Push-Pull fascination (Part 4)

Continuing with this series of blog posts around the 6C4C push-pull design. As suggested my 45 in my previous blog entry, here is the 4P1L-4P1L version:

6C4C PP version 7The first 4P1L driver stage remains unchanged, as does the output stage. The addition of the 4P1L differential pair with CCS tail and LL1660/pp IT is the main change of this design. The LL1660 is configured in ALT M or 2.25+2.25:2+2.  I guess that a different IT could be used instead to get a lower output impedance on the diff pair and improve the performance in A2. The amp has more gain that I need in this configuration as it delivers its maximum power (circa 8W) when input is 1.2V peak.

So how does it performs?

6C4C PP version 7 THD Very low distortion indeed. About 0.06% up to 8W. Mainly odd harmonics dominating the sound of this amp.

Interesting findings…

 

 

Author: Ale Moglia

"A mistake is always forgivable, rarely excusable and always unacceptable. " (Robert Fripp)

6 thoughts on “Push-Pull fascination (Part 4)”

  1. Seems that LL1660 are not well modeled.
    290H is the total primary inductance, this is, two bobbins of 72,5H.
    For secondaries I get two bobbins of 57,28 Hy.

    1. Hi Luis, thanks for pointing it out. I have changed the transformer models in Ltspice anyway as these were not good ones.
      Haven’t had the chance to update this am afraid!
      cheers
      Ale

  2. LL1682 seems to have very high inductance. Lp=400 H
    On the other hand, with this value, secondary must be 0,3636H

    A most real OT would be total Lp = 100H (Lp1 = Lp2 = 25H) and Ls = 91 mH

    1. I found that this model is not ok recently whilst working on Schade feedback designs. Were did you get the 400H of Lp?
      I haven’t got around to measure this am afraid. As per my previous post, I’m working with more accurate Tx models now but need to find the time to post them
      thanks
      Ale

      1. Sorry for delay, is difficult to find my own posts and have no notice of your response.

        I say that Laa = Lp1+Lp2+2k sqrt(Lp1*Lp2) and this is equal to Lp1+Lp2+2Lp1 when k=1, this is 4 Lp1 = 400 H anode to anode inductance (Laa)

        Ouch!! I just see the “Notify me…” checkbox below, excuse my awkwardness.

        Regards.

        1. I also replied your other comment on the Crowbar protection, check it out 🙂
          Got it. I think I need to measure the Lp first given I have now an accurate model in Spice for this. Unfortunately I haven’t had the time to do it yet!
          Thanks! Ale

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