The following is a question from 8.18 in Design of Analog CMOS Integrated Circuit, page 342.
Consider the circuit in this following figure. \$(\frac{W} {L})_{1-4} =\frac{50} {0.5} \$, \$|{I_D}_{1-4}| = 0.5\ \mathrm{mA}\$, \$R_2 = 3\ \mathrm{k\Omega}\$.
Calculate the closed-loop gain and output impedance.
The answer I found from internet (solution link).
$$A \beta = \frac{V_F} {V_t} = \frac{1} {R_2} (R_1 \parallel R_2) \frac{r_{o3}} {(R_1 \parallel R_2) + \frac{1} {g_{m1}}} g_{m2} (r_{o4} \parallel (R_1 + R_2) \parallel r_{o2}) \frac{R_1} {R_1 + R_2}$$
My question is, how can I derive this equation? Any details?
The answer from internet seemed to suggest that
$$\frac{V_x} {V_t} = \frac{1} {R_2} (R_1 \parallel R_2) \frac{r_{o3}} {(R_1 \parallel R_2) + \frac{1} {g_{m1}}}$$
Where does that come from?
Here is my equation.
$$\frac{V_p} {R_1} - \frac{V_t - V_p} {R_2} = -V_p g_{m1} + \frac{V_x - V_p} {r_{o1}} \tag{1}$$
$$-\frac{V_x} {r_{o3}} = -V_p g_{m1} + \frac{V_x - V_p} {r_{o1}} \tag{2}$$
$$\Rightarrow \frac{V_x} {V_t} = \frac{R_1 r_{o3} (1 + g_{m1} r_{o1})} {R_1 r_{o1} + R_1 r_{o3} + R_2 (R_1 + r_{o3} + r_{o1} + R_1 g_{m1} r_{o1}) }$$
Voltage -Voltage feedback
