# How does the signal go through a capacitor

Discussion in 'General Electronics Chat' started by nubi78, Nov 23, 2004.

1. ### nubi78 Thread Starter New Member

Nov 23, 2004
1
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Imagine an amplifier with the output being an AC signal riding on a DC voltage level. All of the text books I have read say you can put a capacitor in series from the output and feed the AC signal from the first amplifier into a second amplifier with no transfer of the DC voltage level. I do not understand how the energy flows across the capacitor in series in such a manner.

In my example, when the DC + AC output voltage from the first stage goes positive, the electrons pull away from that plate closest to the output amplifier. How does that interaction affect the other plate on the input side of the next amplifier?

I always looked at capacitors from a DC perspective where one end is tied to the positive side of a battery and the other is tied to the negative end. The positive end of the battery pulls the electrons away from the plate closest to the positive battery terminal. The negative battery terminal pushes electrons towards the other capacitor plate, giving you the storage capacity of the capacitor.

When you put the capacitor in series, I can see the pull of electrons when the first amplifier goes positive in its output voltage but do not see the push on the input side of the capacitor at the next stage. With that, I do not understand how the DC voltage level is not transferred through the capacitor and the AC voltage does.

Thanks for reading this!! All help is greatly appreciated!

2. ### bodhisatva Member

May 20, 2004
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When you are talking about the capacitor you are talking only about the plates. But you have dielectric between them. So if you charge one of the plates with positive charge you have corresponding negative charge on the other plate.

3. ### mozikluv AAC Fanatic!

Jan 22, 2004
1,437
1
hi,

here's how the coupling cap and a transistor amplifier interact. i'll be using a pnp transistor (Q), an electrolytic cap (CC) and a variable resistor (VR) acting as a volume ctrl.

at the very moment an audio signal is fed to the VR & let's also assume that the negative portion was initially introduced to the CC, electrons (as you already know) will accumulate on the plate directly connected to the VR. due to this electrons on the other plate w/c is connected to the base of Q, will be repelled and forced to move towards the base of Q. in effect increases the forward bias of the base of Q, hence the collector current also increases. as a result a bigger voltage drop develops across the collector load resistor. this voltage drop is the output signal of Q.

after the negative portion has passed thru, the positive portion of the signal enters and this makes the plate connected to the VR, positive. this positive plate will attract electrons at the opposite plate hence pulling electrons away from the base of Q. as a result the forward bias of Q is reduced, collector current also decreases and voltage across the collector load resistor is increased.

as the input signal varies the forward and reverse bias voltage at the base also follows, likewise to the load resistor. this variation of the voltage drop across the collector load resistor is the exact copy of the input signal variation. this is how audio signal flows in the circuit.

between two transistor with a coupling cap, the same principle applies. the output of Q1 acts as a switch for the base signal of Q2. the negative portion of the signal switches on the base of Q2 by providing a forward bias voltage, and vice versa for the positive portion.

to make the long story short the transistor is like a xerox machine duplicating the original signal as long as it is properly biased.

4. ### Martin P Dembrowski Member

Dec 3, 2004
10
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I made this little schematic hopefully clearing this topic up.

[attachmentid=268]

When the voltage swings across the load resistor of Q1 the swing will be reflected in the base circuit of Q2 since it is essentially parallel to the collector route to ground of Q1 through Q1's emitter, so any change across the load of Q1 has to be reflected across the base-emitter junction of Q2, am I right?