Tube DIY Asylum

• Re: I'm confused... did everybody miss this class - dave slagle 13:27:41 12/02/04 (0)
  In Reply to: I'm confused... did everybody miss this class posted by Gordon Rankin on December 2, 2004 at 07:03:25:

  (GR)First off... You cannot look at 1/2 the solution to find the entire argument. You must concentrate on the entire system and how it works.

  exactly, which is why i am confused about all of this discussion about 1/2 the primary. for the class A ac circuit the CT is not there electrically, so the half of the winding that gives you 1/4 a-a is not there either.

  dave

  
  
  

  ---
• Re: I'm not confused :-) - PP model - Damir _the real one 09:40:51 12/02/04 (0)
  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.
  
  
  

  ---
• well then Gordon.... - Sector-7G 08:17:03 12/02/04 (0)
  In Reply to: I'm confused... did everybody miss this class posted by Gordon Rankin on December 2, 2004 at 07:03:25:

  since Mike can't answer this, maybe you can. Iff those Bell Labs Engineers taught you anything of use...

  Take a Class A, PP power stage. It is working into some load a-a, reflected by the secondary. Each of the tubes is working into some load. Pull one valve, what happens to the load on the remaining one?

  up, down, or remains the same will be a sufficient answer.

  core saturation is not an issue for this case.

  Let's see if you have any better grip on this that your Iron supplier does.
  regards,
  Douglas
  !
  mmmmmhhhh...Forbidden Donut!!!
  
  

  ---

This post is made possible by the generous support of people like you and our sponsors:

Sonic Craft

---

[ Top of Page ] [ Contact Us ] [ Support/Wish List ] [ Copyright © 2025, Audio Asylum, all rights reserved ]

---

[ General ] [ Speakers ] [ Tubes ] [ Vinyl ] [ Digital ] [ Hi-Rez ] [ Video Asylum ] [ Cables ] [ Tweaks/DIY ] [ Music ] [ Films ]

---