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PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365520 For a tranistor amplifier, the voltage gain

1 Is low at high and low frequencies and constant at mid frequencies

2 Remains constant for all frequencies

3 Constant at high frequencies and low at low frequencies

4 Is high at high and low frequencies and constant in the middle frequency range

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365521 The $d c$ common emitter current gain of a $n$ - $p$ - $n$ transistor is 50. The potential difference applied across the collector and emitter of a transistor used in $C E$ configuration is $V_{C E} = 2 V$ . If the collector resistance $R_{C} = 4 k Ω$ , the base current $(I_{B})$ and the collector current $(I_{C})$ are

1

I_{B} = 10 μ A, I_{C} = 0.5 m A

2

I_{B} = 0.5 μ A, I_{C} = 10 m A

3

I_{B} = 5 μ A, I_{C} = 1 m A

4

I_{B} = 1 μ A, I_{C} = 0.5 m A

Explanation:

Here, $d c$ current gain $β_{d c} = 50$ .
Voltage across collector and emitter $V_{C E} = 2 V$ .
Collector resistance $R_{C} = 4 k Ω = 4 \times 10^{3} Ω$
As $V_{C E} = I_{C} R_{C},$
$∴ I_{C} = \frac{V_{C E}}{R_{C}} = \frac{2 V}{4 \times 10^{3} Ω} = 0.5 \times 10^{- 3} A = 0.5 m A$
Also, $β_{d c} = \frac{I_{C}}{I_{B}} .$
$∴ I_{B} = \frac{I_{C}}{β_{d c}} = \frac{0.5 \times 10^{- 3} A}{50} = 1 \times 10^{- 5} A = 10 μ A .$

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365522 An amplifier has a voltage gain $A_{V} = 1000$ . The voltage gain in $d B$ is

1

30 d B

2

60 d B

3

3 d B

4

20 d B

Explanation:

Voltage gain, $A_{V} = 1000$
In $d B,$ voltage gain,
$A = 10 \log_{10} 1000 d B$
$= (10 \times 3) \log_{10} 10 d B = 30 d B$
$(∵ \log_{10} 10 = 1)$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365523 Negative feedback

1 Increases stability

2 Decreases stability

3 Produces rectification

4 None of these

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365524 In common emitter amplifier, input resistance is $1000 Ω$ , peak value of input signal voltage is 5 $m V$ and $β = 60$ . The peak value of output current is

1

3 \times 10^{- 4} A

2

0.5 \times 10^{- 4} A

3

2 \times 10^{- 5} A

4

1 \times 10^{- 5} A

Explanation:

Given, input resistance, $R_{i} = 1000 Ω$
Input voltage $V_{i} o r V_{B} = 5 m V = 5 \times 10^{- 3} V$ and $β = 60$
Therefore, input or base current,
$I_{B} = \frac{V_{B}}{R_{i}} = \frac{5 \times 10^{- 3}}{1000} = 5 \times 10^{- 6} A$
As , $β = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{C} = β \times I_{B} = 60 \times 5 \times 10^{- 6} = 3 \times 10^{- 4} A$
Therefore, Peak value of output current
$I_{C} = 3 \times 10^{- 4} A$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365520 For a tranistor amplifier, the voltage gain

1 Is low at high and low frequencies and constant at mid frequencies

2 Remains constant for all frequencies

3 Constant at high frequencies and low at low frequencies

4 Is high at high and low frequencies and constant in the middle frequency range

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365521 The $d c$ common emitter current gain of a $n$ - $p$ - $n$ transistor is 50. The potential difference applied across the collector and emitter of a transistor used in $C E$ configuration is $V_{C E} = 2 V$ . If the collector resistance $R_{C} = 4 k Ω$ , the base current $(I_{B})$ and the collector current $(I_{C})$ are

1

I_{B} = 10 μ A, I_{C} = 0.5 m A

2

I_{B} = 0.5 μ A, I_{C} = 10 m A

3

I_{B} = 5 μ A, I_{C} = 1 m A

4

I_{B} = 1 μ A, I_{C} = 0.5 m A

Explanation:

