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Op-Amp circuit to supply constant power

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data

    see image
    Untitled.png

    2. Relevant equations



    3. The attempt at a solution
    I can make 2 different circuits to solve this. However, Using just op amps and resistors I am not sure how to make a single circuit that could do this
     
  2. jcsd
  3. Jan 22, 2013 #2

    mfb

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    I am not sure if that is intended, but if power just has to be the same for those two specific resistance values, there is an easy solution: Add a specific resistor in series, and find a pair (voltage, resistance) where both resistors get the same power.
     
  4. Jan 22, 2013 #3

    rude man

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    I would consider an op amp circuit set up so that the op amp output voltage is Vo. Then I would run a resistor R from Vo to each of your loads, one at a time of course.

    Given the 10 mW for each load, compute the voltage needed across each load.

    You now have 2 equations and 2 unknowns: R and Vo. Solve for both.

    (You will obviously need about 31.6V across the 100K resistor so figure on an op amp that can handle about 40V between its power supply pins.)
     
  5. Jan 22, 2013 #4
    It is really easy to do with 2 difference circuits. But using one circuit I don't see how changing one resistor could allow it to get 10mW just on the one resistor that has been changed.

    In order for the 100k ohm, it needs ~31V, while the 100 ohm only needs 1V.
     
  6. Jan 22, 2013 #5

    rude man

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    That's what the series resistor is there for! It puts ~ 31V across the 100K but only 1V across the 100 ohm.

    Have you considered doing what I suggested?
     
  7. Jan 22, 2013 #6
  8. Jan 22, 2013 #7

    rude man

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  9. Jan 22, 2013 #8
    32.62 gain, 3162 R
    That way you have 1V over the 100ohm = 10mW
    and 32.62(100000/(3162+100000)=31.62V -> 31.62^2 / 100000 = 9.998mW
     
  10. Jan 22, 2013 #9

    rude man

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    Superb! Congrats!
     
  11. Jan 22, 2013 #10

    NascentOxygen

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  12. Jan 23, 2013 #11
    the link is linked to my account, its forbidden i guess\
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