Semiconductor Electronics Material Devices and Simple Circuits
151098
In n-p-n transistor circuit, the collector current is \(20 \mathrm{~mA}\). If \(\mathbf{9 0 \%}\) of the electrons emitted reach the collector, then the
1 emitter current will be about \(16 \mathrm{~mA}\)
2 emitter current will be \(19 \mathrm{~mA}\)
3 base current will be about \(2 \mathrm{~mA}\)
4 base current will be about \(10 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{I}_{\mathrm{c}}=20 \mathrm{~mA},\) \(\mathrm{I}_{\mathrm{c}}=90 \% \text { of } \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{c}}=\frac{90}{100} \times \mathrm{I}_{\mathrm{E}}\) Then, \(I_E =\frac{100}{90} \times I_c\) \(=\frac{100}{90} \times 20\) \(I_E =22 \mathrm{~mA}\) Now, \(\mathrm{I}_{\mathrm{B}} =\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{c}}\) \(=22-20\) \(=2 \mathrm{~mA}\)Hence, base current will be about \(2 \mathrm{~mA}\)
UPSEE - 2018
Semiconductor Electronics Material Devices and Simple Circuits
151099
Identify the diode which is forward biased in the following set a. B. C. D.
1 Only B
2 A and B
3 A, B and C
4 A, B, C and D
Explanation:
C If \(V_{+}-V_{-}=+v e\) voltage, then it forward biased. \(\mathrm{V}_{+}-\mathrm{V}_{-}=-\mathrm{ve}\) voltage then it reverse biased. Then, For A \(\quad 0-(-5 \mathrm{~V})=5 \mathrm{~V}\), forward biased For B \(5-0=+5 \mathrm{~V}\), forward biased For C \(-2-(-7)=5 \mathrm{~V}\), forward biased. For D \(2-3=-1 \mathrm{~V}\) (negative) reverse biased. So, A, B, and C are forward biased and D is reverse biased.
COMEDK 2011
Semiconductor Electronics Material Devices and Simple Circuits
151103
For the transistor circuit shown in figure, if \(\beta=\) 100, voltage drop across emitter and base is \(0.7 \mathrm{~V}\), then the value of \(\mathrm{V}_{\mathrm{CE}}\) will be-
Semiconductor Electronics Material Devices and Simple Circuits
151104
The input resistance of a common emitter transistor amplifier, if the output resistance is \(500 \mathrm{k} \Omega\), the current gain \(\alpha=0.98\) and the power gain is \(6.0625 \times 10^6\), is-
1 \(198 \Omega\)
2 \(300 \Omega\)
3 \(100 \Omega\)
4 \(400 \Omega\)
Explanation:
B Given, \(\mathrm{R}_0=500 \mathrm{k} \Omega\) \(\alpha=0.98\) Power gain \(=6.0625 \times 10^6\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\) We know, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_0}{R_i}\) \(=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}}\) Also, power gain \(=\) voltage gain \(\times \beta\) \(6.0625 \times 10^6=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}} \times 49\)So, \(\quad \mathrm{R}_{\mathrm{i}}=198 \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
151098
In n-p-n transistor circuit, the collector current is \(20 \mathrm{~mA}\). If \(\mathbf{9 0 \%}\) of the electrons emitted reach the collector, then the
1 emitter current will be about \(16 \mathrm{~mA}\)
2 emitter current will be \(19 \mathrm{~mA}\)
3 base current will be about \(2 \mathrm{~mA}\)
4 base current will be about \(10 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{I}_{\mathrm{c}}=20 \mathrm{~mA},\) \(\mathrm{I}_{\mathrm{c}}=90 \% \text { of } \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{c}}=\frac{90}{100} \times \mathrm{I}_{\mathrm{E}}\) Then, \(I_E =\frac{100}{90} \times I_c\) \(=\frac{100}{90} \times 20\) \(I_E =22 \mathrm{~mA}\) Now, \(\mathrm{I}_{\mathrm{B}} =\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{c}}\) \(=22-20\) \(=2 \mathrm{~mA}\)Hence, base current will be about \(2 \mathrm{~mA}\)
UPSEE - 2018
Semiconductor Electronics Material Devices and Simple Circuits
151099
Identify the diode which is forward biased in the following set a. B. C. D.
