
- For a transistor in common base, the current gain is \[0.95\]. If the load resistance is \[400\;k\Omega \]and input resistance is \[200\Omega \;,\]then the voltage gain and power gain will be
A. \[1900\;and\;1800\]
B. \[1900\;and\;1805\]
C. \[5525\;and\;3591\]
D. \[1805\;and\;1900\]
Answer
594.6k+ views
Hint: In this question here there is common base transistor of which the current gain, load resistance and input resistance is given so first of all we will find the voltage gain with suitable formula in terms of current gain and then we will find the power gain with using formula of powers in terms of current gain and voltage gain.
Using Formula: -
(i) $\alpha = \dfrac{{{I_C}}}{{{I_E}}}\, $
(ii) $\beta = \dfrac{{{V_C}}}{{{V_E}}} $
(iii) ${V_E} = {I_E}{R_E}\,\, $
(iv) ${V_C} = {I_C}{R_L} $
(v) $\dfrac{{{P_C}}}{{{P_E}}} = \alpha \times \beta $
where \[{V_E}\] is the emitter voltage and \[{V_C}\] is the collector voltage. \[{R_E}\] is the input resistance \[{R_L}\] is the load resistance \[{I_E}\] is the emitter current and \[{I_C}\] is the collector current. \[{P_C}\,and\,{P_E}\] are the powers of collector circuit and power of the emitter circuit respectively.
\[\alpha \,and\,\beta \] are current gain and voltage gain respectively.
Given Data:
Current gain \[\alpha = 0.95\,{\rm A}\]
Load resistance \[{R_L} = 400\,k\Omega \]
Input resistance \[{R_E} = 200\,\Omega \]
Voltage gain \[\beta = ?\]
Power gain \[\dfrac{{{P_C}}}{{{P_E}}} = ?\]
Step by step solution:
Voltage gain is the ratio of collector voltage and emitter voltage as follows
\[\beta = \dfrac{{{V_C}}}{{{V_E}}} = \dfrac{{{I_C}{R_L}}}{{{I_E}{R_E}}} = \alpha \dfrac{{{R_L}}}{{{R_E}}} = 0.95 \times \dfrac{{400\, \times 1000\,\Omega }}{{200\,\Omega }} = 1900\]
Power gain is the ratio of power consume in collector circuit and power consume in emitter circuit
\[\dfrac{{{P_C}}}{{{P_E}}} = \dfrac{{{V_C}{I_C}}}{{{V_E}{I_E}}} = \alpha \times \beta = 0.95 \times 1900 = 1805\]………………………… \[({P_C} = {V_C}{I_C}\,and\,{P_E} = {V_E}{I_E})\]
Here from above calculation we found that the voltage gain is equal to \[1900\] and power gain is \[1805\]
Hence option \[(B)\] is correct.
Note: In this question we should be aware of the terms current gain, voltage gain and power gain for the common base transistor it is important to note that we should do the calculation carefully.
Using Formula: -
(i) $\alpha = \dfrac{{{I_C}}}{{{I_E}}}\, $
(ii) $\beta = \dfrac{{{V_C}}}{{{V_E}}} $
(iii) ${V_E} = {I_E}{R_E}\,\, $
(iv) ${V_C} = {I_C}{R_L} $
(v) $\dfrac{{{P_C}}}{{{P_E}}} = \alpha \times \beta $
where \[{V_E}\] is the emitter voltage and \[{V_C}\] is the collector voltage. \[{R_E}\] is the input resistance \[{R_L}\] is the load resistance \[{I_E}\] is the emitter current and \[{I_C}\] is the collector current. \[{P_C}\,and\,{P_E}\] are the powers of collector circuit and power of the emitter circuit respectively.
\[\alpha \,and\,\beta \] are current gain and voltage gain respectively.
Given Data:
Current gain \[\alpha = 0.95\,{\rm A}\]
Load resistance \[{R_L} = 400\,k\Omega \]
Input resistance \[{R_E} = 200\,\Omega \]
Voltage gain \[\beta = ?\]
Power gain \[\dfrac{{{P_C}}}{{{P_E}}} = ?\]
Step by step solution:
Voltage gain is the ratio of collector voltage and emitter voltage as follows
\[\beta = \dfrac{{{V_C}}}{{{V_E}}} = \dfrac{{{I_C}{R_L}}}{{{I_E}{R_E}}} = \alpha \dfrac{{{R_L}}}{{{R_E}}} = 0.95 \times \dfrac{{400\, \times 1000\,\Omega }}{{200\,\Omega }} = 1900\]
Power gain is the ratio of power consume in collector circuit and power consume in emitter circuit
\[\dfrac{{{P_C}}}{{{P_E}}} = \dfrac{{{V_C}{I_C}}}{{{V_E}{I_E}}} = \alpha \times \beta = 0.95 \times 1900 = 1805\]………………………… \[({P_C} = {V_C}{I_C}\,and\,{P_E} = {V_E}{I_E})\]
Here from above calculation we found that the voltage gain is equal to \[1900\] and power gain is \[1805\]
Hence option \[(B)\] is correct.
Note: In this question we should be aware of the terms current gain, voltage gain and power gain for the common base transistor it is important to note that we should do the calculation carefully.
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