Semiconductor Electronics Material Devices and Simple Circuits
151022
In a transistor circuit, the collector current is changed by \(8.9 \mathrm{~mA}\), if the emitter current is changed to \(9.0 \mathrm{~mA}\). The value of current amplification factor \(\beta\) is
1 89
2 92
3 84
4 96
Explanation:
B Given, \(\Delta \mathrm{I}_{\mathrm{C}}=8.9 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{E}}=9 \mathrm{~mA}\) We know that, the current amplification factor \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{E}}-\Delta \mathrm{I}_{\mathrm{C}}}\) \(\beta=\frac{8.9}{9-8.9}=\frac{8.9}{0.1}\) \(\beta=89\)
TS- EAMCET-03.05.2019
Semiconductor Electronics Material Devices and Simple Circuits
151023
In an n-p-n transistor, \(95 \%\) of emitted electrons reach the collector. If the base current is \(2 \mathrm{~mA}\), then collector current is
1 \(1.9 \mathrm{~mA}\)
2 \(38 \mathrm{~mA}\)
3 \(9.5 \mathrm{~mA}\)
4 \(48 \mathrm{~mA}\)
Explanation:
B Given, 95\% emitted electrons reach the collector. \(\mathrm{I}_{\mathrm{B}}=2 \mathrm{~mA}\) \(\text { ie. } \frac{I_C}{I_E}=0.95\) \(I_C=0.95 \times I_E\) We know that, \(\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-0.95 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=.05 \mathrm{I}_{\mathrm{E}}\) \(2=0.05 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{E}}=\frac{2}{0.05}=40 \mathrm{~mA}\) Now, putting the value of \(\mathrm{I}_{\mathrm{E}}\) in equation (i), we get - \(\mathrm{I}_{\mathrm{C}}=0.95 \times 40 \mathrm{~mA}=38 \mathrm{~mA}\)
TS- EAMCET.14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151025
The current gain of a transistor of 0.98 . If the transistor is used in a common emitter arrangement what would be the change in collector current corresponding to a change of \(0.5 \mathrm{~mA}\) in the base current?
1 \(24.5 \mathrm{~mA}\)
2 \(47.5 \mathrm{~mA}\)
3 \(32.5 \mathrm{~mA}\)
4 \(28.5 \mathrm{~mA}\)
Explanation:
B Given, \(\alpha =0.98\) \(\Delta \mathrm{I}_{\mathrm{B}} =0.5 \mathrm{~mA}\) \(\because \quad \beta =\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}\) \(\beta =49\) Hence, \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\) So, collector current is \(\Delta \mathrm{I}_{\mathrm{C}} =\beta \times \Delta \mathrm{I}_{\mathrm{B}}\) \(=49 \times 0.5\) \(\Delta \mathrm{I}_{\mathrm{C}} =24.5 \mathrm{~mA}\)
TS- EAMCET-14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151026
A transistor is used common-emitter configuration. Given its \(\alpha=0.9\), calculate the change in collector current when the base current changes by \(2 \mu \mathrm{A}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Semiconductor Electronics Material Devices and Simple Circuits
151022
In a transistor circuit, the collector current is changed by \(8.9 \mathrm{~mA}\), if the emitter current is changed to \(9.0 \mathrm{~mA}\). The value of current amplification factor \(\beta\) is
1 89
2 92
3 84
4 96
Explanation:
B Given, \(\Delta \mathrm{I}_{\mathrm{C}}=8.9 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{E}}=9 \mathrm{~mA}\) We know that, the current amplification factor \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{E}}-\Delta \mathrm{I}_{\mathrm{C}}}\) \(\beta=\frac{8.9}{9-8.9}=\frac{8.9}{0.1}\) \(\beta=89\)
TS- EAMCET-03.05.2019
Semiconductor Electronics Material Devices and Simple Circuits
151023
In an n-p-n transistor, \(95 \%\) of emitted electrons reach the collector. If the base current is \(2 \mathrm{~mA}\), then collector current is
1 \(1.9 \mathrm{~mA}\)
2 \(38 \mathrm{~mA}\)
3 \(9.5 \mathrm{~mA}\)
4 \(48 \mathrm{~mA}\)
Explanation:
B Given, 95\% emitted electrons reach the collector. \(\mathrm{I}_{\mathrm{B}}=2 \mathrm{~mA}\) \(\text { ie. } \frac{I_C}{I_E}=0.95\) \(I_C=0.95 \times I_E\) We know that, \(\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-0.95 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=.05 \mathrm{I}_{\mathrm{E}}\) \(2=0.05 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{E}}=\frac{2}{0.05}=40 \mathrm{~mA}\) Now, putting the value of \(\mathrm{I}_{\mathrm{E}}\) in equation (i), we get - \(\mathrm{I}_{\mathrm{C}}=0.95 \times 40 \mathrm{~mA}=38 \mathrm{~mA}\)
TS- EAMCET.14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151025
The current gain of a transistor of 0.98 . If the transistor is used in a common emitter arrangement what would be the change in collector current corresponding to a change of \(0.5 \mathrm{~mA}\) in the base current?
