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Circuit to subtract two currents

Discussion in 'Electronic Design' started by owais, May 17, 2007.

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  1. owais

    owais Guest

    Hi,
    I need some guidance or basic starting point on this. I need to
    generate a current that is the difference of two currents. One of the
    currents is constant at ~ 25mA and the other (input current) can vary
    from 0 to 1000mA. My original idea is to do a I-V conversion on the
    two inputs, do a voltage subtraction of the two using a basic opamp
    difference circuit and then an V-I circuit to convert back to
    current...but there has to be a cleaner way.
    Owais
     
  2. Andrew Holme

    Andrew Holme Guest

    Google "current mirror"
     
  3. Jim Thompson

    Jim Thompson Guest

    Compliance? That is, what supply voltages do you have available, and
    what must the difference current drive, impedance-wise?

    ...Jim Thompson
     
  4. John Larkin

    John Larkin Guest

    Currents in parallel add! So connect the 0-1 amp thing to the same
    load as a -25 mA current source.

    John
     
  5. Mike Monett

    Mike Monett Guest

    He used "~' in front of the 25mA, not a minus sign, so you don't know the
    polarity of the 25mA current. If it's positive, it will add instead of
    subtracting.

    Also, connecting the two current sources together means the load voltage
    will appear at both outputs.

    Since you don't know the load resistance, you don't know the max voltage,
    and you don't know if the 25mA current source has sufficient compliance to
    handle the max voltage.

    If the currents are the same polarity, probably the easiest solution is to
    invert the 25mA current, then add it to the 1A as you suggested. The
    conversion circuit can then address any compliance issue with the 1A
    source.

    In any event, it would probably be best to leave the 1A current as is.
    Doing I-V and V-I conversions on a 1A current would dissipate quite a bit
    of power.

    Regards,

    Mike Monett
     
  6. John Larkin

    John Larkin Guest


    OK, I wouldn't hire you.

    John
     
  7. owais

    owais Guest


    Hi,

    The power supply driving the whole solution is rated at 12V, 1A
    continuous, 3.5A PWM; the load is a string of high-power LEDs. The
    difference current output would ideally be 12V at up to
    (InputCurrent-25mA); where InputCurrent is 0->1A

    THanks
     
  8. Mike Monett

    Mike Monett Guest

    [...]
    LOL - too bad. We could have had so much fun!

    Regards,

    Mike Monett
     
  9. owais

    owais Guest

    Thanks to everyone for helping. I now understand how to do this. I
    am going to implement an current sink using a current mirror connected
    to the same point as the 1A supplying the load (which on the led is
    the anode) Seeing how simple the solution is, I'm kind of embarrassed
    at my original question, but really, thanks!

    Owais
     
  10. Mike Monett

    Mike Monett Guest

    Nice of you to let us know you found your solution. Many people ask for
    help, then simply disappear.

    What kind of current mirror do you plan to use - Wilson?

    Regards,

    Mike Monett
     
  11. Fred Bloggs

    Fred Bloggs Guest

    Mmmmmmm....maybe even simpler than that depending on what you're doing.
    If the LED forward drop can be considered constant, then a simple
    parallel resistor can be used to divert the 25mA.
     
  12. owais

    owais Guest

    Actually, I'm using a Widlar and because I don't have bjts on hand,
    I'm going to end up using nmos.
    Thanks!
     
  13. Jasen

    Jasen Guest

    add -25mA to your input current.

    Bye.
    Jasen
     
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