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Op-Amp Integrator

  1. Apr 29, 2005 #1
    I was assigned a lab that requires me to build the attached circuit and show how the output signal is the integral of the input signal. Basically I cant understand what the function of the 10MOhm feedback resistor is. Its giving me a headache when I try to derive my proof for the integrator, and part of the lab is to explain its significance and what would happen if it were removed.

    I dont know if I should be posting this in homework help but this isnt really a math question so I figured I could try here 1st.


    Attached Files:

  2. jcsd
  3. Apr 29, 2005 #2
    Try tackling the problem systematically...

    Convert the circuit to the Laplacian equvilent (i.e. C becomes 1/(Cs) and derive the transfer function. The only special things you have to remember are:

    a. For an ideal op-amp, the currents entering the inverting and non-inverting terminals is zero.
    b. For an ideal op-amp, the voltage between the inverting and non-inverting terminals is zero.

    Clue: In the s domain, a 1/s denotes an intergral.

    Try this and see if it makes more sense to you.
  4. Apr 29, 2005 #3
    Or that the voltage across a cap:

    [tex]Vc = \frac{1}{C} \int i_c dt [/tex]
  5. Apr 30, 2005 #4
    The [tex]R(f)=10M\Omega[/tex] resistor is there to make sure you do not have a small constant voltage on the inverted input (this would be integrated and you don’t want that). You can show that there is a frequency where the impedance of the capacitor equals the impedance of the R(f) resistor. Over this frequency the circuit acts as an integrator, so you can try to neglect that resistor in your calculations
  6. May 3, 2005 #5
    The feedback resistor limis the gain at DC to 100. If it was not there the theoretical gain at DC would be infinit. Any small input offset would cause an integration that would eventually saturate the op amp. Then all the usual assumptions about op amps go to H@$^*.
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