1 \(24.5 \mathrm{~mA}\)
2 \(47.5 \mathrm{~mA}\)
3 \(32.5 \mathrm{~mA}\)
4 \(28.5 \mathrm{~mA}\)
Explanation:
B Given, \(\alpha =0.98\) \(\Delta \mathrm{I}_{\mathrm{B}} =0.5 \mathrm{~mA}\) \(\because \quad \beta =\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}\) \(\beta =49\) Hence, \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\) So, collector current is \(\Delta \mathrm{I}_{\mathrm{C}} =\beta \times \Delta \mathrm{I}_{\mathrm{B}}\) \(=49 \times 0.5\) \(\Delta \mathrm{I}_{\mathrm{C}} =24.5 \mathrm{~mA}\)
TS- EAMCET-14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151026
A transistor is used common-emitter configuration. Given its \(\alpha=0.9\), calculate the change in collector current when the base current changes by \(2 \mu \mathrm{A}\).
Semiconductor Electronics Material Devices and Simple Circuits
151022
In a transistor circuit, the collector current is changed by \(8.9 \mathrm{~mA}\), if the emitter current is changed to \(9.0 \mathrm{~mA}\). The value of current amplification factor \(\beta\) is
1 89
2 92
3 84
4 96
Explanation:
B Given, \(\Delta \mathrm{I}_{\mathrm{C}}=8.9 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{E}}=9 \mathrm{~mA}\) We know that, the current amplification factor \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{E}}-\Delta \mathrm{I}_{\mathrm{C}}}\) \(\beta=\frac{8.9}{9-8.9}=\frac{8.9}{0.1}\) \(\beta=89\)
TS- EAMCET-03.05.2019
Semiconductor Electronics Material Devices and Simple Circuits
151023
In an n-p-n transistor, \(95 \%\) of emitted electrons reach the collector. If the base current is \(2 \mathrm{~mA}\), then collector current is
1 \(1.9 \mathrm{~mA}\)
2 \(38 \mathrm{~mA}\)
3 \(9.5 \mathrm{~mA}\)
4 \(48 \mathrm{~mA}\)
Explanation:
B Given, 95\% emitted electrons reach the collector. \(\mathrm{I}_{\mathrm{B}}=2 \mathrm{~mA}\) \(\text { ie. } \frac{I_C}{I_E}=0.95\) \(I_C=0.95 \times I_E\) We know that, \(\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-0.95 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=.05 \mathrm{I}_{\mathrm{E}}\) \(2=0.05 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{E}}=\frac{2}{0.05}=40 \mathrm{~mA}\) Now, putting the value of \(\mathrm{I}_{\mathrm{E}}\) in equation (i), we get - \(\mathrm{I}_{\mathrm{C}}=0.95 \times 40 \mathrm{~mA}=38 \mathrm{~mA}\)
TS- EAMCET.14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151025
The current gain of a transistor of 0.98 . If the transistor is used in a common emitter arrangement what would be the change in collector current corresponding to a change of \(0.5 \mathrm{~mA}\) in the base current?
