Output signal path - please help me understand

[Deke609]

Please see attached crappy diagram that sets out how I think the signal path works from the output tube.

My thinking is follows - kindly correct anything (everything) that I have wrong:

(1) An amplified signal is produced at the anode, and an opposite "anti-signal" is produced at the cathode. At any given point in time, the voltages of signal and anti-signal are equal but opposite (except at zero crossings, when they are both zero). Accordingly, they comprise opposite voltage potentials that create a current to meet and cancel.

(2) They "meet" via chassis ground. There are two paths available: (a) through the last cap in the power supply (and running through the plate choke); and (b) through the OPT which connects to chassis ground (either before or after the cathode bypass resistor - I have drawn it as connecting before). I have indicated the two paths in dashed light blue lines.

(3) Ohm's Law tells us that at any given moment, the voltage dropped across each path is equal (and results in cancellation of the signal/anti-signal opposing voltage potentials); but the current will vary with the resistance (AC impedance) of each path in the form of the plate choke coil and the OPT.

Do I have this right? If so, I will have some follow-up questions about what this means for power output ... but for now I just want to know whether I am thinking about this correctly.

MTIA, Derek

[Paul Birkeland]

(1) An amplified signal is produced at the anode

Yes.

and an opposite "anti-signal" is produced at the cathode.

There's some backwards ordering here.  The grid controls the current flow from the cathode to the plate.  Your statement would imply that the signal at the plate produces signal at the cathode.

At any given point in time, the voltages of signal and anti-signal are equal but opposite

Well, your cathode has a bypass cap, so there's no signal voltage there (a large capacitor creates an AC short), thus that anti-signal that you're referring to will always be very close to 0V.  You could in fact connect the cathode to ground directly and use a battery or negative bias supply instead of the cathode resistor. 

(2) There are two paths available: (a) through the last cap in the power supply (and running through the plate choke); and (b) through the OPT which connects to chassis ground (either before or after the cathode bypass resistor - I have drawn it as connecting before). I have indicated the two paths in dashed light blue lines.

It helps to draw these as circles to indicate the AC current loops.

(3) Ohm's Law tells us that at any given moment, the voltage dropped across each path is equal (and results in cancellation of the signal/anti-signal opposing voltage potentials)

Are we talking about AC voltage or DC voltage?  Ohms Law has resistance, voltage, current, and power.  Your circuit has caps and inductors.  You can select a frequency and convert these items into a set impedance based on that frequency, then do some analysis for AC signals at that frequency (but not DC).

[Deke609]

Many thanks PB.

There's some backwards ordering here.  The grid controls the current flow from the cathode to the plate.  Your statement would imply that the signal at the plate produces signal at the cathode. ...
Well, your cathode has a bypass cap, so there's no signal voltage there (a large capacitor creates an AC short), thus that anti-signal that you're referring to will always be very close to 0V.  You could in fact connect the cathode to ground directly and use a battery or negative bias supply instead of the cathode resistor.

This I think I get, but I wasn't clear in my OP.  So the grid allows current to flow in proportion to the (varying) voltage at the grid. The source of that current is the DC power supply, which flows non-conventionally from -ve to +ve, so it flows through the tube (via space charge of electrons) from ground via the cathode.  As the current flows through the tube in proportion to the varying voltage at the grid, two opposing signal currents are created simultaneously: one at the anode, and one at the cathode (dc supply current - anode signal current = inverted signal current). But the signal created at the cathode drains to ground. But it can't just disappear - ground can't be a electrical black hole that swallows up voltage potentials, can it? -- so I'm thinking the inverted signal (which is just varying voltage potentials that match (but oppose) those of the anode signal in time) and the anode signal create a current to cancel via ground as conduit.

Are we talking about AC voltage or DC voltage?  Ohms Law has resistance, voltage, current, and power.  Your circuit has caps and inductors.  You can select a frequency and convert these items into a set impedance based on that frequency, then do some analysis for AC signals at that frequency (but not DC).

AC voltages. But my thinking is that it in terms of current paths, it doesn't really matter. At any given point in time, the voltage and frequency of the anode signal will have a fixed values, to which each path to ground (or rather to cancellation via ground) will present a specific fixed impedance (at that particular frequency at that particular moment), and so we can think of the paths as presenting fixed (but almost certainly different) "resistances" to the signal (again at that particular point in time). 

Do I have this wrong?

Many thanks, Derek

[2wo]

 Where I think you're hung up is the two  opposing currents. There is only a single current that flows from B+, to anode to cathode to return, (ground). And is simply controlled by the grid, more or less current flow, not opposing.

Redraw your circuit to just B+, anode, cathode and return, don't even need the grid at this point. Then add the cathode resistor and plot it from there...John

[Paul Birkeland]

The source of that current is the DC power supply cathode

As the current flows through the tube in proportion to the varying voltage at the grid, two opposing signal currents are created simultaneously: one at the anode, and one at the cathode (dc supply current - anode signal current = inverted signal current).

The grid is controlling the flow of current from the cathode to plate.  That seem to be missing from what you're writing.  The cathode does not oppose signal current.

But the signal created at the cathode drains to ground. But it can't just disappear - ground can't be a electrical black hole that swallows up voltage potentials, can it?

A capacitor is a black hole that swallows up AC voltage potential.

-- so I'm thinking the inverted signal (which is just varying voltage potentials that match (but oppose) those of the anode signal in time) and the anode signal create a current to cancel via ground as conduit.

I'm not sure what you're trying to convey, and I am not sure why you have the notion that there needs to be cancellation. 

AC voltages. But my thinking is that it in terms of current paths, it doesn't really matter.

How much DC current flows through the plate choke in your drawing?  How much DC current flows through the output transformer in your drawing?  How much AC current flows through the plate choke in your drawing at 20kHz?  How much AC current flows through the output transformer in your drawing at 20kHz? 

At any given point in time

I would strongly suggest sticking with voltage, impedance, and current for now. 

[Deke609]

Thanks both and apologies: my reference to "opposing currents" was unintentional. I meant opposing voltage potentials at cathode and anode.  But in any event, I'm not making myself clear I'll go study up on electrical theory.

cheers and thanks, Derek

[Paul Joppa]

I will recommend that you look specifically at the Thevenin and Norton analyses.

I think you are grasping the concept, but don't the specific terms in which to express it in the usual ways.

[Paul Birkeland]

I'd also start with a resistively loaded triode rather than a parallel feed output stage. 

[Deke609]

I will recommend that you look specifically at the Thevenin and Norton analyses.

Many thanks PJ. I will do so.

cheers, Derek