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Semiconductor Electronics Material Devices and Simple Circuits

151098 In n-p-n transistor circuit, the collector current is $20 mA$ . If $90 %$ of the electrons emitted reach the collector, then the

1 emitter current will be about

16 mA

2 emitter current will be

19 mA

3 base current will be about

2 mA

4 base current will be about

10 mA

Explanation:

C Given,
$I_{c} = 20 mA,$
$I_{c} = 90 % of I_{E}$
$I_{c} = \frac{90}{100} \times I_{E}$
Then,
$I_{E} = \frac{100}{90} \times I_{c}$
$= \frac{100}{90} \times 20$
$I_{E} = 22 mA$
Now,
$I_{B} = I_{E} - I_{c}$
$= 22 - 20$
$= 2 mA$ Hence, base current will be about $2 mA$

UPSEE - 2018

Semiconductor Electronics Material Devices and Simple Circuits

151099 Identify the diode which is forward biased in the following set
a. original image
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.
$V_{+} - V_{-} = - ve$ voltage then it reverse biased.
Then,
For A $0 - (- 5 V) = 5 V$ , forward biased
For B $5 - 0 = + 5 V$ , forward biased
For C $- 2 - (- 7) = 5 V$ , forward biased.
For D $2 - 3 = - 1 V$ (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 $β =$ 100, voltage drop across emitter and base is $0.7 V$ , then the value of $V_{CE}$ will be-
original image

1 Zero

2

5 V

3

10 V

4

13 V

Explanation:

D Given, $β = 100, R_{C} = 100 k Ω$
$I_{B} = \frac{5 - 0.7}{8.6 \times 10^{3}} = 0.5 mA$
$∴ I_{C} = β \times I_{B}$
$I_{C} = 100 \times 0.5 = 50 mA$
Now,
$V_{C E} = V_{C C} - I_{C} R_{C}$
$V_{C E} = 18 - 50 \times 10^{- 3} \times 100$
$V_{C E} = 18 - 5$ So, $V_{CE} = 13 V$

BCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

151104 The input resistance of a common emitter transistor amplifier, if the output resistance is $500 k Ω$ , the current gain $α = 0.98$ and the power gain is $6.0625 \times 10^{6}$ , is-

1

198 Ω

2

300 Ω

3

100 Ω

4

400 Ω

Explanation:

B Given, $R_{0} = 500 k Ω$
$α = 0.98$
Power gain $= 6.0625 \times 10^{6}$
$∵ β = \frac{α}{1 - α} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
We know,
Voltage gain $(A_{V}) = β \times \frac{R_{0}}{R_{i}}$
$= 49 \times \frac{500 \times 10^{3}}{R_{i}}$
Also, power gain $=$ voltage gain $\times β$
$6.0625 \times 10^{6} = 49 \times \frac{500 \times 10^{3}}{R_{i}} \times 49$ So, $R_{i} = 198 Ω$

BCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151098 In n-p-n transistor circuit, the collector current is $20 mA$ . If $90 %$ of the electrons emitted reach the collector, then the

1 emitter current will be about

16 mA

2 emitter current will be

19 mA

3 base current will be about

2 mA

4 base current will be about

10 mA

Explanation:

C Given,
$I_{c} = 20 mA,$
$I_{c} = 90 % of I_{E}$
$I_{c} = \frac{90}{100} \times I_{E}$
Then,
$I_{E} = \frac{100}{90} \times I_{c}$
$= \frac{100}{90} \times 20$
$I_{E} = 22 mA$
Now,
$I_{B} = I_{E} - I_{c}$
$= 22 - 20$
$= 2 mA$ Hence, base current will be about $2 mA$

UPSEE - 2018

Semiconductor Electronics Material Devices and Simple Circuits

151099 Identify the diode which is forward biased in the following set
a. original image
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.
$V_{+} - V_{-} = - ve$ voltage then it reverse biased.
Then,
For A $0 - (- 5 V) = 5 V$ , forward biased
For B $5 - 0 = + 5 V$ , forward biased
For C $- 2 - (- 7) = 5 V$ , forward biased.
For D $2 - 3 = - 1 V$ (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 $β =$ 100, voltage drop across emitter and base is $0.7 V$ , then the value of $V_{CE}$ will be-
original image

1 Zero

2

5 V

3

10 V

4

13 V

Explanation:

D Given, $β = 100, R_{C} = 100 k Ω$
$I_{B} = \frac{5 - 0.7}{8.6 \times 10^{3}} = 0.5 mA$
$∴ I_{C} = β \times I_{B}$
$I_{C} = 100 \times 0.5 = 50 mA$
Now,
$V_{C E} = V_{C C} - I_{C} R_{C}$
$V_{C E} = 18 - 50 \times 10^{- 3} \times 100$
$V_{C E} = 18 - 5$ So, $V_{CE} = 13 V$

BCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

151104 The input resistance of a common emitter transistor amplifier, if the output resistance is $500 k Ω$ , the current gain $α = 0.98$ and the power gain is $6.0625 \times 10^{6}$ , is-

1

198 Ω

2

300 Ω

3

100 Ω

4

400 Ω

Explanation:

B Given, $R_{0} = 500 k Ω$
$α = 0.98$
Power gain $= 6.0625 \times 10^{6}$
$∵ β = \frac{α}{1 - α} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
We know,
Voltage gain $(A_{V}) = β \times \frac{R_{0}}{R_{i}}$
$= 49 \times \frac{500 \times 10^{3}}{R_{i}}$
Also, power gain $=$ voltage gain $\times β$
$6.0625 \times 10^{6} = 49 \times \frac{500 \times 10^{3}}{R_{i}} \times 49$ So, $R_{i} = 198 Ω$

BCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151098 In n-p-n transistor circuit, the collector current is $20 mA$ . If $90 %$ of the electrons emitted reach the collector, then the

1 emitter current will be about

16 mA

2 emitter current will be

19 mA

3 base current will be about

2 mA

4 base current will be about

10 mA

Explanation:

C Given,
$I_{c} = 20 mA,$
$I_{c} = 90 % of I_{E}$
$I_{c} = \frac{90}{100} \times I_{E}$
Then,
$I_{E} = \frac{100}{90} \times I_{c}$
$= \frac{100}{90} \times 20$
$I_{E} = 22 mA$
Now,
$I_{B} = I_{E} - I_{c}$
$= 22 - 20$
$= 2 mA$ Hence, base current will be about $2 mA$

UPSEE - 2018

Semiconductor Electronics Material Devices and Simple Circuits

151099 Identify the diode which is forward biased in the following set
a. original image
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.
$V_{+} - V_{-} = - ve$ voltage then it reverse biased.
Then,
For A $0 - (- 5 V) = 5 V$ , forward biased
For B $5 - 0 = + 5 V$ , forward biased
For C $- 2 - (- 7) = 5 V$ , forward biased.
For D $2 - 3 = - 1 V$ (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 $β =$ 100, voltage drop across emitter and base is $0.7 V$ , then the value of $V_{CE}$ will be-
original image

1 Zero

2

5 V

3

10 V

4

13 V

Explanation:

D Given, $β = 100, R_{C} = 100 k Ω$
$I_{B} = \frac{5 - 0.7}{8.6 \times 10^{3}} = 0.5 mA$
$∴ I_{C} = β \times I_{B}$
$I_{C} = 100 \times 0.5 = 50 mA$
Now,
$V_{C E} = V_{C C} - I_{C} R_{C}$
$V_{C E} = 18 - 50 \times 10^{- 3} \times 100$
$V_{C E} = 18 - 5$ So, $V_{CE} = 13 V$

BCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

151104 The input resistance of a common emitter transistor amplifier, if the output resistance is $500 k Ω$ , the current gain $α = 0.98$ and the power gain is $6.0625 \times 10^{6}$ , is-

1

198 Ω

2

300 Ω

3

100 Ω

4

400 Ω

Explanation:

B Given, $R_{0} = 500 k Ω$
$α = 0.98$
Power gain $= 6.0625 \times 10^{6}$
$∵ β = \frac{α}{1 - α} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
We know,
Voltage gain $(A_{V}) = β \times \frac{R_{0}}{R_{i}}$
$= 49 \times \frac{500 \times 10^{3}}{R_{i}}$
Also, power gain $=$ voltage gain $\times β$
$6.0625 \times 10^{6} = 49 \times \frac{500 \times 10^{3}}{R_{i}} \times 49$ So, $R_{i} = 198 Ω$

BCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151098 In n-p-n transistor circuit, the collector current is $20 mA$ . If $90 %$ of the electrons emitted reach the collector, then the

1 emitter current will be about

16 mA

2 emitter current will be

19 mA

3 base current will be about

2 mA

4 base current will be about

10 mA

Explanation:

C Given,
$I_{c} = 20 mA,$
$I_{c} = 90 % of I_{E}$
$I_{c} = \frac{90}{100} \times I_{E}$
Then,
$I_{E} = \frac{100}{90} \times I_{c}$
$= \frac{100}{90} \times 20$
$I_{E} = 22 mA$
Now,
$I_{B} = I_{E} - I_{c}$
$= 22 - 20$
$= 2 mA$ Hence, base current will be about $2 mA$

UPSEE - 2018

Semiconductor Electronics Material Devices and Simple Circuits

151099 Identify the diode which is forward biased in the following set
a. original image
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.
$V_{+} - V_{-} = - ve$ voltage then it reverse biased.
Then,
For A $0 - (- 5 V) = 5 V$ , forward biased
For B $5 - 0 = + 5 V$ , forward biased
For C $- 2 - (- 7) = 5 V$ , forward biased.
For D $2 - 3 = - 1 V$ (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 $β =$ 100, voltage drop across emitter and base is $0.7 V$ , then the value of $V_{CE}$ will be-
original image

1 Zero

2

5 V

3

10 V

4

13 V

Explanation:

D Given, $β = 100, R_{C} = 100 k Ω$
$I_{B} = \frac{5 - 0.7}{8.6 \times 10^{3}} = 0.5 mA$
$∴ I_{C} = β \times I_{B}$
$I_{C} = 100 \times 0.5 = 50 mA$
Now,
$V_{C E} = V_{C C} - I_{C} R_{C}$
$V_{C E} = 18 - 50 \times 10^{- 3} \times 100$
$V_{C E} = 18 - 5$ So, $V_{CE} = 13 V$

BCECE-2012

Semiconductor Electronics Material Devices and Simple Circuits

151104 The input resistance of a common emitter transistor amplifier, if the output resistance is $500 k Ω$ , the current gain $α = 0.98$ and the power gain is $6.0625 \times 10^{6}$ , is-

1

198 Ω

2

300 Ω

3

100 Ω

4

400 Ω

Explanation:

B Given, $R_{0} = 500 k Ω$
$α = 0.98$
Power gain $= 6.0625 \times 10^{6}$
$∵ β = \frac{α}{1 - α} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
We know,
Voltage gain $(A_{V}) = β \times \frac{R_{0}}{R_{i}}$
$= 49 \times \frac{500 \times 10^{3}}{R_{i}}$
Also, power gain $=$ voltage gain $\times β$
$6.0625 \times 10^{6} = 49 \times \frac{500 \times 10^{3}}{R_{i}} \times 49$ So, $R_{i} = 198 Ω$

BCECE-2011

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