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Old 11th March 2014, 08:31 PM   #1
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Default 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 author asks to calculate
the value of R2.
He also supplies the answer:
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.
Thanks in advance,
Joe.
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Old 11th March 2014, 10:35 PM   #2
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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.
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Old 11th March 2014, 10:40 PM   #3
DF96 is offline DF96  England
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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.
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Old 12th March 2014, 01:17 AM   #4
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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+
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Old 12th March 2014, 08:35 AM   #5
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Quote:
Originally Posted by nigelwright7557 View Post
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.
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Old 12th March 2014, 12:36 PM   #6
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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
File Type: pdf Opgave12-blz.451-English-I.pdf (122.6 KB, 27 views)
File Type: pdf Opgave12-blz.451-English-II.pdf (112.5 KB, 7 views)
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Old 12th March 2014, 01:03 PM   #7
Merlinb is offline Merlinb  United Kingdom
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Quote:
Originally Posted by DF96 View Post
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.
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Old 12th March 2014, 02:08 PM   #8
DF96 is offline DF96  England
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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.
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Old 12th March 2014, 05:02 PM   #9
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Ok, all theory aside, LTSpice says no-go.
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Old 12th March 2014, 05:13 PM   #10
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Quote:
Originally Posted by funk1980 View Post
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.

How about some comments on my pdf's?
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