Tweedle Dee PI - LNFB and trimmer

[Stephen1966]

(@Ten Over, I had a look at your calculations and I thought it might be useful here to show how your figures work in the calculations.)

If I use the figures for the 5E3 it will be simpler. I am going to stick to my original nomenclature as it will get confusing if I start changing ra to rp and Rk to Rt and so on. I will clarify as I go.

The loaded anode and cathode resistances first. The 56k plate resistor, Ra, can also be called Rp but for semantic reasons, I think I prefer Ra because it indicates the function of the resistor on the anode, much as Rk indicates the function of the resistor on the cathode; unlike Rp which only infers its position in the circuit.

Ra = 56k

When Ra || Rl, and Rl (the V3 power tube grid leak resistor on the anode side) = 220k,

Ra (loaded) = 44.64k (This becomes useful in the input resistance/bootstrapping calculation)

The Rk component in the formulas needs some explanation perhaps because @Ten Over and I consider it differently. You will see by looking back at the schematic at the top of this post that where I call Rb (the bias resistor) and Rk (the tail resistor), @Ten Over calls the sum of Rb + Rt = Rk. I think I would prefer to call my Rk the Rt ... but I made my bed, so I have to lie in it for now! For the bootstrapping formula we need to be able to differentiate between the bias and the tail resistors but it only takes a small adjustment to the formula. For the gain formula generally and the cahtodyne PI specifically, we only need an average of the top and tail resistance so what I call Rb + Rk and @Ten Over calls Rb + Rt is the R(total) resistance of the bias and tail resistor or, what we can now call Rk'.

Rk' = Rb + Rk = 1.5k + 56k = 57.5k

Rk' = 57.5k

Similarly Rk' || Rl (the V4 power tube grid leak resistor on the cathode side) and Rl = 220k,

Rk' (loaded) = 45.59k

For the gain formula we only need a single figure for R and so, summing Ra (loaded) and Rk' (loaded) then dividing by two we have:

R = 45.1k

And so armed with @Ten Over's figures we can proceed to the gain equation.

When R = 45.1k; ra = 60.5k; μ = 104.2

A = μ (R/(ra + R(μ + 2))
A = 100 x (45.1k/60.5k + 45.1k(100 + 2)) = 0.9689 or A = 0.969 rounded up.

So we have a proof that concurs with my earlier calculation using the unloaded resistor values. Mathematically, it is not important whether you use loaded or unloaded values when R is the mean of anode and cathode values and for the gain calculation at least, you can save yourself the bother and just use the unloaded figures without the additional load in parallel calculations. The loaded values come into their own however, when we want to calculate the input resistance with the bootstrapping equation.

Fortunately for us with the 5E3 we have Ra and Rk both at 56k and with a little sleight of hand we can use this feature to convert our previously calculated Ra (loaded) into Rk, and Rk' (loaded) as a substitution of the Rk + Rb denominator in the formula:

PI equations (1) bootstrap formula.jpg

When Rg = 1000k; A = 0.969; Rk = 44.64k; Rk' = 45.59k

Rin = (Rg / 1 - A (Rk / Rk'))
Rin = (1000k / 1 - 0.969 (44.64k / 45.59k)) = 19534.32933k

or about

Rin = 19.5Meg

In the grand scheme of things this is only marginally higher than my unloaded calculation which determined Rin = 18.1Meg, and it remains to be seen whether the derivation of loaded/unloaded values will have much effect on successive calculations for the feedback net in the Tweedle Dee. The 5E3 of course, doesn't have LNFB. One thing is for certain however, with these values, the effect of loading on the previous gain stage is negligible and AC loading for V2a might be simply ignored, with DC loading telling you all you need to know about that stage. It deserves a mention as well, that the values and data I have been using are mostly theoretical whereas @Ten Over's values seem to have been the product of direct observation. The mu of the 12AX7 or any of its equivalent models is considered a theoretical constant: 100. And for testing, this is important because it allows for benchmarks to be established and maintained in controlled and reproducible calculations. In reality, if you have read this far and have even a basic understanding of tube construction, you will know the mu (μ) along with the mutual conductance - or just transconductance these days - (gm) and plate resistance (ra) are valve coefficients that differ from tube to tube. Matching these are what we pay for when we ask for "balanced" tubes. The Radiotron Designer's Handbook 4th Ed. gives a more complete description of these and other coefficients: Section 2.1, page 14.

