Semiconductor Devices: Active Current Limiter

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  • Опубликовано: 18 окт 2024

Комментарии • 10

  • @Hellhound604
    @Hellhound604 3 дня назад

    Hey Prof, wish I had a prof like you when I studied electronics at university 40 years ago, lol… but in my late age I met up with your courses and I enjoy them thoroughly, bringing back elementary stuff I have forgotten ages ago, and I try to re-commit it to my memory. Thanks for it all. You are great for explaining things.👍👍👍

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  3 дня назад

      40 years ago? I was a young teacher back then, only a few years in. Time flies. Glad you're enjoying it.

  • @simonyoungglostog
    @simonyoungglostog 3 дня назад

    Excellent, so far. I'm about halfway through and you've just added the alternate current path and RLimit. This reminds me of a 'ring of two' current source to reduce power supply hum.

  • @aduedc
    @aduedc 3 дня назад

    Thanks for the video
    You can use simple jfet with resistor on its source and gate connected to ground.
    The problem with both circuits is that they waste lots of energy, in your case 0.7xI. When I is large, then it become problematic.

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  3 дня назад +1

      Not when compared to the load power (assuming the circuit uses double digit power supplies, which would be expected for an audio amplifier, and not using tiny loads). Besides, many would argue that "not blowing up" is worth the wasted power ;-)

  • @Stelios.Posantzis
    @Stelios.Posantzis 3 дня назад

    Nice lecture. Won't those flat spots on the crests of the sinewave fry your tweeters though?
    Is there a simple circuit to turn off the drive to the power transistors when such flat spots (i.e. current limiter engaging or the drive voltage is too high) are detected?

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  3 дня назад +2

      The flat spots are a form of clipping and thus produce higher frequency distortion products, but that's probably preferable to throwing in a large current overage. I can imagine special cases where that might not be the case (like overdriving the heck out of a loudspeaker with an extremely low frequency tone), but just remember what you're trying to do here, and that's limit potentially damaging high load currents. Also, you wouldn't want to just "turn off" the output as that would cause large spikes which would be even worse. The best solution (and one that is much more complicated) would be to have a soft knee limiter right before this so the signal *can't* get that high.
      I should also indicate that a circuit like this might be used in non-audio applications, for example, an adjustable lab DC supply.

    • @Stelios.Posantzis
      @Stelios.Posantzis 2 дня назад

      ​@@ElectronicswithProfessorFiore You've got a point there with the soft knee limiter but in addition to the difficulty of implementing it, it also means you must have some power supply voltage margin as you don't want the limiter to engage on every peak, just those over a certain limit. But perhaps there is a way of implementing this without any power supply voltage margin, i.e., allowing the signal to exceed the supply voltage momentarily.
      I saw a really good presentation somewhere explaining why the flat spots burn the tweeters out but I've forgotten where. It also had to do with the presence of crossovers and it essentially showed that the ac energy contained in the harmonics dissipated in the tweeter during the flat portions of the clipped signal was excessive and caused them to burn out.
      Another solution would be a soft turn off once the output signal first crosses the zero voltage. The detection of the zero voltage crossing is not done by a particularly complicated circuit.

  • @simonyoungglostog
    @simonyoungglostog 3 дня назад

    Different question. If we look at the 2 diodes (say 1.4V drop), the Transistors will also drop this plus, potentially, 2 x Limiting resistors. Does this mean the drop across the 10Ohm resistor is zero or does the Diode/Transistor drop vary to accommodate, or something else, I'm thinking crossover distortion? When the overflow transistor is working, that seems to give us 2 x VBE drops against the diode drop. What am I missing?

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  3 дня назад +1

      Remember, the bypass transistor BE is in parallel with the limit resistor. The drop across the 10 ohm will be around 0.7 V when the limiter is activated. That's what turns on the bypass transistor (which shunts current away from the base of the output power transistor). For levels that are below the limit current, the drop across the 10 ohm will be less than 0.7 V (all depending on how large the output current is). It is true that the addition of this circuit will reduce the maximum output compliance by a small amount (0.7 V peak), but that's a small price to pay for the added safety.