1 Only B
2 A and B
3 A, B and C
4 A, B, C and D
Explanation:
C If \(V_{+}-V_{-}=+v e\) voltage, then it forward biased. \(\mathrm{V}_{+}-\mathrm{V}_{-}=-\mathrm{ve}\) voltage then it reverse biased. Then, For A \(\quad 0-(-5 \mathrm{~V})=5 \mathrm{~V}\), forward biased For B \(5-0=+5 \mathrm{~V}\), forward biased For C \(-2-(-7)=5 \mathrm{~V}\), forward biased. For D \(2-3=-1 \mathrm{~V}\) (negative) reverse biased. So, A, B, and C are forward biased and D is reverse biased.
COMEDK 2011
Semiconductor Electronics Material Devices and Simple Circuits
151103
For the transistor circuit shown in figure, if \(\beta=\) 100, voltage drop across emitter and base is \(0.7 \mathrm{~V}\), then the value of \(\mathrm{V}_{\mathrm{CE}}\) will be-
Semiconductor Electronics Material Devices and Simple Circuits
151104
The input resistance of a common emitter transistor amplifier, if the output resistance is \(500 \mathrm{k} \Omega\), the current gain \(\alpha=0.98\) and the power gain is \(6.0625 \times 10^6\), is-
1 \(198 \Omega\)
2 \(300 \Omega\)
3 \(100 \Omega\)
4 \(400 \Omega\)
Explanation:
B Given, \(\mathrm{R}_0=500 \mathrm{k} \Omega\) \(\alpha=0.98\) Power gain \(=6.0625 \times 10^6\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\) We know, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_0}{R_i}\) \(=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}}\) Also, power gain \(=\) voltage gain \(\times \beta\) \(6.0625 \times 10^6=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}} \times 49\)So, \(\quad \mathrm{R}_{\mathrm{i}}=198 \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
151098
In n-p-n transistor circuit, the collector current is \(20 \mathrm{~mA}\). If \(\mathbf{9 0 \%}\) of the electrons emitted reach the collector, then the
1 emitter current will be about \(16 \mathrm{~mA}\)
2 emitter current will be \(19 \mathrm{~mA}\)
3 base current will be about \(2 \mathrm{~mA}\)
4 base current will be about \(10 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{I}_{\mathrm{c}}=20 \mathrm{~mA},\) \(\mathrm{I}_{\mathrm{c}}=90 \% \text { of } \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{c}}=\frac{90}{100} \times \mathrm{I}_{\mathrm{E}}\) Then, \(I_E =\frac{100}{90} \times I_c\) \(=\frac{100}{90} \times 20\) \(I_E =22 \mathrm{~mA}\) Now, \(\mathrm{I}_{\mathrm{B}} =\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{c}}\) \(=22-20\) \(=2 \mathrm{~mA}\)Hence, base current will be about \(2 \mathrm{~mA}\)
UPSEE - 2018
Semiconductor Electronics Material Devices and Simple Circuits
151099
Identify the diode which is forward biased in the following set a. B. C. D.
1 Only B
2 A and B
3 A, B and C
4 A, B, C and D
Explanation:
C If \(V_{+}-V_{-}=+v e\) voltage, then it forward biased. \(\mathrm{V}_{+}-\mathrm{V}_{-}=-\mathrm{ve}\) voltage then it reverse biased. Then, For A \(\quad 0-(-5 \mathrm{~V})=5 \mathrm{~V}\), forward biased For B \(5-0=+5 \mathrm{~V}\), forward biased For C \(-2-(-7)=5 \mathrm{~V}\), forward biased. For D \(2-3=-1 \mathrm{~V}\) (negative) reverse biased. So, A, B, and C are forward biased and D is reverse biased.
