g2 voltage and feedback

[Andy Le Blanc]

After setting the voltage difference between g2 and plate, with the plate being
slightly positive to g2, I applied feed back from the opt secondary to the cathode
of the driving stage of a cathodyne inverter, when I re-checked the difference
between g2 and plate, the screens had gone slightly positive.

To avoid "pentode tone" I reset the difference between g2 and plate.

The g2 / plate difference is easily set with a resistance value where the
choke is located in the circuit. simply match the voltage difference between
the center tap of the opt and tube plate, after you set your desired bias.

It seems desirous to set the operating point, set the plates positive to g2,
apply a feedback ratio, then recheck g2/plate...
or anticipate the application of feedback and set the plates decidedly positive.

[Cliff Schecht]

If my thinking is correct the entire operating point shifts slightly more negative when you apply negative feedback. This slight decrease in current means less voltage dropped at the plate and a higher B+ for a given configuration. But we are talking a few Volts here, not the difference between 50V between P and G2.

I first thought this was a thread about the effects of G2 voltage and negative feedback within the power tube. Lower G2 relates to less negative feedback within the power tube and a brighter sound (or more pentodey).

[Andy Le Blanc]

It is a neat relationship.

It is only a few volts, the power side is a 6l6 50w fender based circuit.
Theres no surprises, w/o feedback the screens were riding .5 volts above the
plate, 470r resistors etc... but the difference is audible.
reversing the relationship to where the plate are slightly positive prevents
the positive plate swing from interacting with the screen.

Degenerative feedback thru an un-bypassed cathode resistor is one thing
global feed back is another, I'm not convinced that the voltage difference
between g2 and plate represents feedback within the tube.
Although it might sound the same...

Thats a good question, It can be observed that there is an interaction of
the g2 voltage with feedback. Does the voltage difference between g2 and
the plate represent "feed back"?, current or voltage feedback?.

[Super_Reverb]

It sounds like the DC operating point is shifting as you increase negative feedback. Could there be something else going on? You could try AC coupling the feedback resistor back to the P/I cathode, because your amp sounds stable at DC, so DC feedback isn't really necessary.

You could also try a much larger feedback resistor in parallel with a capacitor to set a corner at say 20Hz or higher, so that above a specified frequency, your feedback would begin to become active and below the corner frequency, there would be less or near zero f/b.

Many feedback schemes encompass the P/I, outputs, and OT and are strictly resistive, but as many presence controls demonstrate, f/b can be a function of frequency.

The feedback you describe is utilizing a difference amplifier concept, whereby you are feeding back a negative portion of the output signal. With this arrangement, the closed loop voltage gain of the P/I, output, and O/T is determined by the "input" resistor (cathode resistor in the P/I and the feedback resistor and also open loop gain (voltage gain of P/I*output stage w/ O/T)

I'm not convinced that the voltage difference
between g2 and plate represents feedback within the tube.

To be fair, the output tube(s) don't know that you have wrapped feedback around them. The DC point seems to have been affected by application of f/b and from an AC response, the loop will behave differently, but I think the output tubes simply respond to grid bias, O/T center tap voltage, O/T primary characteristics and screen bias voltage and screen limiting resistance as you already described.

Just my $0.02

rob

[Andy Le Blanc]

Its the feed back...

In a nut shell, there is current and voltage feedback...
It can be applied to the circuit either series or shunt...

Both types and both applications have the same effect on gain and distortion...

BUT... depending the type and how it is applied... have different results..
upon the "effective" input resistance and source resistance...

I observed the change of the effective plate resistance of the power tubes,
as it affected the voltage difference between g2 and the plate voltage.

[Firestorm]

Andy-- Wouldn't there have to be some kind of "signal" to be fed back? (Hiss, hum, thermal noise?) Or do you think merely making the connection can somehow alter the plate resistance?

[Cliff Schecht]

Just making the connection will alter the entire output impedance of the amplifier (and therefore the gain and bandwidth as well). All of that stuff gets fed back at a small fraction of the actual amount which, if the noise is low (thermal noise, hum, etc) then it will be further reduced.