1 \(24.5 \mathrm{~mA}\)
2 \(47.5 \mathrm{~mA}\)
3 \(32.5 \mathrm{~mA}\)
4 \(28.5 \mathrm{~mA}\)
Explanation:
B Given, \(\alpha =0.98\) \(\Delta \mathrm{I}_{\mathrm{B}} =0.5 \mathrm{~mA}\) \(\because \quad \beta =\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}\) \(\beta =49\) Hence, \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\) So, collector current is \(\Delta \mathrm{I}_{\mathrm{C}} =\beta \times \Delta \mathrm{I}_{\mathrm{B}}\) \(=49 \times 0.5\) \(\Delta \mathrm{I}_{\mathrm{C}} =24.5 \mathrm{~mA}\)
TS- EAMCET-14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151026
A transistor is used common-emitter configuration. Given its \(\alpha=0.9\), calculate the change in collector current when the base current changes by \(2 \mu \mathrm{A}\).
Semiconductor Electronics Material Devices and Simple Circuits
151022
In a transistor circuit, the collector current is changed by \(8.9 \mathrm{~mA}\), if the emitter current is changed to \(9.0 \mathrm{~mA}\). The value of current amplification factor \(\beta\) is
1 89
2 92
3 84
4 96
Explanation:
B Given, \(\Delta \mathrm{I}_{\mathrm{C}}=8.9 \mathrm{~mA}\) \(\Delta \mathrm{I}_{\mathrm{E}}=9 \mathrm{~mA}\) We know that, the current amplification factor \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{E}}-\Delta \mathrm{I}_{\mathrm{C}}}\) \(\beta=\frac{8.9}{9-8.9}=\frac{8.9}{0.1}\) \(\beta=89\)
TS- EAMCET-03.05.2019
Semiconductor Electronics Material Devices and Simple Circuits
151023
In an n-p-n transistor, \(95 \%\) of emitted electrons reach the collector. If the base current is \(2 \mathrm{~mA}\), then collector current is
1 \(1.9 \mathrm{~mA}\)
2 \(38 \mathrm{~mA}\)
3 \(9.5 \mathrm{~mA}\)
4 \(48 \mathrm{~mA}\)
Explanation:
B Given, 95\% emitted electrons reach the collector. \(\mathrm{I}_{\mathrm{B}}=2 \mathrm{~mA}\) \(\text { ie. } \frac{I_C}{I_E}=0.95\) \(I_C=0.95 \times I_E\) We know that, \(\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-0.95 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{B}}=.05 \mathrm{I}_{\mathrm{E}}\) \(2=0.05 \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{E}}=\frac{2}{0.05}=40 \mathrm{~mA}\) Now, putting the value of \(\mathrm{I}_{\mathrm{E}}\) in equation (i), we get - \(\mathrm{I}_{\mathrm{C}}=0.95 \times 40 \mathrm{~mA}=38 \mathrm{~mA}\)
TS- EAMCET.14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151025
The current gain of a transistor of 0.98 . If the transistor is used in a common emitter arrangement what would be the change in collector current corresponding to a change of \(0.5 \mathrm{~mA}\) in the base current?
1 \(24.5 \mathrm{~mA}\)
2 \(47.5 \mathrm{~mA}\)
3 \(32.5 \mathrm{~mA}\)
4 \(28.5 \mathrm{~mA}\)
Explanation:
B Given, \(\alpha =0.98\) \(\Delta \mathrm{I}_{\mathrm{B}} =0.5 \mathrm{~mA}\) \(\because \quad \beta =\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}\) \(\beta =49\) Hence, \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\) So, collector current is \(\Delta \mathrm{I}_{\mathrm{C}} =\beta \times \Delta \mathrm{I}_{\mathrm{B}}\) \(=49 \times 0.5\) \(\Delta \mathrm{I}_{\mathrm{C}} =24.5 \mathrm{~mA}\)
TS- EAMCET-14.09.2020
Semiconductor Electronics Material Devices and Simple Circuits
151026
A transistor is used common-emitter configuration. Given its \(\alpha=0.9\), calculate the change in collector current when the base current changes by \(2 \mu \mathrm{A}\).