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Old 22nd November 2006, 09:10 AM   #11
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Hi,

Quote:
Originally posted by refference
And more :

Morgan Jones says : “The phase-splitters based on the diffe-
rential pair were all able to provide overall gain , but this was
obtained at the expense of an output balance that is partially
( ... a lot , if you prefer ) dependent on the matching of “mu”
between the valves “ (... or the two halves of them )

Rubbish

Auto balancing LTP:
http://dogstar.dantimax.dk/tubestuf/driver02.htm

My SS CCS equivalent attached.

The imbalance was 5V @ 60V P-P swing without common-mode FB, only 20mV @ 60V P-P swing with

Cheers!
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File Type: gif auto_phase_splitter.gif (10.6 KB, 647 views)
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Old 22nd November 2006, 11:01 AM   #12
SY is offline SY  United States
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The quote from MJ did not, I think, refer to an LTP with a current source as the tail.
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Old 22nd November 2006, 11:22 AM   #13
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Now that you mention it. I think your right...

*goes to get his MJ book*
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Old 22nd November 2006, 04:43 PM   #14
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Quote:
Originally posted by SY
The quote from MJ did not, I think, refer to an LTP with a current source as the tail.
Morgan Jones was referring to LTPs with passive tail loads. The tip-off is the mention of imbalance. VTs are naturally low gain devices. After all, the u's (highest possible theoretical voltage gain) maxes out at 100 for small signal triodes, and doesn't get above 200 even for RF "zero bias" triode finals. Even high gain small signal pentodes have g(m)'s that are pretty low: 12mA/V (12BY7A), 16mA/V (6EW6) and so forth. The g(m)'s of BJTs are limited by how much current you can force through the device before the silicon die melts. Even running at a very modest Ie= 1.0mA, the BJT has a g(m)= 38.46mA/V. How many VTs have g(m)'s that high?

With very limited device gain available, you aren't going to be getting good performance from passively loaded LTPs. You can force output balance by using unequal plate resistors as a "hack". The CMRRs remain relatively poor, and if that's all you had, you'd've been better off with a cathodyne, by far.
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Old 24th November 2006, 12:33 AM   #15
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Hi RayMoth ,


The original design of J.N. Van Scoyoc , ( originally
Published on Radio Electronic Engineering in
1948, November ) uses one 6SN7 and one 6SL7 .
The 6SN7 at the input , acts like cathode-follower ,
coupled to the next tube’s cathodes ( 6SL7 )
that amplifies in push-pull way , with tails
cross-connected to the 6SL7’s grids .

You can use a 12AU7 ( instead 6SN7 ) and 12AX7
( instead 6SL7 ) without change in resistor’s values .

You are correct , the Scoyoc phase-splitter must be
used in the input stage . ( It works with low-level signals )
You are correct too , about the trim-pot between the two
cathodes , to fine-adjust the currents ( working point )

But I have to disagree from you , when you said that :
“this forces the designers to use an all balanced circuit ... “ .
It’s not necessary !! You can use two signals from
two different sources ( works like a mixer ) .
You can use a push-pull balanced or unbalanced signal , or
you can apply the signal at only ONE input , that you
will get at the output two identical signals with inverted
phases .
Bibliography : Eng. Julio Rueda – Audio Amplifiers Circuits
Argentina – 1957

About the Tubes 4e4’s comment , the original design
have only the two output capacitors , as usual in every
phase splitter .

Best Regards ,

Carlos
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Old 24th November 2006, 01:25 AM   #16
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Carlos,

If you put the splitter in the first stage, then the whole amp is going to be a balanced (pp or diff) design. That's all I meant.
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Old 24th November 2006, 06:59 PM   #17
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Hi RayMoth ,

OK ! OK ! Now I understand your point of view . You are correct .
But don't forget , that the output level of a Scoyoc phase splitter
is relatively high , and if you are planning to build a low power
amp ( until 10 - 12 Watts ) , perhaps you won't need another drive stage .

Regards ,

Carlos
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Old 25th November 2006, 01:10 AM   #18
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True, Carlos. With most tube-based splitters using a double triode, you effectively lose half the gain: concertina first stage is normally a voltage amplifer with full gain, feeding the splitter itself which has no gain; LTP splitter, with its single input, gives half the gain of a differential stage with two inputs; and the see-saw gets gain from one half but not the other.

I don't know how the gain of the Scoyoc is calculated, because I can't figure out if the second double triode (e.g. 6SL7, the one that feeds off the cathode followers, e.g. 6SN7) has current NFB or not. In this example the 6SL7 gets its bias from the cathode resistors of the 6SN7, which aren't bypassed (they can't be or there would be no signal), but does this mean that the 6SL7 gain is reduced by current NFB?
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Old 25th November 2006, 10:14 AM   #19
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Default dynamic mu matching

Quote:
Originally posted by refference
Hi all ,
Morgan Jones says : “The phase-splitters based on the diffe-
rential pair were all able to provide overall gain , but this was
obtained at the expense of an output balance that is partially
( ... a lot , if you prefer ) dependent on the matching of “mu”
between the valves “ (... or the two halves of them )
Carlos
Carlos, there is a simple solution to the difference in mu between halves. Even if tightly matched, differences occur. In Electronics Engineering (from the sixties) a cunning solution was presented. In most ECC-series tubes from Philips, both halves have a separate heater for each triode.
--> By supplying each part from a variable voltage source the temperature of the cathode is varied and hence the operating point and mu. The difference in mu can easily be completely taken away with a few components.
albert
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Old 25th November 2006, 11:28 AM   #20
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Now, that is cunning. I daresay, in these days of microelectronic miracles, somebody could dream up a servo arrangement to control the currrent in each half of the heater, so as to correct the mu automatically as th etube ages. It could even be encapsulated within the base of the tube and be invisible to the naked eye

Afterthought: if you have a CCS in the tail of an LTP, there is no need to worry about the mu in each half being different. AC balance is assured, as long as the plate loads are identical. All you have to do is make sure the quiescent current through each triode is the same, which you can correct with resistors in the cathodes.
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