Semiconductor Electronics Material Devices and Simple Circuits
151050
A common emitter amplifier gives an output of \(3 \mathrm{~V}\) for an input of \(0.01 \mathrm{~V}\). If \(\beta\) of the transistor is 100 and the input resistance is \(1 \mathrm{k} \Omega\), then the collector resistance is
1 \(1 \mathrm{k} \Omega\)
2 \(3 \mathrm{k} \Omega\)
3 \(30 \mathrm{k} \Omega\)
4 \(30 \Omega\)
5 \(6 \mathrm{k} \Omega\)
Explanation:
B Given, \(\beta=100\) \(V_{\text {out }}=3 \mathrm{~V} \& \mathrm{~V}_{\text {in }}=0.01 \mathrm{~V}\) \(\mathrm{R}_{\text {in }}=1 \mathrm{k} \Omega\) We know that, Amplifier voltage gain \(=\) current gain \(\times\) resistance gain \(A_V=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{V_{\text {out }}}{V_{\text {in }}}=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{3}{0.01}=100 \times \frac{R_{\text {out }}}{1 \times 10^3}\) \(R_{\text {out }}=\frac{3 \times 10^3}{100 \times 0.01}\) \(R_{\text {out }}=3 \mathrm{k} \Omega\)
COMEDK - 2016
Semiconductor Electronics Material Devices and Simple Circuits
151051
The current gain of a transistor in common base mode is 0.995 . The current gain of the same transistor in common emitter mode is
1 197
2 201
3 198
4 202
5 199
Explanation:
E Given, \(\alpha=0.995\) \(\beta=\text { ? }\) We know that, \(\beta=\frac{\alpha}{1-\alpha}\) \(\beta=\frac{0.995}{1-0.995}=\frac{0.995}{0.005}\) \(\beta=199\)
Kerala CEE 2007
Semiconductor Electronics Material Devices and Simple Circuits
151052
In common emitter amplifier, the current gain is 62. The collector resistance and input resistance are \(5 \mathrm{k} \Omega\) and \(500 \Omega\) respectively. If the input voltage is \(0.1 \mathrm{~V}\), the output voltage is
Semiconductor Electronics Material Devices and Simple Circuits
151053
In a common-emitter amplifier, the load resistance of the output circuit is \(\mathbf{1 0 0 0}\) times the resistance of the input circuit. If \(\alpha=\mathbf{0 . 9 8}\), then voltage gain is :
1 \(49 \times 10^3\)
2 \(2.5 \times 10^2\)
3 \(1.5 \times 10^2\)
4 4.9
5 \(3.5 \times 10^3\)
Explanation:
B Given, \(\alpha=0.98, \mathrm{R}_{\text {out }}=1000 \mathrm{R}_{\text {in }}\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) We know that, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}} \quad\left(\because R_{\text {out }}=1000 R_{\text {in }}\right)\) \(=49 \times 1000\) \(\mathrm{~A}_{\mathrm{V}}=49 \times 10^3\)
Semiconductor Electronics Material Devices and Simple Circuits
151050
A common emitter amplifier gives an output of \(3 \mathrm{~V}\) for an input of \(0.01 \mathrm{~V}\). If \(\beta\) of the transistor is 100 and the input resistance is \(1 \mathrm{k} \Omega\), then the collector resistance is
1 \(1 \mathrm{k} \Omega\)
2 \(3 \mathrm{k} \Omega\)
3 \(30 \mathrm{k} \Omega\)
4 \(30 \Omega\)
5 \(6 \mathrm{k} \Omega\)
Explanation:
B Given, \(\beta=100\) \(V_{\text {out }}=3 \mathrm{~V} \& \mathrm{~V}_{\text {in }}=0.01 \mathrm{~V}\) \(\mathrm{R}_{\text {in }}=1 \mathrm{k} \Omega\) We know that, Amplifier voltage gain \(=\) current gain \(\times\) resistance gain \(A_V=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{V_{\text {out }}}{V_{\text {in }}}=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{3}{0.01}=100 \times \frac{R_{\text {out }}}{1 \times 10^3}\) \(R_{\text {out }}=\frac{3 \times 10^3}{100 \times 0.01}\) \(R_{\text {out }}=3 \mathrm{k} \Omega\)
