A) \[\frac{2}{3}G\]
B) \[1.5\,G\]
C) \[\frac{1}{3}G\]
D) \[\frac{5}{4}G\]
Correct Answer: A
Solution :
| As \[{{A}_{v}}=\beta \frac{{{R}_{L}}}{{{R}_{i}}}\]\[\left[ \because {{g}_{m}}=\frac{\Delta {{l}_{c}}}{\Delta {{V}_{B}}}=\frac{\Delta {{l}_{c}}}{\Delta {{l}_{B}}{{R}_{i}}} \right]\] |
| or \[G=\left( \frac{\beta }{{{R}_{i}}} \right){{R}_{L}}\] \[\left[ \because {{g}_{m}}=\frac{\beta }{{{R}_{i}}} \right]\] |
| \[\Rightarrow \] \[G={{g}_{m}}{{R}_{L}}\] |
| \[\Rightarrow \] \[G\propto {{g}_{m}}\] |
| \[\therefore \] \[\frac{{{G}_{2}}}{{{G}_{1}}}=\frac{{{g}_{{{m}_{1}}}}}{{{g}_{{{m}_{2}}}}}\] |
| \[\Rightarrow \] \[{{G}_{2}}=\frac{0.02}{0.03}\times G\] |
| \[\therefore \] Voltage gain \[{{G}_{2}}=\frac{2}{3}G\] |
You need to login to perform this action.
You will be redirected in
3 sec
