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Modifying the classic Williamson

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Simulation of post #15 based upon 100W Edcor CXPP 5K.
Though I have cheated slightly, fudgin' UL taps at 50%.

20WRMS clean sinewave before it ventures out of AB1.
28W AB2 in short bursts, but you eventually build up a
blocking distortion. 350V was perhaps too conservative?
450V could certainly raise those power figures.
 
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http://www.diyaudio.com/forums/atta...02-modifying-classic-williamson-pwilsoup1.gif

Q1 is our direct coupled phase splitter (no Voltage gain of its own, but
a big gain in drive current) with freebie constant current source across
VBE to load the plate of V1b.

V1b enhances Q1 with all the functionality of a long tail pair of Triodes.
Both -Mu/2 and +Mu/2 Voltage gains of astonishingly equal impedance.
V1b input is cap coupled, as the LTP would have been. +Mu/2 actually
Schade, just scaled to match Mu. Same things if completely different.

One transistor has replaced 2 triodes, a coupling cap, and quite a few
supporting resistors... And this driving stage can abuse a much higher
mu Triode than LTP without plate resistance becoming a limiting factor.
So, open loop gain might be slightly higher than usual for a Williamson.

Note the phases of the PNP splitter outputs are reversed of "normal".
Pretty much had to be P so Schade would be the correct phase for
splitting w. impedances equal. If modding an existing amp, gotta swap
lines at OPT or somewhere else before closing the big loop.

Difference from Williamson? Concertina comes after a virtual LTP of Mu
and Schade. Rather than before a real LTP of Mu and Mu.
 
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Three stages of open loop gain, and equal impedances driving each output pentode.

Concertina by itself is low component count way to split,
but cathode impedance drives with different damping than plate impedance.

Long tail pair fed from one side only, would also not be a perfectly even split.
Teaming up concertina with the long tail pair was the genius of the Williamson.
 
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Three stages of open loop gain, and equal impedances driving each output pentode.

Concertina by itself is low component count way to split,
but cathode impedance drives with different damping than plate impedance.

Long tail pair fed from one side only, would also not be a perfectly even split.
Teaming up concertina with the long tail pair was the genius of the Williamson.

Hi Kenpeter,

I have problems understanding the words youre using I am Dutch.

What do you mean by long tail pair. And teaming up concertina?
 
Three stages of open loop gain, and equal impedances driving each output pentode.

Concertina by itself is low component count way to split,
but cathode impedance drives with different damping than plate impedance.

Long tail pair fed from one side only, would also not be a perfectly even split.
Teaming up concertina with the long tail pair was the genius of the Williamson.


...at the time. You would do it differently these days, probably with sand as your earlier posts show and with far lower parts count.
 
Three stages of open loop gain, and equal impedances driving each output pentode.

Concertina by itself is low component count way to split,
but cathode impedance drives with different damping than plate impedance.

Long tail pair fed from one side only, would also not be a perfectly even split.
Teaming up concertina with the long tail pair was the genius of the Williamson.

Hi Kenpeter,

I have problems understanding the words youre using I am Dutch.

What do you mean by long tail pair. And teaming up concertina?
 
Yes, cathodyne has unequal output impedances, and splits evenly
only when it driving exactly equal loads at all times.

Output pentodes (one in cutoff the other in saturation) not same load...

The differential amp is also a buffer with equal output impedances.
Split not so disturbed by misbehavior of the output Pentodes.
 
The diff.amp that follows the cathodyne functions as a buffer?

No, in fact, the cathodyne has a lower source impedance than the diff amp. And with equal loads (e.g., the output stage), the source impedances are equal and low. See my analysis and experimental proof in the first issue of Linear Audio.

What the diff amp does is adds more open loop gain. Driven from the cathodyne, its bandwidth is quite high. The extra open loop gain is what gives the Williamson its excellent closed loop linearity.
 
No, in fact, the cathodyne has a lower source impedance than the diff amp. And with equal loads (e.g., the output stage), the source impedances are equal and low. See my analysis and experimental proof in the first issue of Linear Audio.

What the diff amp does is adds more open loop gain. Driven from the cathodyne, its bandwidth is quite high. The extra open loop gain is what gives the Williamson its excellent closed loop linearity.

Clear thanks.
 
No, in fact, the cathodyne has a lower source impedance than the diff amp. And with equal loads (e.g., the output stage), the source impedances are equal and low. See my analysis and experimental proof in the first issue of Linear Audio.

What the diff amp does is adds more open loop gain. Driven from the cathodyne, its bandwidth is quite high. The extra open loop gain is what gives the Williamson its excellent closed loop linearity.

Cathode of the cathodyne has much lower impedance than LTP plates.
Anode of the cathodyne has much higher impedance than LTP plates.
Preventing both of these nodes from directly seeing a load is buffering.
Yes I am aware the irony this buffer actually raises one impedance.
Equalization of driving impedances just as important as open loop gain.

Pentode output stage is most certainly not an equal load at all times.
Especially when pushed to the boundaries of A1. Classic unbuffered
cathodyne will misbehave quite differently when challenged by push
than when challenged by pull.

There more ways to fix this problem than you could shake a stick at.
But Williamson is probably the simplest way to do so without another
transformer, given a world where no high voltage P device yet exists.
 
Cathode of the cathodyne has much lower impedance than LTP plates.
Anode of the cathodyne has much higher impedance than LTP plates.
Preventing both of these nodes from directly seeing a load is buffering.

No. Please see my article for the analysis and the experimental verification- with equal load (whether resistive, reactive, or a combination), the source impedances are equal and low.
 
"With Equal Load" yes. The voltages are equal. The impedances are still not.
"With Equal Load" you can't see or hear it, doesn't cause any problem.

This comes back to haunt you when the loads are not equal.

A grid in slight forward conduction, or merely seeing screen voltage begin
to drop due screen current draw is not the same load as other Pentode's
grid that is in deep cutoff.

When the grid connected to the cathode misbehaves, the error will be
forced correct locally, and error will re-appear at the other grid driven
by the anode. Which is fine, cause its in cutoff this half cycle anyway.

When the grid connected to the Anode misbehaves, nothing is corrected.
The error is heard, and no portion of the error is transfered to the cathode
driven side.

OK, not entirely true. There is the 1/Mu portion of the error thats always
on the cathode side...

----

When drives are balanced, half the error is corrected, and half transferred.
and this misbehavior is in symmetry across the middle.

----

Also remember that small errors are aggrivated by the open loop gain.
At that point in the circuit there is no such thing as a small error.
 
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