• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Cathodefollower and strange behaviour with lowering of anode voltage.

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Let's take an ECC83 with two resistors in series on it's cathode.
R1 (directly coupled with the cathode) is 3,3Kohm and in series with it a resistor R2 of 33Kohm, connected to earth. R3 = 1Mohm and runs from gate to the connecting point of R1 and R2.
My question is : why is the voltage on the resistors in series with the cathode getting bigger, while the voltage on the anode gets smaller?
I cannot see this , nor find the theory.
Is the Ri of the tube changing soo much?
 
Is this a simulation or experiment?

If simulation then maybe you are using a bad model, or using a good model outside its domain of applicability. With a 1M grid resistor you will be affected by any grid current which may be present; valve models often get grid current wrong.

If experiment, then have you built the circuit you think you have built? Are you measuring what you think you are measuring? Is parasitic oscillation an issue?
 
I am mainly asking about measurements of several of the same katodefollowers in the book of R. zur Linde. He uses this configuration several times.
In his schematics I observered this tendency in dc. adjustment of ECC83 as cathodefollor. I didn't build them all, so I was quite confused. about this behaviour.
So, I only read about it; it was no measurement of mine. But was eager to try.
That's why I asked if this was normal behaviour of the tube.
 
How about a drawing, gives you something to puzzle :D
Mona
 

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Siepie, you're making a good argument for owning a variable voltage DC bench supply! If you had one, you could breadboard this in a few minutes, then measure (and plot) the voltages you say you're expecting/reading about.

And you don't even need a variable supply… an anode load resistor which is bypassed with a capacitor! (very odd topology to most, but perfectly reasonable) does the trick also. Can't predict exact voltages, but that's what your digital meter is for. Measure 'em.

GoatGuy
 
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