A) 25
B) 50
C) 75
D) 100
Correct Answer: A
Solution :
Since, \[\mu ={{r}_{p}}.{{g}_{m}}\] \[\therefore \] \[{{g}_{{{m}_{1}}}}=\frac{{{\mu }_{1}}}{{{r}_{{{p}_{1}}}}}\] \[=\frac{30}{5}=6\] and \[{{g}_{{{m}_{2}}}}=\frac{{{\mu }_{2}}}{{{r}_{{{p}_{2}}}}}\] \[=\frac{21}{4}\] Since, triodes are connected in parallel, hence effective plate resistance, \[{{r}_{p}}=\frac{{{r}_{{{p}_{1}}}}{{r}_{{{p}_{2}}}}}{{{r}_{{{p}_{1}}}}+{{r}_{{{p}_{2}}}}}\] \[=\frac{5\times 4}{5+4}\] \[=\frac{20}{9}k\Omega \] and transconductance, \[{{g}_{m}}={{g}_{{{m}_{1}}}}+{{g}_{{{m}_{2}}}}\] \[=\left( 6+\frac{21}{4} \right){{(k\Omega )}^{-1}}\] \[\therefore \] Amplification factor \[\mu ={{r}_{p}}.{{g}_{m}}\] \[=\frac{20}{9}\left( 6+\frac{21}{4} \right)\] \[=25\]
You need to login to perform this action.
You will be redirected in
3 sec
