JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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: [ \(\beta \) 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 \({\beta ^2}\frac{{{R_L}}}{{{R_{BE}}}},\frac{{\Delta {I_C}}}{{\Delta {I_E}}},{\beta ^2}\frac{{{R_L}}}{{{R_{BE}}}}\)
- D \(\beta \frac{{{R_L}}}{{{R_{BE}}}},\frac{{\Delta {I_C}}}{{\Delta {I_B}}},{\beta ^2}\frac{{{R_L}}}{{{R_{BE}}}}\)
Answer & Solution
Correct Answer
(D) \(\beta \frac{{{R_L}}}{{{R_{BE}}}},\frac{{\Delta {I_C}}}{{\Delta {I_B}}},{\beta ^2}\frac{{{R_L}}}{{{R_{BE}}}}\)
Step-by-step Solution
Detailed explanation
Curent gain \(\beta=\frac{\Delta I_{C}}{I_{B}}\) Voltage gain \(\mathrm{A}_{\mathrm{v}}=\) Current gain \(\times\) Resistance \(\operatorname{gain}=\beta \frac{R_{L}}{R_{B E}}\) Power gain \(A_{p}=(\text { Current gain })^{2} \times\) Resistance gain…
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