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Semiconductor Electronics Material Devices and Simple Circuits

151067 In CE NPN transistor $10^{10}$ electrons enter the emitter in $10^{- 6} s$ when it is connected to battery. About 5\% electrons recombine with holes in the base. The current gain of the transistor is
$\overset{―}{(e = 1.6} \times 10^{- 19} C)$

1 0.98

2 19

3 49

4 0.95

Explanation:

B Given,
$n = 10^{10}, t = 10^{- 6} s$
$∵$ Emitter current $(I_{E}) = \frac{q}{t} = \frac{ne}{t}$
$= \frac{10^{10} \times 1.6 \times 10^{- 19}}{10^{- 6}}$
$I_{E} = 1.6 mA$
According to equation,
$I_{B} = 5 % I_{E}$
$I_{B} = 1.6 \times \frac{5}{100} = 0.08 mA$
$∴ I_{E} = I_{B} + I_{C}$
$I_{C} = I_{E} - I_{B} = 1.6 - 0.08 = 1.52 mA$
Hence,
$β = \frac{I_{C}}{I_{B}} = \frac{1.52}{0.08}$
$β = 19$

GUJCET 2017

Semiconductor Electronics Material Devices and Simple Circuits

151070 The collector supply voltage in $CE$ transistor amplifier is $10 V$ . The base current is $10 μ A$ in the absence of the signal voltage and the voltage between the collector and the emitter is $4 V$ . The current gain $(β)$ of a transistor is 200 , then the value of the load resistance $R_{L}$

1

1 k Ω

2

2 k Ω

3

3 k Ω

4

4 k Ω

Explanation:

C Given,
$β = 200, V_{CE} = 4 V, V_{CC} = 10 V$
$I_{B} = 10 μ A$
$∵ β = \frac{I_{C}}{I_{B}}$
$I_{C} = β \times I_{B}$
$= 200 \times 10 \times 10^{- 6}$
$I_{C} = 2 \times 10^{- 3} A$
So,
$R_{L} = \frac{Δ V_{CE}}{I_{C}} = \frac{10 - 4}{2 \times 10^{- 3}} = \frac{6}{2 \times 10^{- 3}} = 3 \times 10^{3} Ω = 3 k Ω$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

151073 In a common base circuit, collector base voltage changes by $0.6 V$ and collector current changes by $0.02 mA$ . Then, the output resistance is

1

6 \times 10^{4} Ω

2

3 \times 10^{4} Ω

3

9 \times 10^{4} Ω

4

18 \times 10^{4} Ω

Explanation:

B Given,
$V_{CB} = 0.6 V$
$I_{C} = 0.02 mA = 0.02 \times 10^{- 3} A$
$= 2 \times 10^{- 5} A$
So,
Output Resistance $(R_{out}) = \frac{V_{C B}}{I_{C}}$
$= \frac{0.6}{2 \times 10^{- 5}}$
$R_{out} = 3 \times 10^{4} Ω$

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

151074 For a common emitter amplifier, the audio signal voltage across the collector resistance 2 $k Ω$ is $2 V$ . If the current amplification factor of the transistor is $200$ , and the base resistance is $1.5 k Ω$ , the input signal and base current are

1

0.1 V

and

1 μ A

2

0.15 V

and

10 μ A

3

0.015 V

and

1 A

4

0.0075 V

and

5 μ A

Explanation:

D Given,
$β = 200, R_{C} = 2000 Ω$
$V_{C} = 2 V, R_{B} = 1.5 k Ω$
$∵ I_{C} = \frac{V_{C}}{R_{C}} = \frac{2}{2000} = 10^{- 3} A$
$β = \frac{I_{C}}{I_{B}}$
$∵ I_{B} = \frac{I_{C}}{β} = \frac{10^{- 3}}{200} = 5 \times 10^{- 6} A$
$I_{B} = 5 μ A$
$V_{in} = I_{B} R_{B}$
$= 5 \times 10^{- 6} \times 1.5 \times 10^{3}$
$= 75 \times 10^{- 4} V$
$V_{in} = 0.0075 V$