Here, $d c$ current gain $β_{d c} = 50$ .
Voltage across collector and emitter $V_{C E} = 2 V$ .
Collector resistance $R_{C} = 4 k Ω = 4 \times 10^{3} Ω$
As $V_{C E} = I_{C} R_{C},$
$∴ I_{C} = \frac{V_{C E}}{R_{C}} = \frac{2 V}{4 \times 10^{3} Ω} = 0.5 \times 10^{- 3} A = 0.5 m A$
Also, $β_{d c} = \frac{I_{C}}{I_{B}} .$
$∴ I_{B} = \frac{I_{C}}{β_{d c}} = \frac{0.5 \times 10^{- 3} A}{50} = 1 \times 10^{- 5} A = 10 μ A .$

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365522 An amplifier has a voltage gain $A_{V} = 1000$ . The voltage gain in $d B$ is

1

30 d B

2

60 d B

3

3 d B

4

20 d B

Explanation:

Voltage gain, $A_{V} = 1000$
In $d B,$ voltage gain,
$A = 10 \log_{10} 1000 d B$
$= (10 \times 3) \log_{10} 10 d B = 30 d B$
$(∵ \log_{10} 10 = 1)$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365523 Negative feedback

1 Increases stability

2 Decreases stability

3 Produces rectification

4 None of these

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365524 In common emitter amplifier, input resistance is $1000 Ω$ , peak value of input signal voltage is 5 $m V$ and $β = 60$ . The peak value of output current is

1

3 \times 10^{- 4} A

2

0.5 \times 10^{- 4} A

3

2 \times 10^{- 5} A

4

1 \times 10^{- 5} A

Explanation:

Given, input resistance, $R_{i} = 1000 Ω$
Input voltage $V_{i} o r V_{B} = 5 m V = 5 \times 10^{- 3} V$ and $β = 60$
Therefore, input or base current,
$I_{B} = \frac{V_{B}}{R_{i}} = \frac{5 \times 10^{- 3}}{1000} = 5 \times 10^{- 6} A$
As , $β = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{C} = β \times I_{B} = 60 \times 5 \times 10^{- 6} = 3 \times 10^{- 4} A$
Therefore, Peak value of output current
$I_{C} = 3 \times 10^{- 4} A$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365520 For a tranistor amplifier, the voltage gain

1 Is low at high and low frequencies and constant at mid frequencies

2 Remains constant for all frequencies

3 Constant at high frequencies and low at low frequencies

4 Is high at high and low frequencies and constant in the middle frequency range

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365521 The $d c$ common emitter current gain of a $n$ - $p$ - $n$ transistor is 50. The potential difference applied across the collector and emitter of a transistor used in $C E$ configuration is $V_{C E} = 2 V$ . If the collector resistance $R_{C} = 4 k Ω$ , the base current $(I_{B})$ and the collector current $(I_{C})$ are

1

I_{B} = 10 μ A, I_{C} = 0.5 m A

2

I_{B} = 0.5 μ A, I_{C} = 10 m A

3

I_{B} = 5 μ A, I_{C} = 1 m A

4

I_{B} = 1 μ A, I_{C} = 0.5 m A

Explanation:

Here, $d c$ current gain $β_{d c} = 50$ .
Voltage across collector and emitter $V_{C E} = 2 V$ .
Collector resistance $R_{C} = 4 k Ω = 4 \times 10^{3} Ω$
As $V_{C E} = I_{C} R_{C},$
$∴ I_{C} = \frac{V_{C E}}{R_{C}} = \frac{2 V}{4 \times 10^{3} Ω} = 0.5 \times 10^{- 3} A = 0.5 m A$
Also, $β_{d c} = \frac{I_{C}}{I_{B}} .$
$∴ I_{B} = \frac{I_{C}}{β_{d c}} = \frac{0.5 \times 10^{- 3} A}{50} = 1 \times 10^{- 5} A = 10 μ A .$

