Semiconductor Electronics Material Devices and Simple Circuits
151027
If \(\alpha\) and \(\beta\) are the current gain in the \(C B\) and CE configuration respectively of the transistor circuit, then \(\frac{\beta-\alpha}{\alpha \beta}=\)
1 infinite
2 1
3 2
4 0.5
Explanation:
B We know that, \(\beta=\frac{\alpha}{1-\alpha}\) Multiply both side by \(\alpha\), \(\alpha \beta=\frac{\alpha^2}{1-\alpha}\) Subtract \(\alpha\) from both side in equation (i), \(\beta-\alpha=\frac{\alpha}{1-\alpha}-\alpha\) \(\beta-\alpha=\frac{\alpha^2}{1-\alpha}\) From equation (ii) and equation (iii), we get - \(\beta-\alpha=\alpha \beta\) \(\frac{\beta-\alpha}{\alpha \beta}=1\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
151033
In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is
1 \(0.4 \mathrm{~mA}\)
2 \(0.2 \mathrm{~mA}\)
3 \(0.29 \mathrm{~mA}\)
4 \(0.35 \mathrm{~mA}\)
Explanation:
C Given, Current gain \((\alpha)=0.96\) Emitter current \(\left(\mathrm{I}_{\mathrm{E}}\right)=7.2 \mathrm{~mA}\) Base current, \(\mathrm{I}_{\mathrm{B}}=\) ? We know that, \(\alpha =\frac{I_C}{I_E}=0.96\) \(I_C =\alpha \times I_E\) \(I_C =0.96 \times 7.2\) \(I_C =6.912 m A\) \(I_E =I_C+I_B\) \(I_B =I_E-I_C\) \(=7.2-6.912\) \(I_B =0.288 \approx 0.29 \mathrm{~mA}\)
AP EAMCET (17.09.2020) Shift-II
Semiconductor Electronics Material Devices and Simple Circuits
151034
In the following common-emitter circuit, \(\beta=\) 100 and \(V_{C E}=7 V\). If \(V_{B E}\) is negligible, then the base current is
Semiconductor Electronics Material Devices and Simple Circuits
151035
In a transistor, the value of \(\alpha\) varies between \(\frac{20}{21}\) and \(\frac{100}{101}\). Then the value of \(\beta\) varies between
1 1 and 10
2 0.95 and 0.99
3 20 and 100
4 200 and 300
Explanation:
C Given, \(\alpha=\frac{20}{21} \& \frac{100}{101}\) \(\beta=\frac{\alpha}{1-\alpha}\) Case I. if \(\alpha=\frac{20}{21}\) Then, \(\beta=\frac{20 / 21}{1-20 / 21}=\frac{20}{21-20}=20\) Case II if \(\alpha=\frac{100}{101}\) Then, \(\beta=\frac{100 / 101}{1-100 / 101}=\frac{100}{101-100}=100\)\(\therefore\) Hence, the value of \(\beta\) varies between \(20 \& 100\).
