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(a) By using well matched fransistors, we can get accusate base-emitter voltage and current gain matching and they are fully internally isobted. [well matched means: \beta_{1}=\beta_{2} ](b) D C analysis:\begin{array}{l}V=5 \cdot 9-0.7=5.2 \mathrm{~V} \\\Rightarrow I=\frac{V}{1.2 \mathrm{~K}}=\frac{5.2}{1.2} \mathrm{~mA} \\I=4.33 \mathrm{~mA} \\\therefore I_{c}=\frac{I}{2}=2.167 \mathrm{~mA} . \\V_{c}=15-I_{c} \times 5 \mathrm{~K}=4.167 \mathrm{~V} \\g_{m}=\frac{I_{c}}{V_{T}}=\frac{2.167}{25}=86.68 \mathrm{~ms}\end{array}(c) \gamma_{03}=\frac{1}{\lambda I_{C}} ; A C equivalent diagram is: \gamma_{T 1}=\frac{B}{g_{m}}=865.22(d) Common mode gain!A_{c}=\frac{-5 k}{\frac{1}{9 m}+2 r_{0}}=\frac{-5 k}{11.5+2 r_{0}}ePCommon mode input resistance:\begin{aligned}R_{C M} & =\frac{R_{i n}}{2}=\frac{\gamma_{\pi}+(\beta+1) 2 \gamma_{0}}{2} \\& =\frac{\gamma_{\pi}}{2}+(\beta+1) \gamma_{0} \\R_{i M} & =432.63 \Omega \\R_{\text {out }} & =R_{C}=5 \mathrm{k} \Omega\end{aligned}(f) \gamma_{0}=80 k \Omega \Rightarrow A_{C}=\frac{-5 k}{11.5+2 \times 80 k}=-0.031248 \mathrm{vN}( Prom (d)(g) Differential gain:A_{d}=-g_{m} R_{c} =-86.68 \times 5\therefore A_{d}=-433.4 v / v(h)\begin{array}{l}\text { CMRR }(S)=\frac{\mid A d}{A C} \mid \\\rho=\frac{433.4}{0.031248}=13869 \\\rho_{d B}=20 \log \rho=82.84 d B\end{array}Half circuit(i) zener diode is used to bias the current sourle to achieve the required current. wecan get constant DC potential at the base even though coe have temperature voriantions [EResisfors vary due to the tolexan'es] and Supply vasiations ( \because Zenoracts as volfage regulator) ...