PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365512
The gain of an amplifier without feedback is 90 and the feedback fraction \({\beta}\) is \({\dfrac{1}{10}}\). The gain of the amplifier if negative feedback is introduced is
1 89
2 9
3 8.9
4 900
Explanation:
The gain of an amplifier without feedback is \({A=90}\). Feedback fraction, \({\beta=\dfrac{1}{10}}\) The gain of the amplifier if negative feedback is introduced, \({A_{fb}} = \frac{A}{{1 + A\beta }} = \frac{{90}}{{1 + \frac{{90 \times 1}}{{10}}}} = \frac{{90}}{{1 + 9}} = \frac{{90}}{{10}} = 9\). So correct option is (2)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365513
Consider an \(pn\) transistor amplifier in common emitter configuration. The current gain of the transistor is 100 . If the collector current changes by \(1\,mA\), what will bet he change in emitter current?
1 \(1.1\,mA\)
2 \(1.01\,mA\)
3 \(0.01\,mA\)
4 \(10\,mA\)
Explanation:
Current gain in common emitter mode, \(\beta=\Delta I_{C} / \Delta I_{B}\) But \(\Delta I_{B}=\Delta I_{E}-\Delta I_{C}\) \(\Rightarrow \beta=\dfrac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}\) Given, \(\beta=100\) and \(\Delta {I_C} = 1\;mA\) \(\therefore 100=\dfrac{1}{\Delta I_{E}-1}\) \(\Rightarrow \Delta I_{E}-1=0.01\) \( \Rightarrow \Delta {I_E} = 1 + 0.1 = 1.01\;mA\)
AIIMS - 2005
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365514
If \(\alpha \)- current gain of a transistor is 0.98. What is the value of \(\beta \)- current gain of the transistor?
1 \(4.9\)
2 \(0.49\)
3 \(5\)
4 \(49\)
Explanation:
The relation between \(\alpha \) and \(\beta \) is \(\beta = \frac{\alpha }{{1 - \alpha }}\) Here, \(\alpha = 0.98\) \(\therefore \beta = \frac{{0.98}}{{1 - 0.98}} = \frac{{0.98}}{{0.02}} = 49\)
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365512
The gain of an amplifier without feedback is 90 and the feedback fraction \({\beta}\) is \({\dfrac{1}{10}}\). The gain of the amplifier if negative feedback is introduced is
1 89
2 9
3 8.9
4 900
Explanation:
The gain of an amplifier without feedback is \({A=90}\). Feedback fraction, \({\beta=\dfrac{1}{10}}\) The gain of the amplifier if negative feedback is introduced, \({A_{fb}} = \frac{A}{{1 + A\beta }} = \frac{{90}}{{1 + \frac{{90 \times 1}}{{10}}}} = \frac{{90}}{{1 + 9}} = \frac{{90}}{{10}} = 9\). So correct option is (2)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365513
Consider an \(pn\) transistor amplifier in common emitter configuration. The current gain of the transistor is 100 . If the collector current changes by \(1\,mA\), what will bet he change in emitter current?
1 \(1.1\,mA\)
2 \(1.01\,mA\)
3 \(0.01\,mA\)
4 \(10\,mA\)
Explanation:
Current gain in common emitter mode, \(\beta=\Delta I_{C} / \Delta I_{B}\) But \(\Delta I_{B}=\Delta I_{E}-\Delta I_{C}\) \(\Rightarrow \beta=\dfrac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}\) Given, \(\beta=100\) and \(\Delta {I_C} = 1\;mA\) \(\therefore 100=\dfrac{1}{\Delta I_{E}-1}\) \(\Rightarrow \Delta I_{E}-1=0.01\) \( \Rightarrow \Delta {I_E} = 1 + 0.1 = 1.01\;mA\)
AIIMS - 2005
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365514
If \(\alpha \)- current gain of a transistor is 0.98. What is the value of \(\beta \)- current gain of the transistor?
