(a) For v_{ id }=10 mV , Eq. (9.72) gives the DC drain current i_{ D 1} \text { for transistor } M _{1} as
i_{D1}=\frac{I_Q}{2} +\frac{I_Q}{2} \left(\frac{v_{id}}{V_p} \right) \left[2\left(\frac{I_{DSS}}{I_Q} \right) -\left(\frac{v_{id}}{V_p} \right)^2 \left(\frac{I_{DSS}}{I_Q} \right) ^2\right]^{1/2} (9.72)
\begin{aligned}&i_{ D 1}=\frac{10 m }{2}\left\{1+\frac{100 m }{-4}\left[2\left(\frac{20 m }{10 m }\right)-\left(\frac{100 m }{-4}\right)^{2}\left(\frac{20 m }{10 m }\right)^{2}\right]^{1 / 2}\right\}=5.25 mA \\&i_{ D 2}=I_{ Q }-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. (9.78),
\begin{aligned}g_{ m } &=\left|-\frac{2 I_{ DSS }}{V_{ p }}\left(1-\frac{V_{ GS }}{V_{ p }}\right)\right| \\&=\frac{2}{\left|V_{ p }\right|}\left[\left|I_{ D } I_{ DSS }\right|\right]^{1 / 2}\end{aligned} (9.78)
\begin{aligned}&g_{ m }=\frac{2\left[\left|I_{ D 1} I_{ DSS }\right|\right]^{1 / 2}}{\left|V_{ p }\right|}=\left(\frac{2}{4}\right) \times 2 \overline{5 mA \times 20 mA }=5 mA / V \\&r_{ o 1}=\frac{V_{ M }}{I_{ D }}=\frac{100}{5 m }=20 k \Omega\end{aligned}
From Eq. (9.79), the single-ended differential voltage gain A_{ d } is
A_{ d }=-g_{ m }\left(R_{ D } \| r_{ o 1}\right) (9.79)
A_{ d }=-g_{ m }\left(R_{ D } \| r_{ o 1}\right)=-5 m \times(5 k \| 20 k )=-20 V / V
From Eq. (9.80), the single-ended common-mode voltage gain A_{ C } is
A_{ c }=\frac{-g_{ m } R_{ D }}{1+g_{ m } 2 R_{ SS }} (9.80)
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, 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
v_o=A_{ d } v_{ id }+A_{ c } v_{ ic } (9.10)
v_{ o }=A_{ d } v_{ id }+A_{ c } v_{ ic }=-20 \times 10 mV -0.0399 \times 15 mV =-250.6 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_{\text {id }}=5 / 20=250 mV \text {. } Therefore, V_{ DD } must be greater than I_{ D } R_{ D } in order to allow output voltage swing due to the input voltages.
