# So this looks fun: Matchless Spitfire

Discussion in 'Shock Brother's DIY Amps' started by JuneauMike, Jan 4, 2019.

1. ### petebFriend of Leo's

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thanks

the answer is in there. I still need to understand it better. I will read this answer some more.

I'm getting that there is some 'self balancing' going on in the LTPI an that the balanced load resistors are part of the balancing, but why then do they use 82/100 unbalanced plate load resistors. that's what I am really trying to understand is that it takes unbalance to improve the balance???

2. ### JuneauMikeFriend of Leo's

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The out of phase output is made smaller through resistance to balance the output. The output voltage is related to current across the plates from the input signal. With one input signal, they wouldn't be equal. So the resistor compensates. (If I understand it)

When you use the second stage of the LTP as another input, the output resistors are of equal value.

Last edited: Jan 11, 2019
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3. ### Ten OverTele-Meister

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I think you are jumping around, but the skinny is that the LTP PI in guitar amps is not ideal and some compromises have been made that results in unequal outputs. Some people use unequal plate resistors to balance it out and some use a voltage divider on one plate to equal it out.

There are some Hi Fi designs that actually use a current source in their splitters and they use equal plate resistors.

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4. ### Tom KamphuysTele-Meister

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#### Attached Files:

• ###### Two_Triode_PI_Clare.pdf
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Last edited: Jan 11, 2019
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5. ### noah330Friend of Leo's

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I don't have the schematic I own the map.

I think this is it. I had a Spitfire and they're very, very close sounding boxes. So close that I actually sold my Spitfire because it wasn't worth owning both.

http://www.prowessamplifiers.com/schematics/images/GA_15RV_complete.pdf

6. ### Sean MacTDPRI MemberAd Free Member

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I have been following this thread with great interest and possibly a little confusion for the last few days

It's interesting to think about the theory behind something we know works fine.

The Vox design grounds both sides through 500K pots where the single input Fender style has its unused side straight through a cap to ground.

So the impedance changes as the pots on either side of the Vox PI are adjusted, if I'm thinking about it correctly.

Cool project, I hope it goes well for you.

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7. ### JuneauMikeFriend of Leo's

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Thanks Sean. I won't be building it anytime soon, but it's definitely going to be on my list for a while. If you're interested I've uploaded the schematic.

I've enjoyed this thread too. I seem to have a knack for innocently asking simple questions and then watching them explode into a deeply complicated discussion that I'm barely able to participate in. Ha.

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8. ### SnfoilhatTele-Holic

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Complication exists on different levels or layers, pick your metaphor. Sometimes what is complicated at one level can be conceptualized or modeled or even 'really' concretely works in a simple way, when viewed from another level. Biology is full of this sort of thing.

With a large enough sample or subtle enough statistical tools, one can find differences between two groups of things with a high degree of confidence. The question raised, and a crucial one to try to answer, is whether the discovered difference is large enough to matter in any way, and if so in what ways, under what conditions.

Everything in the thread so far is interesting and valuable in its own right, but no one has plugged into their 1960 Vox AC10 and listened to the Vibrato channel with the Normal channel volume at 0 (PI grid 2 AC grounded) and at 5 (some resistance, looks like 50k||220k) and at 10 (looks like 500k||220k), and told us what the amp sounds like.

Same for the Matchless Chieftain Reverb. What are the results of this listening test? Turning the Reverb up high might make the PI behave crappy, might make the gain drop and take the smile off all out faces. But it might not. And if a 50k pot or a 100k pot doesn't upset the apple cart, then where do we go next? Try to squeeze more gain out of the reverb recovery, try to get away with a 250k? Whatever direction it goes in, the design goes somewhere, instead of being stuck at square one.

