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Old 9th September 2009, 02:55 PM   #1
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Default Miller Capacitance of triode strapped pentode

Ok,

I have searched the forum and found very little definitive info on this subject.

Here is my quandry

I am designing a PP output stage using 2 x 829B's. They will be Triode strapped and paralleled. (don't ask why these tubes..I just like a challenge)

I am having a tough time calculating the Miller Cap of these tubes. It gets complicated because a: the two sections are parallel and B: its a pentode "triode strapped"

Inter-element caps are as follows per section
Cgk 14.5
Cak 7
Cga 0.12


Stage gain will be about 8-9 maybe.

I realize that some of these values will double due to the parallel connection but can someone help me with the method of calculating it?

I need to know how low I need to get the output impedance of the driver stage to keep response up to 25Khz or so.
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Old 9th September 2009, 03:28 PM   #2
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Expect something in range 40-100 pF.
That means, for 50 KHz pole you need at least 30 kOhm.
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Old 9th September 2009, 03:35 PM   #3
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hey-Hey!!!,
There is also the internal g2-k capacitor. As triodes, this now becomes part of the plate circuit.
cheers,
Douglas
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Old 9th September 2009, 03:35 PM   #4
Yvesm is offline Yvesm  France
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Those tubes sometimes integrate an internal bypas capacitor btwn screen and cathode.
Near 200pF IIRC.
But it appears accros the load, not as a "Miller multiplier" !

Yves.
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Old 9th September 2009, 03:36 PM   #5
Yvesm is offline Yvesm  France
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Slower !
Douglas wins !
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Old 9th September 2009, 03:43 PM   #6
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Thanks for the replies but I still don't have a definitive answer.

What is the formula for this type of calc?
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Old 9th September 2009, 03:49 PM   #7
Merlinb is offline Merlinb  United Kingdom
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Quote:
Originally Posted by coldcathode View Post
Thanks for the replies but I still don't have a definitive answer.

What is the formula for this type of calc?
In theory it is:
Cin = Cgk + [(Cg1a+ Cg1g2) * gain]


The problem is in finding a datasheet which gives you a genuine value for Cg1a. Cg1g2 may also be hard to find, but is probably less important.
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Old 9th September 2009, 03:54 PM   #8
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Quote:
Originally Posted by Merlinb View Post
In theory it is:
Cin = Cgk + [(Cg1a+ Cg1g2) * gain]


The problem is in finding a datasheet which gives you a genuine value for Cg1a. Cg1g2 may also be hard to find, but is probably less important.
Cg1-a is very small in pentodes, and Cg1-g2 is typically 20-30% more than the Cg-a of a comparable triode. With a pentode, g2 becomes part of the anode and is multiplied in the Miller effect.
cheers,
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Old 9th September 2009, 04:06 PM   #9
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Quote:
Originally Posted by coldcathode View Post
Thanks for the replies but I still don't have a definitive answer.

What is the formula for this type of calc?
I estimated Cag1 and multiplied it on stage's voltage gain.
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Old 9th September 2009, 04:15 PM   #10
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Well the data I have been able to find is as follows

Cgk = 14.5 (also listed as input C) so I am assuming Grid 1 to cathode
Cak = 7 (internal capacitance of plate to cathode)
Cga = 0.12 grid to plate
I found spec that shows the capacitance of G2 to Cathode including the capacitor is approximately 65pf

G3 is tied to the cathode.

So, with these numbers what do I do?

G2 tied to the plate and is part of the load so where does the 65pf fit in? Ignored correct?
Cak also load it is ignored also?

Do I ignore the G1 to G2 capacitance or make one up?

so if I take the Cga and double it (for parallel) and multiply by 10 (gain 9 +1)
I get 24pf
Then add twice the Cgk = 29pf I get 53pf??
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