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Old 23rd November 2004, 01:52 AM   #1
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Default math quiz- input z triode amp

While in the process of learning tube theory I have come across something in a book I do not understand , simply put the author claims the input Z of a triode amp ( no cathode resistor ) is the
plate to gride capacitance ( 18.4pf) added to the grid to cathode capacitance (1.6pf) this is in parelle with the gride resistor which he calculates by using product over sum and comes up with 81 k .

grid resistor is 100k , miller effect is assumed
freq = 20 khz

I get 97k @ -13 deg

do i need to get some rest ?

jeff
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Old 23rd November 2004, 02:55 AM   #2
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If this is a common cathode triode amp, you haven't taken into account the Miller capacitance.
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Old 23rd November 2004, 03:01 AM   #3
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Default miller

the plate to gride C = 1.6pf gain = 10.5 , acording to author
C = 1.6pf(10.5 + 1) = 18.4 pf + 1.6pf (grid to cathode ) = 20 pf
miller c = c(gain +1)

20 pf @ 20 khz = 398 k ohms

know this is were i believe you cant simply take product over sum
(well you can but it is an equation using complex numbers)
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Old 23rd November 2004, 03:16 AM   #4
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Miller only effects G-P capacitance, although using it for everything added together is going to make little difference.

In the audio range, especially over such a wide frequency range, specifying an impedance is an exercise in futility. Just go by the % attenuation at the highest frequency...

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Old 23rd November 2004, 03:27 AM   #5
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Default miller

the miller effect is only used for the grid to plate c in the equation
author claims this is 1.6 pf as is the grid to cathode c

I agree with youre observation but is the author using bogus math ?


is 100k in parallel with 20 pf(@ 20 khz) = 81 k ohm
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Old 23rd November 2004, 12:28 PM   #6
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Judging by post #3, the capacitance calculation is correct, but you certainly can't just do "product over sum" when the two quantities are at right angles to one another. What you can do is work with admittances instead, and add them as vectors, then convert back to impedances. I tried that, and got a figure of 97k.

On the other hand, why do you need to know the input impedance at a specific frequency? Most people would be happier knowing that it was 100k in parallel with 20pF.
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Old 23rd November 2004, 12:37 PM   #7
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Default why ?

Well I am new to tube learning and the author is using this example to show the effects of the changing impedance at higher frequency .
I gues i just wanterd to know if his calculation was acurate.
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Old 23rd November 2004, 06:05 PM   #8
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Geez, I just plug in total C (miller plus etc.) and see if the reactance at say 20-50-100kHz is comparable to the preceeding stage (Rp || plate R || grid leak). Even with a 12AX7 you honestly won't have any problems with all but the chunkiest tubes.

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