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Semiconductor Electronics Material Devices and Simple Circuits

151027 If $α$ and $β$ are the current gain in the $C B$ and CE configuration respectively of the transistor circuit, then $\frac{β - α}{α β} =$

1 infinite

2 1

3 2

4 0.5

Explanation:

B We know that,
$β = \frac{α}{1 - α}$
Multiply both side by $α$ ,
$α β = \frac{α^{2}}{1 - α}$
Subtract $α$ from both side in equation (i),
$β - α = \frac{α}{1 - α} - α$
$β - α = \frac{α^{2}}{1 - α}$
From equation (ii) and equation (iii), we get -
$β - α = α β$
$\frac{β - α}{α β} = 1$

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

151033 In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is $7.2 mA$ , the base current is

1

0.4 mA

2

0.2 mA

3

0.29 mA

4

0.35 mA

Explanation:

C Given,
Current gain $(α) = 0.96$
Emitter current $(I_{E}) = 7.2 mA$
Base current, $I_{B} =$ ?
We know that,
$α = \frac{I_{C}}{I_{E}} = 0.96$
$I_{C} = α \times I_{E}$
$I_{C} = 0.96 \times 7.2$
$I_{C} = 6.912 m A$
$I_{E} = I_{C} + I_{B}$
$I_{B} = I_{E} - I_{C}$
$= 7.2 - 6.912$
$I_{B} = 0.288 \approx 0.29 mA$

AP EAMCET (17.09.2020) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151034 In the following common-emitter circuit, $β =$ 100 and $V_{C E} = 7 V$ . If $V_{B E}$ is negligible, then the base current is
original image

1

0.01 mA

2

0.04 mA

3

0.02 mA

4

0.03 mA

Explanation:

B Given,
$β = 100, V_{CE} = 7 V$
$V_{CC} = 15 V, R_{C} = 2 k Ω, I_{B} = ?$
$V_{CC} = V_{CE} + I_{C} R_{C}$
$15 = 7 + I_{C} \times 2 \times 10^{3}$
$I_{C} = \frac{8}{2 \times 10^{3}} = 4 \times 10^{- 3} A = 4 mA$
$∴ β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β}$
$I_{B} = \frac{4}{100} = 0.04 mA$

AP EAMCET (21.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151035 In a transistor, the value of $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ . Then the value of $β$ varies between

1 1 and 10

2 0.95 and 0.99

3 20 and 100

4 200 and 300

Explanation:

C Given,
$α = \frac{20}{21} & \frac{100}{101}$
$β = \frac{α}{1 - α}$
Case I. if $α = \frac{20}{21}$
Then,
$β = \frac{20 / 21}{1 - 20 / 21} = \frac{20}{21 - 20} = 20$
Case II if $α = \frac{100}{101}$
Then,
$β = \frac{100 / 101}{1 - 100 / 101} = \frac{100}{101 - 100} = 100$ $∴$ Hence, the value of $β$ varies between $20 & 100$ .

AP EAMCET (20.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151027 If $α$ and $β$ are the current gain in the $C B$ and CE configuration respectively of the transistor circuit, then $\frac{β - α}{α β} =$

1 infinite

2 1

3 2

4 0.5

Explanation:

B We know that,
$β = \frac{α}{1 - α}$
Multiply both side by $α$ ,
$α β = \frac{α^{2}}{1 - α}$
Subtract $α$ from both side in equation (i),
$β - α = \frac{α}{1 - α} - α$
$β - α = \frac{α^{2}}{1 - α}$
From equation (ii) and equation (iii), we get -
$β - α = α β$
$\frac{β - α}{α β} = 1$

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

151033 In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is $7.2 mA$ , the base current is

1

0.4 mA

2

0.2 mA

3

0.29 mA

4

0.35 mA

Explanation:

