Tweedle Dee PI - LNFB and trimmer

[Stephen1966]

I'm currently working up a study of the Tweedle Dee based on the 5E3 circuit and the calculations for the feedback provided by the 3M3, 2n2 LNFB net on the cathodyne PI are not clear. I've been trying to calculate the feedback factor using Blencowe's formulas and with reference to Aiken but I get several derivations of the results depending on how formulas are applied.

This is a simplified schematic of the circuit:

Phase inverter2.jpg

I would much appreciate it if anyone with greater understanding could chime in and settle this once and for all.

The main problem I am having here is how to calculate the input resistance of the cathodyne and this is slightly complicated by the model of an ordinary gain stage that Blencowe uses in Fig.10.3 (page 180) of 'Designing Tube Preamps for Guitar and Bass'.

Blencowe - p180.pdf

Aiken in his whitepaper 'What is Negative Feedback?' https://www.aikenamps.com/index.php/wha ... e-feedback discusses how we can calculate the input resistance when there is no physical resistor present...

In the other case, the feedback is applied to the "signal" side input of the phase inverter. This forms an inverting amplifier configuration, with the gain being set by the ratio of the value of the feedback resistor to the "input" resistor. The input impedance of this type of global negative feedback amplifier is set by the value of the input resistor, so care must be taken in the design of the circuit to avoid loading down the previous stage, as the value of the input resistor usually must be very low in order to get any decent amount of gain with not too large a feedback resistor value. In some cases, there is no input resistor at all, so the effective value of the input resistor is the output impedance of the previous stage. In either case, the overall feedback gain can be set by varying the feedback resistor...

This is in the context of a discussion about global negative feedback with a description of the Fender AB165 Bassman however, and Blencowe treats the calculation of input resistance differently...

In Blencowe's fixed bias model of the cathodyne in https://www.valvewizard.co.uk/cathodyne.html

The input resistance of the fixed-biased version is R1||R2...

or, what we might transpose as Rf || Rg in the simplified Tweedle Dee schematic above. But the fixed bias cathodyne model he uses there is not exactly equivalent either with his R1 attached directly to the HT (above - or before - the plate resistor) and his R2 bypassing Rk all the way to ground. It isn't too far a stretch to imagine the fixed bias model here could be applied in the context of a LNFB net but it's never as straightforward as it seems (it seems!) and in chapter '10.7: Effect on Input and Output Impedance', (page 184) he expresses a formula which demonstrates that when feedback voltage is applied in parallel with the input voltage, that is, when feedback is applied to the input resistance, the input resistance is reduced.

Rin (with feedback) = Rin(1 + Ao) + Rin + Rf/1 + A

When Ao = open loop gain, and A = closed loop gain.

[Edit: I am updating the subject lines of my posts here because Dumble's cathodyne features both a LNFB and variable resistor (trimmer) at the cathode. I could treat the effect of the trimmer separately in its own thread but the NFB and the trimmer are integral to the PI and it makes sense to treat them under the same thread. Please also bear in mind that no-one, to my knowledge, has adequately treated the subject of Tweedle Dee's cathodyne phase splitter before and some of the early mathematical proofs are exploratory. Following the arguments though, the problems are resolved and solutions found.]

[bepone]

this feedback is there to balance outputs from Cathodine PI...it was already discussion..

[rdavy]

Also keep in mind, the 1 Meg grid leak resistor is 10 Meg in the Tweedle Dee.

[Stephen1966]

Looking at Aiken's direction first where 'the effective value of the input resistor is the output impedance of the previous stage',
when the plate resistance of V2a is 50V/0.78mA,
ra is 64k
and the output impedance, Zo
when Ra = 100k
= 100k x 64k/100k+164k = 6400k/164k = 39k

This is wildly different from the "fixed bias" result we get from Blencowe's, R1 || R2 which we apply to the cathodyne (V2b)
when R1 = 3.3M and R2 = 1M
= 1/(1/3300k + 1/1000k) = 767k

And this is different again when we parallel input voltage with feedback voltage to give a reduced input resistance.