Something else worth mentioning in passing, with reference to the idea that the grid leak of the TD was not 1Meg but 10Meg, is that if you factor these values into the input resistance formula the result would be increased by a factor of ten! If we did the same here, with the 5E3 using @Ten Over's tube, the Rin = 195.4 Meg. Again, for the loading on the previous stage, it would hardly matter but now, the high pass, CR filter, with the .02uF coupling cap between V2a and the PI would be the open door to interference on the grid. The fc at the grid of the PI is already very low if you consider the bootstrapped input resistance of the Rg = 1Meg. Make that frequency cutoff ten times smaller with a 10Meg grid stopper. And again, I don't think it is likely the grid leak was 10Meg.

[Stephen1966]

Here are the load lines for the TD cathodyne.

TWEEDLE DEE - V2b - 12AX7 charts.jpg

I've included this just to establish the plate resistance ra which can be cross referenced with the gm and the μ. Reassuringly, the predicted effective plate voltage of 258.53V is very close to the voltage Charlie measured going to the plate, V2b (pin 6), 256.9V

ra = 50V/0.54mA = 92.6k

gm is = 0.00108 mho

[Stephen1966]

We already know the bootstrapped input resistance for the unloaded circuit: Rin = 13.5Meg, so we can roll through the same formula to find the loaded input resistance using the same power tube 220k grid leak resistors in the calculation as before - the TD is like the 5E3 at this stage and beyond.

Recycling some of our previous figures here:

When Rg = 1000k; Rk = 68.3k and Rk' = 99k + 4.7k || 220k = 70.48k; A = 0.97

Rin = (Rg / 1 - A (Rk / Rk'))
Rin = (1000k / 1 - 0.97 (68.3k / 70.48k)) = 16665.87837k or about 16.7Meg

Now, for our feedback formula we just need a few other calculations.

[Stephen1966]

The bootstrapped input resistance aside now, it does not work in the closed loop gain equation if we imbalance one side the equation with a bootstrapped Rin but leave Rf (the feedback resistor) as found. Instead, I am proposing we treat the 1Meg grid leak as Rin = 1000k.

Open loop gain (Ao)

PI equations (1) - open loop gain.jpg

For this, we need the total load Ra' = Ra || (Rf + Rin) = 110k || (3300k + 1000k) = 107.256k

(If we were to add Rk = 99k to this it would not make any significant difference: Rf + Rin is already a large value.)

When Ra' = 107.256k; ra = 92.6k

Ao = 100 x (107.256/(107.256 + 92.6)) = 53.66 (we can ignore the minus sign for mu)

Ao = 53.67

Output impedance before feedback (Zo)

Can be expressed simply as Ra || ra, or - even simpler - R || ra. This perhaps represents the biggest fudge because it's clear from the values chosen that MrD was aiming to create an imbalance in the non-inverting and inverting sides of the PI. It could be argued that really we should be looking at the output impedance of the anode alone, because the NFB is on that side, takes its signal from the anode and feeds it back, from there to the grid. Output impedance at the anode and the cathode are not separately calculated the same though and to paraphrase Blencowe, "when the outputs are equally loaded, the output signals should be perfectly balanced" and one of the ways to achieve this is treat the output impedance as that seen between the non-inverting and inverting outputs. The figure Blencowe uses (39k) is much greater than the modest values his differential output impedance formula (8.3) on page 145 would suggest though and it is the anode in parallel with the internal anode resistance formula (1.6) on page 31 he uses in the feedback equation instead. In order to make it work here we are playing with the idea that R = Ra = Rk.

When R = (Ra + Rk') / 2 = (110k + 103.7k) / 2 = 106.85k; and, ra = 92.6k

Zo = (R x ra) / (R + ra)

Zo = 49.61k

The feedback fraction (B)

This is the gain of the feedback potential divider and why I think it is necessary to treat Rin as an unmolested 1Meg as per the grid leak resistor's actual value.

B = Rin / (Rf + Rin) = 1000k / (3300k + 1000k) = 0.2325

B = 0.23

With these values determined we can now run the formula for the overall closed loop gain.