[Firestorm]

Just making the connection will alter the entire output impedance of the amplifier (and therefore the gain and bandwidth as well). All of that stuff gets fed back at a small fraction of the actual amount which, if the noise is low (thermal noise, hum, etc) then it will be further reduced.

I still don't get it. Sure, when there's signal present, the loop changes output impedance, gain, bandwidth, distortion, etc. But those terms aren't quantifiable in the absence of signal. What I think Andy was seeing (chime in, Andy) was a drop in plate voltage (relative to g2) at idle with no signal. Trying to suss out the electron-level, inside-the-tube reason for that.

I can't remember whether the screen grid was first designed primarily to minimize grid-to-plate capacitance or to "decouple" plate current from plate voltage, but both effects would seem to be tunable along the lines Andy is exploring.

[Andy Le Blanc]

It alters the "effective" plate resistance.

I've been going on about about g2 and the plate, wouldn't have noticed otherwise.

I did some open/closed readings with the FB loop (shorting), it can move
the screens a couple volts in relation to the plate, in static conditions
I observed it move 200-500mv.

[Andy Le Blanc]

Firestorm is correct, but I think the your confusing no signal with no connection.

If there is a load or the output is shorted the feedback loop is effective.

The "internal resistance" of the amp changes.

[Firestorm]

I get the theory; it's the physics that bothers me. Internal resistance to what? There's obviously no DC coupled from the OT secondary via the feedback loop and if there's no signal there should be no AC either. So what is the tube responding too?

Lot's of things alter a tube's plate resistance: relative inter-electrode voltages; inter-electrode capacitances I assume; the electrostatic effect of the screen on the space charge, probably. But I can't my head around how the FB loop can do it with no AC. Unless it's a quantum effect.

Just curious because you were getting such a pronounced effect on tone from a very small change in Vp to Vg2. Maybe it's one of the ingredients that makes some amps sound special compared to other "identical" ones.

[VacuumVoodoo]

Feedback is a reactionary mechanism, it responds to changes in the system. With everything static and constant FB will have nothing to react to.
The small changes you observe are caused by small slow fluctuations in supply voltage that are always present unless it is very heavily regulated. Once you put transconductance and gain sensitivity to fluctuations in supply voltage into the equations, i.e. forgo the simplified signal only analysis and dig deeper, you will see it.

[Cliff Schecht]

So how much effect does the feedback connection to the phase inverter have on the operating point of the power stage?

If my thinking is correct (which it probably isn't at this time of night), making the feedback connection through the phase inverter presents a real, different load to the phase inverter which changes the operating point of this stage. The speaker and secondary are common to ground and connect to the PI eventually. Changing the PI operating point will move a few things, including the current drawn from the power stage and the loading presented to the grids. I think the change in grid loading is negligible for both fixed and cathode bias but the change in operating point in the PI would be enough to cause the G2/P relationship to shift a hair. I think making the feedback connection would drop the gain in the PI and raise the current enough to drop the G2 voltage just a bit more than if there was no feedback connection.

Andy, if you take more measurements make sure to note down whether NFB increases or decreases G2 with relation to the plates.

[Andy Le Blanc]

I think it turns out to be pretty well described...

your need to find : r'a

the output impedance in the presence of feedback.

you divide the plate resistance by 1+BuA

1+[feedback ratio] [amplification factor] [voltage gain of the preceding stage]

then its Rl r'a divided by Rl+r'a..... looks like two resistances parallel

I'm just seeing the slight change as its reflected in the circuit voltages.
I've been eyeballing the voltage difference between g2 and plate,
with "0" as a reference, wouldn't have noticed otherwise.

there's a similar figure for the input resistance...

r'i = ri +(1+AB)

the statements are that there is an increase of the input resistance and a
decrease of the output impedance in the presence of feed back.

It will be fun to fill in the gaps around this one ...

The circuit I've up and running is 6l6, the opt is a 18343 4k/8r, and the
inverter is a 6sn7 split load, the inverting stage is fixed grid bias...