Rin (with feedback) = Rin(1 + Ao) + Rin + Rf/1 + A

Where Ao = open loop gain; μ = 100 (12AX7); Ra' = Ra || (Rf + Rin) = 106.5k; Ra = 110k; Rf = 3300k; Rin = 39k (Aiken); ra = 92.6k
= -μ x Ra'/Ra' + ra = -100 x (106.5k/106.5k + 92.6k) = -53.5

or when Rin = 767k (Blencowe) and Ra' = 107k
= -μ x Ra'/Ra' + ra = -100 x (107k/107k + 92.6k) = -53.6

Nothing too controversial there but the feedback fraction (B) is significantly affected by the different Rin values and this, in turn affects the calculation for the closed loop gain (A) and so on...

When Rin = 39k
B = Rin/(Rf + Rin) = 39k/(3300k + 39k) = 0.011

When Rin = 767k
B = 767k / (3300k + 767k) = 0.19

[Stephen1966]

this feedback is there to balance outputs from Cathodine PI...it was already discussion..

It depends what you mean by "balance" if you mean voltage then this is customarily provided by the Ra (plate) and Rk (tail) resistors along with the effect (loading) of the power tubes' grid leak resistors which are the same in both 5E3 and TD = 220k each. 5E3 has a Ra and Rk of 56k, The TD is slightly imbalanced with Ra = 110k and Rk, the tail (not including the 4.7k bias resistor) = 99k.

I think the jury is still out on whether we want a perfect balance of voltage or a slight imbalance to coax the harmonics out.

Perhaps I am missing something about the discussion you refer to? Could you provide a link?

[Stephen1966]

Also keep in mind, the 1 Meg grid leak resistor is 10 Meg in the Tweedle Dee.

You have a point and I'm familiar with the discussion regarding the original photos. I'm erring towards the grid leak being just 1 Meg because the bootstrapped impedance is already very high - over 13 Meg in my calculations - and it's already complicated enough with at least three different approaches to calculating the input resistance of the feedback loop. The point is, with the bootstrapped impedance being so high with a 1 Meg grid leak, the AC load is practically the same as the DC load already, and I'm not sure what difference a bootstrapped 10 Meg would make, if any.

[Stephen1966]

I thought about the problem and I think I have (a part of) the answer.

On reflection, the best option for calculating the closed loop gain would be to use the equivalent universal feedback formula described on page 180 of Blencowe's book mentioned earlier.

PI equations (1) - equivalent feedback equation.jpg

The figures for the open loop gain (Ao), the feedback fraction (B), and the differential output impedance (Zo) can all be derived in a relatively straightforward way and the LNFB resistor (Rf) is known so it only comes down to how we arrive at the input resistance (Rin) when there is no physical "input resistor" in circuit. The problem caught me off guard because Blencowe's 'Practical circuit with local feedback', Fig 10.3 (page 180) shows Rin and Rg (in faint) but this is an ordinary gain stage with local feedback, not a cathodyne PI or even a cathode follower where the grid leak resistor (Rg) is bootstrapped to magnify the input resistance to AC signals many times the value of the grid leak, Rg. This, I think gives us a virtual Rin or input resistance which can be used in the equivalent feedback formula.

I'm looking at the schematic for the AB165 Aiken referred to and I think it's less appropriate to derive the input resistance from the output impedance illustrated in that circuit; the 7025 is an ordinary/mixer gain stage feeding a LTP with global negative feedback. The resulting output impedance is so relatively small that it can in no way measure up to the bootstrapped input resistance of the cathodyne circuit.

[bepone]

And.. how much is it after so much math?
You can recheck your math simple with one coupling cap, so math is no needed after all

Btw. why you are collecting those data? I saw that there is some commercial background on everything with Dumble amps--> on your signature , so all help what you can get on this forum is doubtful in the start like with everybody what trying to sell something and asking all the answers / solutions on the forum in the same time

[Stephen1966]

Before we can determine the bootstrapped value of Rg and hence the effective input resistance (Rin) using the bootstrap formula below we need to find the value of the differential gain (A)

PI equations (1) bootstrap formula.jpg

The formula for differential gain takes the 'total output voltage, measured between the two outputs, and divide(s) by the input voltage to get a sort of overall measure of gain', Blencowe, p.144.

