A) \[{{A}_{V}}=150,\,\,{{V}_{out}}=1.8\,V,\,\,and\,\,power\,\,gain=9000\]
B) \[{{A}_{V}}=20,\,\,{{V}_{out}}=1V,\,\,and\,\,power\,\,gain\,\,=\,\,2000\]
C) \[{{A}_{V}}=150,\,\,{{V}_{out}}=1.5V,\,\,and\,\,power\,\,gain\,\,=\,\,8500\]
D) \[{{A}_{V}}=20,\,\,{{V}_{out}}=1.5V,\,\,and\,\,power\,\,gain\,\,=\,\,2000\]
Correct Answer: A
Solution :
Voltage gain \[{{A}_{v}}=\beta \frac{{{R}_{C}}}{{{R}_{B}}}=60\times \,\,\frac{5\times {{10}^{3}}}{2\times {{10}^{3}}}=\,\,150\] Power gain \[=\,\,{{\beta }^{2}}\frac{{{R}_{C}}}{{{R}_{B}}}=\,\,60\times \,\,60\times 2.5\,\,=\,\,9000\] \[\operatorname{Base} current = {{I}_{B}}=\,\,\frac{12\times {{10}^{-3}}}{2\times {{10}^{3}}}\,\,=6\times {{10}^{-\,6}}A\] \[{{\operatorname{I}}_{C}}=\,\,\beta {{I}_{B}}=60\times 6\times {{10}^{-\,6}}\,=3.6\times {{10}^{-\,4}}\,A\] \[\operatorname{Output}=\,\,{{I}_{C}}{{R}_{L}}=3.6\times {{10}^{-\,4}}\,\,\times \,\,5\times 1{{0}^{3}}\Omega =\,\,1.8\,V\]
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