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#1 |
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diyAudio Member
Join Date: Sep 2001
Location: Finland, Ostrobothnia
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Hi everyone
It's about 15 years since I studied electrical engineering and I am trying to remember some basic things again. I have some questions about the selection of output transformers for a tube amplifier and I would appreciate if someone could aid me in the right direction. The circuit I am looking at is a Williamson type parallel push-pull ultra-linear with 6L6GC tubes. Most similar designs that I have seen use one pair of tubes in push-pull UL, which with a pair of 6L6GC tubes makes about 6k6 output impedance. In this design there are two pairs, or so called parallel push-pull. A similar circuit by Acrosound is here: http://oestex.com/tubes/acrowill_1.gif The Acrosound amp uses a 3k3 output transformer. Instead of using an output transformer designed for 3k3 primary impedance, the circuit author uses a 6k transformer. I'm now asking, what is the benefit of using a 6k OT here? Maybe to make it easy to switch to other tubes? If I were to build this amplifier, wouldn't it be better to use a 3k3 transformer? I am trying to understand how it would affect power output and characteristics. I would appreciate some inputs on this. I have been looking at some available transformers and for instance Edcor has many transformers in their CXPP series, but none with 6k impedance. I would have to choose between 6k6, 5k or lower (there are 4k2, 4k, 3k8, 3k5, 3k3 etc.). Here is the circuit in question: http://img855.imageshack.us/img855/1221/parts100dpi.png Note. The author of the above circuit is Jukka Tolonen, which has published articles in Glass Audio Magazine and a Finnish Hifi magazine. His homepage is here: Jukka Tolonen Home Page Last edited by johan; 6th March 2011 at 11:30 AM. |
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#2 |
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diyAudio Member
Join Date: Apr 2007
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Loadlines, loadlines, loadlines! education+diy or Steve's Tube Pages
After digging that through you'll be able to figure out yourself. For triodes it is pretty straight forward, higher load gives lower output power, lower distortion and lower output impedance. For pentodes there is a pretty narrow optimal load span which gives reasonable THD and power out, output impedanece is always high and needs to be treated with nfb. UL is somewhere inbetween depending on UL percentage. It is a little trickier since there is seldom plate curves avilable in the datasheets. But if I remember correct there should be an example in the GE 6550 datasheet. /Olof
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I'm not allowed to do magic, union rules... |
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#3 |
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diyAudio Member
Join Date: Sep 2001
Location: Finland, Ostrobothnia
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#4 |
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diyAudio Member
Join Date: Apr 2007
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Yes, that's all the work done already.
So according to this the optimal loading is somewhere btwn 7-10 k(a-a) (depending on how you value thd vs Pout)for a single pair. In your parallell PP 3k5 to 5k should be good. I would go close to 5K but more seems to be giving up power for no good reasons. /Olof
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I'm not allowed to do magic, union rules... |
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#5 |
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diyAudio Member
Join Date: Sep 2001
Location: Finland, Ostrobothnia
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Thank you.
The Edcor 100W 5k transformer CXPP100-8-5K is probably the cheapest but still a good option. |
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#6 |
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diyAudio Member
Join Date: Sep 2001
Location: Finland, Ostrobothnia
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Another question. Suppose I'd like to temporarily remove one pair of output tubes and connected the 8 ohm speaker to the 4 ohm tap on the secondary (assuming I had such an OT). Would this load the output transformer in some weird way or affect bandwidth or something?
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#7 |
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diyAudio Member
Join Date: Apr 2007
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Should not be any problem. BW will be less than with all tubes in place since your drive impedance will be higher. But a reasonably decent OPT will cope.
/Olof
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I'm not allowed to do magic, union rules... |
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#8 |
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diyAudio Member
Join Date: Sep 2001
Location: Finland, Ostrobothnia
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Thank you. Good to know. Now I can proceed buying the transformers.
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