I'm using a 6AU6 in triode mode but with g2 acting as the anode; g3 and plate are both grounded. I intend using a 1M log pot as a violume control immediately before the tube, so the input capacitance is important because of the HF cut-off at low volume settings.
I think the effective input capacitance of a 6AU6 used in this way will be = mu x g1-to-g2 capacitance (Miller) + g1-k capacitance + g1-plate capacitance + g1-g3 capacitance.
The last term (g1 to g3) is probably small enough to ignore. I don't know the g1-g2 capacitance but I would guess around 1.5pF. Any ideas please?
I think the effective input capacitance of a 6AU6 used in this way will be = mu x g1-to-g2 capacitance (Miller) + g1-k capacitance + g1-plate capacitance + g1-g3 capacitance.
The last term (g1 to g3) is probably small enough to ignore. I don't know the g1-g2 capacitance but I would guess around 1.5pF. Any ideas please?
My printed GE data sheet shows triode mode (screen, suppressor and plate tied together) as G1-P = 2.6pF, input (G1-K) = 3.2pF, output (P-K) = 8.5pF with shield, 1.2pF without. So for a voltage gain of say 30, your input C will be around 2.6 * 30 + 3.2 = 81.2pF. Grid-screen capacitance (the only active combination) is the same, plate grounded or not so I doubt your situation will be very different.
But you shouldn't have to ask this question anyway - *throws a quarter at ray_moth* - here, go out and buy some pot! 😱
Tim
But you shouldn't have to ask this question anyway - *throws a quarter at ray_moth* - here, go out and buy some pot! 😱
Tim
Thank you: with that quarter, I can afford to go out and buy a 1M pot., as you suggest 😉
My reason for asking the question is: I thought that, since the plate is grounded, its capacitance to g1 would not be multiplied by mu; only the g1-g2 capacitance would be subjected to the Miller effect. Further, the g1-g2 capacitance should be prettyy low, certainly lower than the g1-plate capacitance, since g2 has a much smaller surface area (although it's also much closer to g1). Am I barking up the wrong tree?
My reason for asking the question is: I thought that, since the plate is grounded, its capacitance to g1 would not be multiplied by mu; only the g1-g2 capacitance would be subjected to the Miller effect. Further, the g1-g2 capacitance should be prettyy low, certainly lower than the g1-plate capacitance, since g2 has a much smaller surface area (although it's also much closer to g1). Am I barking up the wrong tree?
No, because g2 controls the electric field in the area between g1 and g3. It (and g3) shield g1 from P, however.
Tim
Tim
Sure, that's true in pentode mode. However, the way I am using the 6AU6 is not pentode mode at all. It's a triode, with a skinny anode (g2) surrounded by a couple of grounded structures (g3 and plate).
No, it applies EXACTLY the same. It defines the ELECTRICAL FIELD in SPACE. Thing lines of force and stuff. That electric field. The only thing that has changed are the electrical parameters, in this case g1-g2 is now Miller capacitance and g3 and p act as shields.
Tim
Tim
I agree, sorry if I gave the wrong impression. So, what I'm trying to ask is, how can I calculate the Miller capacitance for the way I'm using the tube?
Find g1-g2 C, lol....
Like I said, it won't be much different from the datasheet version, as g1-g3, g1-p and g1-shield are all low capacitance figures, only contributing to static load (like g1-k) anyway, not miller.
Tim
Like I said, it won't be much different from the datasheet version, as g1-g3, g1-p and g1-shield are all low capacitance figures, only contributing to static load (like g1-k) anyway, not miller.
Tim
Hi,
The formula for finding Cin is:
Cin = Cgk + Cgp*(A+1)
In this case with the "trioded" 6AU6 and an assumed gain of 30, Miller effect will be very low and will have little effect on bandwidth provided source impedance is reasonably low.
If you have a capacitance meter you can measure g1 - g2 capacitance (it will be slightly less once the tube is operating).
The -3dB cut off point can be found by using this formula:
f = 1/(2*pi*R*C)
If you're really picky you can augment the value of C by adding stray capacitance but you'll find that bandwidth will be more than ample for all things audio.
Cheers, 😉
The formula for finding Cin is:
Cin = Cgk + Cgp*(A+1)
In this case with the "trioded" 6AU6 and an assumed gain of 30, Miller effect will be very low and will have little effect on bandwidth provided source impedance is reasonably low.
If you have a capacitance meter you can measure g1 - g2 capacitance (it will be slightly less once the tube is operating).
The -3dB cut off point can be found by using this formula:
f = 1/(2*pi*R*C)
If you're really picky you can augment the value of C by adding stray capacitance but you'll find that bandwidth will be more than ample for all things audio.
Cheers, 😉
Thanks for the ecouragement, Frank. I'm using the tube as a preamp stage, preceded by a James (a.k.a "passive Baxandall") tone network and a volume control. This volume control is presently a 225k pot but it really needs to be at least 1M to get optimal results from the tone network. I was wondering if replacing it with 1M would cause unwanted HF rolloff at low volume settings (e.g. with ~800k or more in series with the 6AU6 grid). I haven't noticed such a problem with the present 225k pot at low volume settings.
If that's the case, can you use a CF instead, with like a 10k pot? Though I suppose you need the gain, and that would add another stage...
Tim
Tim
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