C Given,
Current gain $(α) = 0.96$
Emitter current $(I_{E}) = 7.2 mA$
Base current, $I_{B} =$ ?
We know that,
$α = \frac{I_{C}}{I_{E}} = 0.96$
$I_{C} = α \times I_{E}$
$I_{C} = 0.96 \times 7.2$
$I_{C} = 6.912 m A$
$I_{E} = I_{C} + I_{B}$
$I_{B} = I_{E} - I_{C}$
$= 7.2 - 6.912$
$I_{B} = 0.288 \approx 0.29 mA$

AP EAMCET (17.09.2020) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151034 In the following common-emitter circuit, $β =$ 100 and $V_{C E} = 7 V$ . If $V_{B E}$ is negligible, then the base current is
original image

1

0.01 mA

2

0.04 mA

3

0.02 mA

4

0.03 mA

Explanation:

B Given,
$β = 100, V_{CE} = 7 V$
$V_{CC} = 15 V, R_{C} = 2 k Ω, I_{B} = ?$
$V_{CC} = V_{CE} + I_{C} R_{C}$
$15 = 7 + I_{C} \times 2 \times 10^{3}$
$I_{C} = \frac{8}{2 \times 10^{3}} = 4 \times 10^{- 3} A = 4 mA$
$∴ β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β}$
$I_{B} = \frac{4}{100} = 0.04 mA$

AP EAMCET (21.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151035 In a transistor, the value of $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ . Then the value of $β$ varies between

1 1 and 10

2 0.95 and 0.99

3 20 and 100

4 200 and 300

Explanation:

C Given,
$α = \frac{20}{21} & \frac{100}{101}$
$β = \frac{α}{1 - α}$
Case I. if $α = \frac{20}{21}$
Then,
$β = \frac{20 / 21}{1 - 20 / 21} = \frac{20}{21 - 20} = 20$
Case II if $α = \frac{100}{101}$
Then,
$β = \frac{100 / 101}{1 - 100 / 101} = \frac{100}{101 - 100} = 100$ $∴$ Hence, the value of $β$ varies between $20 & 100$ .

AP EAMCET (20.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151027 If $α$ and $β$ are the current gain in the $C B$ and CE configuration respectively of the transistor circuit, then $\frac{β - α}{α β} =$

1 infinite

2 1

3 2

4 0.5

Explanation:

B We know that,
$β = \frac{α}{1 - α}$
Multiply both side by $α$ ,
$α β = \frac{α^{2}}{1 - α}$
Subtract $α$ from both side in equation (i),
$β - α = \frac{α}{1 - α} - α$
$β - α = \frac{α^{2}}{1 - α}$
From equation (ii) and equation (iii), we get -
$β - α = α β$
$\frac{β - α}{α β} = 1$

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

151033 In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is $7.2 mA$ , the base current is

1

0.4 mA

2

0.2 mA

3

0.29 mA

4

0.35 mA

Explanation:

C Given,
Current gain $(α) = 0.96$
Emitter current $(I_{E}) = 7.2 mA$
Base current, $I_{B} =$ ?
We know that,
$α = \frac{I_{C}}{I_{E}} = 0.96$
$I_{C} = α \times I_{E}$
$I_{C} = 0.96 \times 7.2$
$I_{C} = 6.912 m A$
$I_{E} = I_{C} + I_{B}$
$I_{B} = I_{E} - I_{C}$
$= 7.2 - 6.912$
$I_{B} = 0.288 \approx 0.29 mA$

AP EAMCET (17.09.2020) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151034 In the following common-emitter circuit, $β =$ 100 and $V_{C E} = 7 V$ . If $V_{B E}$ is negligible, then the base current is
original image

1

0.01 mA

2

0.04 mA

3

0.02 mA

4

0.03 mA

Explanation:

B Given,
$β = 100, V_{CE} = 7 V$
$V_{CC} = 15 V, R_{C} = 2 k Ω, I_{B} = ?$
$V_{CC} = V_{CE} + I_{C} R_{C}$
$15 = 7 + I_{C} \times 2 \times 10^{3}$
$I_{C} = \frac{8}{2 \times 10^{3}} = 4 \times 10^{- 3} A = 4 mA$
$∴ β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β}$
$I_{B} = \frac{4}{100} = 0.04 mA$

