| In a common emitter configuration with suitable bias, it is given than \[{{R}_{L}}\] is the load resistance and \[{{R}_{BE}}\] is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: [JEE Online 15-04-2018] |
| \[\beta \] is current gain, \[{{I}_{B}},{{I}_{C}},{{I}_{E}}\] are respectively base, collector and emitter currents: |
A) \[\beta \frac{{{R}_{L}}}{{{R}_{BE}}},\frac{\Delta {{I}_{E}}}{\Delta {{I}_{B}}},{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}}\]
B) \[{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}},\frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}},\beta \frac{{{R}_{L}}}{{{R}_{BE}}}\]
C) \[{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}},\frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}},{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}}\]
D) \[\beta \frac{{{R}_{L}}}{{{R}_{BE}}},\frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}},{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}}\]
Correct Answer: D
Solution :
| [d] Voltage gain = |
| \[\frac{{{V}_{CE}}}{{{V}_{BE}}}=\beta \frac{{{R}_{L}}}{{{R}_{BE}}}\] |
| Current gain \[=\beta =\frac{{{I}_{C}}}{{{I}_{B}}}\] |
| \[\text{Power gain=voltage gain}\times \text{current gain=}{{\beta }^{2}}\frac{{{R}_{L}}}{{{R}_{BE}}}\] |
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