Semiconductor Electronics Material Devices and Simple Circuits
151027
If \(\alpha\) and \(\beta\) are the current gain in the \(C B\) and CE configuration respectively of the transistor circuit, then \(\frac{\beta-\alpha}{\alpha \beta}=\)
1 infinite
2 1
3 2
4 0.5
Explanation:
B We know that, \(\beta=\frac{\alpha}{1-\alpha}\) Multiply both side by \(\alpha\), \(\alpha \beta=\frac{\alpha^2}{1-\alpha}\) Subtract \(\alpha\) from both side in equation (i), \(\beta-\alpha=\frac{\alpha}{1-\alpha}-\alpha\) \(\beta-\alpha=\frac{\alpha^2}{1-\alpha}\) From equation (ii) and equation (iii), we get - \(\beta-\alpha=\alpha \beta\) \(\frac{\beta-\alpha}{\alpha \beta}=1\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
151033
In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is
1 \(0.4 \mathrm{~mA}\)
2 \(0.2 \mathrm{~mA}\)
3 \(0.29 \mathrm{~mA}\)
4 \(0.35 \mathrm{~mA}\)
Explanation:
C Given, Current gain \((\alpha)=0.96\) Emitter current \(\left(\mathrm{I}_{\mathrm{E}}\right)=7.2 \mathrm{~mA}\) Base current, \(\mathrm{I}_{\mathrm{B}}=\) ? We know that, \(\alpha =\frac{I_C}{I_E}=0.96\) \(I_C =\alpha \times I_E\) \(I_C =0.96 \times 7.2\) \(I_C =6.912 m A\) \(I_E =I_C+I_B\) \(I_B =I_E-I_C\) \(=7.2-6.912\) \(I_B =0.288 \approx 0.29 \mathrm{~mA}\)
AP EAMCET (17.09.2020) Shift-II
Semiconductor Electronics Material Devices and Simple Circuits
151034
In the following common-emitter circuit, \(\beta=\) 100 and \(V_{C E}=7 V\). If \(V_{B E}\) is negligible, then the base current is
Semiconductor Electronics Material Devices and Simple Circuits
151035
In a transistor, the value of \(\alpha\) varies between \(\frac{20}{21}\) and \(\frac{100}{101}\). Then the value of \(\beta\) varies between
1 1 and 10
2 0.95 and 0.99
3 20 and 100
4 200 and 300
Explanation:
C Given, \(\alpha=\frac{20}{21} \& \frac{100}{101}\) \(\beta=\frac{\alpha}{1-\alpha}\) Case I. if \(\alpha=\frac{20}{21}\) Then, \(\beta=\frac{20 / 21}{1-20 / 21}=\frac{20}{21-20}=20\) Case II if \(\alpha=\frac{100}{101}\) Then, \(\beta=\frac{100 / 101}{1-100 / 101}=\frac{100}{101-100}=100\)\(\therefore\) Hence, the value of \(\beta\) varies between \(20 \& 100\).
Semiconductor Electronics Material Devices and Simple Circuits
151027
If \(\alpha\) and \(\beta\) are the current gain in the \(C B\) and CE configuration respectively of the transistor circuit, then \(\frac{\beta-\alpha}{\alpha \beta}=\)
1 infinite
2 1
3 2
4 0.5
Explanation:
B We know that, \(\beta=\frac{\alpha}{1-\alpha}\) Multiply both side by \(\alpha\), \(\alpha \beta=\frac{\alpha^2}{1-\alpha}\) Subtract \(\alpha\) from both side in equation (i), \(\beta-\alpha=\frac{\alpha}{1-\alpha}-\alpha\) \(\beta-\alpha=\frac{\alpha^2}{1-\alpha}\) From equation (ii) and equation (iii), we get - \(\beta-\alpha=\alpha \beta\) \(\frac{\beta-\alpha}{\alpha \beta}=1\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
151033
In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is
1 \(0.4 \mathrm{~mA}\)
2 \(0.2 \mathrm{~mA}\)
3 \(0.29 \mathrm{~mA}\)
4 \(0.35 \mathrm{~mA}\)
Explanation:
C Given, Current gain \((\alpha)=0.96\) Emitter current \(\left(\mathrm{I}_{\mathrm{E}}\right)=7.2 \mathrm{~mA}\) Base current, \(\mathrm{I}_{\mathrm{B}}=\) ? We know that, \(\alpha =\frac{I_C}{I_E}=0.96\) \(I_C =\alpha \times I_E\) \(I_C =0.96 \times 7.2\) \(I_C =6.912 m A\) \(I_E =I_C+I_B\) \(I_B =I_E-I_C\) \(=7.2-6.912\) \(I_B =0.288 \approx 0.29 \mathrm{~mA}\)
