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
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
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)
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)
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...
Tim
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...
Tim
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.
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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