(a) I_{ EE }=1 mA \text {. Since } v_{ B 2}=0 , we can write
R_{ EE } I_{ EE }+V_{ BE }=V_{ EE }=15 V
Thus,
R_{ EE }=\frac{15-0.7}{1 m }=14.3 k \Omega
We have
i_{ Cl }=i_{ C 2}=\frac{I_{ EE }}{2}=\frac{1 mA }{2}=0.5 mA
and g_{ m }=\frac{i_{ Cl }}{V_{ T }}=\frac{0.5 mA }{26 mV }=19.23 mA / V
Since v_{ B 2}=V_{ B 2}+v_{ b 2}=0, v_{ b 2}=0. Then
v_{ id }=v_{ b 1}-v_{ b 2}=v_{ b 1}
and v_{ ic }=\frac{v_{ b 1}+v_{ b 2}}{2}=\frac{v_{ b 1}}{2}
From Eq. (9.10), we get
v_o=A_{ d } v_{ id }+A_{ c } v_{ ic } (9.10)
v_{ ol }=A_{ c } v_{ ic }+\frac{A_{ d } v_{ id }}{2}=A_{ c } \frac{v_{ b 1}}{2}+\frac{A_{ d } v_{ b 1}}{2}=\frac{v_{ b 1}}{2}\left(A_{ c }+A_{ d }\right) (9.151)
Substituting for A_{ d } \text { and } A_{ c } from Eqs. (9.141) and (9.146) gives the voltage gain A_{1} as
A_{ d }=\frac{v_{ od }}{v_{ id }}=-g_{ m } R_{ C } (9.141)
\begin{aligned}A_{ c } =\frac{-g_{ m } R_{ C }}{1+2 g_{ m } R_{ EE }\left(1+1 / \beta_{ F }\right)}\end{aligned} (9.146)
A_{1}=\frac{v_{ o1}}{v_{ b 1}}=\frac{1}{2}\left(A_{ c }+A_{ d }\right)=-\frac{1}{2}\left[\frac{g_{ m } R_{ C }}{1+2 g_{ m } R_{ EE }\left(1+1 / \beta_{ F }\right)}+g_{ m } R_{ C }\right] (9.152)
Substituting A_{1}=-250, g_{ m }=19.23 mA / V , R_{ EE }=14.3 k \Omega, \text { and } \beta_{ F }=100 into Eq. (9.152) gives R_{ C }=25.95 k \Omega .
(b) From Eq. (9.141),
A_{ d }=-g_{ m } R_{ C }=-19.23 m \times 25.95 k =-499.01
From Eq. (9.146),
A_{ c }=\frac{-19.23 mA / V \times 25.95 k \Omega}{1+2 \times 19.23 mA / V \times 14.3 k \Omega \times(1+1 / 100)}=-0.897
Thus, CMRR =\left|A_{ d } / A_{ c }\right|=499.01 / 0.897=556.3 \text { (or } 54.91 dB \text { ) } .