AP EAMCET (21.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151035 In a transistor, the value of $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ . Then the value of $β$ varies between

1 1 and 10

2 0.95 and 0.99

3 20 and 100

4 200 and 300

Explanation:

C Given,
$α = \frac{20}{21} & \frac{100}{101}$
$β = \frac{α}{1 - α}$
Case I. if $α = \frac{20}{21}$
Then,
$β = \frac{20 / 21}{1 - 20 / 21} = \frac{20}{21 - 20} = 20$
Case II if $α = \frac{100}{101}$
Then,
$β = \frac{100 / 101}{1 - 100 / 101} = \frac{100}{101 - 100} = 100$ $∴$ Hence, the value of $β$ varies between $20 & 100$ .

AP EAMCET (20.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151027 If $α$ and $β$ are the current gain in the $C B$ and CE configuration respectively of the transistor circuit, then $\frac{β - α}{α β} =$

1 infinite

2 1

3 2

4 0.5

Explanation:

B We know that,
$β = \frac{α}{1 - α}$
Multiply both side by $α$ ,
$α β = \frac{α^{2}}{1 - α}$
Subtract $α$ from both side in equation (i),
$β - α = \frac{α}{1 - α} - α$
$β - α = \frac{α^{2}}{1 - α}$
From equation (ii) and equation (iii), we get -
$β - α = α β$
$\frac{β - α}{α β} = 1$

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

151033 In a p-n-p transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is $7.2 mA$ , the base current is

1

0.4 mA

2

0.2 mA

3

0.29 mA

4

0.35 mA

Explanation:

C Given,
Current gain $(α) = 0.96$
Emitter current $(I_{E}) = 7.2 mA$
Base current, $I_{B} =$ ?
We know that,
$α = \frac{I_{C}}{I_{E}} = 0.96$
$I_{C} = α \times I_{E}$
$I_{C} = 0.96 \times 7.2$
$I_{C} = 6.912 m A$
$I_{E} = I_{C} + I_{B}$
$I_{B} = I_{E} - I_{C}$
$= 7.2 - 6.912$
$I_{B} = 0.288 \approx 0.29 mA$

AP EAMCET (17.09.2020) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151034 In the following common-emitter circuit, $β =$ 100 and $V_{C E} = 7 V$ . If $V_{B E}$ is negligible, then the base current is
original image

1

0.01 mA

2

0.04 mA

3

0.02 mA

4

0.03 mA

Explanation:

B Given,
$β = 100, V_{CE} = 7 V$
$V_{CC} = 15 V, R_{C} = 2 k Ω, I_{B} = ?$
$V_{CC} = V_{CE} + I_{C} R_{C}$
$15 = 7 + I_{C} \times 2 \times 10^{3}$
$I_{C} = \frac{8}{2 \times 10^{3}} = 4 \times 10^{- 3} A = 4 mA$
$∴ β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β}$
$I_{B} = \frac{4}{100} = 0.04 mA$

AP EAMCET (21.04.2019) Shift-II

Semiconductor Electronics Material Devices and Simple Circuits

151035 In a transistor, the value of $α$ varies between $\frac{20}{21}$ and $\frac{100}{101}$ . Then the value of $β$ varies between

1 1 and 10

2 0.95 and 0.99

3 20 and 100

4 200 and 300

Explanation:

C Given,
$α = \frac{20}{21} & \frac{100}{101}$
$β = \frac{α}{1 - α}$
Case I. if $α = \frac{20}{21}$
Then,
$β = \frac{20 / 21}{1 - 20 / 21} = \frac{20}{21 - 20} = 20$
Case II if $α = \frac{100}{101}$
Then,
$β = \frac{100 / 101}{1 - 100 / 101} = \frac{100}{101 - 100} = 100$ $∴$ Hence, the value of $β$ varies between $20 & 100$ .

AP EAMCET (20.04.2019) Shift-II

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