Back to basics: Practical capacitor charging currents by LTspice demonstration

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  • Опубліковано 7 лют 2024
  • Note that an inductance at the input to a closed loop switch mode system may induce instability.
    See:
    • Instability and oscill...
    • Correction to:Instabil...

КОМЕНТАРІ • 23

  • @PavolFilek
    @PavolFilek 5 місяців тому +1

    Thank you very much. I need this to my MPPT DC-DC BUCK 2-phase.

  • @krish2nasa
    @krish2nasa 5 місяців тому +1

    Very informative, Thank you very much.

  • @zaikindenis1775
    @zaikindenis1775 5 місяців тому +2

    Thank you for the video. So, the maximum current will no be higher than Uin/sqrt(L/C). And the same effect one can see in gate drivers, where we need high gate current in hard switching aplication.

    • @sambenyaakov
      @sambenyaakov  5 місяців тому +1

      Yes, correct. 👍. Hold on to my next, eye opener video.

  • @nikolaivic9987
    @nikolaivic9987 5 місяців тому +1

    Very nice.
    Keep in mind that LTSpice by defualt adds Rser=1mOhm to every inductor in the model to improve convergency. This can be changed to 0Ohmin Control Panel -> Hacks -> Always default inductors to Rser=0. Not sure though what is applied when L=0, whcih may be the rootcause of the difference in Energy you see :)

    • @sambenyaakov
      @sambenyaakov  5 місяців тому

      All this is known. But...you missed the REAL question: is the loss with L dependent on R?

    • @zaikindenis1775
      @zaikindenis1775 5 місяців тому +1

      ​@@sambenyaakov Interesting question, thank you. Intuitively, not. Battery provide energy E*q. And capacitor is charged finally to E*q/2 energy. So we always lose E*q/2 in heat, we assume charge save in the circuit.

    • @sambenyaakov
      @sambenyaakov  5 місяців тому

      @@zaikindenis1775 Hold on to my next video on losses. Coming soon😊

    • @nikolaivic4480
      @nikolaivic4480 5 місяців тому +1

      ⁠@@sambenyaakovI mean that the total energy is not measured only over R2 if Rser=1mOh, hence the difference between L=0 and L!=0 energies. L is irrelevant for the E 🙂

    • @sambenyaakov
      @sambenyaakov  5 місяців тому

      @@nikolaivic4480 OK, Got you now. You are of course correct.

  • @KratosGhostofSparta-pp1cr
    @KratosGhostofSparta-pp1cr 2 місяці тому +1

    Hi sir. I have a question about LTspice.
    Is there a way that add TI models in to ltspice and use it?
    Thank you

    • @sambenyaakov
      @sambenyaakov  2 місяці тому

      You may be lucky and the model might be compatible, but if it includes non generic sources or syntax it will take an expert to convert.

  • @biswajit681
    @biswajit681 5 місяців тому +1

    Great insight...Sir could you please make video on smps input filter design with respect to voltage mode and current mode control and input filter impact on loop characteristics

  • @yinan8719
    @yinan8719 5 місяців тому +1

    Thanks Dr Sam Ben-Yaakov. Is that also concluded that no matter how large the inductance is the energy dissipation is the same? Also seems it may takes a bit longer to dissipate the same amount of energy than no inductance at all.

    • @sambenyaakov
      @sambenyaakov  5 місяців тому

      Indeed. Hold on to my next video an eye opener.

  • @petarsimic4097
    @petarsimic4097 5 місяців тому

    🙏❤

  • @tamaseduard5145
    @tamaseduard5145 5 місяців тому +2

    👍🙏❤️

  • @richardandrews573
    @richardandrews573 5 місяців тому

    At 4:46 I calculated 28mJ not 38mJ?

    • @sambenyaakov
      @sambenyaakov  5 місяців тому

      75/2= 28? Or you used another method?