Calculating mid-band gain of an inverting operational amplifier with a capacitor in its feedback network

[Holden97]

Hello everyone! This is a topic I'm feeling very confused about 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 (by hand) the bode diagram of the transfer function and calculate the 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 (sorry but it gives me an error if I try to use the math editor for some reason):

• Vo(s)/Vs(s) = - Zf/Zi = -(jwRC + 1)/(jw*Rs* C) = - (100jw + 1)/jw

I'm very confused regarding the mid band gain because I have the open loop gain (74dB) and the capacitor in the feedback network is throwing me off. I know that the closed loop gain is equal to Aol/(1 + β*Aol) and without the capacitor β would be equal to Rs/(Rs + R), however replacing R and C with an impedance (sRC+1)/sC the feedback factor depends on frequency, correct? This is where I become lost, how am I supposed to calculate the feedback factor if it varies for all frequencies? Especially if I'm mostly interested in mid-band gain?

Thank you very much for your help and sorry for the confusion!

 

[barry]

Why don’t you think feedback factor can be a function of frequency?

 

[Holden97]

Why don’t you think feedback factor can be a function of frequency?

I do! Just a bit confused as to what to do to calculate mid-band gain. I was thinking of calculating feedback factor for high frequencies (ω → inf) and low frequencies (ω → 0) which should make calculating gain at both high and low frequencies easy enough, but I'm still unsure about mid-band gain since it's what I'm actually supposed to find

 

[std_match]

You now have a kind of integrator. You may want to put C in parallel with R, instead of a series connection.

 

[barry]

I do! Just a bit confused as to what to do to calculate mid-band gain. I was thinking of calculating feedback factor for high frequencies (ω → inf) and low frequencies (ω → 0) which should make calculating gain at both high and low frequencies easy enough, but I'm still unsure about mid-band gain since it's what I'm actually supposed to find

"mid-band" is an indeterminate value as you've presented it. It's the value between specific upper and lower frequencies (which you haven't stated). It's not the value between zero and infinity.

 

[danadakk]

Its odd you are given Aol but not its intrinsic frequency dependency :

Here is open loop Aol dependency (single pole compensated in most OpAmps) :

Is the 74 db Gain value you are given considered the Aol G at midband of the Aol curve ?
Reason I ask most OpAmps these days their Aol at DC is >= 100 db.

https://www.ti.com/lit/slyt367 General analysis considering Aol


Regards, Dana.

 

[FvM]

The usual way to handle the problem is to sketch a bode diagram with ideal OP (infinite gain and GBW, zero Rout). Then determine how real transfer characteristic is affected by real OP parameters. You don't specify finite GBW, thus closed loop gain above RC corner frequency is R/Rs = 10 with effectively no Aol influence. Corner frequency is very low with 0.0016 Hz, so any "mid frequency" is most likely in constant gain branch of bode diagram.

 

[BradtheRad]

1.
The single capacitor creates a rolloff curve whose range of action covers a wide band of frequencies. At one end of this band is slight attenuation. At the other end is extreme attenuation. We can point to some middle frequency and call that the midband. It might be the arithmetic mean or it might be a geometric mean. By that reasoning it's arbitrary.

2.
Another way of looking at maximum gain is to think in terms of power (VxA), that is, find where the greatest power (VxA) goes through the capacitor. This just happens to be at the rolloff frequency which is the point where the amplitude is .707 times the non-attenuated frequency. This is the conventional definition arrived at by a mathematical (non-arbitrary) formula.

 

[danadakk]

The usual way to handle the problem is to sketch a bode diagram with ideal OP (infinite gain and GBW, zero Rout). Then determine how real transfer characteristic is affected by real OP parameters. You don't specify finite GBW, thus closed loop gain above RC corner frequency is R/Rs = 10 with effectively no Aol influence. Corner frequency is very low with 0.0016 Hz, so any "mid frequency" is most likely in constant gain branch of bode diagram.

An interesting overview of AC G error :

https://www.ti.com/lit/pdf/slyt383

Regards, Dana.