Integrator op amp-derivation help

Discussion in 'Homework Help' started by u-will-neva-no, Apr 2, 2012.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    2
    Hey everyone, I have attached the question and solution to this problem but do not arrive at the same solution. My problem is that I do not know how to use the Ib_2 current so my approach was just to say that another current would flow through the capacitor and that is equal to the current flowing through the resistor:

    I_r_e_s = I_c_a_p


     \frac{v_i - 0}{R} = -c\frac{dV_o}{dt}


    -\frac{v_o}{RC} = \frac{dv_o}{dt}
    V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t) + V_o(0)

    so V_o(t) = -\frac{1}{RC} \int^t_0 V_i(t)

    Any advice why the above is not applicable to this question and how I should go about it? Thanks!
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Ib2 does not flow in the integrator circuit, so it is a red herring.
    You have totally ignored Ib1, which is in the integrator circuit. Therefore, Ires≠Icap.
     
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  3. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Hey Ron, thanks for the reply. So Ib1 = Icap?
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You're not thinking this out. Use KCL.
     
  5. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Okay I think its Ires= Ic1+Icap but I have always learnt that a current flowing into the inverting op-amp is 0; thus Ic1=0 (ideally). This results in what I had previously using KCL..
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    THe key word here is "ideally". Not all real-world op amps have insignificant input bias current. I do think this is a poorly formulated question, because you still have to assume that open loop gain=∞ and input offset voltage=0 in order to solve it. If the op amp is not ideal, then all parameters that are relevant to the problem should be listed, IMHO.
     
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