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Does a cathod follower invert the phase?

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  • #16
    Originally posted by Satamax View Post
    But is it in relation with the rotation of the cow's tail or definately not? Or is there a Π fraction of the rotation of the cow's tail in relation with it???
    OK,let's get a little deeper in the subject:if we're talking cows and if it's a matter of "Π"(English read:PEE) than the 90 degrees angle issue refers clearly to the cow's tail as being bent upwards in a fixed position (not rotating).As you already know,a cow having such a tail's position is diagnosed as "mad" (unstable,self-oscilating) and has to be somehow "neutralised".Despite of the EC regulations,a mad cow doesn't have to be cruely executed but,according to an ancient wisdom,"...one can get rid of the Mad Cow Syndrome by divorcing her".
    If we're talking bulls,we have a more clearly situation here:The best example is Ferdinand The Bull (yes,the flower smeller from the cartoon).He's enjoying smelling daisies wich are round (360 degrees) flowers.Hence,the daisies don't rotate the phase at all.
    Q.E.D.!
    (Oh Gawd,please don't provoke me anymore... )

    Regards,

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    • #17
      Originally posted by Arthur B. View Post
      Since no explanation of grounded cathode and grounded anode operation is given, I'll give my own.

      In the grounded cathode configuration, the input is the grid, the anode is the output and the cathode is the reference for both, hence grounded cathode.

      The gain of this stage without the degenerative feedback of a cathode resistor is:

      dv(a-k)/dv(g-k) = -mu (note the sign) or dv(a-k) = -mu*dv(g-k)

      In the grounded anode configuration, the input is the grid, the cathode is the output and the cathode is the reference for both, hence grounded anode.

      Before we go on, we need the anode to grid voltage. That is simply:

      dv(a-g) = dv(a-k)-dv(g-k) = -mu*dv(g-k)- dv(g-k) = -(mu+1)*dv(g-k)

      The gain of this stage is:

      dv(k-a)/dv(g-a) = (-mu*dv(g-k))/(-(mu+1)*dv(g-k)) = mu/(mu+1)

      Note that there is always a loss with the “cathode follower”.

      Questions? Comments? Flames?
      Absolutely correct!!!
      Nico
      Kindest Regards

      Nico Verduin
      http://www.verelec.com
      http://home.wanadoo.nl/nico.verduin

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      • #18
        maybe...?

        Maybe you got the +/- 90* thing from a phase shift discussion. IIRC +/- 90* shift is where feedback loops become more trouble than help.

        Chuck

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        • #19
          Originally posted by Enzo View Post
          OK, while I learned electronics in an RF environment, it was 40-50 years ago, so I forgot a lot. Didn't want someone coming back with "No no no, at UHF there is a 4 degree phase shift..."

          Thanks
          I don't think you anything to worry about. If there should be any slight shift in phase at audio frequencies, it would not be enough for any concern given the application.

          -g
          ______________________________________
          Gary Moore
          Moore Amplifiication
          mooreamps@hotmail.com

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          • #20
            Oh I wasn't worried about audio, I was just heading any potential nitpickers off at the pass.
            Education is what you're left with after you have forgotten what you have learned.

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            • #21
              Originally posted by JGR View Post
              Hey all,

              Can anyone tell me if a cathode follower in a typical Marshall style circuit inverts the phase of the signal?

              Thanks!

              JR
              It does not.

              -g
              ______________________________________
              Gary Moore
              Moore Amplifiication
              mooreamps@hotmail.com

              Comment

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