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365522 An amplifier has a voltage gain $A_{V} = 1000$ . The voltage gain in $d B$ is

1

30 d B

2

60 d B

3

3 d B

4

20 d B

Explanation:

Voltage gain, $A_{V} = 1000$
In $d B,$ voltage gain,
$A = 10 \log_{10} 1000 d B$
$= (10 \times 3) \log_{10} 10 d B = 30 d B$
$(∵ \log_{10} 10 = 1)$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365523 Negative feedback

1 Increases stability

2 Decreases stability

3 Produces rectification

4 None of these

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365524 In common emitter amplifier, input resistance is $1000 Ω$ , peak value of input signal voltage is 5 $m V$ and $β = 60$ . The peak value of output current is

1

3 \times 10^{- 4} A

2

0.5 \times 10^{- 4} A

3

2 \times 10^{- 5} A

4

1 \times 10^{- 5} A

Explanation:

Given, input resistance, $R_{i} = 1000 Ω$
Input voltage $V_{i} o r V_{B} = 5 m V = 5 \times 10^{- 3} V$ and $β = 60$
Therefore, input or base current,
$I_{B} = \frac{V_{B}}{R_{i}} = \frac{5 \times 10^{- 3}}{1000} = 5 \times 10^{- 6} A$
As , $β = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{C} = β \times I_{B} = 60 \times 5 \times 10^{- 6} = 3 \times 10^{- 4} A$
Therefore, Peak value of output current
$I_{C} = 3 \times 10^{- 4} A$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365520 For a tranistor amplifier, the voltage gain

1 Is low at high and low frequencies and constant at mid frequencies

2 Remains constant for all frequencies

3 Constant at high frequencies and low at low frequencies

4 Is high at high and low frequencies and constant in the middle frequency range

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365521 The $d c$ common emitter current gain of a $n$ - $p$ - $n$ transistor is 50. The potential difference applied across the collector and emitter of a transistor used in $C E$ configuration is $V_{C E} = 2 V$ . If the collector resistance $R_{C} = 4 k Ω$ , the base current $(I_{B})$ and the collector current $(I_{C})$ are

1

I_{B} = 10 μ A, I_{C} = 0.5 m A

2

I_{B} = 0.5 μ A, I_{C} = 10 m A

3

I_{B} = 5 μ A, I_{C} = 1 m A

4

I_{B} = 1 μ A, I_{C} = 0.5 m A

Explanation:

Here, $d c$ current gain $β_{d c} = 50$ .
Voltage across collector and emitter $V_{C E} = 2 V$ .
Collector resistance $R_{C} = 4 k Ω = 4 \times 10^{3} Ω$
As $V_{C E} = I_{C} R_{C},$
$∴ I_{C} = \frac{V_{C E}}{R_{C}} = \frac{2 V}{4 \times 10^{3} Ω} = 0.5 \times 10^{- 3} A = 0.5 m A$
Also, $β_{d c} = \frac{I_{C}}{I_{B}} .$
$∴ I_{B} = \frac{I_{C}}{β_{d c}} = \frac{0.5 \times 10^{- 3} A}{50} = 1 \times 10^{- 5} A = 10 μ A .$

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365522 An amplifier has a voltage gain $A_{V} = 1000$ . The voltage gain in $d B$ is

1

30 d B

2

60 d B

3

3 d B

4

20 d B

Explanation:

Voltage gain, $A_{V} = 1000$
In $d B,$ voltage gain,
$A = 10 \log_{10} 1000 d B$
$= (10 \times 3) \log_{10} 10 d B = 30 d B$
$(∵ \log_{10} 10 = 1)$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365523 Negative feedback

1 Increases stability

2 Decreases stability

3 Produces rectification

4 None of these

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365524 In common emitter amplifier, input resistance is $1000 Ω$ , peak value of input signal voltage is 5 $m V$ and $β = 60$ . The peak value of output current is