Closed loop gain (A)

PI equations (1) - equivalent feedback equation.jpg

When Ao = 53.67; B = 0.23; Rf = 3300k; Zo = 49.61k and Rin = 1000k

A = (Ao / (1 + AoB)) x ((Rf + Zo) / (Rin + Rf + Zo))

A = (53.67 / 1 + (53.67 x 0.23)) x ((3300k + 49.61k) / (1000k + 3300k + 49.61k))
A = 4.022 x 0.77 = 3.097

A = 3.1

The feedback factor (Ao/A)

So with these figures we reduce the gain by the factor of 53.67 / 3.1 = 17.3

or, expressed in dB

20 x log(10) (17.3) = 24.76 dB

Plugging in the loaded values we have:

Ao = 43.77
Zo = 40.48
B = 0.23 (unchanged)
A = 2.38

This provides us with a feedback factor of

Ao/A = 18.39
25.29 dB of feedback

In both cases, the amount of feedback seems high!

[Ten Over]

I'm not really following much of that, so I took off in my own direction.

The cathodyne already has 50% NFB, 30.2dB of gain reduction, before the 3.3M local NFB thing.

CF 11 PNG.png

So I used the gain without the 3.3M local NFB as the open loop gain.

I used this cheat sheet for the closed loop gain.

NFB Local Equations.png

This is how I figured the cathodyne output impedance.

CF 40 PNG.png

CF 41 PNG.png

There is no physical Rg, but there is the output impedance of V2a which I assigned a value of 38k to.

EPSON643.JPG

[Ten Over]

So I built the thing and got an open loop gain of 1.824 with no regard for significant figures. Using this in the closed loop equation above, I calculated a closed loop gain of 1.766. The observed closed loop gain was 1.774, pretty close.

The observed gain reduction was 1.774/1.824 = 0.973 = 0.24dB.

This small amount of gain reduction from NFB is not surprising when you have 3300k working against 38k. It strikes me as an exercise in the silly.

[Stephen1966]

So I built the thing and got an open loop gain of 1.824 with no regard for significant figures. Using this in the closed loop equation above, I calculated a closed loop gain of 1.766. The observed closed loop gain was 1.774, pretty close.

The observed gain reduction was 1.774/1.824 = 0.973 = 0.24dB.

This small amount of gain reduction from NFB is not surprising when you have 3300k working against 38k. It strikes me as an exercise in the silly.

Thank you for the feedback

I think the glaring error with my last proof is the open loop gain - against every fibre of common sense which told me the open loop gain was 0.97 I dug away with the equation for an ordinary gain stage. It was downhill from there!

I'm starting to buy into the idea that the closed loop gain is dealing with some very small numbers but I'm not quite where you are yet. As for the silly, hmm, I think that's a pretty fair comment but I'm also looking at this from the design perspective and there's an argument to be made that such small changes are all part of a master plan. Everything from the decoupled cathodes in the first stage, the tighter filtering between the early stages, the increased voltages, the stiffer rectifier, all this appears to be aimed at addressing one of the shortcomings of the typical 5E3 which was the tendency to break up early on and the flabbier bass. It's interesting to me that he left the power stage virtually untouched but the cleaner, greater headroom from the preamp reins in the bass without sacrificing the mids. I do think the LNFB is problematic though, when it clips, it clips abruptly and hard and that's partly due to the imbalance he deliberately engineered - it seems - to encourage the higher order harmonics. Personally, I think because of the already immense level of negative feedback in the cathodyne, the circuit would be better served with a warmer bias and no NFB would mean a smoother transition into overdrive and softer clipping.

I have the amp right here beside me, so I will be clipping some leads into it shortly. These cheat sheets you have posted are gold though. Thanks again.

[Stephen1966]

The bug in my ointment was the formula for closed loop gain. I ran all the numbers again with some slight differences to @Ten Over's and it came out, pretty much the same.

[Edit: I was thinking I might write up the results of my calculations for you here but the only real differences to @Ten Over's calculations were how I arrived at the coefficients of mu, the internal plate resistances of V2, and the output impedances. These do figure in the final calculations - specifically, as they compare with the observed results - but the method he used, and the closed loop gain formula yielded such similar results that they hardly bear repeating here.]

After this I went to the amp and hooked up the probes.

• Yellow traces - probe between the coupling cap of V2a and the grid of the cathodyne input, before the NFB resistor; roughly equivalent to the open loop gain: Ao.