COMEDK 2017

Semiconductor Electronics Material Devices and Simple Circuits

151067 In CE NPN transistor $10^{10}$ electrons enter the emitter in $10^{- 6} s$ when it is connected to battery. About 5\% electrons recombine with holes in the base. The current gain of the transistor is
$\overset{―}{(e = 1.6} \times 10^{- 19} C)$

1 0.98

2 19

3 49

4 0.95

Explanation:

B Given,
$n = 10^{10}, t = 10^{- 6} s$
$∵$ Emitter current $(I_{E}) = \frac{q}{t} = \frac{ne}{t}$
$= \frac{10^{10} \times 1.6 \times 10^{- 19}}{10^{- 6}}$
$I_{E} = 1.6 mA$
According to equation,
$I_{B} = 5 % I_{E}$
$I_{B} = 1.6 \times \frac{5}{100} = 0.08 mA$
$∴ I_{E} = I_{B} + I_{C}$
$I_{C} = I_{E} - I_{B} = 1.6 - 0.08 = 1.52 mA$
Hence,
$β = \frac{I_{C}}{I_{B}} = \frac{1.52}{0.08}$
$β = 19$

GUJCET 2017

Semiconductor Electronics Material Devices and Simple Circuits

151070 The collector supply voltage in $CE$ transistor amplifier is $10 V$ . The base current is $10 μ A$ in the absence of the signal voltage and the voltage between the collector and the emitter is $4 V$ . The current gain $(β)$ of a transistor is 200 , then the value of the load resistance $R_{L}$

1

1 k Ω

2

2 k Ω

3

3 k Ω

4

4 k Ω

Explanation:

C Given,
$β = 200, V_{CE} = 4 V, V_{CC} = 10 V$
$I_{B} = 10 μ A$
$∵ β = \frac{I_{C}}{I_{B}}$
$I_{C} = β \times I_{B}$
$= 200 \times 10 \times 10^{- 6}$
$I_{C} = 2 \times 10^{- 3} A$
So,
$R_{L} = \frac{Δ V_{CE}}{I_{C}} = \frac{10 - 4}{2 \times 10^{- 3}} = \frac{6}{2 \times 10^{- 3}} = 3 \times 10^{3} Ω = 3 k Ω$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

151073 In a common base circuit, collector base voltage changes by $0.6 V$ and collector current changes by $0.02 mA$ . Then, the output resistance is

1

6 \times 10^{4} Ω

2

3 \times 10^{4} Ω

3

9 \times 10^{4} Ω

4

18 \times 10^{4} Ω

Explanation:

B Given,
$V_{CB} = 0.6 V$
$I_{C} = 0.02 mA = 0.02 \times 10^{- 3} A$
$= 2 \times 10^{- 5} A$
So,
Output Resistance $(R_{out}) = \frac{V_{C B}}{I_{C}}$
$= \frac{0.6}{2 \times 10^{- 5}}$
$R_{out} = 3 \times 10^{4} Ω$

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

151074 For a common emitter amplifier, the audio signal voltage across the collector resistance 2 $k Ω$ is $2 V$ . If the current amplification factor of the transistor is $200$ , and the base resistance is $1.5 k Ω$ , the input signal and base current are

1

0.1 V

and

1 μ A

2

0.15 V

and

10 μ A

3

0.015 V

and

1 A

4

0.0075 V

and

5 μ A

Explanation:

D Given,
$β = 200, R_{C} = 2000 Ω$
$V_{C} = 2 V, R_{B} = 1.5 k Ω$
$∵ I_{C} = \frac{V_{C}}{R_{C}} = \frac{2}{2000} = 10^{- 3} A$
$β = \frac{I_{C}}{I_{B}}$
$∵ I_{B} = \frac{I_{C}}{β} = \frac{10^{- 3}}{200} = 5 \times 10^{- 6} A$
$I_{B} = 5 μ A$
$V_{in} = I_{B} R_{B}$
$= 5 \times 10^{- 6} \times 1.5 \times 10^{3}$
$= 75 \times 10^{- 4} V$
$V_{in} = 0.0075 V$

