You'll find a derivation of closed loop transfer function in text books as Gray Meyer Analysis and Design... I'll remember, that I calculated similar circuits at times, also transfer functions of higher order active filters.
I admit, that I'm content in most cases with a qualitative analysis: How does the output resistance affect the transfer function? How does the transfer function modify at the GBW limit? In some cases, a simple pole-zero gain expression can be estimated.
In other cases, particularly when higher orders of the amplifier gain function must be considered, a simulation is necessary.
I mean if the opamp is not ideal(for most cases, it is the truth), and it's gain is A, output resistance is Ro, what's the transfer function?
Found it's very complicated.
Do you really want to include the output resistor ? Why not also the input impedance ?
I propose only to consider the finite gain A, since the contribution of Ro is really negligible.
Then, the transfer function is
G=AHf/(1+AHr) with the return function Hr=R1/(R1+Xp) and Xp=R2||(1/jwC)
and the forward function Hf=Xp/(R1+Xp)
The ideal opamp's output resistance is zero, while actural opamp's ouput resistance is very large, for example, several million ohms, especially for the OTA. And if you want to minimize the output resistance, you must increase either the output transitor size or the output stage bias current, or use the output buffer stage. All these ways will introduce bargain with the die size, power consumption, and circuit complication , and sometimes may be not applicable.
On the other hand, the input resistance is normally very large for CMOS process due to the gate dielectric. So it can be ignored.
The ideal opamp's output resistance is zero, while actural opamp's ouput resistance is very large, for example, several million ohms, especially for the OTA.
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Correction: The "actual opamps output resistance" is rather low. The figure from walker5678 shows an opamp and NOT an OTA. An OTA is no opamp but a transconductance amp. These terms should not be mixed in order to avoid misunderstandings.
For the simple two stage amplifier, if the Ids of output stage is 1mA(which is very large for CMOS IC), and lambda is 0.01, then the output resistance is about 0.5/lambda*Ids = 50K.
So is the simple two stage amplifier an operational amplifier or an OTA?
OK, so most of the onchip amplifier is essentially OTA, which has large output impedence, thus if you want to add a pole or zero in the control loop, maybe it's hard to estimate the precise location of these poles/zeros, ... unless use the simulation method.
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For the simple two stage amplifier, if the Ids of output stage is 1mA(which is very large for CMOS IC), and lambda is 0.01, then the output resistance is about 0.5/lambda*Ids = 50K.
So is the simple two stage amplifier an operational amplifier or an OTA?
Neither - nor.
But, of course, it´s a matter of definition.
I would call your two-stage amplifier simply "amplifier".
The term opamp applies - for my opinion (and so it is used worldwide) - to amplifiers which have a very high gain and input impedance and a very low output impedance.
Only in this case you can create mathematical "OPERATIONS" which are determined solely by the feedback network.
Conversely, an OTA is an amplifier which can be seen as a voltage controlled current source. But it is clear that this is true only for a limited range of load impedances.
OK, I see. It´s really necessary to find an agreement on terms and conventions.
For me, an ideal opamp has really "ideal" parameters: A infinite (or sometimes 1E6), rin infinite, rout=0.
And the "real" opamp is something between ideal and real:
A finite and frequency dependent; rin and rout are considered as far as it makes sense.