Charging a capacitor with an AC source

Status
Not open for further replies.
P

PhaseShift

Senior Member
With a circuit consisting of an AC source and a capacitor the capacitor will charge and discharge with the alternating source voltage. Is this correct?

What is a situation where the capacitor can build up a voltage field higher then the supply voltage? I have read information where the capacitor voltage field can keep increasing until insulation or dieletric breaks down?
 
Can you be more specific about what you read - it will not charge higher than voltage applied.

If you apply higher than rated voltage you will have insulation breakdown.
 
There is a circuit that uses caps called a "charge pump" which is used to "pump" the voltage to a higher or lower voltage than the source. This is done by charging the cap to the source voltage level, then disconnecting it and reconnecting it so that the lower voltage terminal of the cap is connected to the higher voltage terminal of the source. This effectively doubles the voltage at the higher voltage terminal of the cap vs. that of the source. By cascading such circuits fairly high voltages can be achieved. This is commonly used in integrated circuits where higher (or lower) voltages with low current requirements are needed (such as with flash memories). The "pump" circuits can also be designed to use an AC source, though the output is still generally DC.
 
100628-2107 EST

A series or parallel resonant circuit when appropriately excited can have larger voltages on the capacitor and inductor than the excitation voltage.

An automotive ignition coil is a simple example if you consider the 12 V battery as the excitation source.

If you take a low impedance AC source of a low voltage and connect across this source a series circuit of an inductor and capacitor tuned to the frequency of the voltage source and use low loss components for the inductor and capacitor, then the voltages across the capacitor and inductor will be greater than the source voltage.

Using a Signal Transformer 241-6-20 primary as the inductor and a 1 mfd series capacitor I get resonance at about 170 Hz with 0.083 excitation voltage resulting in about 0.33 V across the capacitor, and about 0.35 across the inductor. Thus, the inductance is about 0.85 H. A transformer like this is fairly lossy. On my LRC bridge the inductor measures about 0.6 H at 1 kHz. Also the inductance varies considerably with voltage level. With the oscillator as the signal source I adjusted the its voltage to approximate the voltage on the coil when connected to the bridge.

.
 
sometimes scratching a AC lead across a cap can cause it to charge to couple thousand volts, I have done this many times leaving them on my work bench to stop other workers from messing with my stuff, but this is a bi-poler cap, not a polarized cap, a polarized cap can be used in a series of caps and diodes to produce a doubling and tripling effect at low currents, this is how most all CRT HV supply's worked, buy taking the 9kv output of the fly-back transformer, and tripling it up to 25kv or quadrupling it to 35kv. as to why a bi-poler will react like this, I have long forgot this answer but if I'm thinking right it has to do with something like inductive kickback, but it will work, hook up one lead then just scratch the other lead on the other terminal with a DVM or a 20k ohm per volt meter, set to 1kv range and it will max out the meter, I have gotten one as high as 3kv before, most caps have a bled resistor and wont allow this without removing it.
 
Last edited:
If you have a series LC circuit, being driven by an AC source at it's resonant frequency, the voltage in the capacitor can rise above the supply voltage. Here is a very basic description:
http://www.allaboutcircuits.com/vol_2/chpt_6/3.html

Unwanted resonant voltage rise can occur in circuits with series capacitors and inductors (or wiring or transformers with sufficient inductance), if the values of each cause the circuit to be resonant at 60hz.

Resonant rise is used to great effect in radio transmitter circuits, Tesla coils, pulsed lasers, etc.
 
There is also the basic power supply. You take AC, fully rectify it to DC, and the peak to peak voltage is 1.4 time the RMS value. So a 120VAC source could easily be turned into a 170V DC source.

The AC waveform, however, spends very little time at 170V (it is the peak of the wave). So use much current and the capacitor will quickly drain. But it will recharge quickly as the voltage drops since more and more of the waveform is at or above the voltage of the capacitor. This is what makes the output voltage ripple. A huge capacitor helps, but can only do so much.
 
You might also be thinking of a 'restriking fault' in an ungrounded system.

In this situation, an arc conducts for only part of the AC cycle, and the fault in combination with the inductance of the wires themselves can charge the capacitance of the wires relative to grounded metal (eg the conduit). Eventually this can lead to insulation failures.

The key in all of these is that you have more than a capacitor and an AC voltage source. You also need inductors or switching elements or something else.

-Jon
 
Status
Not open for further replies.
Top