COMEDK 2017

Semiconductor Electronics Material Devices and Simple Circuits

151067 In CE NPN transistor $10^{10}$ electrons enter the emitter in $10^{- 6} s$ when it is connected to battery. About 5\% electrons recombine with holes in the base. The current gain of the transistor is
$\overset{―}{(e = 1.6} \times 10^{- 19} C)$

1 0.98

2 19

3 49

4 0.95

Explanation:

B Given,
$n = 10^{10}, t = 10^{- 6} s$
$∵$ Emitter current $(I_{E}) = \frac{q}{t} = \frac{ne}{t}$
$= \frac{10^{10} \times 1.6 \times 10^{- 19}}{10^{- 6}}$
$I_{E} = 1.6 mA$
According to equation,
$I_{B} = 5 % I_{E}$
$I_{B} = 1.6 \times \frac{5}{100} = 0.08 mA$
$∴ I_{E} = I_{B} + I_{C}$
$I_{C} = I_{E} - I_{B} = 1.6 - 0.08 = 1.52 mA$
Hence,
$β = \frac{I_{C}}{I_{B}} = \frac{1.52}{0.08}$
$β = 19$

GUJCET 2017

Semiconductor Electronics Material Devices and Simple Circuits

151070 The collector supply voltage in $CE$ transistor amplifier is $10 V$ . The base current is $10 μ A$ in the absence of the signal voltage and the voltage between the collector and the emitter is $4 V$ . The current gain $(β)$ of a transistor is 200 , then the value of the load resistance $R_{L}$

1

1 k Ω

2

2 k Ω

3

3 k Ω

4

4 k Ω

Explanation:

C Given,
$β = 200, V_{CE} = 4 V, V_{CC} = 10 V$
$I_{B} = 10 μ A$
$∵ β = \frac{I_{C}}{I_{B}}$
$I_{C} = β \times I_{B}$
$= 200 \times 10 \times 10^{- 6}$
$I_{C} = 2 \times 10^{- 3} A$
So,
$R_{L} = \frac{Δ V_{CE}}{I_{C}} = \frac{10 - 4}{2 \times 10^{- 3}} = \frac{6}{2 \times 10^{- 3}} = 3 \times 10^{3} Ω = 3 k Ω$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

151073 In a common base circuit, collector base voltage changes by $0.6 V$ and collector current changes by $0.02 mA$ . Then, the output resistance is

1

6 \times 10^{4} Ω

2

3 \times 10^{4} Ω

3

9 \times 10^{4} Ω

4

18 \times 10^{4} Ω

Explanation:

B Given,
$V_{CB} = 0.6 V$
$I_{C} = 0.02 mA = 0.02 \times 10^{- 3} A$
$= 2 \times 10^{- 5} A$
So,
Output Resistance $(R_{out}) = \frac{V_{C B}}{I_{C}}$
$= \frac{0.6}{2 \times 10^{- 5}}$
$R_{out} = 3 \times 10^{4} Ω$

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

151074 For a common emitter amplifier, the audio signal voltage across the collector resistance 2 $k Ω$ is $2 V$ . If the current amplification factor of the transistor is $200$ , and the base resistance is $1.5 k Ω$ , the input signal and base current are

1

0.1 V

and

1 μ A

2

0.15 V

and

10 μ A

3

0.015 V

and

1 A

4

0.0075 V

and

5 μ A

Explanation:

D Given,
$β = 200, R_{C} = 2000 Ω$
$V_{C} = 2 V, R_{B} = 1.5 k Ω$
$∵ I_{C} = \frac{V_{C}}{R_{C}} = \frac{2}{2000} = 10^{- 3} A$
$β = \frac{I_{C}}{I_{B}}$
$∵ I_{B} = \frac{I_{C}}{β} = \frac{10^{- 3}}{200} = 5 \times 10^{- 6} A$
$I_{B} = 5 μ A$
$V_{in} = I_{B} R_{B}$
$= 5 \times 10^{- 6} \times 1.5 \times 10^{3}$
$= 75 \times 10^{- 4} V$
$V_{in} = 0.0075 V$

