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How determine p-g1 capacitance in trioded pentode

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I have a small signal pentode of which only the pentode-connected capacitances are given in the data-sheet:

Input 3.6 pF
Output 2.95 pF
anode-grid1 less than 0.008 pF
anode-cathode less than 0.025 pF

How can I determine (measure, calculate, estimate ...) the respective capacitances when triode connected, especially the plate-grid1 which should be much higher then ?
 
The effective grid-plate capacitance of a pentode connected fro triode operation is essentially a tad more than the grid-screen capacitance. So without the grid-screen capacitance listed in teh datasheet, you can't calcutate it from the datasheet values given.

You could ROUGHLY estimate the grid-screen capacitance if you can see the grid rods poking through the micas. Assume the grids are sheets for this purpose, and use the standard formula for capacitance between two concentric cylinders.

The best way to measure it is by using the "Q-Meter" method. Failing that, set the tube up as a colpits oscillator at about 10 MHz or so. Measur the oscillation frequency with two different values of tuning capacitance. By algebraic manipulation, you can, from the two values of tuning capacitance, derive the extra capacitance contributed by the tube in order to get the ration of frequencies that you measured. Frequency is inversely proportional to capacitance squared. Due to the capaciatnce of teh tube, the frequency ratio will be slightly less.
 
When triode connected the g1 - c capacitance remains the same 3.6 pF, but Cg1-a increases essentially due to Miller effect and the change in connection.
You can estimate that capacitance by putting a relatively big series resistance between signal source and g1 and measuring the frequency where gain of the stage drops 6 dB.
At this frequency the capacitive impedance of the tube input is equal to used series resistance.
Then simply calculate the capacitance (C = 1/2pi x Xc). Remember to deduct the 3.6 pF g1 - a. The rest is Cg1-a including Miller effect.
The real Cg1-a can be calculated by dividing the above capacitance by A (amplification) of the stage.
 
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Thank you Keit and artosalo for your inputs.
I didn't think of actually calculating c from geometry, but why not; although in my case this could be tricky because it's a sovjet miniature rod tube (they are not often used and nobody seems to have any substantial c-data) ; it has 4 shield, 4 g3, 4g2, 2g1, 2 anode -rods and the filament, some of the spacings are sub millimeter ...
I guess, I will try both, the oscillator and the gain method and see whether they give similar results.
In the meantime I found another method on the web, but not shure about that http://www.radiomuseum.org/forumdat...ture_Tubes/1Zh37B_Capacitance_measurement.GIF
 
Don't forget that the C's will be a little bigger when there is space charge present during operation.

I have just used a digital C meter before, with the lead wires spread out on an insulating surface and a long stick on one to just touch the lead to the g1 pin, or not. So as to read the delta C, without disturbing the wire positions. Seemed to give readings close to the datasheet.

It is the extraneous lead wire capacitances that will cause accuracy issues, so the delta test at the pin is helpful. Of course, the real circuit will add those lead wire capacitances right back in.
 
2nd thoughts:
since the tubes are actually amplifying in both methods mentioned above, they seem to measure "Miller-effected" capacitances rather than just bare c, something like (cag1)*(1 + effective Av);
which brings up the question what effective Av is in case of the oscillator ...

unfortunately my c-meter is not good enough, just 10pF resolution ...
 
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I finally fell for the method described by Joe Sousa in http://www.radiomuseum.org/forumdat...ture_Tubes/1Zh37B_Capacitance_measurement.GIF.
It basically uses a capacitive devider between the tbd cap and a known cap, a signal generator and a scope.
I took a 47pF+/-1pF styroflex as reference, 100kHz square wave at 25v pk-pk together with a 15pF probe.
To validate the setup I first measured another 47+/-1pF cap and it came out as 46.8.
Then a 4+/-1pF which tested as 4.1pF, finally a 10pF as 10.2, so it seems to work quite well.
Now for the tube, a 1j24b (1Ж24Б), pins not under test grounded:
ca-g1 = 2.4pF (trioded) vs. 0.15pF (pentode)
cg1-k = 2.6pF (trioded)
ca-k = 2.0pF (trioded) vs. 0.3pF (pentode)
I am not too surprised that 0.008 and 0.025 from the data sheet came out as 0.15 and 0.3, it just shows that my error is on the order of 0.2pF.
 
The beauty of the oscillator method is that it does NOT measure the miller capacitance, and there is NO need to ascertain Av. It measures the true capacitance - that's why I suggested it. It's main weakness is that the calculation involves subtracting two mumbers to get a small number - this sort of calc always magnifies measurement error. However these days with frequency counters we can measure frequencies very accurately far more accurately than measuring an sort of capacitance directly. Also, the Cak and Cgk capacitances, considered in series, also add to the effective capacitance.

"Miller capacitance" is not a physical capacitance - it is merely a notional capacitance that has the same effect on input current as a capacitance to earth in the input of an ideal inverting amplifying stage. Since the current thru Cga is dreiven by a anode signal Av times the input, the input current is then Cga x (1 + Av).

In a colpits oscillator, where the tank circuit capacitance is between grid and anode, Cga is merely part of the tank circuit capacitance (parallel) and is not multiplied by Av.

The increase in tube capacitances due to space charge is very small. I've measured them by the Q-meter method and for small receiving tubes it is of the order of 0.03 to 0.1 pF for Cgk. Cga and Cak are affected much less.
 
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