Hiwatts..

[eschertron]

The CF holds its cathode voltage nearly constant due to its configuration derived feedback, yes, but it must sink/supply the current that allows this to happen. Since its bias current is low its runs out of current swing, going into shut off or saturation well before the previous stage runs out of voltage swing. When this happens, its output impedance is high, and so the loading on the previous stage is decreased and the gain into the PI increases.

I'd love to see someone demonstrate this with a scope and a video! I'm interpreting what you're saying as to sound something like a NFB loop that suddenly opens when the headroom is reached. Is that what you have in mind with regard to the sudden decrease in loading on the previous stage?

[Dave H]

I'd love to see someone demonstrate this with a scope and a video!

What do you expect to see? I don't think it will be dramatic. I calculated that the gain only changes from about 57 to 67 for the part of the waveform where the CF is driven off.

[Malcolm Irving]

The CF holds its cathode voltage nearly constant due to its configuration derived feedback, yes, but it must sink/supply the current that allows this to happen. Since its bias current is low its runs out of current swing, going into shut off or saturation well before the previous stage runs out of voltage swing. When this happens, its output impedance is high, and so the loading on the previous stage is decreased and the gain into the PI increases.

But doesn’t the calculated output impedance of the CF tell us precisely about its ability to source or sink current (provided it stays within the range of validity of that calculated impedance)?

I’ve done some (rough) estimates as follows:

Not sure about the B+ here, so let’s say it is 300V.
Assuming no grid current into the CF, the voltage at its grid is 300 x 220k / (1M +220k) = 54V
Considering the ECC83 plate characteristics with a load line of 220k and B+ of 300V, I estimated the following quiescent conditions:
Vgk = -2.7V approx. (which justifies the no grid current assumption). Plate current Ia = 0.3mA approx.
The voltage at the grid stays constant. If we now source some output current from the CF cathode (we go off the load line but stay with the plate characteristics):
If the cathode voltage goes down by 1V, Vgk becomes -1.7 and Ia goes up to about 1.5 mA.
Current through the cathode resistor is now (54+1.7)/220k = 0.25 mA, so 1.5-0.25 = 1.25mA is sourced to the external load.
Looking at this the other way around: if the cathode is required to source an additional 1.25mA to the external load it can do so easily with a variation of only 1V at the cathode.
Estimating the output impedance from the above, we get 1V/1.25mA = 800 ohms (which is close to the calculated value of the CF output impedance).

If the cathode voltage goes up by 1V, Vgk becomes -3.7V and Ia goes down to zero, the tube is cut-off. The CF can now sink approx. 0.3mA from the external load and this current goes through the 220k cathode resistor.
Looking at this the other way around: the cathode can only sink about 0.3mA before it goes into cut-off.
Estimating the output impedance again: 1V/0.3mA = 3k3 (getting higher because we are going into cut-off).

Summarising the above, the CF is running pretty cold, is capable of sourcing additional current quite easily and I don’t think is likely to reach saturation. On the other hand, it is not far from cut-off and does not have so much ability to sink current.

[Malcolm Irving]

The CF can source 1.25mA without saturating. This current has to go through the 100k resistor to the driving stage, i.e. a voltage drop of 1.25x100 = 125V. So a negative going swing (from quiescent) of 125V at the plate of the driving stage can easily be accommodated.

The CF can only sink about 0.3mA before cutting off, so a positive going swing of only 30V at the plate of the driving stage can be accommodated before the impedance seen at the cathode of the CF rises to 220k.

[Mike Sulzer]

The CF can source 1.25mA without saturating. This current has to go through the 100k resistor to the driving stage, i.e. a voltage drop of 1.25x100 = 125V. So a negative going swing (from quiescent) of 125V at the plate of the driving stage can easily be accommodated.

The CF can only sink about 0.3mA before cutting off, so a positive going swing of only 30V at the plate of the driving stage can be accommodated before the impedance seen at the cathode of the CF rises to 220k.

The current through the 100K resistor goes through the coupling capacitor as well, and so average current in both directions must be the same, limited by how much the CF can sink. (not sure about the details of how this works) The PI is a 12AT7 running on low current. I suspect that the gain of this stage is not very high, and so quite a bit of drive is required.