• Red traces - between the coupling cap and grid leak (220k) on the anode side - closed loop gain on the anode side: A(anode).

• Green traces - between the coupling cap and grid leak (220k) on the cathode side - closed loop gain on the cathode side: A(cathode).

You can see that the input signal (yellow) and output signal on the cathode (green) are (in phase) non-inverting but the output signal on the anode (red) is (anti-phase) inverting. The Vp-p for each are displayed in the snapshots.

This first set of traces is looking at the input and outputs using a 1kHz sine.

SDS00049.jpg

However, to mitigate any capacitive reactance by a reasonable degree, from circuit wiring or the tube's internal capacitance, this next set of traces was made using a 20kHz sine.

SDS00050.jpg

Treating these observations as simple gain-in/gain-out (Ao/A) voltages we have a NFB factor in the order of:

Inverting (anode side):

Ao = 32.4V; A(anode) = 28.8V; f = 1kHz

Ao/A(anode) = 32.4/28.8 = 1.125
20 x log_10 (1.125) = 1 dB

Ao = 11.2V, A(anode) = 10.2V; f = 20kHz

Ao/A = 11.2/10.2 = 1.098
20 x log_10 (1.098) = 0.8 dB

Non-inverting (cathode side):

Ao = 32.4V; A(cathode) = 28.4V; f = 1kHz

Ao/A(cathode) = 32.4/28.4 = 1.141
20 x log_10 (1.141) = 1.1 dB

Ao = 11.2V, A(cathode) = 10.0V; f = 20kHz

Ao/A(cathode) = 11.2/10.0 = 1.12
20 x log_10 (1.12) = 0.98 dB

The numbers don't lie!

[Stephen1966]

The results of the calculations and the observations taken together, really bring into question the efficacy of the NFB on the PI. All the advantages of NFB such as augmented headroom, diminished distortion, extended bandwidth and so on are rendered practically ineffective when the feedback factor is so small. An "exercise in the silly" to paraphrase. Worse yet, I believe there is another negative impact on the performance of the circuit which is discussed in this section from the Radiophonic Designer's Handbook 4th Edition (RDH4):

RDH4 - Voltage feedback from plate.jpg

The resistance (Rf), 'acts as a shunt path...' and 'behaves as though an additional shunt resistance... were connected from grid to earth'.

The cumulative effect is to negate any benefit of the bootstrapped grid leak resistor in the PI and weigh down the ac load on V2a's gain stage. So, it seems the LNFB doesn't just end up doing very little, but because it constitutes a parallel ac shunt as well, it reduces the headroom of the driver in V2a and may actually be contributing to greater distortion much earlier in the shift from low to high gain. I mentioned earlier, the NFB also tends to clip more abruptly and harder and this is hardly usable no matter how delicate you are with the guitar's volume knob.

In conclusion, I think we owe @Ten Over a beer for shining a light on the complexities of NFB in the cathodyne. There is very little in the way of any useful information on this kind of circuit out there and even Blencowe, who is usually a go-to source for demystifying such things doesn't help us to penetrate the darkness around the subject. As much as we be loathe to admit it, we may need to concede that MrD had a similar problem sorting the proverbial "wheat from the chaff" with this as well, and while it may not have hurt the tone of the amp very noticably - or at all - I think we can safely assume, it can't have helped it much either.

I really wanted the results to be wrong, the allure of NFB benefits and what they mean for headroom, distortion and noise, bandwidth and also input and output impedance are something we can all appreciate... when they work. Some experiments though don't give us the results we want. Rather, the results we need.

[Charlie Wilson]

OK, so I am envious of being able to do all those calculations but did you, at some point, simply disconnect the LNFB and listen to the amp turned up? Do you hear a difference with it in and out of the circuit? If you don't hear a difference, then all the calculating may show why but if you do hear a difference then you may be making assumptions on why it is there. An example of this is the assumption that the trim pot is there to balance the PI. I have seen and heard several real deal Tweedle Dees and I am convinced the trimmer is there to do the opposite. Also, after seeing several of these amps I am convinced that the PI grid resistor is 1 meg not 10 meg.
CW

[ElectronAvalanche]

Hi Charlie,

Now I am a bit confused. Is it a 10M or 1M?

