Topology for Schade

[TheGimp]

I'm working on my Webcor amp that I pulled from a console and now that I've got it completely rebuilt I'm thinking about Schade feedback.

Today was the anual Tree Street Yard Sale event

The topology is basically an input triode gain stage with global feedback to a cathode resistor, followed by a concertina phases plitter driving the output tubes (12AB5).

It seems to me that this topology is not suitable for Schade feedback, and either I need to follow the concertina splitter with cathode followers, or I need to use an LTP to allow me to feedback to the same impedance for both output tubes.

Is there any way to use Schade feedback with a concertina splitter directly driving the output tubes?

[DF96]

As I understand it Schade feedback is simply a new name for the anode follower, which in turn is the valve forerunner of the inverting opamp configuration. For a P-P pair each must see a signal coming from the same source impedance. LTP phase splitter is the simplest way to provide this. If you use a pair of CF then you will need a largish build-out resistor for each so you are almost throwing away the benefit of a CF.

[smoking-amp]

One way to preserve the concertina splitter would be to connect it's plate load resistor to a P type current mirror (connected to B+). Then a matched load resistor from the mirror drain back down to Gnd. Then take the two Hi-Z outputs off the concertina plate and the mirror drain. Should be almost matched Z for the Schade feedbacks. The concertina (triode) will have some finite plate impedance, so some lowering of the mirror load resistor needed to compensate, or replace the concertina itself with an N Mosfet to get hi Zout there too.

Hmmm, on second thought, this would introduce twice the neg. feedback to one side (mirror out) as to the other (concertina plate out) since the current input to the mirror gets affected.

All you really need is to just use one Schade feedback to the INPUT of the concertina.

[mashaffer]

Ooo.. Smoking amp, that is an interesting idea. I will watch with interest.

[kenpeter]

Schade IMHO: is the transformation of high impedance Pentode
curves to lower impedance Triode curves, by local nfeedback.

Of course, you need to make sure your nfeedback is of the sort
that helps linearize output voltage, rather than output current.
And should be of lowest phase shift, preferably direct coupled.

Its a little uncertain what is Schade and what is not in a push
pull amplifier? Cause opposing plate voltages are also coupled
by the output transformer. The nfeedback to each Pentode is
never pure like Schade's infamous "Fig35", a completely single
ended example. Then again, push-pull blended plate feedback
might still be a valid transform to triodes with similarly coupled
plates...

I don't see a problem with GNF as Schade as long as its kept
less than one pole of phase shift. Throw two or more poles at
the loop, and you got potential problems of a Williamson...

I agree with Post #2, new name for same old anode follower.
However, Schade proved and made obvious exactly why this
should be equivalent to a Triode. And strongly hinted to the
method by which Mu is formed inside a real triode.

[Ian444]

I've been pondering how effective an unbypassed cathode resistor on an output tube is for local feedback (PP or SE). If the grid 1 signal was say 15VRMS and at that input level you have say 1VRMS on the (unbypassed) cathode of the output tube, are you achieving some useful local feedback? Of course to get that scenario you would need a combination of fixed bias and cathode bias, since the cathode resistor would be small. I feel this may be relevant to the topic in that it may be an alternative to Schade for amps where Schade feedback is difficult to implement, other than using GNFB. I could be totally on the wrong track too, hence the question.

[kenpeter]

Unbypassed cathode resistor increases current linearity and plate resistance,
decreases Gm, does nothing to externally reconnect Mu broken by the screen.
Indeed, if any Mu still exists, actively fights against it having any influence.

If you stand that cathode upon a transformer output, and tolerate some very
small amount of DC: Parasitic cathode winding resistance creates exactly the
same situation given above. Plate resistance increases, no Triode, no Schade.
The small resistance here definately not helping us Schadewise.

But check voltage divider from the plate to the cathode created by inductance!
Now here we got Schade, bigtime. A positive voltage swing fed back to cathode
is exactly equivalent to a negative swing fed back to grid... Way more than the
winding resistance is fighting. With Mu effectively reconnected, Triode happens.
Plate resistance decreases dramatically. Schade used transformer voltage divider
(coupling to the grid) in his example...

Examine output of ARC ST-70-C3. The 4 ohm tap is grounded, 0 and 16 taps
abused for cathode windings. I don't know how well DC will balance into 4 or
8 ohm loads? But ought to be in very good cancellation for the full 16 out.
You gotta have an output transformer anyway, basically a freebie, take it!

This scheme is more or less direct coupled. One pole of phase shift at worst...
Unless'n you got some really weird transformer parasitics. And we ain't talkin'
like Williamson, where you add one or two more caps of shift, and magnify the
chance for problems by an entire triode worth of gain added to that long loop.

[smoking-amp]

On the idea of keeping the Concertina in place, both Schade feedbacks (from output plates) could be implemented, I think. One S fdbk would be sent to the Concertina input (grid), and the other sent to the bottom of the Concertina cathode load resistor. Ie, that Schade feedback divider would have its low value resistor inserted under the Concertina cathode resistor to ground. The other Schade feedback (to the Concertina grid) would depend on the previous input tube's plate resistor for dividing down feedback. The Concertina plate resistor might need a little tweek to get the two Concertina output levels matched then. The Schade feedback resistors will have an apparent assymetry, due to the low resistance needed in the Concertina cathode circuit, versus the high resistance needed in the C grid circuit. But functionally, they both provide the same attenuation ratio of the two feedbacks.

This would have the advantage of using both Schade feedbacks from the output tube plates equally, no feedback having to pass through the OT then (as for the single fdbk to the Concertina grid case), so OT xfmr frequency anomalies would be kept out of the loops. Seems workable, anyone see problems?

Edit: I think I see a minor issue that can be fixed. The small value feedback divider resistor placed below the Concertina cathode load is effectively adding in additional neg. feedback to that output side due to its series connection. (beside the required neg. fdbk to the Concertina grid-cathode V diff.). This can be fixed by increasing the Schade fdbk attenuation a little (ie, raising the Schade feedback resistor value on that side) so the net neg. fdbk is equal for both Concertina outputs. Experiments probably required. (or Sim. it)

[smoking-amp]

Correction:
Seems the series Concertina cathode feedback resistor is actually REDUCING the neg. feedback for that side's output effectively, so the Schade feedback resistor needs some decreasing in value on that side. This needs some Simulation now for sure to see if this can work out for equal neg. feedback to each side. I guess if it doesn't work out, we would have another way to get SE sound from P-P.

[Wavebourn]

You can borrow mine.

The first tube (triode-pentode) is ECF200, LTP driver tubes are 6J51P (may be substituted by 6P15P), output tubes are GU-50