Unusual Cathode bias resistor config...need some help

[reubster]

I came across this preamp stage the other day and am curious as to what it does and how I apply calculations for the gain and bias.
I would assume that the grid (bias) voltage can be calculated from the voltage across the 1.8K resistor and that the addition of the 22K resistor will create greater neg feedback .......so therefore the overall gain of this stage will be lower than for a typical cathode resistor value (eg 1.8K).

Lets assume B+ = 300VDC
So for the DC load line the cathode resistor is (1.8 + 22K) and the overall current is 300 / (100K +23.8K) = 2.42mA
But this gives me a grid voltage of 1.8K x 2.42Ma = -4.3V which seems too low/close to cut off.

I'm sure doing something wrong.

Any comments would be much appreciated

[es345]

Looking to your schematic:

the grid is referenced to the cathode resistor of 1,8 K. Take this resistor only for your calculation.
The 22K resistor is an additional NFB. The amplication of this gain stage will be something like 4 (my estimation without calculation)

[reubster]

Thanks ES

I'm just wondering if you could elaborate on your answer.....How did you estimate the gain?

I'm trying to do this from a loadline, but am having some chicken & egg ground hog moments: do I start with the anode current and work out the grid voltage or do I pick a grid voltage and make it fit onto the load line?
I see how I can ignore the 22K resistor because, the grid voltage is referenced across the 1.8K, but would it not come into play when calculating the anode current?

Too many questions make my head hurt

[es345]

I'm just wondering if you could elaborate on your answer.....How did you estimate the gain?

It is a rough estimation methode working if you have a cathode resistor without bypass cap.

Take your schematic:

The current through the Cathode resistor(s) is the same as the one going through the plate resistor.
Let us assume we would have an ideal triode: unlimited amplification

The cathode Voltage signal will then follow exactly the Voltage change at the grid injected (cathode follower),
The Cathode current will change accordingly. As the Cathode current is the same as the anode current the plate voltage will change in the opposite direction based on the size of the plate resistor. If now the plate resistor is e.g. 5 times higher than the cathode resistor, you have a resulting amplification of 5. That means: the ratio - Plate resistor/cathode resistor) - equals to the amplification of the circuit with an ideal tube.

In reality the tube amplification is limited. Therefore this estimation is an upper limit. But the closer the ratio of Plate to Cathode resistor is coming to 1 the better the estimation is. Remember the Cathodyn inverter where the Cathode resistor and the Plate resistor are basically the same (omitting the Cathode bias resistor). There you have an amplification of one with 2 outputs (at Plate and Cathode) but phase inverted.

[reubster]

Thanks ES

GREAT EXPLANATION