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Finding an expression for the transfer function of an op-amp with a finite open loop gain

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mtg97

Student
Aug 15, 2024
1
Hello everyone! I'm studying for an exam and I was practicing op-amps. This specific one popped up at one of my professor's exams but I'm very unsure as to how I'm supposed to proceed.
circuit_jrfv5i.png

Besides the values the resistors and capacitor in the picture, I know that A[sub]OL[/sub] = 74dB and Z[sub]out[/sub] = 5 ohms.

I'm supposed to find an expression for the transfer function, plot the Bode diagram and calculate mid-band gain. Normally if I were given an op-amp with ideally infinite open loop gain I'd just use the virtual short method to calculate the transfer function, or even just divide the feedback impedance by the input impedance (in this case, (R[sub]1[/sub] + R[sub]2[/sub] + Z[sub]C[/sub])/R[sub]1[/sub] however the open loop gain and output impedance threw me off. I'm assuming that they both have an effect on the transfer function, so I was thinking of working my way to it using A[sub]V[/sub] = (A[sub]OL[/sub]/(1 + βA[sub]OL[/sub]). In this case, feedback factor β would be (R[sub]1[/sub]/(R[sub]1[/sub] + R[sub]2[/sub] + Z[sub]C[/sub]). Replacing Z[sub]C[/sub] with 1/sC this is the expression for β that I got:
beta_fx6dwc.png


And plugging it into A[sub]V[/sub] = (A[sub]OL[/sub]/(1 + βA[sub]OL[/sub]) this is what I ended up with, kinda got messy but it should be correct (not sure if it is)
av1_ufrfas.png

rearranged:
av2_jshhwf.png


Now, I was wondering if my approach up until here is correct, and which one is. Is the last expression for Av, the one that takes in consideration A[sup]OL[/sup] correct, or am I just supposed to get an expression for the transfer function with V[sub]O[/sub]/V[sub]S[/sub] = -Z[sub]F[/sub]/Z[sub]IN[/sub] (Z[sub]F[/sub] being the impedance in the feedback network).

Thank you very much!
 
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