In computing the gain in the two states, you must consider the output impedance of the previous stage as well, adding maybe 40K to the 100K resistor. That is the series leg of the voltage divider. The shunt leg changes from about 100K to about 320K. It looks like the gain increases by about a factor of 1.7.

[Malcolm Irving]

The current through the 100K resistor goes through the coupling capacitor as well, and so average current in both directions must be the same ...

Yes. Good point.

In a transient from 'no signal' to continuous large signal conditions, the DC voltage across the coupling cap would shift so that it neither charges or discharges further over a signal cycle (i.e. the average current through the coupling cap becomes zero in the steady state, as you say).

[Malcolm Irving]

However, the way that a time-average current of zero is achieved (in the overdriven steady state) in these types of circuits is not usually that the forward peak current becomes equal to the reverse peak current, rather the 'duty cycle' of the signal is changed.
A large current in one direction is balanced by a smaller current in the other direction which lasts for a longer time.

[mhuss]

Guys, it's just a voltage reference (for the PI). In this variation, there's no signal going through it at all, the triode is just a buffer/impedance converter for the 1M/220k voltage divider.

The older (early 70's circuit) actually had signal going through it as well, so that analysis is a little more complicated. (http://hiwatt.org/Schematics/DR_Pre4Input_v1a.pdf)

[Mike Sulzer]

Guys, it's just a voltage reference (for the PI). In this variation, there's no signal going through it at all, the triode is just a buffer/impedance converter for the 1M/220k voltage divider.

The older (early 70's circuit) actually had signal going through it as well, so that analysis is a little more complicated. (http://hiwatt.org/Schematics/DR_Pre4Input_v1a.pdf)

Really? How do you know that? What is the purpose using a tube to provide dc to one side of the PI through a 100K resistor, and the other side through a 1 Meg resistor? Why do you need a tube to do that? If it is just a voltage reference, why is the 100K resistor load the ac to the grid of the pi? And why run the tube off such a low current that it cannot sink the ac current over the full linear range of the stage that drives it?

I am perfectly willing to accept that you are right and it is just a stupid circuit, but I would like to see some evidence. Until then, I have to think that maybe the designer had a purpose in mind, and maybe the circuit is clever.

[mhuss]

It's a fixed voltage reference for the PI. Instead of being self-biasing, like the standard LTP (and 98% of all tube amp stages), this variation uses a fixed reference voltage to set the operating point. The low-impedance voltage source acts like a virtual ground for signal, but the fixed DC level holds the grids at a more or less fixed point - plus any signal applied.

Scope it out, there's no AC signal to speak of on any of the VRef triodes' three terminals.

[frus]

So do you have a theory why they went from the CF circuit to just the voltage reference?
I mean, they are essentially "wasting" another triode and could have saved a tube, so it doesn't make sense from cost cutting perspective

FWIW, I worked on a reissue recently, and they chose the earlier circuit with the CF

[Malcolm Irving]

If it was just a voltage reference for the PI, you would expect it to be connected to the PI grid by a 1M resistor similar to the other PI grid. The 100k resistor is crucial to the way it operates. My guess is that the designer played about with different values for that resistor and chose one that sounded 'right'.

[Malcolm Irving]

Scope it out, there's no AC signal to speak of on any of the VRef triodes' three terminals.

Try scoping across the 100k resistor to see how much current is flowing through it. (Connection into the scope needs to be floating.)

[d95err]

they are essentially "wasting" another triode and could have saved a tube, so it doesn't make sense from cost cutting perspective

That's what I find so fascinating with this circuit. If they'd had a leftover triode and used it for this purpose it would have made sense, but actually adding an extra tube for it...

The circuit designer must have had a really good explanation of the benefits of having fixed bias for the PI, and be able to "sell" it internally. I would really like to know what it was

[SoulFetish]

Wait, i have an idea. If this could otherwise be achieved via a voltage divider from HT to ground, maybe this was done for noise supression. Any ripple present at the HT would also show up at the grid via the divider, correct? Cathode followers generally have good PSRR. Maybe this was enough of a reason to choose this topology. I would think the performance would be more consistent over the life of the tube versus the life of an EXTRA rc filtering stage stage. Its just an idea, what do you guys think?