Test: The Bipolar Junction Transistor

Question 1

A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is N_E = 10⁸ cm^-3, N_B= 5.2 x 10¹⁶ cm^-3 and N_C = 10¹⁵ cm^-3 .For V_EB = 0.6 V, the p_B at x =0 is

Accepted Answer

5.2 x 10^-13 cm^-3

Answer

5.2 x 10^-19 cm^-3

Answer

5.2 x 10¹⁶ cm^-3

Answer

5.2 x 10¹¹ cm^-3

Question 2

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x" = 0 is

Q. The minority carrier concentration at x = x_B is

Accepted Answer

2.6 x 10¹⁴ cm^-3

Answer

4.5 x 10¹⁴ cm^-3

Answer

2.6 x 10¹² cm^-3

Answer

3.9 x 10¹⁴ cm^-3

Question 3

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x" = 0 is

Q. The minority carrier concentration at x" = 0 is

Accepted Answer

4.5 x 10¹⁴ cm^-3

Answer

3.9 x 10¹⁴ cm^-3

Answer

2.7 x 10¹² cm^-3

Answer

2.7 x 10¹⁴ cm^-3

Question 4

An pnp bipolar transistor has uniform doping of N_E = 6 x 10¹⁷cm^-3, N_B= 2 x 10¹⁶ cm^-3 and N_C= 5 x 10¹⁴cm^-3. The transistor is operating is inverse-active mode. The maximum V_CB voltage, so that the low injection condition applies, is

Accepted Answer

0.48 V

Answer

0.86 V

Answer

0.32 V

Answer

0.60 V

Question 5

The following currents are measured in a uniformly doped npn bipolar transistor:

I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA

Q.

The α is

Accepted Answer

0.787

Answer

0.667

Answer

0.733

Answer

0.8

Question 6

The following currents are measured in a uniformly doped npn bipolar transistor:

I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA

Q.

The β is

Accepted Answer

3.69

Answer

0.44

Answer

2.27

Answer

8.39

Question 7

The following currents are measured in a uniformly doped npn bipolar transistor:

I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA

Q.

The γ is

Accepted Answer

0.923

Answer

0.816

Answer

1.083

Answer

0.440

Question 8

A silicon npn bipolar transistor has doping concentration of N_E = 2 x 10¹⁸cm^-3, N_B =10¹⁷cm^-3 and N_C = 15 x 10¹⁶ cm^-3. The area is 10^-3 cm² and neutral base width is 1 μm. The transistor is biased in the active region at V_BE = 0.5 V. The collector current is
(D_B = 20 cm²/s)

Accepted Answer

17&mu;A

Answer

9 &mu;A

Answer

22 &mu;A

Answer

11 &mu;A

Question 9

A uniformly doped npn bipolar transistor has following parameters:

N_E = 10¹⁸ cm^-3 N_B = 5 x 10¹⁶ cm^-3,
N_c = 2 x 10¹⁹ cm^-3,
D_E = 8 cm² /s , D_B = 15 cm² /s , D_c = 14 cm² /s
x_E = 0.8 μm, x_B = 0.7 μm

The emitter injection efficiency γ is

Accepted Answer

0.977

Answer

0.999

Answer

0.982

Answer

0.934

Question 10

In bipolar transistor biased in the forward-active region the base current is I_B = 50 μA. and the collector currents is I_C= 27 μA. The α is

Accepted Answer

0.982

Answer

0.949

Answer

54

Answer

0.018

Question 11

For the transistor in fig., I_S = 10¹⁵ A, β_F = 100, β_R = 1. The current I_CBO is

Accepted Answer

1.01 x 10^-15A

Answer

1.01 x 10^-14A

Answer

2 x 10^-14A

Answer

2 x 10^-15A

Question 12

Determine the region of operation for the transistor shown in circuit in question

Accepted Answer

Cutoff

Answer

Forward-Active

Answer

Reverse-Active

Answer

Saturation

Question 13

Determine the region of operation for the transistor shown in circuit in question

Accepted Answer

Forward-Active

Answer

Reverse-Active

Answer

Saturation

Answer

Cutoff

Question 14

Determine the region of operation for the transistor shown in circuit in question.

Accepted Answer

Reverse-Active

Answer

Forward-Active

Answer

Saturation

Answer

Cutoff

Question 15

Determine the region of operation for the transistor shown in circuit in question.

Accepted Answer

Saturation

Answer

Forward-Active

Answer

Reverse-Active

Answer

Cutoff

Question 16

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A

Q. The base-emitter voltage is

Accepted Answer

0.84 V

Answer

0.53 V

Answer

0.7 V

Answer

0.98 V

Question 17

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A

Q. The current I₁ is

Accepted Answer

- 12.75 mA

Answer

12.75 mA

Answer

12.5 mA

Answer

-12.5 mA

Question 18

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA

Q. The value of β is

Accepted Answer

79

Answer

81

Answer

80

Answer

None of the above

Question 19

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA

Q. The value of I_C is

Accepted Answer

2.77 mA

Answer

2.4 mA

Answer

2.34 mA

Answer

1.97 mA

Question 20

For a BJT, I_C = 5 mA, I_B = 50 μA and I_CBO = 0.5μA.