Best, Electron

Well, six years later and I think a made a mistake with regards to one of the component values in the Tweedle Dee. After talking to Jelle a while ago and staring at my own photos I now believe that the grid leak resistor on the PI is a 10 meg rather than a 1 meg.
CW

[Stephen1966]

OK, so I am envious of being able to do all those calculations but did you, at some point, simply disconnect the LNFB and listen to the amp turned up? Do you hear a difference with it in and out of the circuit? If you don't hear a difference, then all the calculating may show why but if you do hear a difference then you may be making assumptions on why it is there. An example of this is the assumption that the trim pot is there to balance the PI. I have seen and heard several real deal Tweedle Dees and I am convinced the trimmer is there to do the opposite. Also, after seeing several of these amps I am convinced that the PI grid resistor is 1 meg not 10 meg.
CW

Thanks Charlie, the maths can be pretty tedious but when they pan out I think they can reveal the mechanism in more granular detail. Testing the assumptions is what it's all been about.

In fact, I was messing around with the LNFB in the amp but my first impressions were that it was just a bit noisier without the feedback. If anyone wants to try this themselves, you only need to lift the bottom leg of the 3M3 resistor where it connects up with the grid wire and the grid leak resistor. Using clips on a lead then you listen to it with and without feedback without any major surgery. Just be careful! The little cap on top of the feedback resistor is all that separates you from B+3 and a fancy new hairstyle. You don't need a scope to do this either - you can measure the voltages with a DMM.

This is where we get the reality check. I should have trusted my earlier instincts!

Here are the calculations along with a note about the observed data at the bottom:

Feedback calculations.jpg

And here are the traces; first without feedback and then with.

Signal without feedback.jpg

Signal with feedback.jpg

It was seeing it like this which sent me back to the calculations because if my latest findings using the configuration presented by @Ten Over were correct, I should have seen a much smaller difference between the open and closed loop gain. But this was greater than predicted and performing a rough feedback factor calculation on the measured results in which I didn't change anything - other than add/remove the LNFB from circuit - I got: Ao/A = 1.5, roughly speaking (the calculations above show the more accurate figures). So, I'm sorry @Ten Over but we miscalculated in a couple of places.

The first error was in determining the input resistance. Aiken proposed using the output impedance of the preceding triode as the Rin but in fact it is much too small a value to work in this configuration. Taking my cue from Merlin again, I considered Rf and Rg in parallel, to be Rin. It's a quite approximate figure because the paths to earth return through the cathode and the anode and I haven't taken any detailed account of the plate or cathode resistors but we know they are pretty close and so unlikely to affect the reciprocal by much. AC doesn't really care about any capacitors in the return to earth. And so when I re-estimated Rin using Rf and Rg I was then able to resurrect Blencowe's closed loop gain formula and running the revised numbers in it this time, it gives us the feedback factor which tallies up with the observed results.

[Stephen1966]

Hi Charlie,

Now I am a bit confused. Is it a 10M or 1M?

Best, Electron

Well, six years later and I think a made a mistake with regards to one of the component values in the Tweedle Dee. After talking to Jelle a while ago and staring at my own photos I now believe that the grid leak resistor on the PI is a 10 meg rather than a 1 meg.
CW

I can save you a bit of maths here...

With a 10M grid leak the feedback factor is increased to 2.48 for 7.9dB of feedback.

[Stephen1966]

Who was it who said: 'There's no such thing as a free lunch!'

Same here if the observations are anything to go by.

First off, you will see the (yellow) input voltage which is electrically equivalent to the signal at the grid of V2b before the feedback/grid leak divider is unchanged: 44Vp-p regardless of whether the LNFB is in circuit or not. What we don't see and which is kind of trivial anyway, is that when the LNFB is not in circuit, the grid leak is bootstrapped and so presents almost no ac load to the previous stage, V2a. However, when the LNFB is in circuit, the once bootstrapped grid leak now connects to a voltage divider, Rf and so first the bootstrap effect is mitigated and Rg is just Rg again (1Meg) and the ac load on V2a = Rf || Rg = 767k. Not terrible. I was hoping to capture the difference in the break up of signal at the grid of the PI but it's only a few volts out of the total headroom and for our purposes, we might ignore it.