1

3 \times 10^{- 4} A

2

0.5 \times 10^{- 4} A

3

2 \times 10^{- 5} A

4

1 \times 10^{- 5} A

Explanation:

Given, input resistance, $R_{i} = 1000 Ω$
Input voltage $V_{i} o r V_{B} = 5 m V = 5 \times 10^{- 3} V$ and $β = 60$
Therefore, input or base current,
$I_{B} = \frac{V_{B}}{R_{i}} = \frac{5 \times 10^{- 3}}{1000} = 5 \times 10^{- 6} A$
As , $β = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{C} = β \times I_{B} = 60 \times 5 \times 10^{- 6} = 3 \times 10^{- 4} A$
Therefore, Peak value of output current
$I_{C} = 3 \times 10^{- 4} A$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365520 For a tranistor amplifier, the voltage gain

1 Is low at high and low frequencies and constant at mid frequencies

2 Remains constant for all frequencies

3 Constant at high frequencies and low at low frequencies

4 Is high at high and low frequencies and constant in the middle frequency range

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365521 The $d c$ common emitter current gain of a $n$ - $p$ - $n$ transistor is 50. The potential difference applied across the collector and emitter of a transistor used in $C E$ configuration is $V_{C E} = 2 V$ . If the collector resistance $R_{C} = 4 k Ω$ , the base current $(I_{B})$ and the collector current $(I_{C})$ are

1

I_{B} = 10 μ A, I_{C} = 0.5 m A

2

I_{B} = 0.5 μ A, I_{C} = 10 m A

3

I_{B} = 5 μ A, I_{C} = 1 m A

4

I_{B} = 1 μ A, I_{C} = 0.5 m A

Explanation:

Here, $d c$ current gain $β_{d c} = 50$ .
Voltage across collector and emitter $V_{C E} = 2 V$ .
Collector resistance $R_{C} = 4 k Ω = 4 \times 10^{3} Ω$
As $V_{C E} = I_{C} R_{C},$
$∴ I_{C} = \frac{V_{C E}}{R_{C}} = \frac{2 V}{4 \times 10^{3} Ω} = 0.5 \times 10^{- 3} A = 0.5 m A$
Also, $β_{d c} = \frac{I_{C}}{I_{B}} .$
$∴ I_{B} = \frac{I_{C}}{β_{d c}} = \frac{0.5 \times 10^{- 3} A}{50} = 1 \times 10^{- 5} A = 10 μ A .$

KCET - 2018

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365522 An amplifier has a voltage gain $A_{V} = 1000$ . The voltage gain in $d B$ is

1

30 d B

2

60 d B

3

3 d B

4

20 d B

Explanation:

Voltage gain, $A_{V} = 1000$
In $d B,$ voltage gain,
$A = 10 \log_{10} 1000 d B$
$= (10 \times 3) \log_{10} 10 d B = 30 d B$
$(∵ \log_{10} 10 = 1)$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365523 Negative feedback

1 Increases stability

2 Decreases stability

3 Produces rectification

4 None of these

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365524 In common emitter amplifier, input resistance is $1000 Ω$ , peak value of input signal voltage is 5 $m V$ and $β = 60$ . The peak value of output current is

1

3 \times 10^{- 4} A

2

0.5 \times 10^{- 4} A

3

2 \times 10^{- 5} A

4

1 \times 10^{- 5} A

Explanation:

Given, input resistance, $R_{i} = 1000 Ω$
Input voltage $V_{i} o r V_{B} = 5 m V = 5 \times 10^{- 3} V$ and $β = 60$
Therefore, input or base current,
$I_{B} = \frac{V_{B}}{R_{i}} = \frac{5 \times 10^{- 3}}{1000} = 5 \times 10^{- 6} A$
As , $β = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{C} = β \times I_{B} = 60 \times 5 \times 10^{- 6} = 3 \times 10^{- 4} A$
Therefore, Peak value of output current
$I_{C} = 3 \times 10^{- 4} A$

MHTCET - 2020

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