AP EAMCET (17.09.2020) Shift-II
Semiconductor Electronics Material Devices and Simple Circuits
151034
In the following common-emitter circuit, \(\beta=\) 100 and \(V_{C E}=7 V\). If \(V_{B E}\) is negligible, then the base current is
Semiconductor Electronics Material Devices and Simple Circuits
151035
In a transistor, the value of \(\alpha\) varies between \(\frac{20}{21}\) and \(\frac{100}{101}\). Then the value of \(\beta\) varies between
1 1 and 10
2 0.95 and 0.99
3 20 and 100
4 200 and 300
Explanation:
C Given, \(\alpha=\frac{20}{21} \& \frac{100}{101}\) \(\beta=\frac{\alpha}{1-\alpha}\) Case I. if \(\alpha=\frac{20}{21}\) Then, \(\beta=\frac{20 / 21}{1-20 / 21}=\frac{20}{21-20}=20\) Case II if \(\alpha=\frac{100}{101}\) Then, \(\beta=\frac{100 / 101}{1-100 / 101}=\frac{100}{101-100}=100\)\(\therefore\) Hence, the value of \(\beta\) varies between \(20 \& 100\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
151027
If \(\alpha\) and \(\beta\) are the current gain in the \(C B\) and CE configuration respectively of the transistor circuit, then \(\frac{\beta-\alpha}{\alpha \beta}=\)
1 infinite
2 1
3 2
4 0.5
Explanation:
B We know that, \(\beta=\frac{\alpha}{1-\alpha}\) Multiply both side by \(\alpha\), \(\alpha \beta=\frac{\alpha^2}{1-\alpha}\) Subtract \(\alpha\) from both side in equation (i), \(\beta-\alpha=\frac{\alpha}{1-\alpha}-\alpha\) \(\beta-\alpha=\frac{\alpha^2}{1-\alpha}\) From equation (ii) and equation (iii), we get - \(\beta-\alpha=\alpha \beta\) \(\frac{\beta-\alpha}{\alpha \beta}=1\)
MHT-CET 2020
Semiconductor Electronics Material Devices and Simple Circuits
151033
In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is
1 \(0.4 \mathrm{~mA}\)
2 \(0.2 \mathrm{~mA}\)
3 \(0.29 \mathrm{~mA}\)
4 \(0.35 \mathrm{~mA}\)
Explanation:
C Given, Current gain \((\alpha)=0.96\) Emitter current \(\left(\mathrm{I}_{\mathrm{E}}\right)=7.2 \mathrm{~mA}\) Base current, \(\mathrm{I}_{\mathrm{B}}=\) ? We know that, \(\alpha =\frac{I_C}{I_E}=0.96\) \(I_C =\alpha \times I_E\) \(I_C =0.96 \times 7.2\) \(I_C =6.912 m A\) \(I_E =I_C+I_B\) \(I_B =I_E-I_C\) \(=7.2-6.912\) \(I_B =0.288 \approx 0.29 \mathrm{~mA}\)
AP EAMCET (17.09.2020) Shift-II
Semiconductor Electronics Material Devices and Simple Circuits
151034
In the following common-emitter circuit, \(\beta=\) 100 and \(V_{C E}=7 V\). If \(V_{B E}\) is negligible, then the base current is
Semiconductor Electronics Material Devices and Simple Circuits
151035
In a transistor, the value of \(\alpha\) varies between \(\frac{20}{21}\) and \(\frac{100}{101}\). Then the value of \(\beta\) varies between
1 1 and 10
2 0.95 and 0.99
3 20 and 100
4 200 and 300
Explanation:
C Given, \(\alpha=\frac{20}{21} \& \frac{100}{101}\) \(\beta=\frac{\alpha}{1-\alpha}\) Case I. if \(\alpha=\frac{20}{21}\) Then, \(\beta=\frac{20 / 21}{1-20 / 21}=\frac{20}{21-20}=20\) Case II if \(\alpha=\frac{100}{101}\) Then, \(\beta=\frac{100 / 101}{1-100 / 101}=\frac{100}{101-100}=100\)\(\therefore\) Hence, the value of \(\beta\) varies between \(20 \& 100\).