COMEDK 2017

Semiconductor Electronics Material Devices and Simple Circuits

151067 In CE NPN transistor $10^{10}$ electrons enter the emitter in $10^{- 6} s$ when it is connected to battery. About 5\% electrons recombine with holes in the base. The current gain of the transistor is
$\overset{―}{(e = 1.6} \times 10^{- 19} C)$

1 0.98

2 19

3 49

4 0.95

Explanation:

B Given,
$n = 10^{10}, t = 10^{- 6} s$
$∵$ Emitter current $(I_{E}) = \frac{q}{t} = \frac{ne}{t}$
$= \frac{10^{10} \times 1.6 \times 10^{- 19}}{10^{- 6}}$
$I_{E} = 1.6 mA$
According to equation,
$I_{B} = 5 % I_{E}$
$I_{B} = 1.6 \times \frac{5}{100} = 0.08 mA$
$∴ I_{E} = I_{B} + I_{C}$
$I_{C} = I_{E} - I_{B} = 1.6 - 0.08 = 1.52 mA$
Hence,
$β = \frac{I_{C}}{I_{B}} = \frac{1.52}{0.08}$
$β = 19$

GUJCET 2017

Semiconductor Electronics Material Devices and Simple Circuits

151070 The collector supply voltage in $CE$ transistor amplifier is $10 V$ . The base current is $10 μ A$ in the absence of the signal voltage and the voltage between the collector and the emitter is $4 V$ . The current gain $(β)$ of a transistor is 200 , then the value of the load resistance $R_{L}$

1

1 k Ω

2

2 k Ω

3

3 k Ω

4

4 k Ω

Explanation:

C Given,
$β = 200, V_{CE} = 4 V, V_{CC} = 10 V$
$I_{B} = 10 μ A$
$∵ β = \frac{I_{C}}{I_{B}}$
$I_{C} = β \times I_{B}$
$= 200 \times 10 \times 10^{- 6}$
$I_{C} = 2 \times 10^{- 3} A$
So,
$R_{L} = \frac{Δ V_{CE}}{I_{C}} = \frac{10 - 4}{2 \times 10^{- 3}} = \frac{6}{2 \times 10^{- 3}} = 3 \times 10^{3} Ω = 3 k Ω$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

151073 In a common base circuit, collector base voltage changes by $0.6 V$ and collector current changes by $0.02 mA$ . Then, the output resistance is

1

6 \times 10^{4} Ω

2

3 \times 10^{4} Ω

3

9 \times 10^{4} Ω

4

18 \times 10^{4} Ω

Explanation:

B Given,
$V_{CB} = 0.6 V$
$I_{C} = 0.02 mA = 0.02 \times 10^{- 3} A$
$= 2 \times 10^{- 5} A$
So,
Output Resistance $(R_{out}) = \frac{V_{C B}}{I_{C}}$
$= \frac{0.6}{2 \times 10^{- 5}}$
$R_{out} = 3 \times 10^{4} Ω$

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

151074 For a common emitter amplifier, the audio signal voltage across the collector resistance 2 $k Ω$ is $2 V$ . If the current amplification factor of the transistor is $200$ , and the base resistance is $1.5 k Ω$ , the input signal and base current are

1

0.1 V

and

1 μ A

2

0.15 V

and

10 μ A

3

0.015 V

and

1 A

4

0.0075 V

and

5 μ A

Explanation:

D Given,
$β = 200, R_{C} = 2000 Ω$
$V_{C} = 2 V, R_{B} = 1.5 k Ω$
$∵ I_{C} = \frac{V_{C}}{R_{C}} = \frac{2}{2000} = 10^{- 3} A$
$β = \frac{I_{C}}{I_{B}}$
$∵ I_{B} = \frac{I_{C}}{β} = \frac{10^{- 3}}{200} = 5 \times 10^{- 6} A$
$I_{B} = 5 μ A$
$V_{in} = I_{B} R_{B}$
$= 5 \times 10^{- 6} \times 1.5 \times 10^{3}$
$= 75 \times 10^{- 4} V$
$V_{in} = 0.0075 V$

COMEDK 2017

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