A) \[\frac{1}{\alpha }\]
B) \[\frac{\alpha }{1+\alpha }\]
C) \[\frac{\alpha }{1-\alpha }\]
D) \[\alpha -\frac{1}{\alpha }\]
Correct Answer: C
Solution :
Current gain in common-base configuration is. |
\[\alpha ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}} \right)}_{{{V}_{CB}}}}\] |
Current gain in common-emitter configuration is, |
\[\beta ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}} \right)}_{{{V}_{CE}}}}\] |
Also, \[{{i}_{B}}={{i}_{E}}-{{i}_{C}}\] |
or \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] |
\[\therefore \] \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{B}}}\] |
or \[\beta =\alpha \times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}\] |
or \[B=\alpha \times \frac{1}{1-\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}}\] |
or \[\beta =\frac{\alpha }{1-\alpha }\] |
Note: \[\beta \] is always greater than \[\alpha \]. Also \[\alpha <1\] and \[\beta >1\]. |
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