COMEDK - 2016
Semiconductor Electronics Material Devices and Simple Circuits
151051
The current gain of a transistor in common base mode is 0.995 . The current gain of the same transistor in common emitter mode is
1 197
2 201
3 198
4 202
5 199
Explanation:
E Given, \(\alpha=0.995\) \(\beta=\text { ? }\) We know that, \(\beta=\frac{\alpha}{1-\alpha}\) \(\beta=\frac{0.995}{1-0.995}=\frac{0.995}{0.005}\) \(\beta=199\)
Kerala CEE 2007
Semiconductor Electronics Material Devices and Simple Circuits
151052
In common emitter amplifier, the current gain is 62. The collector resistance and input resistance are \(5 \mathrm{k} \Omega\) and \(500 \Omega\) respectively. If the input voltage is \(0.1 \mathrm{~V}\), the output voltage is
Semiconductor Electronics Material Devices and Simple Circuits
151053
In a common-emitter amplifier, the load resistance of the output circuit is \(\mathbf{1 0 0 0}\) times the resistance of the input circuit. If \(\alpha=\mathbf{0 . 9 8}\), then voltage gain is :
1 \(49 \times 10^3\)
2 \(2.5 \times 10^2\)
3 \(1.5 \times 10^2\)
4 4.9
5 \(3.5 \times 10^3\)
Explanation:
B Given, \(\alpha=0.98, \mathrm{R}_{\text {out }}=1000 \mathrm{R}_{\text {in }}\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) We know that, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}} \quad\left(\because R_{\text {out }}=1000 R_{\text {in }}\right)\) \(=49 \times 1000\) \(\mathrm{~A}_{\mathrm{V}}=49 \times 10^3\)
Semiconductor Electronics Material Devices and Simple Circuits
151050
A common emitter amplifier gives an output of \(3 \mathrm{~V}\) for an input of \(0.01 \mathrm{~V}\). If \(\beta\) of the transistor is 100 and the input resistance is \(1 \mathrm{k} \Omega\), then the collector resistance is
1 \(1 \mathrm{k} \Omega\)
2 \(3 \mathrm{k} \Omega\)
3 \(30 \mathrm{k} \Omega\)
4 \(30 \Omega\)
5 \(6 \mathrm{k} \Omega\)
Explanation:
B Given, \(\beta=100\) \(V_{\text {out }}=3 \mathrm{~V} \& \mathrm{~V}_{\text {in }}=0.01 \mathrm{~V}\) \(\mathrm{R}_{\text {in }}=1 \mathrm{k} \Omega\) We know that, Amplifier voltage gain \(=\) current gain \(\times\) resistance gain \(A_V=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{V_{\text {out }}}{V_{\text {in }}}=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{3}{0.01}=100 \times \frac{R_{\text {out }}}{1 \times 10^3}\) \(R_{\text {out }}=\frac{3 \times 10^3}{100 \times 0.01}\) \(R_{\text {out }}=3 \mathrm{k} \Omega\)
COMEDK - 2016
Semiconductor Electronics Material Devices and Simple Circuits
151051
The current gain of a transistor in common base mode is 0.995 . The current gain of the same transistor in common emitter mode is
1 197
2 201
3 198
4 202
5 199
Explanation:
E Given, \(\alpha=0.995\) \(\beta=\text { ? }\) We know that, \(\beta=\frac{\alpha}{1-\alpha}\) \(\beta=\frac{0.995}{1-0.995}=\frac{0.995}{0.005}\) \(\beta=199\)
Kerala CEE 2007
Semiconductor Electronics Material Devices and Simple Circuits
151052
In common emitter amplifier, the current gain is 62. The collector resistance and input resistance are \(5 \mathrm{k} \Omega\) and \(500 \Omega\) respectively. If the input voltage is \(0.1 \mathrm{~V}\), the output voltage is