PI equations (1) - gain when Ra and Rk equal.jpg

When Rk and Ra are the same, Ra = Rk = R = 56k, μ = 100 and using the plate resistance ra = 70.4k from the 5E3 because Ra and Rk are equal in that amp.

A = μ (R/(ra + R(μ + 2))
A = 100 x (56k/70.4k + 56k(100 + 2)) = 0.968 or 0.97 rounded up.

This result also tells us that we are very close to unity gain for the stage. The differential gain formula works well when the plate and tail resistors are equivalent but when there is a deliberate imbalance built in, as in the TD, separate formulas, for the gain to the anode and gain to the cathode need to be used. They should both add up to less than 2 and if we took a mean it should look very similar to the result of the differential gain formula.

With A settled, we can now figure out the bootstrapped value of Rg... Again, using 5E3 values for simplicity.

Rin = (Rg / 1 - A (Rk / Rk + Rb))
Rin = (1000k / 1 - 0.97 (56k / 56k + 1.5k)) = 18081.76095k or about 18.1Meg

[Stephen1966]

And.. how much is it after so much math?
You can recheck your math simple with one coupling cap, so math is no needed after all

Btw. why you are collecting those data? I saw that there is some commercial background on everything with Dumble amps--> on your signature , so all help what you can get on this forum is doubtful in the start like with everybody what trying to sell something and asking all the answers / solutions on the forum in the same time

You can never have too much maths

[Stephen1966]

For the TD I'm also interested in the differential gain because I am only trying to find the equivalent input resistance, Rin. So, by taking the resistances of the plate and tail and balancing them, rendering them equal, I have a "close enough" factor I can use in the object of the exercise, which is to find the bootstrapped input resistance for the feedback calculation.

When R = Ra + Rk / 2 = 110k + 99k / 2 = 104.5k; ra = 64k

A = μ (R/(ra + R(μ + 2))
A = 100 x (104.5k/64k + 104.5k(100 + 2)) = 0.9745 or 0.97 rounded down.

And for the bootstrap equation:

Rin = (Rg / 1 - A (Rk / Rk + Rb))
Rin = (1000k / 1 - 0.97 (99k / 99k + 4.7k)) = 13520.2086k or about 13.5Meg

So in both the 5E3 and the TD, the bootstrapped input resistance to AC signals is so high it presents very little load to the previous gain stage (V2a), providing that stage with an AC load virtually the same and in parallel with the DC load. That maximises output swing coupled with a higher B+ in the TD for much better clean headroom. It was a happy accident with the 5E3 perhaps but even so, it still had a reputation for being quick to break up. At least in the TD, the point where it breaks up occurs later, but that isn't just about the operating point of its cathodyne; the PI still has lots of grunt to drive the power tubes into distortion.

[Ten Over]

Before we can determine the bootstrapped value of Rg and hence the effective input resistance (Rin) using the bootstrap formula below we need to find the value of the differential gain (A)


PI equations (1) bootstrap formula.jpg


The formula for differential gain takes the 'total output voltage, measured between the two outputs, and divide(s) by the input voltage to get a sort of overall measure of gain', Blencowe, p.144.


PI equations (1) - gain when Ra and Rk equal.jpg


When Rk and Ra are the same, Ra = Rk = R = 56k, μ = 100 and using the plate resistance ra = 70.4k from the 5E3 because Ra and Rk are equal in that amp.

A = μ (R/(ra + R(μ + 2))
A = 100 x (56k/70.4k + 56k(100 + 2)) = 0.968 or 0.97 rounded up.

That is not the formula for differential gain. It's the formula for the absolute value of the gain at either output when both outputs are loaded the same. The loaded plate resistance is 44.64k and the loaded cathode resistance is 45.59k in the 5E3. The average of the two is 45.1k and that is a pretty good fudged number to use for the gain equation. I get an internal anode impedance of 60.5k and a mu of 104.2 for a gain of 0.969 at either output.

CF 36 PNG.png

[Ten Over]

For the TD I'm also interested in the differential gain because I am only trying to find the equivalent input resistance, Rin. So, by taking the resistances of the plate and tail and balancing them, rendering them equal, I have a "close enough" factor I can use in the object of the exercise, which is to find the bootstrapped input resistance for the feedback calculation.

