Howland, Deboo and the non inverting Op Amp integrator

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  • Опубликовано: 21 окт 2024

Комментарии • 42

  • @px6436
    @px6436 2 года назад +9

    babe wake up new sam ben-yaakov video just dropped.

  • @Chris_Grossman
    @Chris_Grossman 2 года назад +1

    I have never seen that circuit. Thank you for introducing it to me.

  • @hamidk4772
    @hamidk4772 2 года назад +2

    Outstanding Job.

  • @aduedc
    @aduedc 2 года назад +3

    Thanks, very interesting presentation.
    Of course, Howland current source is well known, but the stability criterion is not well documented.
    So from your presentation we gather R4 > Real (Z*R1*R3/(R1*R3+Z*R2)) in order to have stable operation.
    @8:56 change Z with non-ideal LC Tank. Let R1=R2=R3=R4=R now the current through the Z is Iz=Vin/R and if we make the circuit unstable using mismatch we will have oscillation.
    By adjusting Vin, we can get very clean oscillation with low phase noise.
    Also, if you want to use it as integrator, all the real integrators suffer from high frequency noise integration, so you need to add a low pass circuit ( a differentiator) to the circuit to prevent saturation.

  • @frunsebischkek1050
    @frunsebischkek1050 2 года назад +5

    Thank you! A very interesting way for deriving of the function. There are some types of improved Howland current pumps. For example it's possible to split R3 in two or introduce a buffer (another opamp) between the load and the positiv pin of the opamp. 👉 "improved Howland current pump".
    Maybe the circuit could be used as a differential integrator, when not grounding the resistor R1.
    By the way, could you please explain the function of a transconductance amplifier and the variety of its applications? Many thanks in advance!

  • @CircuitShepherd
    @CircuitShepherd 2 года назад +3

    Very interesting! I'll refer to this video in one of my future tutorials if you don't mind.

  • @doublewishbone312
    @doublewishbone312 2 года назад +1

    Another great topic. I would say the instability described while true does necessarily mean instability of the overall system. Another way is to look at offset current being injected due to resistive network mismatch that happens to also be a function of the state of the integrator and if this circuit is in a bigger loop and this offset (bounded by bigger loop or w.c. supply while main loop starts up) can be tolerated this circuit could still be used even with the mismatch in the "runaway" direction.

    • @sambenyaakov
      @sambenyaakov  2 года назад

      For same offset, with mismatch the other direction there is no drift.

  • @byronwatkins2565
    @byronwatkins2565 2 года назад +3

    I solved that circuit three times two different ways and breadboarded it myself before I would believe it would work.

  • @ats89117
    @ats89117 2 года назад +6

    Very interesting. TI has put out a few papers on Howland Current Pump inspired circuits recently (shown below). I don't remember them warning about the instability criteria though...
    SBOA437 - OCTOBER 2020
    SNOA474A-January 2008-Revised April 2013
    Okay, 2013 isn't that recent, but at my age, it seems that way...

    • @aduedc
      @aduedc 2 года назад +1

      Thanks for references. I could find SBOA437 "Analysis of Improved Howland Current Pump Configurations" However, I could not find SNOA474A

    • @sambenyaakov
      @sambenyaakov  2 года назад +1

      www.ti.com/lit/an/snoa474a/snoa474a.pdf

    • @sambenyaakov
      @sambenyaakov  2 года назад +1

      👍🙏😊

    • @aduedc
      @aduedc 2 года назад +1

      @@sambenyaakov
      Thank you professor Ben-Yaakov.
      I know people use this circuit (Howland Current Source or Pump) for Linear V2I convertor. I guess a chip with all the resistors matched inside a chip (R1=R2=R3=R4=R) is in order. One important factor is the maximum frequency of operation of this circuit. The other is Maximum current that it can support.
      One possible application I am contemplating is to replace Z with a transmission line, inside a chip and outside a chip. Such transmission line is more immune to noise and interference than voltage transmission line. Specially, If we could modify this circuit so it have differential output instead of single ended output. One possible solution is to use two such circuit, but matching is an issue. Please notice for ground noise, which is common in a chip, even a single ended current signal can compensate for ground noise.

