Frequency Response of Amplifiers -1

Let us analyze the following amplifier "differential pair with current mirror" and guess the frequency response  

First we find the nodes of interest in the small signal / ac equivalent circuit of the above. The source nodes of $M_{1}$ and $M_{2}$ are connected to ac ground. The remaining nodes are X and Y. 

Impedance @ node X is $$\frac{1}{g_{m3}}||r_{o1}||r_{o3} =\frac{1}{g_{m3}} $$

Impedance @ node Y is $$r_{o2}||r_{o4}$$

From this we can see node X has lower impedance compared to that of node Y. Therefore the pole at the output node (Y) is lower than the pole at node X.

Dominant pole is at output node $$ \boxed{\omega_{p1} = \frac{1}{(r_{o2}||r_{o4})C_{Y}}}$$ 

Non-dominant pole is at node X $$ \boxed{\omega_{p2} = \frac{g_{m3}}{C_{X}}}$$ 

[Note: $^{\dagger}$"Operational Transconductance Amplifier(OTA) can be defined as an amplifier where all nodes are low impedance except the input and output nodes". By this definition our amplifier is an OTA ]

From the circuit topology we can see that, from the input $v_{in}$ to $v_{out}$ there are two signal paths in parallel. One is fast path ($M_{1},M_{2}$) and second is slow path ($M_{1},M_{3},M_{4}$).

$$ \frac{V_{out}}{V_{in}} = \frac{A_{0}}{1+\frac{s}{\omega_{p1}}} + \frac{A_{0}}{(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})} $$
$$\boxed{\frac{V_{out}}{V_{in}} =\frac{A_{0}(2+\frac{s}{\omega_{p2}})}{(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})}}$$ Where $A_{0}$ is dc gain.
From the transfer function we see a left-hand side zero.  The following is the desired magnitude - frequency response of our amplifier. We want to push the non-dominant pole as far as possible after unity frequency - frequency where the magnitude reaches 0dB.

desired frequency response - magnitude

$\omega_{u}$ is unity frequency

Reference: Design of Analog CMOS Integrated Circuits,Razavi, B.
$\dagger$ CMOS: Circuit Design, Layout, and Simulation, R. Jacob Baker