Question 9.15: Analyzing BiCMOS amplifiers The DC biasing current of a BiCM...

(a) We have

\begin{aligned}r_{ o 2} &=\frac{2 V_{ M }}{I_{ Q }}=\frac{2 \times 20 V }{10 \mu A }=4 M \Omega \\r_{ o 4} &=\frac{2 V_{ A }}{I_{ Q }}=\frac{2 \times 50 V }{10 \mu A }=10 M \Omega \\R_{ o } &=r_{ o 2}\left\|r_{ o 4} \equiv 4 M \Omega\right\| 10 M \Omega=2.86 M \Omega \\g_{ m 2} &=\sqrt{2 K_{ n } I_{ Q }}=\sqrt{2 \times 25 \mu \times 10 \mu}=22.36 \mu A / V\end{aligned}

Thus, the differential voltage becomes

A_{ d }=-g_{ m 2} R_{ o }=-22.36 \mu \times 2.86 M =-63.9

(b) We have

\begin{aligned}&r_{ o 4}=r_{ o 6}=\frac{2 V_{ A }}{I_{ Q }}=\frac{2 \times 50 V }{10 \mu A }=10 M \Omega \\&R_{ o }^{\prime}=\beta_{ F } r_{ o 6}=40 \times 10 M \Omega=400 M \Omega \\&R_{ o }=r_{ o 4}\left\|R_{ o }^{\prime}=10 M \right\| 400 M =10 M \Omega \\&g_{ m 2}=\sqrt{2 K_{ n } I_{ Q }}=\sqrt{2 \times 25 \mu \times 10 \mu}=22.36 \mu A / V\end{aligned}

Thus, the differential voltage becomes

A_{ d }=-g_{ m 2} R_{ o }=-22.36 \mu \times 10 M =-223.6 V / V

(c) From Eq. (9.187),

\begin{aligned}R_{ o } &=R_{ D }^{\prime}\left\|R_{ S }^{\prime}=\left(r_{ o 6} g_{ m 6} \beta_{ F 4} r_{ o 4}\right)\right\|\left(\beta_{ F 8} r_{ o 8}\right) \\&=\beta_{ F } r_{ o } \quad\left(\text { for } r_{ o 4}=r_{ o 8}=r_{ o }, \beta_{ F 8}=\beta_{ F }\right)\end{aligned}                           (9.187)

\begin{gathered}R_{ o }=\beta_{ F } r_{ o 4}=40 \times 10 M =400 M \Omega \\g_{ m 2}=\frac{I_{ Q }}{2 V_{ T }}=\frac{10 \mu A }{2 \times 25.8 mV }=193.8 \mu A / V\end{gathered}

Thus, the differential voltage gain becomes

A_{ d }=-g_{ m 2} R_{ o }=-193.8 \mu A / V \times 400 M \Omega=-77,520 V / V

which is considerably larger than for the other two configurations.