normal diode, all they do is pass a constant current regardless of the voltage across them. I believe internally they are the same electrically as JFETs with the gate and source pins joined together through a small resistance. As the current increases, the voltage across the internal resistor rises and chokes he FET off by making the gate more negative compared to the source.

I warn you they can be expensive and it may be difficult to obtain them for the current you need. This is why it is more normal to use other methods to generate constant current.

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