How to figure output Z of SRPP?
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 10th January 2007, 12:39 PM #1 smbrown   diyAudio Member     Join Date: Jun 2004 Location: Portland, OR How to figure output Z of SRPP? Can someone give me the process for computing the output impedance of a SRPP preamp both when the output is taken from the cathode of the upper tube, and when the output is taken from the plate of the lower tube? Thanks!!
 10th January 2007, 04:01 PM #2 first   diyAudio Member   Join Date: Nov 2006 I want to know too
 10th January 2007, 04:53 PM #3 tubetvr diyAudio Member   Join Date: Jan 2003 Location: Sweden Given that both tubes are identical the Zout at the cathode of the upper tube is: rp * (rp + R)/ (2*rp+(u+1)*R) where rp is internal anode resistance and R is the cathode resistor value. For a 12AX7 with 1.2k cathode resistor this gives ~16k The output impedance at the lower tube anode is higher and will be (rp + (u+1) * R) / 2 if the lower tube cathode is not decoupled and will be: ((rp+(u+1)*R)*Rp)/(2rp+(u+1)*R) if the lower athode resistor is decoupled. Typical values for a 12A7 will be ~92k with no deoupling and 47k with decoupled cathode resistor Regards Hans PS, this is described in chapter 11 of "Vacuum tube amplifiers " by Valley and Wallman where the SRPP is first described and analysed.
 10th January 2007, 06:23 PM #4 smbrown   diyAudio Member     Join Date: Jun 2004 Location: Portland, OR Thanks! In fact I'm using a 6072 and had an opportunity to measure it at lunch. The bottom is not bypassed. The output at the bottom of the upper tube cathode is about 15k, which seems to work out okay. Kind of a higher output Z than I'd like, but sure sounds nice as a preamp.
 11th January 2007, 05:03 AM #5 tubetvr diyAudio Member   Join Date: Jan 2003 Location: Sweden Yes, 15k with a 6072 sounds about right. A mu follower will give lover output impedance and higher gain than a SRPP but it doesn't have the same output current capability and the distortion rises very quickly when a mu follower is loaded too hard, (e.g. when driving an output stage which clips and start to draw grid current) the SRPP is much better when loaded hard. Regards Hans
 11th January 2007, 08:42 AM #6 Merlinb   diyAudio Member     Join Date: May 2006 Location: Lancashire Here's a tutorial on the SRPP: http://www.valvewizard.co.uk and go to SRPP
Geek
diyAudio Member

Join Date: Sep 2004
I have this sitting on my HD for a reference. Can't remember where I got it from.
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tubetvr
diyAudio Member

Join Date: Jan 2003
Location: Sweden
Quote:
 Here's a tutorial on the SRPP:http://www.valvewizard.co.uk and go to SRPP
Interesting site with a lot of useful information but unfortunately it contain a common error when describing the output impedance of the cathodyne phase splitter, it gives output impedance values that are different for cathode and anode when in reality these are equal when both outputs are loaded equally.

Regards Hans

 11th January 2007, 03:10 PM #9 smbrown   diyAudio Member     Join Date: Jun 2004 Location: Portland, OR Thanks for all the great feedback, very appreciated. My little project was to just try a 6072 in a SRPP as a preamp. I used 2k R's and this resulted in about 1.2ma of current, and 15k Zout. In listening last night a friend and I thought it sounded very nice until pushed a bit harder, then it started to get that over driven sound to it. I'm thinking that the fault is likely a combination of the lower current and highish Zout. The input to the power amp is 100k, but the capacitance of the cables and the input tube on the power amp comes in at about 280pf, which I think is challenging the slew capability of the preamp SRPP stage. I really like the sound of the 6072, but not very much grunt there. Maybe switching to 1k and around 2ma would do it. I'll play some more.
plovati
diyAudio Member

Join Date: Mar 2005
Location: near Milan
Quote:
 Originally posted by tubetvr Interesting site with a lot of useful information but unfortunately it contain a common error when describing the output impedance of the cathodyne phase splitter, it gives output impedance values that are different for cathode and anode when in reality these are equal when both outputs are loaded equally. Regards Hans
Please could You explain a little bit?
Thank You
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