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Designing an op amp integrator

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Design an op-amp-based integrator with the dc gain equal to -10 V/V (use R1 = 200 Ω and R2 = 2 kΩ). Knowing the gain-bandwidth product of the μA741 op amp, find the proper value of a capacitance C of the integrator so that its 3dB frequency is 50 kHz.

    2. Relevant equations

    bg7QK.png (1)

    3. The attempt at a solution

    I have found out that the gain-bandwidth product of the amp is 1 MHz
    And this is my guess how the amplifier looks like.
    zBK1s.png


    I think I could use formula (1) but i would think I had to replace R2 with this formula :
    dcgd6.png (2)

    because the Op amp has a capacitor.


    Am I on the right track for this?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 17, 2012 #2

    rude man

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    Gold Member

    Re: Designing an op amp

    The canonical approach to this is to model the op amp as a gain block ω0/s

    where ω0 is your 1 MHz*2π. (So that the gain at 1 MHz = 1).

    Then write the amplitude response of the combined network using your R2 and C chosen such that the new 3dB-down frequency is 50 KHz.

    I might as well warn you that the result is a second-order transfer function, not just a nice
    K/(Ts+1) first-order one. Of course, it's still a lowpass filter with a definite 50-KHz 3db-down cutoff frequency. 2nd-orders are just bitchier than 1st to plot & calculate, but it's straight-forward.

    Your text/instructor might have given you a short-cut approximation to this problem, yielding another 1st-order transfer function. I don't know.

    You could try approximating the 2nd order xfr fn by a 1st order one by factoring the 2nd-order polynomial, starting tentatively with C assuming an infinite-gain op amp, then throwing out the less significant pole.
     
    Last edited: Feb 17, 2012
  4. Feb 18, 2012 #3
    It is also known as a low pass filter because at 'low' frquencies Xc is large and the DC(and AC) gain is given by -R2/R1.
    At 'high' frequencies Xc is small and 'shorts' R2. The gain falls with increasing frequency.
    An important point (from a practical point of view) is when Xc = R2. Can you see when this occurs the DC gain will have halved?
     
  5. Feb 18, 2012 #4

    rude man

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    1. The OP is asked to include the effects of the gain-bandwidth product of the 741.
    2. Gain when XC = R2 is reduced to 1/√2 of the DC gain, not 1/2.
     
  6. Feb 19, 2012 #5
    Correction noted 1/√2 not 1/2
     
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