Capacitor coupling question

"jimi"

1. The capacitor has no charge initially.

2. When connected into a circuit at two points, it acquires a charge voltage equal to that found of across the same two points in the circuit before the cap was fitted.

1. AC voltages appearing on one side of the the cap are transferred to the other as if there is no impediment to them passing.

........ Phil

You seem to have a reasonable understanding there. Notably you understand that the signal voltage on the FET gate in your example will swing below ground potential as well as above ground.

Because the capacitor isn't charged by an AC signal. Once the bias conditions of such an amplifier have stabilised after being turned on, the voltage across a coupling cap is constant (aside from the very small AC voltage across it caused by any load current).

See above.

Graham

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Dear friends,

We always seem to know what is happening but are unable to express it. I too travel in the same boat. I try to explain what i understand.

First, capacitor doesnot short the ac completly. it will maintain some dc out of the ac signal. Friends please help to make this statement more precise.

What happens is : time zero: no voltage is there. a dc voltage is applied. capacitor resistance is zero right now. it starts charging to the dc voltage applied. the time it takes to charge depends on actual value of capacitor, the resistance and inductance it sees.

time x: a ac signal is sitting on dc. the capacitor resistance to the rising signal (or falling signal) is very low. so the rising / falling component seems to see lower resistance than the dc component and infact passes through the capacitor.

I think there is a better explaination than what i gave so i hope someone will make it more precise.

with regards

manoj

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I was shocked when I realised the signal went below ground! Even when it's a single ended supply, right? I still don't get how it centres itself though. I get that there's a steady voltage accross the capacitor with no signal present. Something like a music signal looks like random wiggles. After it passes through the capacitor how does it know where 'centre' is to go up and down from, if you know what I mean?

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I think I might have it :) Both sides of the capacitor are sitting at a different DC voltage like you said. For example lets say the input side of the capacitor is at 5 volts. A signal comes along that makes the voltage rise to 5.5v then back to 5v. The signal is fast enough that the capacitor passes it with little resistance. The other side is at 3 volts. What appears there is a signal going in a positive direction raising 3 volts to 3.5 and back to 3. The same happens in a negative direction so the signal appears centered on 3 volts or at whatever voltage is on the other side. Yes?

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Yes.

If you charged it up to some supply and then disconnected the "ground" end and moved the "supply" end to ground, then the disconnected end would be at *minus* supply volts.

Robin

"jimi"

1. An AC signal has no DC component.

2. An AC signal averages to zero.

Not all varying signals are genuine AC.

...... Phil

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You've got it.

The reason it works like that is because the coupling capacitor has a large value, so the signal cannot change the charge (and hence the voltage) on the capacitor appreciably.

In single supply applications like your example, the large capacitor is the reason we get 'turn-on/off 'thumps' from the circuit as the initial charge on the capacitor is established (and removed at turn-off).

Graham

And indeed that is exactly how some simple low current negative supply generator ICs work.

Graham

generator ICs work.

Thanks, what you said kept running around my head until it popped out and I got it. What a relief! LOL I just had to get this. It seems so simple but I just couldn't see it. Jimi