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23rd May 2006, 12:45 PM  #21 
diyAudio Member
Join Date: Sep 2003
Location: quebec

hi shoog
90ma for bias don t forget pa max=13 watt (26 watt for one 6080) and 70% of pa max is better for tube life what is your idle watt / triode bye 
23rd May 2006, 01:55 PM  #22  
diyAudio Member
Join Date: Jul 2004
Location: Ardeche

Quote:
http://www.audioxpress.com/resource/...ss/ga300ac.pdf and http://www.audioxpress.com/resource/...ss/ga400ac.pdf Yves. 

23rd May 2006, 03:37 PM  #23 
diyAudio Member
Join Date: Aug 2002
Location: Eire

I am running 102mA against 100V plate voltage (taking the 30V bias). This gives me 100x0.1=10W or 20W per valve. It is a little less than one of the suggested operating points on the datasheet.
Shoog 
23rd May 2006, 11:07 PM  #24  
diyAudio Member
Join Date: Oct 2003
Location: Finger Lakes, NY

Quote:
If you look at class A PP as a balanced source then it might make sense. jlsem is right; it's not really that simple because rp varies with signal, so they don't make a truly balanced source. But it's a good simplification because it makes the analysis A LOT simpler. Anyway, consider a 1:1 PP transformer. Put 1000 ohm resistors across each half of the secondary. Since it's a 1:1 transformer the plate to plate load seen at the primary is 2000 ohms. Now ask yourself the question: what load is seen across each half of the primary? Seems obvious by inspection that it must be 1000 ohms. It's a 1:1 transformer; the same load that's across each half of the secondary must appear across each half of the primary. That's 1/2 the plate to plate load. If you drive only one side of the primary then you have a 1:2 voltage ratio which is a 1:4 impedance ratio. In that case the load seen would be 1/4 * 2000 ohms = 500 ohms. 1/4 plate to plate. Incidentally, it's not hard to prove using conservation of energy. Since it's a 1:1 transformer the same currents and voltages must exist at both the primary and secondary. If the tubes each saw 1/4 the plate to plate load in class A then there would be a different amount of power going into the transformer than coming out. In fact, IIRC (it's been a while since I proved it to myself) you'll get different calculated powers depending on whether you use i*v, v^2/R, or i^2*R. OTOH, when driving only one side (pure class B, if such a thing existed) then the power calculations work out ok at 1/4 plate to plate load.  Dave 

24th May 2006, 10:24 AM  #25 
diyAudio Member
Join Date: Aug 2002
Location: Eire

Thanks for all the information. Another jump on the learning curve.
Heres a related question. Assuming we stay in pure class A PP, what can we expect as max efficiency. Would it be 50%. Shoog 
24th May 2006, 01:19 PM  #26 
diyAudio Member
Join Date: Aug 2002
Location: Eire

I just wanted to correct some statements I have made.
The drive to the 6080's is 60Vpp which gives aplate output of 55Vpp. The measurements I stated before were down to problems with I have with interpreting my scope. I apologies for any misunderstanding. Shoog 
24th May 2006, 03:48 PM  #27 
diyAudio Member
Join Date: Nov 2005

Thanks Dave,
We were really talking apples and oranges... I didn't realize at first they we were talking about an "effective impedance" as a result of a compound analysis rather than a superposition type. I'm itching to build a PP tube rig... if I can design or purchase OPT's that truely thrill me. Funny you mention energy calcs... I am always using energy calcs to solve stuff. Crunch tricky stuff in a hurry and avoid all that Simulation and Laplace ****. One of these days I'll fork out for Mathematica and let it crunch instead... in the meantime, spreadsheets... 
25th May 2006, 02:02 AM  #28  
diyAudio Member
Join Date: Jun 2003
Location: Dallas,TX

Quote:
Nice try, John 

25th May 2006, 04:54 AM  #29 
diyAudio Member
Join Date: Nov 2005

jlsem,
Thanks... 
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