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2nd May 2011, 12:26 PM  #1 
diyAudio Member
Join Date: Jan 2010
Location: Germany

Determining Zout for SRPP
Dear folks,
I want to use a SRPP with 6N1P to drive a low impedance ( ~ 10K ). This nice Triode has mu=35, rp = 4400 After reading The Tube CAD Journal,SRPP Decoded http://http://valvewizard2.webs.com/SRPP_Blencowe.pdf http://www.euronet.nl/~mgw/background/tubes/uk_tubesinfo_1.html I´m a little bit confused about how to determine Zout. The optimum value for Rak seemed to be 750Ohms at Rload=10K for fully bypassed Rk. Using formular at tubecad I calculate 632Ohm Using formular at valvewizard I calculate 725Ohm And using it from euronet.nl I calculate 2250Ohm If I measure the Output impedance using 1Khz signal i calculate 4889Ohm What is the "right" formular for a fully bypassed SRPP?? Thanks a lot for any help.. Karsten 
2nd May 2011, 12:59 PM  #2 
diyAudio Member
Join Date: May 2007

You may be confusing output impedance with optimum load. They are not the same thing. To add confusion, different people seem to give different formulas. I have been doing an analysis of the SRPP which one day I might publish. My results are:
output impedance Zout is (Ra1+(mu+1)Rk1 + Rk2)Ra2/(Ra1+(mu+1)Rk1+Ra2+(mu+1)Rk2) where Ra is anode impedance, Rk is cathode resistor (which may be zero if bypassed), suffix 1 refers to lower valve and 2 to upper valve. For identical valves and lower cathode resistor bypassed (and smallish Rk), this simplifies to (Ra^2)/(2Ra+(mu+1)Rk); in the typical case this will be about Ra/3. The optimum load, Zopt, set by requiring equal currents from each valve, is approximately (mu Rk2  Ra2)/2  note that this is often quite small (could even be negative!) unless a large value for Rk2 (upper cathode resistor) is used. This is best done using a mu follower, which is a closely related circuit with different bias arrangements. Assuming bypassed lower cathode, Rk2=750, Ra=4400, mu=35 I get Zout = 633 (full formula) or 541 (simple formula) Zopt = 10925 Bear in mind that you need to use the actual Ra and mu at the chosen bias point, not the quoted values for the valve in general. My guess is that 750 ohms at the cathode will bias the valve back, so Ra will be larger and mu smaller, but I am not familiar with the valve. Check the data sheet. 
2nd May 2011, 02:51 PM  #3 
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Join Date: Jan 2010
Location: Germany

Dear DF96,
Excelent! Thank you so much for providing proved formulars. I will try to reproduce this in the next evenings ( Thanks a lot again to give me some reasons to vanish in my basement ) So the hope is real now to drive a Zload of ~10K using a SRPP stage. Deep bow about your knowledge.. regards Karsten 
2nd May 2011, 03:11 PM  #4 
diyAudio Moderator

Dave, is that any different than the Amos and Birkenshaw equation?
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"Pity, wrath, heroism, filled them, but the power of putting two and two together was annihilated." EM Forster 
2nd May 2011, 04:06 PM  #5 
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Join Date: May 2007

I'm not sure, I haven't seen that one  do you have a reference? My results are slightly different from Blencowe (and Broskie, IIRC). Also, Morgan Jones gives a formula which is different from mine but his numerical result agrees with mine.
So, Karsten, please don't bow too low  I'm smart, but not infallible! If anyone else wishes to have a go themselves, I will summarise my analysis. The SRPP starts as a common cathode amp with an active load. The complication is that the output is taken from the top cathode instead of the bottom anode. This means that the open circuit voltage gain is reduced a little because of the signal voltage drop across the upper cathode resistor. The short circuit current output can be worked out from the lower valve simply driving the upper cathode resistor, then this voltage is multiplied by gm. The ratio between open circuit voltage gain and short circuit transconductance is the output impedance. (Note that this is circuit transconductace, not valve transconductance). The optimum load is approximately given by both valves contributing equal signal currents. There is a bit of slack here, which may be why different people get slightly different formulas. The differences are small, though, the distortion has a broad minimum and valve characteristics will not be precisely known so any formula should only be taken as a rough guide. 
2nd May 2011, 05:49 PM  #6 
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Join Date: May 2007

I now realise that Stuart is referring to the book on TV engineering by Amos and Birkenshaw. Was this where the SRPP first appeared in print? Unfortunately I don't have access to a copy.

2nd May 2011, 06:12 PM  #7 
diyAudio Member
Join Date: Jan 2010
Location: Germany

Well, I search for "Amos and Birkenshaw" but don´t find appropiate books/links. If there is a onlinecopy available and it´s readable for a non educated mathematic maniac I´m interested in.
Enjoy the evening and "Carpe Diem" Karsten 
2nd May 2011, 07:49 PM  #8 
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Join Date: May 2007

I could not find an online copy. Maybe one will turn up on ebay.

2nd May 2011, 11:30 PM  #9 
diyAudio Moderator

Yes, they derive the equations in Volume 2. I'll check the reference when I'm home tonight and compare your result to theirs.
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"Pity, wrath, heroism, filled them, but the power of putting two and two together was annihilated." EM Forster 
3rd May 2011, 04:01 AM  #10 
diyAudio Member
Join Date: Nov 2007
Location: Dallas

Split the SRPP resistor in half and tap your output at the middle.
Will make both triodes work equally into the load. Zout increases, this is true. But Zopt decreases to lowest possible for that SRPP, as the slew limits become symmetrical. Of course you must include cathode follower impedance as part of "the SRPP resistor" when deciding where the true middle is... Pretend the top cathode is loaded with a CCS, and what would that impedance be? Then add SRPP resistor, and divide by two. Thats the point where you can best drive a challenging load... Strangely enough, plate resistance of the lower triode doesn't figure into finding the middle. It does however, figure into Zout. (Plate resistance + Cathode resistance + SRPP resistor) /2 Last edited by kenpeter; 3rd May 2011 at 04:14 AM. 
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