Hello everyone! I'm extremely confused about this and I was looking for advice, I'm sorry if this is the wrong place to ask.
I have the following inverting operational amplifier with these values:

- Rs = 1k ohms
- R = 10k ohms
- C = 10 mF
- Aol = 74dB
- Zin = inf
- Zout = 5 ohms

I need to find the transfer function, plot the bode diagram of the transfer function and calculate mid-band gain. Got no issues with the first two, I found an expression for the transfer function dividing the impedance on the feedback network by the input impedance, this is what I got:


\[ \frac{V_o}{V_s} = -\frac{sRC+1}{sR_sC} = - \frac{100s+1}{s} \]

I'm very confused regarding mid band gain. I know that the open loop gain is 74dB but the capacitor in the feedback network is throwing me off. The closed loop gain is equal to:\[ Av = \frac{A_{OL}}{1 + \beta A_{OL}} \] If I replace the resistor R and capacitor C in the feedback network with one impedance, I get that beta is equal to this:

\[ \beta = \frac{R_s}{R_s + \frac{jwRC+1}{jw}} \] which is dependent on frequency, this is where I'm lost. How can I find the mid-band gain if it's dependent on frequency? I can find it easily for low frequencies (ω → 0) and high frequencies (ω → inf), however I really don't know what to do when it comes to mid-band.

Thanks for the help and sorry for the confusion!

 

Sorry but it won't let me edit the post. I think I got it, actually. I'm supposed to calculate mid-band gain by plotting the bode diagram, mid-band gain in this case is where the diagram stays flat, right? If this is the case, I'm still confused about the open loop gain. Where does it come into play? What would I use it for? Should it be used to calculate bandwidth limits?

 

I'm still confused about the open loop gain. Where does it come into play?

The open-loop gain is given as a gain-bandwidth product, which determines how close the closed-loop gain is to the ideal (as calculated by your second equation) at a given frequency.
Since Aol of a normal op-amp rolls off with frequency at a nominal 6dB/octave (20dB/decade), it determines the maximum frequency response for a given closed-loop gain.
See example below:

 

The open-loop gain is given as a gain-bandwidth product, which determines how close the closed-loop gain is to the ideal (as calculated by your second equation) at a given frequency.
Since Aol of a normal op-amp rolls off with frequency at a nominal 6dB/octave (20dB/decade), it determines the maximum frequency response for a given closed-loop gain.
See example below:

View attachment 329278

That's extremely useful, thank you very much!