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In Reply to: P.S. posted by Tre' on August 22, 2011 at 10:14:52:
" Posted by Damir _the real one on December 2, 2004 at 09:40:51
In Reply to: I'm confused... did everybody miss this class posted by Gordon Rankin on December 2, 2004 at 07:03:25:
Well, I mathematicaly described 30:1 turns ratio SE OPT with CT secondary, see below. In short, you can connect one 4 Ohms speaker on the half of the sec., one 16 Ohms speaker across the whole sec., or two 8 Ohms speakers across both halves. In all cases, our tube on the primary will "see" the same reflected impedance and "gave" the same power. Well, let`s reverse the thing:
1.) We have, say 300 turns CT primary (150+150 turns), and 10 turns secondary. Turns ratio from whole primary to secondary is 30:1, or from halves of the primary 15:1.
2.) Let`s connect the speaker of 8 Ohms across the secondary. Then our reflected impedance to the whole primary is (30^2)*8 = 7200 Ohms.
Reflected impedance across the one or another half of the primary is (15^2)*8=1800 Ohms, or 1/4 of the whole primary.
3.) Let`s connect the tube, or better - for simplicity AC generator that gives 100Vrms no matter of load. We can connect the generator through the whole primary, Raa=7200 Ohms. Our 100 Vrms will give the power Ppr`=U^2/Raa=100^2/7200=1,3889W. That power is "transferred" to the secondary load. Usec is Upr/turns ratio, or Usec`=Upr/30=3,333Vrms, and power is Psec=Usec`^2/Rsp=3,33^2/8=1,3889W.
We can connect our generator across only the half primary, our generator "see" Raa/4=1800 Ohms, and gives power Ppr``=100^2/1800=5,555W. On the speaker - Usp``=100/15=6,667Vrms, Psp``=6,667^2/8=5,555W.
4.) Now, we feel like goin` PP, and connect two identical generators with 100Vrms output, but in antiphase. Connection - one wire from the each generator on the one side of the primary, then remaining two wires connect together and grounding (the same like with tubes). We can now say that we have two generators in series, in antiphase, connected on the primary, and each one gives 100Vrms.
When we measured AC voltage across our primary, we measured 200Vrms!
Now, Ppr```=200^2/7200=5,555W, and our Usec```=200/30=6,667Vrms, and Psp```=6,667^2/8=5,555W.
5.) The question is - what load each tube/generator in PP A class (both tubes working) "see"?
The Law of power conservation - Ppr=Psec. Every tube/generator gives half of the total voltage and half of the total power on the primary (and transferred to the secondary load).
Usec```^2/Rsp = Ugen1^2/Rload1 + Ugen2^2/Rload2
We know that Ugen1=Ugen2=100Vrms, we know Usec```^2/Rsp=5,555W, and we know that Rload1 = Rload2 = Rload1,2 = unknown resistances of the one and another halves of the primary in PP case.
Then we have: 5,555 = 2*(100^2)/Rload1,2 and
Rload1,2 = 20000/5,5555 = 3600 Ohms (half of the total Raa=7200 Ohms).
We can express this with turn ratios, the result is the same :
1/Rsec = 1/(Rload1*(Nsec/Nhalf pr1.)^2)+1/(Rload2*(Nsec/Nhalf pr2)^2)
1/8=1/(Rload1,2*(10/150)^2) + 1/(Rload1,2*(10/150)^2)
and from that Rload1,2 = 3600 Ohms.
6.) Summary and conclusion:
Reflected secondary load across the whole primary is the square of the turns ratio times Rsec, or in this case Raa=7200 Ohms. This load (7200 Ohms) "see" the tube connected across the whole primary.
When we connect the tube across the only half of primary, our tube "see" Raa/4, or in this case, 1800 Ohms.
When we connect two tubes in PP A class, each tube "see" half of the primary resistance, Raa/2, or in this case 3600 Ohms. "
Tre'
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- And now for the math - Tre' 10:47:17 08/22/11 (0)
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