Op Amps: Current Boosting

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  • Опубликовано: 4 окт 2024
  • Sometimes we can obtain sufficient voltage from an op amp but do not have sufficient current. Examples include driving low impedance loads or loads with high capacitance. While it is possible to use ICs which boast higher output current and power, another approach is to add a current boosting stage within the op amp's feedback loop, as shown in this video.
    References: Operational Amplifiers: Theory and Application; Chapter 4, section 4.
    My free texts and lab manuals are available for download at my college web site www.mvcc.edu/j... and at my personal site www.dissidents.com
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Комментарии • 15

  • @NTMMFTS
    @NTMMFTS 5 месяцев назад +1

    Thanks, this is exactly what I needed to learn for a project with a 5 ohm antenna.

  • @andymouse
    @andymouse 5 месяцев назад

    Excellent tutorial ! Dave Jones was talking about ideal diodes and the moved on to talk about the composite amp, so thanks for the TINA demo....cheers

  • @doggwoggle
    @doggwoggle 5 месяцев назад

    composite amps are clever.
    thanks for all the great videos, professor 🙂

  • @JonDeth
    @JonDeth 4 месяца назад +1

    I'll have to try this with my currently developing transistor fuzz-stortion. I've been avoiding clipping diodes as much as possible until I'm fully satisfied with what I'm getting from overdriving my transistors. I fully grasp that voltage gain is the most significant factor when we're trying to generate clipping by exceeding a devices forward voltage, but I've also noticed there's a richness and texture in the final product when there is a nice boost in current included.
    The simple way I did this in the past was running a ground resistor off the output, and forcing the op amp into Class-A or nearly by doing it, but I imagine there are advantages to adding this current gain stage on the feedback path. With the simple method, it's just a resistor and a bypass capacitor to shunt the noise the method also introduces.
    I'm just in the practice of establishing as many traits as I can for the project before introducing clipping diodes because 1, I still add other passive components to alter the available voltage that switches them on and 2, in doing so have found using a capacitor so I'm getting a specific corner frequency up/down clipped or bandpass allows for you to get those extra harmonics in the most desired bands and not so obviously reducing output to your load.

  • @bobdoritique7347
    @bobdoritique7347 5 месяцев назад

    Merci for this nice video. Very interesting, pragmatic and efficient, as far as necessary!

  • @mahmouda1920
    @mahmouda1920 5 месяцев назад

    Thank you

  • @MsFireboy2
    @MsFireboy2 5 месяцев назад

    Professor I’m using a JRC 4556 high current op-amp to drive the primary side of a transformer to sweep the Collector Emitter of a transistor.

  • @leeaho437
    @leeaho437 5 месяцев назад

    Good explanation. What software package are you using?

  • @rolandoaguilera3114
    @rolandoaguilera3114 5 месяцев назад

    Great content! Is there any textbook for reference on op amp current boosting? Thanks!

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  5 месяцев назад +1

      Yes. My free textbook. See links in the video description above for download as well as inexpensive print versions.

  • @simonyoungglostog
    @simonyoungglostog 5 месяцев назад

    This is great.
    Would this feedback setup negate thermal runaway on the transistors?
    Is it still worth putting small emitter resistors on the final stage transistors?
    Would we still need the biasing diodes?
    Thanks again, I hope all is well with you.

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  5 месяцев назад

      Feedback doesn't get rid of thermal runaway here, and yes, include the diodes for lowest distortion.

  • @steveh1273
    @steveh1273 5 месяцев назад

    what about the gain change from adding the push pull circuit in the feedback loop? Does the p-p circut have very low impedance? I'm assuming it does since the output voltage changed very little.

    • @ElectronicswithProfessorFiore
      @ElectronicswithProfessorFiore  5 месяцев назад

      It's inside the feedback loop. With this much negative feedback, the feedback defines the system gain.