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!

 

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.

 

Hey Ron, thanks for the reply. So Ib1 = Icap?

 

Hey Ron, thanks for the reply. So Ib1 = Icap?

You're not thinking this out. Use KCL.

 

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..

 

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..

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.