Switched-Resonator Converter

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  • Опубликовано: 14 ноя 2024

Комментарии • 13

  • @babylonfive
    @babylonfive 8 лет назад +2

    YAY! Glad to see you back on here, Sam.

  • @solenskinerable
    @solenskinerable 2 года назад +2

    can the 3rd phase be simply implemented in a half-bridge 2-stage cockcroft walton capacitive multiplier for higher efficiency, e.g. with a zener or such?

    • @sambenyaakov
      @sambenyaakov  2 года назад

      Hi, can you please indicate to which minute in the video are you referring?

    • @solenskinerable
      @solenskinerable 2 года назад

      @@sambenyaakov can i send you an ltspice schematic?

    • @sambenyaakov
      @sambenyaakov  2 года назад

      @@solenskinerable Yes, sby@bgu.ac.il

  • @alexanderkushnerov4556
    @alexanderkushnerov4556 8 лет назад

    Shmuel, I happy to see your returning to the audience!!! You remember, I like to dig (sometimes off to the side :) In the case with this gyrator, I dug out the old good gyrator, described in the book of Erickson and Maksimovic. It is also resonant and uses a diode bridge as shown in Fig. 4.1 here
    ecee.colorado.edu/~ecen5817/notes/ch4.pdf
    From the point of view of waveforms, the output current in the presentation is a rectified version of the inductor current in Fig. 4.34(a), page 31.

    • @sambenyaakov
      @sambenyaakov  8 лет назад

      Hi Alex,
      Nice to hear from you. Thanks for the good comment.
      Please write to my email.

    • @alexanderkushnerov4556
      @alexanderkushnerov4556 8 лет назад

      Thank you Shmuel ! In the second part of my comment "From the point of view of waveforms..." I apparently made a mistake. The circuit I simulated was a bit different. I connected an inductor to the SCC of 1/2 and made the 3rd topology as a short-circuited resonant tank. In any case, all these things are very interesting...

  • @pooviravalli9816
    @pooviravalli9816 6 лет назад

    Hello Professor, Thank you for the explanation. As with some resonant converters, I was wondering if the switching frequency is changed for load and line variations to accommodate zero voltage switching.

  • @mikeguitar9769
    @mikeguitar9769 6 лет назад

    @0:52 Doesn't a transformer satisfy those equations?

    • @sambenyaakov
      @sambenyaakov  6 лет назад +1

      Not really. The relationships in a transformer are between I&I and V&I and not V&I.

    • @mikeguitar9769
      @mikeguitar9769 6 лет назад +1

      @@sambenyaakov
      You're right. I was just noticing that the following equation is apparently valid for both a gyrator and a transformer.
      -Io/Iin = 1/ (Vo / Vin)
      (In a transformer it is the turns ratio, and in a gyrator the G term cancels.)
      But apparently this equation follows directly from conservation of power (so I guess it doesn't say much about the device except that it doesn't accumulate energy in steady-state and it's a lossless/"ideal" device).
      Dividing the 2 equations from the definition of the gyrator given indicates "conservation of power", (same as with an ideal transformer). But that operation reduces the 2 equation system to 1. If the second equation is then written, presumably in terms of the impedance relationship, then this indicates that in the Gyrator, Zout and Zin are inversely proportional to each other, whereas in a transformer Zout and Zin are proportional to each other.
      ------------------------------------------
      Derivations:
      Derivation of Conservation of power:
      Gyrator definition:
      -Io = G Vin
      Iin = G Vo
      Divide one equation by the other to get:
      -Io/Iin = Vin/Vo
      Rearranging:
      -Io Vo = Vin Iin
      (indicates Conservation of power)
      Derivation of Impedance Relationship:
      From Gyrator definition:
      Vo = Iin/G
      -Io = G Vin
      Dividing these:
      Vo/-Io = (1/G^2) (1/(Vin/Iin))
      (Indicates Impedance Relationship is inversely proportional)
      ===================================
      Transformer
      Conservation of Power:
      -Iout/Iin = Vin/Vo = n
      -Io Vo = Vin Iin
      Impedance relationship:
      Zout = Vo/-Iout
      Zin = Vin/Iin
      Zin = n^2 Zout
      =================================================================