capacitorpower supply
Just got a (220<–>12-0-12) transformer hooked up with a bridge rectifier and it measured 13 volts DC output from rectifier, but when I added a 1uF capacitor it just jumped up to 20 volts, and same reading(20 volts) from a 0.1uF capacitor, how is that even possible!?
Note: Nothing is connected in the circuit more than a transformer and a bridge rectifier in the first case, and only a capacitor added in the second case along with the voltmeter.
Appropriate bleeder resistors are the usual solution. They aren't usually switched, although they can be.
The value depends on the time you require to discharge the capacitors. The formula is
$$ V_{t} = V_{0} \, e^{ -t / RC } $$
where \$V_{t}\$ is the voltage at time t and \$V_{0}\$ is the initial voltage at time 0. It's an exponential function, so I'd just assume 1/10 of the initial voltage.
It isn't a power function, as someone edited it!
You should find that the power taken by the bleeder resistors is negligible compared to the 120W capability of the supply.
Lets look at this another way. What is the voltage at the terminals of this circuit:
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!).
Best Answer
\$V_{rms}\$ vs. \$V_{peak}\$. The peak voltage is \$\sqrt{2}\approx 1.4\$ times larger than the average (RMS) voltage.
If you put a capacitor on a rectified AC waveform it will smooth out the supply. If there is no load, it will smooth it out to around the peak voltage of the supply.
A 220V AC supply is 220V RMS, which is equivalent to 311V peak. If you bring that down to 12V AC, that is equivalent to about 17V peak.
As to how you got 20V, either your meter is dodgy (unlikely) or the supply voltage is higher than you thought, or the transformer is not the ratio you think it is.
If you add a load, the voltage will drop because the average voltage supplied is the RMS value - the capacitor can't sustain a current at the peak voltage because that would require it to deliver more power to the load than is being delivered by the transformer.
That the rectified supply is varying between 0 and \$V_{peak}\$ with the average voltage being \$V_{rms}\$. If you are drawing a current, the capacitor will smooth this out to be around \$V_{rms}\$, but if you aren't drawing any current, then it will keep getting "topped up" with charge until it reaches the peak voltage. As a demonstration of this, try putting, say, a 1k resistor at the output of your supply, you should see the voltage drop.