math quiz- input z triode amp

tenderland · 23rd November 2004, 01:52 AM

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

SY · 23rd November 2004, 02:55 AM

If this is a common cathode triode amp, you haven't taken into account the Miller capacitance.

tenderland · 23rd November 2004, 03:01 AM

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)

Sch3mat1c · 23rd November 2004, 03:16 AM

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

tenderland · 23rd November 2004, 03:27 AM

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

EC8010 · 23rd November 2004, 12:28 PM

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.

tenderland · 23rd November 2004, 12:37 PM

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.

Sch3mat1c · 23rd November 2004, 06:05 PM

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.

Tim

23rd November 2004, 01:52 AM	#1
tenderland diyAudio Member Join Date: Jul 2004 Location: us	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

23rd November 2004, 02:55 AM	#2
SY Tetsujin diyAudio Moderator Join Date: Oct 2002 Location: Austin, TX	If this is a common cathode triode amp, you haven't taken into account the Miller capacitance. __________________ "We all feel like that, Reggie, now and then, especially when Spring is upon us, but few of us would care to put it on our cards." — Sir Archibald Clerk-Kerr

23rd November 2004, 03:16 AM	#4
Sch3mat1c diyAudio Member Join Date: Jan 2003 Location: Milwaukee, WI	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 __________________ See my Electronics webpage -- the home of Vacuum Tube Drag Racing. The key to being a successful Audiophile: "I reject your reality and substitute my own!"

23rd November 2004, 03:27 AM	#5
tenderland diyAudio Member Join Date: Jul 2004 Location: us	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

23rd November 2004, 12:28 PM	#6
EC8010 diyAudio Moderator Join Date: Jan 2003 Location: Near London. UK	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. __________________ The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...


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23rd November 2004, 03:01 AM	#3
tenderland diyAudio Member Join Date: Jul 2004 Location: us	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)

23rd November 2004, 12:37 PM	#7
tenderland diyAudio Member Join Date: Jul 2004 Location: us	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.

23rd November 2004, 06:05 PM	#8
Sch3mat1c diyAudio Member Join Date: Jan 2003 Location: Milwaukee, WI	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. Tim __________________ See my Electronics webpage -- the home of Vacuum Tube Drag Racing. The key to being a successful Audiophile: "I reject your reality and substitute my own!"

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