When R = Ra + Rk / 2 = 110k + 99k / 2 = 104.5k; ra = 64k

A = μ (R/(ra + R(μ + 2))
A = 100 x (104.5k/64k + 104.5k(100 + 2)) = 0.9745 or 0.97 rounded down.

And for the bootstrap equation:

Rin = (Rg / 1 - A (Rk / Rk + Rb))
Rin = (1000k / 1 - 0.97 (99k / 99k + 4.7k)) = 13520.2086k or about 13.5Meg

So in both the 5E3 and the TD, the bootstrapped input resistance to AC signals is so high it presents very little load to the previous gain stage (V2a), providing that stage with an AC load virtually the same and in parallel with the DC load. That maximises output swing coupled with a higher B+ in the TD for much better clean headroom. It was a happy accident with the 5E3 perhaps but even so, it still had a reputation for being quick to break up. At least in the TD, the point where it breaks up occurs later, but that isn't just about the operating point of its cathodyne; the PI still has lots of grunt to drive the power tubes into distortion.

Ra=73.3k
Rk=70.5k
ra=92k
u=95.4
Ak=0.9665
Rin=12.9M

Presuming 220k grid leaks.

[Stephen1966]

Before we can determine the bootstrapped value of Rg and hence the effective input resistance (Rin) using the bootstrap formula below we need to find the value of the differential gain (A)


PI equations (1) bootstrap formula.jpg


The formula for differential gain takes the 'total output voltage, measured between the two outputs, and divide(s) by the input voltage to get a sort of overall measure of gain', Blencowe, p.144.


PI equations (1) - gain when Ra and Rk equal.jpg


When Rk and Ra are the same, Ra = Rk = R = 56k, μ = 100 and using the plate resistance ra = 70.4k from the 5E3 because Ra and Rk are equal in that amp.

A = μ (R/(ra + R(μ + 2))
A = 100 x (56k/70.4k + 56k(100 + 2)) = 0.968 or 0.97 rounded up.

That is not the formula for differential gain. It's the formula for the absolute value of the gain at either output when both outputs are loaded the same. The loaded plate resistance is 44.64k and the loaded cathode resistance is 45.59k in the 5E3. The average of the two is 45.1k and that is a pretty good fudged number to use for the gain equation. I get an internal anode impedance of 60.5k and a mu of 104.2 for a gain of 0.969 at either output.

CF 36 PNG.png

Thank you. I didn't go so far as to consider the plate or cathode under load and 'differential' is probably a poor choice of words. In this case, I was only trying to get to the feedback calculation so I only needed an averaged unbypassed, gain figure. I will look at the maths again, working in the loaded resistances and see if that takes me anywhere unexpected. So far, considering we got there by different paths,we seem to be mostly in agreement.

[Stephen1966]

For the TD I'm also interested in the differential gain because I am only trying to find the equivalent input resistance, Rin. So, by taking the resistances of the plate and tail and balancing them, rendering them equal, I have a "close enough" factor I can use in the object of the exercise, which is to find the bootstrapped input resistance for the feedback calculation.

When R = Ra + Rk / 2 = 110k + 99k / 2 = 104.5k; ra = 64k

A = μ (R/(ra + R(μ + 2))
A = 100 x (104.5k/64k + 104.5k(100 + 2)) = 0.9745 or 0.97 rounded down.

And for the bootstrap equation:

Rin = (Rg / 1 - A (Rk / Rk + Rb))
Rin = (1000k / 1 - 0.97 (99k / 99k + 4.7k)) = 13520.2086k or about 13.5Meg

So in both the 5E3 and the TD, the bootstrapped input resistance to AC signals is so high it presents very little load to the previous gain stage (V2a), providing that stage with an AC load virtually the same and in parallel with the DC load. That maximises output swing coupled with a higher B+ in the TD for much better clean headroom. It was a happy accident with the 5E3 perhaps but even so, it still had a reputation for being quick to break up. At least in the TD, the point where it breaks up occurs later, but that isn't just about the operating point of its cathodyne; the PI still has lots of grunt to drive the power tubes into distortion.

Ra=73.3k
Rk=70.5k
ra=92k
u=95.4
Ak=0.9665
Rin=12.9M

Presuming 220k grid leaks.

Cool! Thanks again. Yeah, 220k.