Semiconductor Electronics Material Devices and Simple Circuits
151053
In a common-emitter amplifier, the load resistance of the output circuit is \(\mathbf{1 0 0 0}\) times the resistance of the input circuit. If \(\alpha=\mathbf{0 . 9 8}\), then voltage gain is :
1 \(49 \times 10^3\)
2 \(2.5 \times 10^2\)
3 \(1.5 \times 10^2\)
4 4.9
5 \(3.5 \times 10^3\)
Explanation:
B Given, \(\alpha=0.98, \mathrm{R}_{\text {out }}=1000 \mathrm{R}_{\text {in }}\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) We know that, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}} \quad\left(\because R_{\text {out }}=1000 R_{\text {in }}\right)\) \(=49 \times 1000\) \(\mathrm{~A}_{\mathrm{V}}=49 \times 10^3\)
Semiconductor Electronics Material Devices and Simple Circuits
151050
A common emitter amplifier gives an output of \(3 \mathrm{~V}\) for an input of \(0.01 \mathrm{~V}\). If \(\beta\) of the transistor is 100 and the input resistance is \(1 \mathrm{k} \Omega\), then the collector resistance is
1 \(1 \mathrm{k} \Omega\)
2 \(3 \mathrm{k} \Omega\)
3 \(30 \mathrm{k} \Omega\)
4 \(30 \Omega\)
5 \(6 \mathrm{k} \Omega\)
Explanation:
B Given, \(\beta=100\) \(V_{\text {out }}=3 \mathrm{~V} \& \mathrm{~V}_{\text {in }}=0.01 \mathrm{~V}\) \(\mathrm{R}_{\text {in }}=1 \mathrm{k} \Omega\) We know that, Amplifier voltage gain \(=\) current gain \(\times\) resistance gain \(A_V=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{V_{\text {out }}}{V_{\text {in }}}=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\) \(\frac{3}{0.01}=100 \times \frac{R_{\text {out }}}{1 \times 10^3}\) \(R_{\text {out }}=\frac{3 \times 10^3}{100 \times 0.01}\) \(R_{\text {out }}=3 \mathrm{k} \Omega\)
COMEDK - 2016
Semiconductor Electronics Material Devices and Simple Circuits
151051
The current gain of a transistor in common base mode is 0.995 . The current gain of the same transistor in common emitter mode is
1 197
2 201
3 198
4 202
5 199
Explanation:
E Given, \(\alpha=0.995\) \(\beta=\text { ? }\) We know that, \(\beta=\frac{\alpha}{1-\alpha}\) \(\beta=\frac{0.995}{1-0.995}=\frac{0.995}{0.005}\) \(\beta=199\)
Kerala CEE 2007
Semiconductor Electronics Material Devices and Simple Circuits
151052
In common emitter amplifier, the current gain is 62. The collector resistance and input resistance are \(5 \mathrm{k} \Omega\) and \(500 \Omega\) respectively. If the input voltage is \(0.1 \mathrm{~V}\), the output voltage is
Semiconductor Electronics Material Devices and Simple Circuits
151053
In a common-emitter amplifier, the load resistance of the output circuit is \(\mathbf{1 0 0 0}\) times the resistance of the input circuit. If \(\alpha=\mathbf{0 . 9 8}\), then voltage gain is :
1 \(49 \times 10^3\)
2 \(2.5 \times 10^2\)
3 \(1.5 \times 10^2\)
4 4.9
5 \(3.5 \times 10^3\)
Explanation:
B Given, \(\alpha=0.98, \mathrm{R}_{\text {out }}=1000 \mathrm{R}_{\text {in }}\) \(\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\) We know that, Voltage gain \(\left(A_V\right)=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}} \quad\left(\because R_{\text {out }}=1000 R_{\text {in }}\right)\) \(=49 \times 1000\) \(\mathrm{~A}_{\mathrm{V}}=49 \times 10^3\)
