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I connected a voltage dropping capacitor in series with 220v AC supply and I measured the output voltage and it was 220 v too. I was expecting to get a voltage that is less than 220 volts. Did I make a mistake or the capacitor doesn't work ? I just want to test the capacitor before making the rectifier and load circuit.

One more thing, It is written on the capacitor 822j does that mean it is 8200 pF ?

It is about 0.9 uF when I measure it.

I know it is a dangerous circuit and I should use a transformer but I would like to make a transformer less power supply.

Thank you very much,

enter image description here

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    \$\begingroup\$ I VERY STRONGLY DISCOURAGE people from making their own transformer-less power supplies. They can be very dangerous, especially if you don't really know what you're doing. Why not use a transformer? You'd be MUCH safer! \$\endgroup\$ – DerStrom8 Aug 21 '15 at 13:20
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    \$\begingroup\$ Stop before you hurt yourself. \$\endgroup\$ – Spehro Pefhany Aug 21 '15 at 14:40
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Lets look at this another way. What is the voltage at the terminals of this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Will the voltage drop? Simple answer: no. Long answer: Well there is no load on it, so no current flowing, so by Ohm's law, no voltage drop across the resistor.

The same is true for your capacitor, there is no current flow, so no voltage drop across it's reactance.

Now what a capacitor will do in an AC circuit if you have a load on it, is to limit the current flowing, which it can do without dissipating large amounts of power like a normal resistor would - it stores and releases charge. That doesn't mean to say it won't heat up, there are losses in it which limit how much power it can safely transfer. It will also change the power factor a long way from unity which can be bad for whatever is supplying your mains voltage (increased losses in transformers and transmission lines, etc.).

Using a capacitor in this way means you would be building a potential divider circuit with whatever you connect, so it won't be a very good regulator. With little load the voltage will go up (no load = open circuit = full mains voltage), and if you put a large load the voltage will drop as more and more voltage is dropped over the capacitor (so high enough rated X series capacitor!).

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A capacitor "drops voltage" only if it is used as a voltage divider in conjuction with another impedance. You would have gotten the same result if you had used a resistor in this configuration. In this case, the second impedance is the input impedance of your voltmeter, which is probably 10MΩ or more.

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First of all: BE CARFEUL WHAT YOU DO HERE! 220 VAC (I assume 50 Hz) are no fun to get in touch with. You dont have a closed circuit here, cause your cap isnt connected to ground, so the 220 Volts are dropping all over the open clamps (infinite resistance --> maximum voltage drop). Please dont connect it this way because you're gonna blow it up and it might get nasty.

Edit: The calculation of the reactance is rather irrelevant, since as Dave Tweed mentions, the only power dissipated are for parasitic losses.

Maybe we can help you more, if you show the rest of your schematic.

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    \$\begingroup\$ A reactance does not "dissipate" power. Since the current is out of phase with the voltage, the net power flow is essentially zero, except for parasitic losses. \$\endgroup\$ – Dave Tweed Aug 21 '15 at 13:35
  • \$\begingroup\$ The output voltage depends on the load impedance, and he probably connected to a multimeter (high impedance), so the cap doesn't filter anything at all. \$\endgroup\$ – MdxBhmt Aug 21 '15 at 13:37
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In order to drop voltage, first the capacitor has to charge up. And in order to charge up it needs a current, which clearly that circuit doesn't have.

Just connecting the open ends would be a bad idea; put a resistor in series.

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