Secondly, you will see that there is a fourth voltage (blue) which represents the power going to the speaker, measured at the passive 8 Ohm load (8.65 Ohm, actual). It's quite easy to see here that just as the output of the cathodyne is reduced, so too is the output power and we go from 9.8 W(rms) without feedback to 5.1 W(rms) with the feedback in place, no other changes to the channel gain being made. On the other hand, we do see an improvement in the headroom of the PI amounting to a relatively modest (theoretical) 6Vp-p where it might have been in the region of 4Vp-p previously. And, similarly, there is a modest extension of the bandwidth and linearity of the circuit when noise and distortion are effectively being cancelled out by the negative feedback signal. That's something of a double-edged sword as well, because when the PI is driven to cutoff, the LNFB is starved into non-existence and so there's a rapid transition (abrupt) into even greater (hard clipped) cutoff as the full gain of the triode is unleashed, unrestrained by the normally moderating effect of the negative feedback. It's at this point as well, we see the voltage at the cathode rising sharply. These are effects that you don't often see in a normal gain stage; effects you don't normally get when the tube soft-clips around cut off, and where the transition is smoother and - we might add - more controllable.

There's a trade-off with the addition of the LNFB network and the dispensing of the original 5E3 circuit architecture. Pros and cons. As choices go though, it's an easy decision to reverse if you don't like the results.

[Stephen1966]

I want to look in more detail at the cathode circuit of the Tweedle Dee here and perhaps dispel the notion that it is some way, the trimmer is meant to balance the plate and cathode voltages. With the formulas I will use below, it is a relatively straightforward matter to prove that different plate and voltages are all but guaranteed when we look at the circuit. I am working on a different hypothesis regarding the trimmer but to get there, we need to review and establish the basic principles first. For this I am going to refer to this article by Abert Preisman which is a repost from Aiken Amps https://www.google.co.uk/url?sa=t&sourc ... dg0XjcP5LN:

cathodyne-1.pdf

What we want the formula(s) to show is that for a given input voltage (Vg) we get one output voltage from the in-phase, non-inverting, cathode side and another voltage from the inverting plate side and we need two formulas for this. In Preisman's article, the assumption is that we want a perfectly balanced pair of voltages from the plate and the cathode and in Hi-Fi, where frequency response and minimum THD are at a premium, a balance is more strictly required but in guitar amps and for all practical purposes where a bias resistor might be included in the calculation (or not), the bias resistor itself is usually a much smaller value than the tail resistor and so hardly affects the outcome. In real terms it might be ignored altogether and so Eq.11 in Preisman's article would be the one to use for both plate and cathode when R = Ra = Rk or 'Z_k = Z_L = Z' as he puts it. Substituting the circuit desigations into his formulas for cathode and plate output voltages, we have, respectively:

Non-inverting output voltage.jpg

and

Inverting output voltage.jpg

The output voltages are an expression of gain and are given as Ap (plate voltage output) and Ak (cathode voltage output) accordingly. In these calculations, you will see that when the resistors in the plate and the cathode parts of the circuit are the same, that is, when Ra = Rk (in this case = 56k) then the output at Ap and Ak is the same. Also, that when we add the bias resistor into the calculation we get a fractionally different output voltage at the cathode. As before, we are using a 12AX7 (or equivalent tube) with a µ = 100. Using the same input signal voltage throughout (Vg = 10V) we have:

Ap and Ak output voltages.jpg

Normally we would not bother calculating the result to so many decimal places but it was necessary here to show that even for a small change in the value of the resistors on either side of the triode there is an accompanying small difference in the output voltage between the inverting and non-inverting sides. The differences here, with the model of the Fender 5E3 circuit values, it's easy to see that for all intents and purposes, with an input signal voltage of 10V the output at the plate and the cathode would be 9.7V. Which, agrees with our earlier calculations for the open loop gain where the formula:

Open loop gain formula.jpg

You may recall this formula gave us an open loop gain factor of 0.97.

This, as Preisman remarks, is 'in spite of the fact that one is driven by a high impedance source and the other by a low-impedance source.' Even if we don't dig any deeper into the mathematics to validate that last statement, we can accept Preisman's opening remarks:

It is stated in various texts that the impedance seen looking in the plate circuit is different from that looking into the cathode circuit, yet for equal loads, the output voltages and frequency response are identical.