PSU Design Fundamentals: Experimental Demonstration of Inductor Back EMF

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  • čas přidán 11. 10. 2023
  • In one of our previous videos we explained the action of an inductor in a switch mode power supply. This is the second part of our introductory videos about inductors. In this video, we will demonstrate just how big an inductor back EMF can get even when energised with a tiny battery.
    By simply breaking the current path of an inductor we light up, albeit momentarily, a 230V lightbulb with a single PP3 9V Battery.
    Download more detailed lecture notes as a PDF: www.biricha.com/psudf-inducto...
    Part 1 of the this video is here:
    • PSU Design Fundamental...
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Komentáře • 5

  • @waterfuel
    @waterfuel Před 5 měsíci

    I need to know how to get minimum 100v or higher from 12v battery, switch and 1.75 Millihenry wire coil as about .66 ohms wire?? The inductor circuit must have a parallel outside diode, so that the balance of coil energy released fast, >100v, goes ALSO forward to the resistor load. (Coil Polarity + and - changes around so as - and +) Does the circuit require capacitor to store energy and release through load in same direction? Current flow I assumed was only positive to negative through anode. Or is that negative reverse current through cathode diode?

  • @calholli
    @calholli Před 3 měsíci

    So why are we not capturing that back EMF and using it to recharge the battery to prolong is lifespan? (such as when using the coil in a motor, etc). Of course it would still have loses, but if you're recouping even a little bit each time, that would add up quite a bit over the span of the batteries total discharge; extending the usable time for each charge. Basically increasing efficiency

  • @doit.20
    @doit.20 Před 8 měsíci

    Hi, how to figure out Rds ON of an IGBT with the datasheet?

  • @syedzameerbashas4217
    @syedzameerbashas4217 Před 8 měsíci

    @biricha how does the back emf goes to -300v bcz the battery is only 9v

    • @gaynzz6841
      @gaynzz6841 Před 3 měsíci

      v(t) = L * di(t)/dt
      It's a differential equation. The faster the switch is opened, the higher the voltage