(a) For v_{ id }=10 mV , Eq. (9.42) gives the DC drain current i_{ D 1} for transistor M _{1} as
i_{ D 1}=\frac{I_{ Q }}{2}+\sqrt{2 K_{ n } I_{ Q }}\left(\frac{v_{ id }}{2}\right)\left[1-\frac{\left(v_{ id } / 2\right)^{2}}{\left(I_{ Q } / 2 K_{ n }\right)}\right]^{1 / 2} (9.42)
\begin{aligned}&i_{ D 1}=\frac{10 m }{2}+2 \overline{2 \times 1.25 m \times 10 m } \times\left(\frac{10 m }{2}\right)\left[1-\frac{(10 m / 2)^{2}}{10 m /(2 \times 1.25 m )}\right]^{1 / 2}=5.25 mA \\&i_{ D 2}=I_{ D }-i_{ D 1}=10 mA -5.25 mA =4.75 mA\end{aligned}
(b) We know that I_{ D 1}=I_{ D 2}=I_{ Q } / 2=10 mA / 2=5 mA . From Eq. (9.54),
g_{ m }=2 K_{ n }\left(V_{ GS }-V_{ t }\right) =2 \overline{2 K_{ n } I_{ Q }} (9.54)
g_{ m }=2 \overline{2 \times K_{ n } I_{ Q }}=2 \overline{2 \times 1.25 m \times 10 m }=5 mA / V
r_{ ol }=\frac{V_{ M }}{I_{ D }}=\frac{100}{5 m }=20 k \Omega
From Eq. (9.55), the single-ended differential voltage gain A_{ d } is
A_{ d }=\frac{v_{ od } / 2}{v_{ id } / 2}=-g_{ m }\left(r_{ ol } \| R_{ D }\right) (9.55)
A_{ d }=-g_{ m }\left(R_{ D } \| r_{ ol }\right)=-5 m \times(5 k \| 20 k )=-20 V / V
From Eq. (9.57), the single-ended common-mode voltage gain A_{ c } is
A_{ c }=\frac{v_{ oc }}{v_{ ic }}=\frac{-g_{ m } R_{ D }}{1+g_{ m } 2 R_{ SS }} (9.57)
A_{ c }=\frac{-g_{ m } R_{ D }}{1+g_{ m } 2 R_{ SS }}=\frac{-5 m \times 5 k }{1+5 m \times 2 \times 50 k }=-0.0399 V / V
Thus, \text { CMRR }=\left|A_{ d } / A_{ c }\right|=20 / 0.0399=501(\text { or } 54 dB ) .
From Eq. (9.59),
R_{ ic }=R_{ id }=\infty (9.59)
R_{ id }=R_{ ic }=\infty
We know that
v_{ id }=v_{ g 2}-v_{ g 1}=20 mV -10 mV =10 mV
and v_{ ic }=\frac{v_{ g 1}+v_{ g 2}}{2}=\frac{10 mV +20 mV }{2}=15 mV
Using Eq. (9.10), we have
\begin{aligned}v_{0} &=A_{1}\left(v_{ ic }+\frac{v_{ id }}{2}\right)+A_{2}\left(v_{ ic }-\frac{v_{ id }}{2}\right) \\&=\left(\frac{A_{1}-A_{2}}{2}\right) v_{ id }+\left(A_{1}+A_{2}\right) v_{ ic } \\&=A_{ d } v_{ id }+A_{ c } v_{ ic }\end{aligned} (9.10)
v_{ o }=A_{ d } v_{ id }+A_{ c } v_{ ic }=-20 \times 10 mV -0.0399 \times 15 mV =-201 mV
(c) The DC drain voltage at the drain terminal of a transistor is
V_{ D }=V_{ DD }-I_{ D } R_{ D }=30 V -5 mA \times 5 k \Omega=5 V
Thus, for A_{ d }=-20, the maximum differential voltage will be v_{ id }=5 / 20=250 mV . Therefore, V_{ DD } must be greater than I_{ D } R_{ D } in order to allow output voltage swing due to the input voltages.
