
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Tubes / Valves All about our sweet vacuum tubes :) Threads about Musical Instrument Amps of all kinds should be in the Instruments & Amps forum 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
10th December 2014, 09:57 AM  #1 
diyAudio Member
Join Date: Apr 2014
Location: US

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=i1i2 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. 
10th December 2014, 10:31 AM  #2 
diyAudio Moderator

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.
__________________
And while they may not be as strong as apes, don't lock eyes with 'em, don't do it. Puts 'em on edge. They might go into berzerker mode; come at you like a whirling dervish, all fists and elbows. 
10th December 2014, 10:32 AM  #3 
diyAudio Member
Join Date: Mar 2002
Location: Glasgow

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 commongrid side gets lost as variations in the current in Rk. This small signal current is i1i2. 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.
__________________
"Some of the arguments conjure images of whitecoated engineers with putty in their ears, designing audio equipment and not caring how it sounds, only how it measures. I have never met such a person"http://scopeboy.com/amps 
10th December 2014, 10:48 AM  #4 
diyAudio Member
Join Date: May 2007

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. 
10th December 2014, 05:46 PM  #5 
diyAudio Member
Join Date: Apr 2014
Location: US

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. 
10th December 2014, 05:52 PM  #6  
diyAudio Moderator

Quote:
Quote:
__________________
And while they may not be as strong as apes, don't lock eyes with 'em, don't do it. Puts 'em on edge. They might go into berzerker mode; come at you like a whirling dervish, all fists and elbows. 

10th December 2014, 06:17 PM  #7  
diyAudio Member
Join Date: Apr 2014
Location: US

Quote:
The one with resistor tail will work. 

10th December 2014, 06:21 PM  #8 
diyAudio Member
Join Date: Feb 2010
Location: Helsinki

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. 
10th December 2014, 06:22 PM  #9 
diyAudio Moderator

Try building it.
__________________
And while they may not be as strong as apes, don't lock eyes with 'em, don't do it. Puts 'em on edge. They might go into berzerker mode; come at you like a whirling dervish, all fists and elbows. 
10th December 2014, 09:28 PM  #10 
diyAudio Member
Join Date: Apr 2014
Location: US

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. 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Question about CCS used in Morgan Jones' book.  Alan0354  Tubes / Valves  24  11th December 2014 04:07 AM 
Morgan Jones Kapacitors  Nick F  Tubes / Valves  30  4th January 2013 06:59 AM 
Morgan Jones' CCS's  Kay Pirinha  Construction Tips  10  15th June 2012 07:41 AM 
Morgan Jones book to swap  pingfloid  Tubes / Valves  3  8th May 2004 05:19 AM 
MY Morgan Jones headamp.  digi01  Tubes / Valves  9  25th April 2003 02:19 AM 
New To Site?  Need Help? 