The transistor parameters are: $K_{n1} = K_{n2} = 0.1 mA/V^2, K_{n3} = K_{n4} = 0.3 mA/V^2$ and for all transistors, $\lambda = 0$ and $V_{TN} = 1V$. Determine the maximum range of common-mode input voltage.

The reference current can be determined from,

$I_1 = \dfrac {20 - V_{GS4}}{R_1}$

and from,

$I_1 = K_{N3}(V_{GS4} - V_{TN})^2$

Combining these two equations and substituting the parameter values, we obtain

$9V_{GS4}^2 - 17 V_{GS4} - 11 = 0$

which yields

$V_{GS4} = 2.40V$ and $I_1 = 0.587 mA$

Since $M_3$ and $M_4$ are identical, we also find

$I_Q = 0.587 mA$

The quiescent drain currents in $M_1$ and $M_2$ are $I_{D1} = I_{D2} = I_Q /2 \cong 0.293 mA$

The gate-to-source voltages are then

$V_{GS1} = V_{GS2} = \sqrt {\dfrac{I_{D1}}{K_{n1}}} + V_{TN} = \sqrt {\dfrac{0.293}{0.1}} + 1 = 2.71 V$

The quiescent values of $v_{o1}$ and $v_{o2}$ are

$v_{o1} = v_{o2} = 10 - I_{D1}R_D = 10 - (0.293)(16) = 5.31V$

The maximum common-mode input voltage is the value when $M_1$ and $M_2$ reach the transition point, or

$V_{DS1} = V_{DS2} = V_{DS1}(sat) = V_{GS1} - V_{TN} = 2.71 - 1 = 1.17 V$

Therefore,

$v_{CM}(max) = v_{o1} - V_{DS1}(sat) + V_{GS1} = 5.31 - 1.71 + 2.71 = 6.31 V $

The minimum common-mode input voltage is the value when $M_4$ reaches the transition point, or

$V_{DS4} = V_{DS4}(sat) = V_{GS4} - V_{TN} = 2.4 - 1 = 1.4 V$

Therefore,

$v_{CM}(min) = V_{GS1} + V_{DS4}(sat) - 10 = 2.71 + 1.4 - 10 = -5.89 V $

For this circuit the maximum range for the common-mode input voltage is $-5.89 V \leq v_{CM} \leq 6.31 V$