    • @sambenyaakov
      @sambenyaakov  2 года назад +1

      @@aduedc Good points, thanks. An intersting point that I dodged in video is the dependence of the operation frequency on the open loop transfer function of the Op Amp.

  • @Theineluctable_SOME_CANT
    @Theineluctable_SOME_CANT 2 года назад +2

    I wish I was a Howland.
    But I fear I am just another DeBoo...

    • @sambenyaakov
      @sambenyaakov  2 года назад +1

      It's OK as long as you are not deja vu

  • @Blanke94
    @Blanke94 Год назад +1

    Prof. Ben Yaakov, I enjoy your videos a lot! you make things "as simple as they can be, but not any less". This circuit seems simple but produces interesting insights.
    Can the feedbacks be realized matched in practice? Can it be made to track the matched configuration somehow with active feedback (and still be worth it)?

    • @sambenyaakov
      @sambenyaakov  Год назад +1

      Thanks for note. In many applications, absolute matching is not necessary since a small error can be tolerated..

    • @Blanke94
      @Blanke94 Год назад

      @@sambenyaakov thank you. would you consider making a video comparing the typical SMPS topologies (i.e Flyback, half bridge, forward..) and considerations for choosing between them?

  • @imetr8r
    @imetr8r 4 месяца назад +1

    This was a great explanation of a single-supply, non-inverting integrator. Could you make a single-supply, non-inverting derivative by exchanging C and R4?

    • @sambenyaakov
      @sambenyaakov  4 месяца назад

      Good question. I will have to think about that.

    • @imetr8r
      @imetr8r 3 месяца назад +1

      @@sambenyaakov I have implemented your version of the Deboo integrator into a PI closed servo loop for a cheap Chinese mini lathe which only has a 5 volt supply for logic. It works fairly well, but I had to place the encoder feedback device where I could find room and that was directly on the lathe spindle rather than the motor (for lack of space around the motor). With the spindle drive belt between the motor and encoder it induces some compliance into the system resulting in a low frequency oscillation when the loop is closed and a load applied. I am hoping a derivative will dampen this oscillation. I wish I could find a motor with an encoder (or tachometer) build in that will fit the restricted space, but lacking that I'll attempt to modify the Deboo integrator to a Sam Ben-Yaakov derivative. Or maybe an I-M-E-Trader derivative as this single rail function does not seem to have yet been invented.

    • @sambenyaakov
      @sambenyaakov  3 месяца назад

      @@imetr8r Thanks for sharing I hope you reach an optimal solution.

  • @Theineluctable_SOME_CANT
    @Theineluctable_SOME_CANT 2 года назад +2

    LOVE IT...

  • @tamaseduard5145
    @tamaseduard5145 2 года назад +4

    👍🙏

  • @jjmcrosbie
    @jjmcrosbie 2 года назад +1

    Isn't that what we used to call a "bootstrap" integrator?

  • @rikka0_059
    @rikka0_059 Год назад +1

    If the input has a very high slew rate (e.g. a square wave), will the OpAmp be able to catch up?

    • @sambenyaakov
      @sambenyaakov  Год назад +1

      Op Amps have inheret slew rate limitation that ca not be exceeded no matter the feedback network.

    • @sambenyaakov
      @sambenyaakov  Год назад +1

      See ruclips.net/video/B9sIAbdfIyE/видео.html

    • @rikka0_059
      @rikka0_059 Год назад +1

      @@sambenyaakov Thank you! In case of an integrator, if the input is a square wave and the output's slew rate does not exceed the limitation, will the GBP of the OpAmp affect the operation? e.g. using a 1.5MHz OpAmp to make an integrator to integrate a 100kHz square wave? (Assuming the gain is sufficient small that does not cause saturation or slew rate limitation)

    • @sambenyaakov
      @sambenyaakov  Год назад +1

      @@rikka0_059 No it will not