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Old 26th November 2006, 03:44 PM   #1
jayme is offline jayme  United States
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Default calculating HF rolloff at 20kHz from input impedance

How do I calculate the corner frequency for HF rolloff, and the amount of dB rolloff that I have at 20kHz, due to the volume control imput impedance and the Miller capacitance of the driver tube?

I have a shunt-style volume control arrangement: a 100k series resistor shunted by a 100k pot. (Thus, the max shunt value is 100k, but I do most of my listening around the 2k shunt level, with a max around 5k. Anything more than that is just too loud).

The circuit is an Aikido linestage with all 6SN7s. The first stage of the Aikido is more or less a grounded cathode 6SN7 with a 6SN7-based CCS on top. I believe this makes the input capacitance equal to mu * Miller capacitance, or 20*3.8pF = 76pF.

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Old 27th November 2006, 03:56 PM   #2
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Here are the relevant equations, assuming I am understanding your attenuator circuit correctly.

R1 is series input resistance, R2 and C2 in parallel shunt R1 to ground, and the signal is then taken from junction of R1 and R2. In your case R1 = 100K, R2 = 2K, C2 = 76p. Note that in practical terms, with such a low value of R2 compared to R1, you dont really need to worry about frequency response. But here are the equations :

3 dB frequency = 2 * pi * (R1 + R2) / (R1 * R2 * C2 )

dB at a frequency F = 10 * log( (R1 + R2)^2 / ( (R1 + R2)^2 + (2 * pi * R1 * R2 * C2)^2)

So for your example, with R2 = 2K, 3dB is at 1.07 MHz,
and dB at 20Khz is -.0015. In other words, forget about it, you have bigger things to worry about.

At R2 = 5K, you have 3dB = 440KHz, and 20KHz down -.009 dB

At full volume, R2 = 100K, then 3dB is at 41.9 KHz, and 20 KHz is down to -.89 dB.

But note that this is just playing with the math, in reality that C2 may be much higher than 76p due to wiring capacitance and who knows what else. On the other hand a few pF of wiring capacitance in parallel with R1 will balance out the 76pF in parallel with R2, and give a perfectly flat response, so with a low C2, and a high ratio of R1 to R2, you wont really know unless you measure it.
Robert McLean
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Old 27th November 2006, 04:38 PM   #3
jayme is offline jayme  United States
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Thanks! That is very helpful.

Those were the equations I was looking for...and your analysis clears up a lot of the questions I had.
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Old 27th November 2006, 04:44 PM   #4
kevinkr is offline kevinkr  United States
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Miller capacitance is actuallly Cgp x (mu + 1) or in this case about 80pF, add to that the Cgk and any strays you can measure. (or estimate) Since you are using a tube based ccs the 6SN7 mu should closely approach the theoretical value of 20. (otherwise you need to know the effective mu in the circuit in question to make this calculation)

In any case it won't make much of a difference, you've got plenty of bandwidth.

See AikenAmps: for a very good treatise on this subject.

"To argue with a person who has renounced the use of reason is like administering medicine to the dead." - Thomas Paine
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