coupling cap .....

[trem]

.....OH Boy.
I Really Wish it was not always ME that asked these dumb-shit questions.
But I do not know the Reason/Answer, so then I wonder.....and I end up here.
The books I have never address electronics at a level that is low enough for me to understand. The knowledge that the author(s) assume the reader will have is always greater than my own.

On a simple gain stage (two of them I guess) with no tone stack or anything in between...just the cap and that 1M grid leak...what happens with the current.?
I understand (I think) that when the plate side charges, the grid side of the cap evacuates electrons.
What happens when the phase reverses.?
Does the grid side of the cap "suck" current from the grid, and create some kind of grid current.?
And what about the plate side...if the grid side of the cap is "charging" and the plate side of the cap is losing electrons.....where do Those Go.?
Thank You

[Pdavis68]

I'll take a shot at this in the hopes that I understand your question correctly.

The point of the coupling cap is to block the DC from the plate of the first stage from going to the grid of the second stage while allowing the AC audio signal to pass through. Otherwise it would mess up the biasing of the grid of the following stage, which presumably is a few volts below the cathode. So it's not really related to plate current and grids don't pass current.

[trem]

I'll take a shot at this in the hopes that I understand your question correctly.

The point of the coupling cap is to block the DC from the plate of the first stage from going to the grid of the second stage while allowing the AC audio signal to pass through. Otherwise it would mess up the biasing of the grid of the following stage, which presumably is a few volts below the cathode. So it's not really related to plate current and grids don't pass current.

Thanks for the reply, but no...not what I am asking.
I Understand the purpose of the cap.
I am wondering where the flow of electrons is going to and coming from when the grid side of the cap charges.
Thanks

[DRH1958]

I'll take a stab at it. When an AC signal is present, the plate of that 100k resistor that has the 290v on it(pin 6) is considered at ground potential as far as AC is concerned. So the signal input charges the .022 cap and this voltage change appears at the grid of the next stage on pin 7. Then when the signal reverses, the cap discharges through the 100k resistor to ground(which is really the point where it gets the 290v) and this signal also appears at the grid of pin 7 as the opposite polarity of the first signal, thus driving the tube in the other direction. It really is a loop from the 290v supply up through the 100k resistor to the .022 cap, down through the 1 Meg resistor, through the 56k resistor to ground. Since the 290 supply is at ground as far as AC is concerned, this completes the path to ground at the other end. From what I understand it does this because it is in series with the filter caps and these provide the path to ground. That .022 cap charging and discharging is how the AC signal appears at the grid of the phase inverter(pin7).

[Jazz P Bass]

I think of it more as 'manipulating the charge'.

The signal itself is what is being passed on to the grid.

[trem]

I'll take a stab at it. When an AC signal is present, the plate of that 100k resistor that has the 290v on it(pin 6) is considered at ground potential as far as AC is concerned. So the signal input charges the .022 cap and this voltage change appears at the grid of the next stage on pin 7. Then when the signal reverses, the cap discharges through the 100k resistor to ground(which is really the point where it gets the 290v) and this signal also appears at the grid of pin 7 as the opposite polarity of the first signal, thus driving the tube in the other direction. It really is a loop from the 290v supply up through the 100k resistor to the .022 cap, down through the 1 Meg resistor, through the 56k resistor to ground. Since the 290 supply is at ground as far as AC is concerned, this completes the path to ground at the other end. From what I understand it does this because it is in series with the filter caps and these provide the path to ground. That .022 cap charging and discharging is how the AC signal appears at the grid of the phase inverter(pin7).

I am going over this now.
I THINK it answers most of my question.
And Sorry.....I forgot this schem was a voltage amplifier into a PI. I had meant to post a schem of one "simple" amplifier into another. But I think the explanation would be about the same.
I think my biggest wonder was when the grid side "charges". Do electrons flow from the grid, or the 1M resistor to the cap.?
Anyway, let me study this for a bit.
Thank You

[DRH1958]

Yep, I would agree with that. It is probably more correct to say reversing the charge rather than charging and discharging like I said in my explanation.. Because of this action, the signal is passed to the grid.
To answer your earlier question trem, no the current flow does not come from the grid. Any major grid current flow in preamps does not happen. Minor amounts maybe. Power amps are another story.

[trem]

....... That .022 cap charging and discharging is how the AC signal appears at the grid of the phase inverter(pin7).

Yes, I absolutely do understand the purpose of the cap, and how it works.
I was wondering about the path of travel for current, into and out of the cap, during each phase flip.
I think the rest of your post has set me straight on that.....Thank YOU.

[Jazz P Bass]

Man, you are entering some heavy territory.

Maxwell's equations nailed it when he united electricity, magnetism & light.

The charge on a capacitor is transfered by it's leads, from one side to the other side.
Remember, there is a dielectric separating the two plates.
What gets set up between the two plates is an electrostatic field.

In a circuit as you are talking about, with a Vac signal, the E Field becomes an electric displacement field.
EE's tend to think of the flow as a diplacement current.

I would imagine none of this helps, but you really entered a realm with that question.

Displacement current - Wikipedia, the free encyclopedia

[eschertron]

I think we're taking it farther than what the OP intended. Not an information theory analysis, or a first law analysis, but simply 'what is going on with the electrons' on the grid side of the cap kind of question.

If we accept that for preamp tubes under discussion that there is no current flowing in or out of the grid (as mentioned above), then the only other path for current is across the grid leak between the 'downstream' side of the cap and the ground wire. The electrostatic field (above, thanks JazzP) attracts and repels electrons that must come from the ground wire (actually sourced from the neg terminal of the preamp's power supply cap) through the grid leak. I haven't - recently, anyway - calculated the actual charge that accumulates in the coupling cap during any representative half-cycle; but it must be small. The current through the grid leak resistor is going to be uA.
JM2C - your charge capacity may vary

[trem]

JPB.....Thanks.
I had never heard that term before..."displacement current".
It is quite a bit above My Pay Grade, but I will peruse that link with interest.
Thanks Again

[nsubulysses]

This may be useful to you - https://www.youtube.com/watch?v=ppWBwZS4e7A

Especially starting at 8:11

You could study theory for years but really the best way to grasp what's going on is just to get a displacement current meter and probe the circuit

[trem]

Yeah.....will do.
Thanks Again Guys
I Appreciate It

[Chuck H]

Way to go eschertron That's the description "I" would have expected (except I didn't have the term "displacement current" in my cranium ). But this is how I've always conceptualized it. And a nice write up directed toward the basic question. Though I couldn't have exactly described it, I sort of got the concept after being introduced to how a "grid leak" triode stage works and it's pro's and con's. I'll venture to say that the grid leak (or grid load in an elevated cathode arrangement) is sort of the neutral end of the AC because it's impedance is much lower than the grid itself. Is that about right? This would also explain the predicaments that occur when the grid leak (or grid load) is omitted, because then the grid DOES become manipulated as an electron source.

EDIT: P.S. This thread is shaking hands with the reason "grid stop" resistors work to prevent blocking distortion.