Finding an expression for the transfer function and mid-band gain of a real operational amplifier with a capacitor in its feedback network

Question

I'm studying for an exam and I was practicing on this amplifier, however I'm a bit confused when it comes to a non-ideal operational amplifier.

It's not ideal because it has an open loop gain \$ A_{OL} = 74dB \approx 2.51 \times10^{-7} \$ and a finite output impedance \$ Z_{out} = 5\Omega \$.

I'm asked to find the transfer function, plot the Bode diagram and calculate mid-band gain. As for mid-band gain, I'd analyze the Bode diagram since mid-band gain is supposed to be the gain value at which the diagram is constant, correct? However I need the transfer function in order to be able to plot the Bode diagram.

Normally I'd find an expression for the transfer function by using the virtual short method, however that requires that the OpAmp is ideal which is not the case here so I'm assuming that having a finite open loop gain and an output impedance greater than 0 will have an effect on the transfer function. I know that when an OpAmp has finite open loop gain, the expression of the closed loop gain is the following: $$ A_v = \frac{A_{OL}}{1 + \beta A_{OL}} $$ where \$ \beta \$ is the feedback factor which in this case is given by: $$ \beta = \frac{R_1}{R_1 + R_2 + Z_C}$$ with \$ Z_C = \frac{1}{sC} \$, ending up with these expression for \$ \beta \$: $$ \beta = \frac{sCR_1}{sC(R_1 + R_2)+1} $$

The final expression for the closed loop gain knowing that \$ A_{OL} \approx 2.51 \times 10^{-7} \$ is the following:

$$ A_v = \frac{A_{OL}}{1 + \frac{sCR_1}{sC(R_1 + R_2)+1}A_{OL}}$$ Tried to rearrange it a tiny bit and it got messy without plugging in the values but this should be it: $$A_v = \frac{A_{OL}(sC(R_1 + R_2) + 1)}{sC(R_1+R_2)+sCR_1A_{OL} + 1} $$

I'm a bit lost here. Assuming up until here it's correct, is \$ A_v \$ the transfer function I'm looking for? If I plug in the numbers and plot the Bode diagram of this transfer function will I get the correct mid-band gain looking at where the diagram is flat?

If this is all correct, I'm still lost about the output impedance. How does it come into play with the transfer function? Because so far I haven't considered it. Thanks for the help!

Answer 1

The best and fastest way to obtain the transfer function in a clear and ordered form - read low-entropy type of expression - is to resort to the fast analytical circuits techniques or FACTs. You have plenty of examples in my last book on the subject and let's add another one with your circuit.

First, you start with \$s=0\$, which is the dc gain of your circuit determined with the capacitor open. It's simple, you have the inverting open-loop gain of your op-amp:

Then, you determine the pole by zeroing the excitation, \$V_{in}=0\$, and then you install a test generator \$I_T\$ across the capacitor (which is temporarily disconnected). By determining the resistance \$R\$ offered by the connecting terminals of the capacitor (or an inductance), you will find the time constant of the circuit equal to \$\tau=RC_1\$. And, for a first-order circuit, the pole is immediately found as \$\omega_p=\frac{1}{\tau}\$:

A quick sanity check with SPICE shows our calculation is correct:

For the zero, several options exist but the null double injection is quite simple here and offers the best format in the end. The principle is to null the response (the output) by bringing the input bias back in the process. The current source \$I_T\$ is then adjusted to bring the response to 0 V. This is what I did with a SPICE simulator and I can check that my calculation is correct. This sanity check is easy and invaluable to chase any mistake during the calculation process:

There you go, you have the pole and the zero, you can assemble the transfer function. However, the leading term \$H_0\$ has the dimension of a gain - and that is what we want - but does not tell much about the design goal. Remember, you write transfer expressions for design purposes, this is a design-oriented analysis or D-OA, as forged by Dr. Middlebrook years ago. By revealing an inverted pole and zero, you have a leading term now representative of the high-frequency gain. I would not call it a "mid-band" gain personally, but more a high-frequency or asymptotic gain in this case:

Time to test the analysis versus a SIMetrix plot and import the data in Mathcad (see here how to do it):

And this ends my contribution for today! : )