Question 9.14: Analyzing a BJT differential amplifier with a current mirror...

We have

I_{ Cl }=I_{ C 2}=I_{ C 4}=\frac{20 \mu A }{2}=10 \mu A

I_{ B 6}=\frac{I_{ Q }}{\beta_{ F 7} \beta_{ F 8}}=\frac{20 \mu A }{100 \times 100}=2 nA

\begin{aligned}&g_{ m }=\frac{I_{ C 2}}{V_{ T }}=\frac{10 \mu A }{26 mV }=387.6 \mu A / V \\&r_{ o 2}=r_{ o 3}=r_{ o }=\frac{V_{ A }}{I_{ C }}=\frac{100 V }{10 \mu A }=10 M \Omega \\&r_{\pi}=\frac{\beta_{ F } V_{ T }}{I_{ C }}=\frac{100 \times 26 mV }{10 \mu A }=260 k \Omega\end{aligned}

From Eq. (9.155),

A_{ d }=-\left(\frac{I_{ C }}{V_{ T }}\right)\left(\frac{V_{ A }}{2 I_{ C }}\right)=-\frac{V_{ A }}{2 V_{ T }}                                    (9.155)

A_{ d }=-\frac{V_{ A }}{2 V_{ T }}=-\frac{100 V }{2 \times 26 mV }=-1923 V / V

From Eq. (9.156),

R_{ id }=\frac{v_{ id }}{i_{ d }}=2 r_{\pi} \quad\left(\text { for } I_{ C }>0\right)                             (9.156)

R_{ id }=2 r_{\pi}=2 \times 260 k \Omega=520 k \Omega

and      R_{ o }=r_{ o 2} \| r_{ o 4}=\frac{r_{ o }}{2}=\frac{10 M \Omega}{2}=5 M \Omega

Since Q_{6} \text { and } Q_{7} form a compound transistor, its effective B-E voltage is that of two B-E junctions in series. Thus,

R_{ L }=2 r_{\pi 6}=\frac{2 V_{ T }}{I_{ B 6}}=\frac{2 \times 26 mV }{2 nA }=26 M \Omega

The effective transconductance is G_{ m }=g_{ m }=387.6 \mu A / V . Thus, the overall voltage gain with load A_{ d ( load )} is

A_{ d ( load )}=-G_{ m }\left(R_{ o } \| R_{ L }\right)=-387.6 \mu A / V \times(5 M \Omega \| 26 M \Omega)=-1625 V / V