COMEDK 2011
Semiconductor Electronics Material Devices and Simple Circuits
151103
For the transistor circuit shown in figure, if \(\beta=\) 100, voltage drop across emitter and base is \(0.7 \mathrm{~V}\), then the value of \(\mathrm{V}_{\mathrm{CE}}\) will be-
Semiconductor Electronics Material Devices and Simple Circuits
151104
The input resistance of a common emitter transistor amplifier, if the output resistance is \(500 \mathrm{k} \Omega\), the current gain \(\alpha=0.98\) and the power gain is \(6.0625 \times 10^6\), is-
1 \(198 \Omega\)
2 \(300 \Omega\)
3 \(100 \Omega\)
4 \(400 \Omega\)
Explanation:
B Given, \(\mathrm{R}_0=500 \mathrm{k} \Omega\) \(\alpha=0.98\) Power gain \(=6.0625 \times 10^6\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\) We know, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_0}{R_i}\) \(=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}}\) Also, power gain \(=\) voltage gain \(\times \beta\) \(6.0625 \times 10^6=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}} \times 49\)So, \(\quad \mathrm{R}_{\mathrm{i}}=198 \Omega\)
Semiconductor Electronics Material Devices and Simple Circuits
151098
In n-p-n transistor circuit, the collector current is \(20 \mathrm{~mA}\). If \(\mathbf{9 0 \%}\) of the electrons emitted reach the collector, then the
1 emitter current will be about \(16 \mathrm{~mA}\)
2 emitter current will be \(19 \mathrm{~mA}\)
3 base current will be about \(2 \mathrm{~mA}\)
4 base current will be about \(10 \mathrm{~mA}\)
Explanation:
C Given, \(\mathrm{I}_{\mathrm{c}}=20 \mathrm{~mA},\) \(\mathrm{I}_{\mathrm{c}}=90 \% \text { of } \mathrm{I}_{\mathrm{E}}\) \(\mathrm{I}_{\mathrm{c}}=\frac{90}{100} \times \mathrm{I}_{\mathrm{E}}\) Then, \(I_E =\frac{100}{90} \times I_c\) \(=\frac{100}{90} \times 20\) \(I_E =22 \mathrm{~mA}\) Now, \(\mathrm{I}_{\mathrm{B}} =\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{c}}\) \(=22-20\) \(=2 \mathrm{~mA}\)Hence, base current will be about \(2 \mathrm{~mA}\)
UPSEE - 2018
Semiconductor Electronics Material Devices and Simple Circuits
151099
Identify the diode which is forward biased in the following set a. B. C. D.
1 Only B
2 A and B
3 A, B and C
4 A, B, C and D
Explanation:
C If \(V_{+}-V_{-}=+v e\) voltage, then it forward biased. \(\mathrm{V}_{+}-\mathrm{V}_{-}=-\mathrm{ve}\) voltage then it reverse biased. Then, For A \(\quad 0-(-5 \mathrm{~V})=5 \mathrm{~V}\), forward biased For B \(5-0=+5 \mathrm{~V}\), forward biased For C \(-2-(-7)=5 \mathrm{~V}\), forward biased. For D \(2-3=-1 \mathrm{~V}\) (negative) reverse biased. So, A, B, and C are forward biased and D is reverse biased.
COMEDK 2011
Semiconductor Electronics Material Devices and Simple Circuits
151103
For the transistor circuit shown in figure, if \(\beta=\) 100, voltage drop across emitter and base is \(0.7 \mathrm{~V}\), then the value of \(\mathrm{V}_{\mathrm{CE}}\) will be-
Semiconductor Electronics Material Devices and Simple Circuits
151104
The input resistance of a common emitter transistor amplifier, if the output resistance is \(500 \mathrm{k} \Omega\), the current gain \(\alpha=0.98\) and the power gain is \(6.0625 \times 10^6\), is-
1 \(198 \Omega\)
2 \(300 \Omega\)
3 \(100 \Omega\)
4 \(400 \Omega\)
Explanation:
B Given, \(\mathrm{R}_0=500 \mathrm{k} \Omega\) \(\alpha=0.98\) Power gain \(=6.0625 \times 10^6\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\) We know, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_0}{R_i}\) \(=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}}\) Also, power gain \(=\) voltage gain \(\times \beta\) \(6.0625 \times 10^6=49 \times \frac{500 \times 10^3}{\mathrm{R}_{\mathrm{i}}} \times 49\)So, \(\quad \mathrm{R}_{\mathrm{i}}=198 \Omega\)