Even if we are very pro theory and very skeptical of 'proven' designs (because the music instrument industry has had some amateurs working in it, that seems sure, and music history has made some questionable electronics into legends), we can't answer @JuneauMike 's question until there are some real design constraints to set boundaries for the design. What is the budget, what is the chassis size, what are the sound/tone/behavioral needs of the player? Why talk about LTP PI performance at all (other than as a very cool discussion of electronics background) if we can ditch the LTP and install a mixing network and an additional gain stage, like Fender did in the Princeton Reverb? Everything is tradeoffs. I suppose knowing the ideals may let you make the best trades. But the end goals of a guitar player are not simple or obvious--the holistic behavior of the amp. This is an opposing phenomenon to that which I first mentioned. The error of reductionism might lead one to believe that a great sounding amp is a bunch of well-tuned parts. Many circuit blocks are relatively simple, but the speaker-OT-power amp-preamp-instrument-player system dynamically +interactively working across full frequencies and all the typical signal amplitudes (that is to say, playing guitar) is really complicated! So these very technical seeming iterative steps in design elements may not have the most black and white ideals either.

Last edited: Jan 12, 2019
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10. ### petebFriend of Leo's

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So it turns out that the LTPI is not self balancing, it’s just two common cathode triodes and they can be un balanced.

I believe there is another splitter called the self balanced phase inverter used on even older fender amps.

So the triodes need to be balanced in order to be balanced.

Not only the out puts need to be balanced, but everything needs to be balanced. The entire table needs to be set.

Essential to the operation of The LTPI is that the signal on the combined cathode must be half way inbetween the signal on the two plates. That’s why they have balanced grid leak resistors. If this balance was upset, say one was changed from 1 M to 500 K then the LTPI balance would be lost.

11. ### JuneauMikeFriend of Leo's

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I think I read it differently. They are self balancing in that the ratio of the two outputs are consistant across an array of voltage input levels. But the are structurally flawed, and the output resistors correct that measurable and predictable flaw. When both inputs are used to send a signal, the output resistors don't need different values to correct output voltage and current.

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12. ### Ten OverTele-Meister

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The grid leak resistors don't need to be the same. There are different considerations for each one.

Grid Leak 1 is the load that the previous stage sees and we don't want to load it down too much. Most other places we want the biggest grid leak we can have but not so big that grid current starts to offset the DC to an unacceptable level. It is the same for the LTP input, but this time the input impedance is about twice the nominal value of the grid leak. So we could use a 500K and still have the usual 1M load for the previous stage. The balance actually improves with a smaller Grid Leak 1 because there will be more AC current through it and that current is sourced through the tail resistor. More current through the tail resistor means a greater voltage drop, so the stages don't need as much current imbalance to get the necessary cathode voltage. Part of the cathode voltage has already been supplied by the greater voltage drop across the tail resistor.

Grid Leak 2 is in parallel with the tail resistor for AC purposes. Lower values for Grid Leak 2 would be the same thing as lowering the value of the tail resistor for AC, but the DC stays the same. Lowering the resistance of the tail causes a greater imbalance between the outputs. You would be better off leaving the grid leak high and lowering the value of the tail resistor since that would at least improve headroom.

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13. ### Ten OverTele-Meister

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The problem with dissimilar plate resistors when both inputs are used is that it will balance the output of one side while throwing the other side further out of balance. A compromise must be made.

14. ### petebFriend of Leo's

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If they are self balancing. What exactly is causing the balance?

15. ### petebFriend of Leo's

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Thanks for the excellent explanation.

What you said above about grid leak 1 surprised me. I did not notice that the previous stage did not have its own plate load resistor. Often a tube will work both into its own plate load and into the next stage’s grid leak. That’s an interesting detail about the LTPI.

Ten over, I am not disagreeing with you, your explanation stands. I am just going to continue a little further to explain what I see. Please don’t feel the need to correct me, but if you do, that’s fine too.

From kirschoffs voltage drop rule, the signal where the two 1 M grid leaks meet must be half way between the signal on the two grids. Like I said, this is essential to the balance on the LTPI, but this signal is not the same as is on the cathode. And the signal on the cathode is the one that really needs to be mid way between the signal on the two grids. The signal on the cathode is different than the signal where the two grid leaks meet by the amount of signal voltage increase over the cathode resistor, which is the smaller of the two cathode resistors, so this voltage change over the cathode resistor is small, leaving the signal close but not equal to the signal where the two grid leaks meet.

Could this shift in signal from what it should be, be the reason for the imbalanced load resistors? It looks like it is not.

