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Cathode-coupled Phase Splitter Formula

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I have the following formula for calculating Plate resistor value of V1
Rl1= (amplication factor+1) x Rk x Rl2 / Rl2 + (amp factor+1) x Rp x Rk
Somehow I do not think this is right? Can not get a proper answer for a 6SN7. I have Rl2 = 26k amplication factor = 20 Rp = 7.7k Rk = 8.5k
What should Rl1 be?? Thanks Mel
 
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There isn't a single best value for plate load for a voltage amplification stage. As SY says, draw a load-line on the triode plate voltage vs plate current curves.

Within reason, the higher the value of load resistor the better: it'll give you a more horizontal load-line and better linearity. However, you also tend to get better linearity and a more useful bias point if your plate current is higher; unfortunatley, this tends to be in conflict with using a higher plate load resistor, because of the higher voltage drop across it. As with most things, it's a compromise.

I'm operating 6SN7s happily with plate current of 3.6mA and plate load of 47K. The load is dropping approv 170v and, with B+ of 400v, this leaves me with 230v across the tube - cathode resistor combination. After playing around and asking questions, I think this is a reasonable operating point, considering the B+ that I have available. Some might argue that I should allow more current and reduce the load resistor.
 
I'm putting together a 300B PP. Going to use the 6SN7 as a phase-splitter and to drive the 300B's. I'll need a gain of 3 to 4 for that section using my favorite EF86(triode connected) as input tube. From what I have read I need 4.1mA minimum going through each section of the 6SN7's to reach 20,000Hz on the 300B's. I'll balance the output of the 6SN7's with my o-scope...just wondered about that formula. The Author of the book (Bruce Rozenblit) says it gives VERY accurate results.......but as I said seems to be something wrong with the equation. I just wanted to check and compare the equation with my scope...;) Thanks Mel Of course I may end up with something totally different by the time I'm done playing.
 

PRR

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> for a 6SN7. I have Rl2 = 26k amplication factor = 20 Rp = 7.7k Rk = 8.5k What should Rl1 be??

"26K, adjust on test"

First-approximation, equal plate resistors gives equal output voltages. So pencil 26K.

Correction: V2 gets less drive because Rk is typically smaller than the value that gives optimum balance. But in this case the error is around 5%. The amp will work fine with 5% unbalance. If you want to get fussy, Rozenblit's formula looks like the right form for best small-signal equality. But it relies on small-signal approximations like Mu. With real tubes, Mu is not a constant. With the high voltages you seem to want, Mu is not a constant. SY would draw load-lines, which is as valid as the tube-plot you have (and in this case at least you are up in the area where the plots are pretty correct). But the long-tail amp is hard to plot both sides at once. And it is a hell of a lot quicker to tack in a 27K resistor, measure unbalance, and fudge to equal outputs. So Rl1 might be 25K, +/-10% for tube and other tolerances. "26K, adjust on test"

I don't see how a 300B needs any 10mA in a 6SN7 to make 20KHz. How bad is the Miller effect? 300pFd? The output impedance of a 6SN7 with plate resistor at 10mA is close to 5K. 300pFd in 5K is around 100KHz. There may be other advantages to working the 6SN7 very rich, but treble does not demand that much.

There is such a thing as too-low load resistance. For voltage-swing, rule-of-thumb, I would not make Rl less than 2X Rp. If I was not hurting for load current, 5X is a safer and lower-distortion value. Over most of a rich-running 6SN7's range, that suggests around 40K-50K, as Ray is using. If Miller capacitance is not a big problem (pentodes or NFB) then 100K is often a good Rp for driving self-bias output grids.
 
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