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Old 10th December 2014, 09:57 AM   #1
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Default Question about LTP in Morgan Jones' book

I have question in page 400 of Morgan Jones 3rd edition book that I cannot follow. It is regarding to LTP calculation proving that you need Rk to be very high in order to make the plate resistor R1=R2.

v1=i1R1, v2=i2R2.

It said if V1 matching V2 and in balance condition ( no input signal), then

v1=v2 => i1 X R1 = i2 X R2. (1)

Gain of V2 is A2 => v2=vgk A2 => vgk=v2/A2. (2)

For V2 being common grid, vgk= ik X Rk where ik is the AC cathode current.

A) This is where I don't understand: I thought ik should equal i1 + i2, not i1 - i2 as the book said.

Let's just continue on using assumption ik=i1-i2 in the book.

i1 X Rk X A2 =i2 X (R2 + Rk X A2) (3)

B) The next step the book want to substitute (1) into (3)!!! I thought (1) is for balance condition, then how can you substitute into (3) that derived from imbalance condition that vgk exist? What am I missing?

Thanks

Last edited by Alan0354; 10th December 2014 at 10:23 AM.
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Old 10th December 2014, 10:31 AM   #2
SY is offline SY  United States
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Remember the convention that the lower case signifies signal current (or voltage). The signal current in the cathode resistor is the difference between the signal currents in each half (since they are 180 degrees opposite phase). Intuitively, you can see that this is correct since as the current in the tail tends toward constant, the signal current through it has to tend to zero, i.e., the signal currents in each half are equal and opposite.
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Old 10th December 2014, 10:32 AM   #3
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Intuitively, what is happening is that one side of the LTP operates as a cathode follower, driving the other side which operates as a common grid amplifier.

If Rk is less than infinity, you can see that some of the signal current coming out of the cathode follower that should drive the common-grid side gets lost as variations in the current in Rk. This small signal current is i1-i2. You can think of it as an unbalance current, or a differential mode signal, as opposed to i1+i2 which is the common mode signal.
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Old 10th December 2014, 10:48 AM   #4
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It would help if he had a labelled diagram, instead of relying on us guessing exactly what he means by V1 etc. It would also help if he had defined precisely what he means by A2 - I think he means A2 = mu2 x R2 / (R2 + ra2) i.e. the gain a grounded cathode stage would have, or a grounded grid but ignoring the source impedance.

The signal voltage developed at the cathode arises from the difference in the signal currents, because he has already slipped in a minus sign somewhere.

The equation you have labelled (1) is true when output voltage balance occurs, and is approximately true otherwise. I suspect you may be confusing output voltage balance with current balance - they are not the same thing unless Rk is infinite. It is perfectly OK to combine voltage balance with current imbalance.

I would put the formula for A2 into his last equation, so the the result is expressed in terms of valve parameters and circuit values. I would also present the result the other way round, so that the degree of voltage imbalance arising from equal anode resistors can be seen too - it is not as bad as it might seem.

Anyway, his basic result stands: you need high Rk or high gain (i.e. high mu) for good balance with equal anode resistors. Having said that, there is little point in striving for balance which far exceeds the balance of the output stage.
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Old 10th December 2014, 05:46 PM   #5
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Thanks all in replying. The problem with Jones is he does not label anything. I am sure the result is correct, but how does he get there. I want to go through the equation because I feel this is one of the more important equation. Also, in equation, I cannot go by intuition, it has to be exact. This why I spent a lot of time typing the formulas out here.

Yes, I understand lower case is AC signal. Therefore i1 and i2 can be positive or negative and they have to be opposite polarity from each other. I use ik = i1 + i2 because I have to assume one is negative. That's how every other text book I used before.

I even have question with (1) because R1 is NOT defined to be equal to R2. In triode, if R1 not equal to R2, you create imbalance voltage and you change the current through V1 and V2 because of ra. v1 will not equal to v2 if R1 not equal to R2. That's the reason I specified in my post R1=R2. You just cannot have i1R1=i2R2 unless both tubes are balance AND R1=R2.

I just don't see how he can substitute (1) into (3) because (1) is for balance where i1R1=i2R2.


I really don't like this book, I just cannot find a better one for hifi amps, In text books, if you derive a formula, it has to be precise, well labeled. I definitely have a better book in tube guitar amp, problem is it does not cover much in lowering distortion as guitar amp goes the opposite route. I just feel formula is not precise in Jones' book. There is still an open question on using CCS for LTP in my other thread. Two people came out and said they have problem with the circuit described by Jones.

Thanks

Last edited by Alan0354; 10th December 2014 at 06:14 PM.
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Old 10th December 2014, 05:52 PM   #6
SY is offline SY  United States
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Quote:
Originally Posted by Alan0354 View Post
I just don't see how he can substitute (1) into (3) because (1) is for balance.
Since he's solving for what the resistance values have to be in order to achieve balance, that's the constraint.

Quote:
Two people came out and said they have problem with the circuit described by Jones.
And yet, it worked exactly as advertised when I built it. And the Bulwer-Lytton amp in VA4 works with that same circuit.
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Old 10th December 2014, 06:17 PM   #7
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Quote:
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And yet, it worked exactly as advertised when I built it. And the Bulwer-Lytton amp in VA4 works with that same circuit.
I don't see the one with CCS on top and bottom can achieve balance. Because the tube has no control on the current no matter how you use the ra. The balance totally relies on the two CCS to be exactly equal which it is not likely.

The one with resistor tail will work.
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Old 10th December 2014, 06:21 PM   #8
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If the tail is a couple of orders of magnitude stronger, then it works just fine. Then the top CCS is "in the same category as the tube" in terms of current control; but when looking at the top CCS & tube, the CCS wins so it controls the loadline.

Say, put a darlington cascoded CCS in the tail, and a single MOSFET CCS on the plates. Or, like I do, gyrators.
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Old 10th December 2014, 06:22 PM   #9
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Quote:
Originally Posted by Alan0354 View Post
I don't see the one with CCS on top and bottom can achieve balance.
Try building it.
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Old 10th December 2014, 09:28 PM   #10
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Originally Posted by SY View Post
Try building it.
I specifically refer to Fig 2.49 in page 136. This is my theory. Maybe you have other reason why it work and I like to hear it as it is still open. This is my theory:

The grid of the tubes do not conduct current. So what goes into the two cathodes has to come out from the two plates. This means whatever current sourced from the two current sources of MPSA92 HAVE to be sink by the CCS at the bottom. There is no other path. The only way the circuit can operate is the bottom current is EXACTLY equal to the top two CCS combined. Or else, the plate voltage is going to drift.

Engineering is not about making one work. If you tweak the current, you might be able to make it work. But it will drift with temperature, device variation and resistor variation. It is not going to be stable.

The one in Fig.2.51 works because you have only the top current source, the tube can adjust the plate voltage to adjust the current to match the CCS. this is because ra comes into play. This is a solid design. Not the one in Fig. 2.49.
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