Op-Amp Constant Current Source

Discussion in 'Analog & Mixed-Signal Design' started by enggricha, Oct 11, 2016.

  1. enggricha

    Thread Starter Member

    May 17, 2014
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    [​IMG]

    I am trying to do something like this. My load will be up to 3K ohms and I want to be able to drive up to 10mA current through it. Which would mean I would need a supply rail of +/- 35V to be comfortable. Looks like there are very few Op-Amps that can handle 70V R-R voltages. Is there a simpler way of doing this?

    The load has be on the low side, it cannot be a high side / floating load.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    How accurate and stable does the current need to be with varying load resistances?
     
  3. wayneh

    Expert

    Sep 9, 2010
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    Why do you need -35V? Do you need -10mA?
     
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    You could used matched voltage dividers to reduce the voltages at the inputs of the opamp, and use a level-shifter on the opamp output. That would allow use of a 'normal' opamp (albeit with reduced current resolution related to the division ratio).
     
  5. ci139

    Member

    Jul 11, 2016
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    Random_CC32.png Random_CC32h.png nowwith3shldetect -- if such for any improvement
    No credits of any kind ((the feedback here is not exactly what it should be -- i just tested if i manage to fetch a mosfet 3shld - which now i approx can imagine is to figure out minimum conductance OpAmp circuit . . . ))
    +more fq. tolerant v.
    Random_10mAcc.png ((nothing to do))
     
    Last edited: Oct 12, 2016
  6. crutschow

    Expert

    Mar 14, 2008
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    Here's a simple circuit that should do what you want, using a dedicated constant-current IC.
    The LT3092 requires only a minimum of 1.2V headroom.
    The LTspice simulation is shown below for the pot U2 rotation of 0 to 100%.

    upload_2016-10-11_16-39-43.png
     
    Last edited: Oct 11, 2016
  7. Bordodynov

    Well-Known Member

    May 20, 2015
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    Hi crutschow.
    Iout = (R(U2) / R2+1) * 10uA
     
  8. crutschow

    Expert

    Mar 14, 2008
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    By that you mean ((R(U2) / R2) + 1) * 10uA to be totally precise and add the 10uA reference current to the total current (which can typically be ignored unless you are trying to generate very low currents).
     
    Last edited: Oct 12, 2016
  9. enggricha

    Thread Starter Member

    May 17, 2014
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    I do not need -10mA, but I need to be be able to get down to 0uA. Dont I need -35V to be able to get close to 0uA.
     
  10. enggricha

    Thread Starter Member

    May 17, 2014
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    I want to be able to go below 0.5mA, apart from that this was a great idea. Will probably use this sometime in the future.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    That requirement would have been nice to know up front. :rolleyes:

    The simulation shows a minimum of about 0.3 mA.
    The worst-case minimum current @ room ambient would be about 2/3 mA minimum due to the offset voltage.
     
  12. ci139

    Member

    Jul 11, 2016
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  13. enggricha

    Thread Starter Member

    May 17, 2014
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    Ur right its too big a parameter to be missed out. Anyways thanks for the suggestion as I mentioned, I am sure this will come in handy someday.
     
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