1 \(4.9\)
2 \(0.49\)
3 \(5\)
4 \(49\)
Explanation:
The relation between \(\alpha \) and \(\beta \) is \(\beta = \frac{\alpha }{{1 - \alpha }}\) Here, \(\alpha = 0.98\) \(\therefore \beta = \frac{{0.98}}{{1 - 0.98}} = \frac{{0.98}}{{0.02}} = 49\)
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365512
The gain of an amplifier without feedback is 90 and the feedback fraction \({\beta}\) is \({\dfrac{1}{10}}\). The gain of the amplifier if negative feedback is introduced is
1 89
2 9
3 8.9
4 900
Explanation:
The gain of an amplifier without feedback is \({A=90}\). Feedback fraction, \({\beta=\dfrac{1}{10}}\) The gain of the amplifier if negative feedback is introduced, \({A_{fb}} = \frac{A}{{1 + A\beta }} = \frac{{90}}{{1 + \frac{{90 \times 1}}{{10}}}} = \frac{{90}}{{1 + 9}} = \frac{{90}}{{10}} = 9\). So correct option is (2)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365513
Consider an \(pn\) transistor amplifier in common emitter configuration. The current gain of the transistor is 100 . If the collector current changes by \(1\,mA\), what will bet he change in emitter current?
1 \(1.1\,mA\)
2 \(1.01\,mA\)
3 \(0.01\,mA\)
4 \(10\,mA\)
Explanation:
Current gain in common emitter mode, \(\beta=\Delta I_{C} / \Delta I_{B}\) But \(\Delta I_{B}=\Delta I_{E}-\Delta I_{C}\) \(\Rightarrow \beta=\dfrac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}\) Given, \(\beta=100\) and \(\Delta {I_C} = 1\;mA\) \(\therefore 100=\dfrac{1}{\Delta I_{E}-1}\) \(\Rightarrow \Delta I_{E}-1=0.01\) \( \Rightarrow \Delta {I_E} = 1 + 0.1 = 1.01\;mA\)
AIIMS - 2005
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365514
If \(\alpha \)- current gain of a transistor is 0.98. What is the value of \(\beta \)- current gain of the transistor?
1 \(4.9\)
2 \(0.49\)
3 \(5\)
4 \(49\)
Explanation:
The relation between \(\alpha \) and \(\beta \) is \(\beta = \frac{\alpha }{{1 - \alpha }}\) Here, \(\alpha = 0.98\) \(\therefore \beta = \frac{{0.98}}{{1 - 0.98}} = \frac{{0.98}}{{0.02}} = 49\)
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365512
The gain of an amplifier without feedback is 90 and the feedback fraction \({\beta}\) is \({\dfrac{1}{10}}\). The gain of the amplifier if negative feedback is introduced is
1 89
2 9
3 8.9
4 900
Explanation:
The gain of an amplifier without feedback is \({A=90}\). Feedback fraction, \({\beta=\dfrac{1}{10}}\) The gain of the amplifier if negative feedback is introduced, \({A_{fb}} = \frac{A}{{1 + A\beta }} = \frac{{90}}{{1 + \frac{{90 \times 1}}{{10}}}} = \frac{{90}}{{1 + 9}} = \frac{{90}}{{10}} = 9\). So correct option is (2)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365513
Consider an \(pn\) transistor amplifier in common emitter configuration. The current gain of the transistor is 100 . If the collector current changes by \(1\,mA\), what will bet he change in emitter current?
1 \(1.1\,mA\)
2 \(1.01\,mA\)
3 \(0.01\,mA\)
4 \(10\,mA\)
Explanation:
Current gain in common emitter mode, \(\beta=\Delta I_{C} / \Delta I_{B}\) But \(\Delta I_{B}=\Delta I_{E}-\Delta I_{C}\) \(\Rightarrow \beta=\dfrac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}\) Given, \(\beta=100\) and \(\Delta {I_C} = 1\;mA\) \(\therefore 100=\dfrac{1}{\Delta I_{E}-1}\) \(\Rightarrow \Delta I_{E}-1=0.01\) \( \Rightarrow \Delta {I_E} = 1 + 0.1 = 1.01\;mA\)
AIIMS - 2005
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365514
If \(\alpha \)- current gain of a transistor is 0.98. What is the value of \(\beta \)- current gain of the transistor?
1 \(4.9\)
2 \(0.49\)
3 \(5\)
4 \(49\)
Explanation:
The relation between \(\alpha \) and \(\beta \) is \(\beta = \frac{\alpha }{{1 - \alpha }}\) Here, \(\alpha = 0.98\) \(\therefore \beta = \frac{{0.98}}{{1 - 0.98}} = \frac{{0.98}}{{0.02}} = 49\)
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
