cap voltage

[J Rindt]

This has bothered me for a while.......
I can look at a schem, and I See/Understand how a cathode cap charges and discharges...same thing for the power supply caps. I see how they charge and discharge, and can follow the path through the schem.
But the coupling caps between gain stages...I do not get it. I can see where one side would see the plate voltage rise and fall as guitar signal hits the grid. But I do not see how the other side ever sees any AC.
If you understand what I am asking.....what is the charge and discharge path for each side of a coupling cap...like in this Princeton.....the two gain stages that follow the Tone Stack.
Where/How is that 0.022 charging and discharging.?
Thank You

[eleventeen]

It's not so much charge/discharge, it's more (a different functionality/use of caps, in general) "pass AC/block DC". Frankly I believe this would a better description of bypass caps under a cathode. If I feed a sine wave to one side of a coupling cap and charge it up to +10 volts on a voltage peak, then the discharge path is right back where those 10 volts came from when the sine wave reverses, falls below 10 volts, and then actually goes negative (assuming for this mental picture that the "zero crossing" is zero volts) Standing on a 12AX7 preamp plate in a Fender, I might have 180 volts of DC, then with a 10 volt peak-to-peak signal being produced at the plate, 185 volts, followed by 175 volts.

[J Rindt]

I understand what you are saying (i think) but how does THAT translate to AC changing on The Other Side of the cap.
If the plate side charges up 10 volts, and then discharges as the plate voltage falls, how does that "pass" AC on the other side, the grid side, of the cap.?
Can you humor me.?.......Explain it like I am 17 years old, and taking my first electronics class.
Thank You

[PRR]

> a cathode cap charges and discharges...

No it doesn't.

It charges-up when you turn-on. It discharges when you turn off. In between it holds the "same" voltage, not varying. (OK, if you pass bass tones too small for the cap, there is some charge/discharge.)

Same for the coupling cap. When you turn-on, one side goes up to 200V, the other side tends to zero V. In a second or so, the cap is holding a charge of 200V, the difference. It holds that all the time you are playing.

When you play, the plate swings *rapidly* up to 201V, down to 199V, and back (in emulation of the fine tones you play). This is passed-on, minus the 200V steady charge, to the next grid as a +1V/-1V swing.

Actually the cap voltage can change but *slowly*. If the wall-voltage sags 120V to 114V, the plate may drop from 200V to 190V. The cap voltage will drop to 190V. It will slowly accommodate changes of supply voltage, tube condition, resistor drift, so-that the output side tends to be zero V (as urged by grid resistor).

[shooter]

The non-engineer explanation I was taught for *seeing* the AC *thru* the coupling cap.  look at the schematic drawing of a cap   -----|  |------
The DC hits the vertical wall and stops, the AC swings Over the vertical wall, falls thru the gap, then swings back up from under the other vertical wall (sin wave).  The value of the *gap* (cap value) determines the frequency that can swing thru the gap to the otherside

Explained by a Navy Chief when I kept pestering him for the  *why?*
 

[2deaf]

Every time an electron shows up at one plate of a capacitor, an electron leaves from the other plate.  It take a change in voltage to convince an electron to move.  At DC, there is no voltage change and all of the electrons that are on one plate stay put so that no electrons leave or arrive on the other plate.  If voltage changes, electrons leave or arrive at one plate causing electrons to arrive or leave at the other plate.

[J Rindt]

Every time an electron shows up at one plate of a capacitor, an electron leaves from the other plate.  It take a change in voltage to convince an electron to move.  At DC, there is no voltage change and all of the electrons that are on one plate stay put so that no electrons leave or arrive on the other plate.  If voltage changes, electrons leave or arrive at one plate causing electrons to arrive or leave at the other plate.

Yeah.....sorry for my stupidity. I do not think the other guys understood the fundamental lack of understanding I was experiencing.
While the plate side charges with the falling plate voltage (its grid going more positive) the grid side of the cap is discharging, making it grid more positive and causing more current to flow through that tube.
That is how it mimics the phase of the preceding tube.
Thank You

[sluckey]

Something that always helps me see the charge and discharge paths for coupling caps is to first mentally take the tube out of linear analog operation and consider it as a digital switch, ie, it's either cut off or hard on. Fig. 1 shows the charge path with the tube cut off (like an open switch) and Fig. 2 shows the discharge path with the tube switched on hard (like a closed switch). When the tube is switched off the cap will charge to the full value of the B+ battery as shown in Fig.1. And when the tube is switched on the cap will completely discharge through the tube.

Now shake your head and go back to linear analog mode. The tube will neither be totally cut off or totally switched on. With no AC signal applied to the grid the tube will bias up to some steady state of conduction and the plate to cathode will appear as a fixed resistor. This tube resistance is in series with the plate load resistor R_A across the B+ battery. This creates a voltage divider and some steady voltage will be dropped across the resistance of the tube. Now the cap can only charge to the voltage level across the tube resistance. The cap can never charge to the full B+ voltage. This is the quiescent state for the circuit and the cap charge will not change as long as no signal is applied to the tube grid.

Now let's put an AC signal into the grid. During the positive portion of the AC signal the tube will conduct more, it's resistance will decrease, causing the voltage across the tube to decrease. The cap will discharge down from the quiescent level to this new level. During the negative portion of the AC signal the tube will conduct less, it's resistance will increase, causing the voltage across the tube to increase. The cap will charge up to this new level. This charge/discharge will continue as long as the AC signal is present on the grid. (Hope you noticed the phase inversion, ie, increase in grid voltage causes a decrease in plate voltage.)

Since R_G is in series with the cap, an AC voltage will be developed across it as the cap charges and discharges at the frequency of the AC signal applied to the grid. This is how the cap actually blocks the DC voltage and passes only the AC voltage. Whenever the cap is setting at a quiescent state, there is no charge/discharge current, therefore no voltage is dropped across R_G.

I know this is a simplistic way to look at this, but it works for me. I left out the cathode bias resistor just for simplicity. The important things to remember about this is that the resistance of the tube changes IAW the AC signal applied to the control grid and this causes the DC voltage on the plate to change. It is only this change in DC voltage that gets coupled to R_G.