Trying to solve a Plate Follower Problem
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diyAudio Member

Join Date: Feb 2011
Trying to solve a Plate Follower Problem

Hello,
In trying to revive my knowledge of tube audio amplifier theory, I am currently solving the problems in a book dealing with vacuum tube fundamentals.
At this moment I am trying to solve problems in a chapter about phase splitters for driving a push-pull stage.
There is one particular problem which I did solve but, am uncertain if my answers are correctly executed. It concerns a so called 'plate follower' or 'anode follower' stage in a two stage phase splitter, where the first stage is a normal grounded cathode amplifier stage driving the second one which has this 'plate follower' configuration. Here is the circuit diagram and the data:

μ= 100; (gain factor)
Ri= 20kΩ;
Ra= 80kΩ;
R1= 400kΩ;
R3= 20kΩ;
v2=-v1

the value of R2.
R2= 467Ω, but not how to calculate this.

Before I am bothering the forum with my solution of this problem I want to ask you if someone is willing to help me with checking my
solution... If so, then I will translate my solution using the English language.
Joe.
Attached Images
 PlateFollower1.jpeg (19.6 KB, 215 views)

 11th March 2014, 10:35 PM #2 diyAudio Member     Join Date: Sep 2011 Location: Groningen Are you sure those numbers are correct?? The only way I can see it working is by reducing R1 to 4k instead of 400k. The massive feedback from R3 would swamp it otherwise. But the 4k would make it a horrible stage. The first triode gives an amplification of 80, but with such a load, it's reduced to about 3 or 4! Or the given answer is wrong and should be multiplied by 1000, but than R3 should be altered to maintain balance between outputs. So... something's fishy.
 11th March 2014, 10:40 PM #3 diyAudio Member   Join Date: May 2007 Either you have misunderstood, or the author is wrong. The circuit with the component values you have given will not work. For a plate/anode follower as a phase splitter you would need R1 and R3 to be about the same value, and R2 to be much larger than 467R.
 12th March 2014, 01:17 AM #4 diyAudio Member     Join Date: Apr 2008 Location: Carlisle, England I just use a single triode phase splitter as it is easy to understand. The anode and cathode resistors are the same and the grid is set to 1/4 B+ __________________ http://www.murtonpikesystems.co.uk PCBCAD50 pcb design software.
diyAudio Member

Join Date: Sep 2011
Location: Groningen
Quote:
 Originally Posted by nigelwright7557 I just use a single triode phase splitter as it is easy to understand. The anode and cathode resistors are the same and the grid is set to 1/4 B+
Yes the 'cathodyne' or 'split load' PI is a nice stage, but not always applicable since is has no gain of its own (you'll need another triode) and looses its balance when unevenly loaded. Horrible when overdriven as well.

I don't think the OP's intention is to settle with a certain PI, but to fundamentally understand the workings, and to mathematically solve, a given tube circuit.

diyAudio Member

Join Date: Feb 2011
Hello forum members,
Thanks for your responses and suggestions!
After having solved this problem for the first time I was also astonished about the extreme low value of the calculated R2. Which I did not expect.
Then I tried to solve it in a somewhat different way arriving at the same result. But I still I have the unpleasant feeling that 'something is wrong' here but cannot detect what is it.
So here the two pdf files with my solutions. Hope you guys can detect where I went wrong.
Now I got the idea to let LTspice calculate this R2 value but did not had the time to do so..
Joe.
Attached Files
 Opgave12-blz.451-English-I.pdf (122.6 KB, 27 views) Opgave12-blz.451-English-II.pdf (112.5 KB, 7 views)

diyAudio Member

Join Date: May 2006
Location: Lancashire
Quote:
 Originally Posted by DF96 Either you have misunderstood, or the author is wrong. The circuit with the component values you have given will not work.
Actually the book is correct, a value of 467 ohms would give unity gain. Admittedly, it is a peculiar way to arrange the circuit.

I used a very quick 'n' dirty calculation:

First estimate the open-loop gain, noting that the load on the valve is roughly Ra||R3 = 16k.
A = 100 * 16 / (16 + 20) = 44
Now assume +44V at the anode, -1V at the grid and -44V input.
Current in R3 = 45/20k = 2.25mA
Current in R1 = 43/400k = 108uA
Therefore current in R2 = 2.25 - 0.108 = 2.142mA
Since there is 1V across it, its value must be 1/2.142mA = 467 ohms.

Last edited by Merlinb; 12th March 2014 at 01:09 PM.

 12th March 2014, 02:08 PM #8 diyAudio Member   Join Date: May 2007 Think again. R1 provides the input signal current for the inverter. 400k feeding around 467R gives an attenuation worse than 0.001. Times the stage gain of -44 you get around 0.04 gain from anode to anode, before you add the negative feedback. You should have checked your arithmetic for plausibility. A plate/anode follower will have similar signal currents in the input resistor and feedback resistor - yours does not. It will also have similar values for the input and feedback resistor - yours does not (n.b. if gain is required then the feedback resistor will be larger than the input resistor - yours is much smaller!). Assuming a circuit works, and then doing the arithmetic to get component values, is a valid way of analysing provided that the assumption is correct. You have mixed up feedback current and input current.
 12th March 2014, 05:02 PM #9 diyAudio Member     Join Date: Sep 2011 Location: Groningen Ok, all theory aside, LTSpice says no-go.
diyAudio Member

Join Date: Feb 2011
Quote:
 Originally Posted by funk1980 Ok, all theory aside, LTSpice says no-go.
funk1980,
Look at this LTspice diagram, especially the second voltage source.
Then run this and look at the results.

Attached Images
 PlateFollower.jpg (206.4 KB, 114 views)

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