Output Stage & calculating halfboost freq. can you doublecheck my results?

[SoulFetish]

Can you double check my results?
I'm calculating the halfboost frequency for a given cathode bypass capacitor value in an output stage. Most of the discussions seem to center around implementation in a preamp stage. In order to achieve accurate results, I'm using this formula:

*
The formula was found in this short article:
http://valvewizard.co.uk/ChoosingBypassCaps.pdf
Here is the datasheet for an EL84:
EL84-2.pdf
*
the parameters for a pair of EL84s operating in a class AB push-pull output stage are as follows:
10k p-p load impedance
220Ω Rk(per tube)
20µF Ck(per Rk)
internal resistance ≈ 38k
µ = 19

solving for f during class B operation I came up with f_(halfboost) ≈ 37Hz (37.125908162)
Is this correct?

[Mike Sulzer]

That equation is meant to be used with triodes, not pentodes. But I guess you can; the mu for a pentode is high. Using the plate resistance you give, 38K, and a transonducdtkance of 11.3 mA/V, I get 429.4.

[pdf64]

From the linked info, the 10k loads are for triode connected EL84; is that your arrangement?
8k is more usual for a pentode p-p arrangement.

Without bothering to calculate the formula, 20uF seems a bit low to fully bypass a 220 ohm individual cathode resistor.

[Malcolm Irving]

The mu value of 19 that you can find in datasheets for EL84 is the ‘inner amplification factor: mu g1 - g2’.
The usual mu value varies so much over the operating characteristics that a value is not usually quoted.
More details about this are given here:
Amplification and Valve Characteristics

[Mike Sulzer]

But that equation for half boost frequency becomes insensitive to mu when mu is large, and so any large value should give nearly the same answer. This also means that a simple equation without mu could be used.

[Mike Sulzer]

So it looks like that square root term just becomes the square root of three when mu gets very large. That makes a very simple equation if I have not served up the math again!

[Helmholtz]

Can you double check my results?
I'm calculating the halfboost frequency for a given cathode bypass capacitor value in an output stage. Most of the discussions seem to center around implementation in a preamp stage. In order to achieve accurate results, I'm using this formula:

*
The formula was found in this short article:
http://valvewizard.co.uk/ChoosingBypassCaps.pdf
Here is the datasheet for an EL84:
[ATTACH]47627[/ATTACH]
*
the parameters for a pair of EL84s operating in a class AB push-pull output stage are as follows:
10k p-p load impedance
220Ω Rk(per tube)
20µF Ck(per Rk)
internal resistance ≈ 38k
µ = 19

solving for f during class B operation I came up with f_(halfboost) ≈ 37Hz (37.125908162)
Is this correct?

For most practical purposes (where µ>10 and Ra>>Rk) it is completely sufficient to estimate the -3dB frequency from the time RC time constant alone, especially if you consider that electrolytics can have huge tolerances of e.g. +20%/-50%.
This means replacing the square root term by 1. Thus I find f=36.2Hz, which is close enough.

[Mike Sulzer]

For most practical purposes (where µ>10 and Ra>>Rk) it is completely sufficient to estimate the -3dB frequency from the time RC time constant alone, especially if you consider that electrolytics can have huge tolerances of e.g. +20%/-50%.
This means replacing the square root term by 1. Thus I find f=36.2Hz, which is close enough.

But the paper referred to in the first post defines the half boost point in terms of voltage gain, not power. So if we have sqrt(1 + (omega*R*C)**2) (that is, the voltage frequency response of a first order high pass filter), then omega = sqrt(3)/(R*C) gives 2. Thus the sqrt(3) term in their equation when it is simplified by letting mu get very large. sqrt(3) = 1.7320508075688772, so this is large enough to be significant.

[SoulFetish]

So it looks like that square root term just becomes the square root of three when mu gets very large. That makes a very simple equation if I have not served up the math again!

(...I'm afraid to even ask)

So, when you refer to the square root term, are you referring to this entire expression

where as using the figures I was going by, I got 1/(2πRkCk)*sqrt(1.05346294) = 1/(2πRkCk)*1.026383427388

you would simplify it to say 1/(2πRkCk)*sqrt(3) = 1/(2πRkCk)*1.7320508075688772
??
go easy on me.

[Helmholtz]

But the paper referred to in the first post defines the half boost point in terms of voltage gain, not power. So if we have sqrt(1 + (omega*R*C)**2) (that is, the voltage frequency response of a first order high pass filter), then omega = sqrt(3)/(R*C) gives 2. Thus the sqrt(3) term in their equation when it is simplified by letting mu get very large. sqrt(3) = 1.7320508075688772, so this is large enough to be significant.

Hello Mike,

you are right, the -6dB or half-gain-frequency differs from the -3dB frequency by a factor of (3)^0.5. But the -6dB frequency of an ideal first order high pass filter must be lower than the -3dB frequency, whereas the square root term in the formula from the article can only increase the frequency value.
For the ECC83 example in the paper the square root term yields a factor of 1.17.
I don't think the formula in the paper can be used for pentodes also. Pentodes do have higher µ values as well as higher ra values, which would partly compensate in the formula. µ is actually a theoretical number, meaning the intrinsic open-loop-voltage-gain. It is not typically found in pentode specs as it strongly depends on operating conditions. It is certainly much higher than the value of 17 given in Post 1.
With all these uncertainties I would go by the simple -3dB formula, which at least gets you in the ballpark, even if the frequency may be a little too low.
If it has to be precise, nothing beats a measurement of the stage's frequency response.

[nickb]

I posted this link earlier but I see the post never made it. Here it is again, now updated to use the more useful half boost freq instead of the -3dB point

Cap Value Calculator

[SoulFetish]

I posted this link earlier but I see the post never made it. Here it is again, now updated to use the more useful half boost freq instead of the -3dB point

Cap Value Calculator

This is a great utility, nick. One thing I noticed was that when I input a Ck value lower than 1k(ie. as a decimal value), I get error messages. Does this have to do with how how it links to the numerical value-
"rko= rk*1000;"

[nickb]

This is a great utility, nick. One thing I noticed was that when I input a Ck value lower than 1k(ie. as a decimal value), I get error messages. Does this have to do with how how it links to the numerical value-
"rko= rk*1000;"

The Rk is in K ohms, so for example for 470 ohms enter 0.47. I don't see any error messages.

[Helmholtz]

I posted this link earlier but I see the post never made it. Here it is again, now updated to use the more useful half boost freq instead of the -3dB point

Cap Value Calculator

Very nice tool indeed. The µ-range seems to be limited to 200 max though. This is certainly not enough for pentodes.

[nickb]

True. The slider has to have an upper limit. I've changed it to 500.