The little bit of signal voltage added by the cathode resistor would move the cathode voltage up, closer to the signal voltage on grid 1 and away from the signal on grid 2, which is a form of negative feedback, lowering the gain on triode 1, meaning triode one would need the larger plate load to compensate, but that’s not how it is, triode 1 has the smaller plate load instead of the larger one.

This question I would like an answer to.

Can any one explain (sorry if I missed it) why does triode 1 use the smaller plate load? It seems that if this were explained, then one could better see what causes the balance in the first place.

I might be wrong, but it seems that the explanation for the unbalanced plate load resistors is that “there is an unbalance” but I don’t think I have read (at least recently) what causes the unbalance and why it gets unbalanced in a certain direction. This I want to know.

16. ### Ten OverTele-Meister

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In post #92, I spoke of increasing the AC voltage across the tail resistor so that the triodes wouldn't need as much imbalance in order to get the required cathode voltage. The typical NFB insertion has a similar strategy except the entire tail resistor is riding on an AC voltage.

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17. ### petebFriend of Leo's

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Ten over,

Did I stumble upon the reason for the unbalance?

I think I did.

I will read yours closer again.

Thank you

18. ### Tom KamphuysTele-Meister

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The reason can be found in the twin triode PDF with some effort. I haven't put the effort in to tell you what it is (yet...).

Why the triode 1 plate resistor is lower (and 82k) is simple using eq. 9:

R2/R1=1+(Ra2+R2)/(uRk)

= 1 + (65k + 100k)/(100 * 10k)
= 1.165

100k/R1 = 1.165
R1=100k/1.165=85k

Edit: changed some values to their actual value. Loading by the power tube grid leak resistors will make the formulas a bit more complicated, but results in a very similar value for R1.

Last edited: Jan 14, 2019
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19. ### Tom KamphuysTele-Meister

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My take on the reason for the imbalance:

Second grid is grounded for AC. Signal applied to first grid.
For output signals to be balanced with equal plate resistors, the cathode should be .5x signal. This can only be achieved by a change in current through the tail resistor. But to get a current change, the outputs have to be imbalanced, otherwise the increased current of one tube is counteracted by the decreased current through the other.
Both situations can thus not coexist.
The circuit 'resolves' this by a bit of imbalance.

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20. ### petebFriend of Leo's

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agreed

did you see my point that the two 1 M grid leak resistors are going to evenly split the signal in half. the junction of the two grid leak resistors have the exact signal needed at the cathode.

can we agree on this?

the problem is that this signal is going to change as it goes thru the cathode resistor before it gets to the cathode. that means the signal on the cathode is not half way between the two grids, but slightly towards one or the other. I think it will rise and become slightly closer to the signal on grid 1. thus, it is the signal voltage developed across the cathode resistor that makes the LTPI unbalanced, until the different plate load resistors re balance it.

it seems pretty straight forward to me.

did i make a mistake in my reasoning?

I see where this comes from: (Blencoe)

Balance:
Because this circuit does not use a constant-current sink in the tail, the two outputs will not be perfectly balanced if both triodes have the same value anode resistors (although balance will be pretty good all the same). The difference in gain between the two outputs is given by:
A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))]
Where Rk is the total tail resistance Rb+Rt.
In this case we can see from the characteristics graph that ra = 40k, mu=45:
A1/A2 = 1 + [(82k+40k)/(47k(45+1))]
= 1.06

And I like Blencoe's next part:

So the inverting output will be 6% higher than the non-inverting output. This could be corrected by making Ra1 6% smaller in value, but in practice it is not necessary; a slightly unbalanced phase inverter is often quite benificial to guitar tone, due to the additional 2nd harmonic it introduces. Nevertheless, this is why the 'traditional' version of the circuit also uses mis-matched anode resistors.

this matches what I was thinking and fixes my problem, or why I thought I was wrong.

I said the signal on the cathode is going to get closer to the signal on grid 1, losing gain here. that's also what Blencoe says, the inverted side has a higher output. now for the fix. triode one or the non inverting triode has less gain. how to balance? I thought to balance this, triode 1 would need more plate resistance, but Blencoe says otherwise, he says the side with lower output, side 1, needs a smaller plate load, which is what is in place on the AB763 pre amps.

Last edited: Jan 14, 2019
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