Q. The value of β is

Accepted Answer

103

Answer

91

Answer

83

Answer

51

GATE ECE (Electronics) Test: The Bipolar Junction Transistor Free Online

MCQ Practice Test & Solutions: Test: The Bipolar Junction Transistor (20 Questions)

A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is N_E = 10⁸ cm^-3, N_B= 5.2 x 10¹⁶ cm^-3 and N_C = 10¹⁵ cm^-3 .For V_EB = 0.6 V, the p_B at x =0 is

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x = x_B is

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x" = 0 is

An pnp bipolar transistor has uniform doping of N_E = 6 x 10¹⁷cm^-3, N_B= 2 x 10¹⁶ cm^-3 and N_C= 5 x 10¹⁴cm^-3. The transistor is operating is inverse-active mode. The maximum V_CB voltage, so that the low injection condition applies, is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The α is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The β is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The γ is

A silicon npn bipolar transistor has doping concentration of N_E = 2 x 10¹⁸cm^-3, N_B =10¹⁷cm^-3 and N_C = 15 x 10¹⁶ cm^-3. The area is 10^-3 cm² and neutral base width is 1 μm. The transistor is biased in the active region at V_BE = 0.5 V. The collector current is
(D_B = 20 cm²/s)

A uniformly doped npn bipolar transistor has following parameters:
N_E = 10¹⁸ cm^-3 N_B = 5 x 10¹⁶ cm^-3,
N_c = 2 x 10¹⁹ cm^-3,
D_E = 8 cm² /s , D_B = 15 cm² /s , D_c = 14 cm² /s
x_E = 0.8 μm, x_B = 0.7 μm
The emitter injection efficiency γ is

In bipolar transistor biased in the forward-active region the base current is I_B = 50 μA. and the collector currents is I_C= 27 μA. The α is

For the transistor in fig., I_S = 10¹⁵ A, β_F = 100, β_R = 1. The current I_CBO is

Determine the region of operation for the transistor shown in circuit in question

Determine the region of operation for the transistor shown in circuit in question

Determine the region of operation for the transistor shown in circuit in question.

Determine the region of operation for the transistor shown in circuit in question.

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A
Q. The base-emitter voltage is

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A
Q. The current I₁ is

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA
Q. The value of β is

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA
Q. The value of I_C is

For a BJT, I_C = 5 mA, I_B = 50 μA and I_CBO = 0.5μA.
Q. The value of β is

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GATE ECE (Electronics) Test: The Bipolar Junction Transistor Free Online

MCQ Practice Test & Solutions: Test: The Bipolar Junction Transistor (20 Questions)

A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is NE = 108 cm-3, NB = 5.2 x 1016 cm-3 and NC = 1015 cm-3 .For VEB = 0.6 V, the pB at x =0 is

An npn bipolar transistor having uniform doping of NE = 1018 cm-3 NB = 1016 cm-3 and Nc = 6 x 1015 cm-3 is operating in the inverse-active mode with VBE = - 2 V and VBC = 0.6 V. The geometry of transistor is shown in figQ. The minority carrier concentration at x = xB is

An npn bipolar transistor having uniform doping of NE = 1018 cm-3 NB = 1016 cm-3 and Nc = 6 x 1015 cm-3 is operating in the inverse-active mode with VBE = - 2 V and VBC = 0.6 V. The geometry of transistor is shown in figQ. The minority carrier concentration at x" = 0 is

An pnp bipolar transistor has uniform doping of NE = 6 x 1017 cm-3, NB = 2 x 1016 cm-3 and NC = 5 x 1014 cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, is

The following currents are measured in a uniformly doped npn bipolar transistor:InE = 1.20 mA, IpE = 0.10 mA, InC = 1.18 mA IR = 0.20 mA, IG = 1 μA, IpC0 = 1 μAQ. The α is

The following currents are measured in a uniformly doped npn bipolar transistor:InE = 1.20 mA, IpE = 0.10 mA, InC = 1.18 mA IR = 0.20 mA, IG = 1 μA, IpC0 = 1 μAQ.The β is

The following currents are measured in a uniformly doped npn bipolar transistor:InE = 1.20 mA, IpE = 0.10 mA, InC = 1.18 mA IR = 0.20 mA, IG = 1 μA, IpC0 = 1 μAQ. The γ is

A silicon npn bipolar transistor has doping concentration of NE = 2 x 1018cm-3, NB =1017cm-3 and NC = 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5 V. The collector current is (DB = 20 cm2/s)

A uniformly doped npn bipolar transistor has following parameters:NE = 1018 cm-3 NB = 5 x 1016 cm-3, Nc = 2 x 1019 cm-3, DE = 8 cm2 /s , DB = 15 cm2 /s , Dc = 14 cm2 /s xE = 0.8 μm, xB = 0.7 μmThe emitter injection efficiency γ is

In bipolar transistor biased in the forward-active region the base current is IB = 50 μA. and the collector currents is IC = 27 μA. The α is

For the transistor in fig., IS = 1015 A, βF = 100, βR = 1. The current ICBO is

Determine the region of operation for the transistor shown in circuit in question

Determine the region of operation for the transistor shown in circuit in question

Determine the region of operation for the transistor shown in circuit in question.

Determine the region of operation for the transistor shown in circuit in question.

For the circuit shown in fig., let the value of βR =0.5 and βF = 50. The saturation current is 10-16 AQ. The base-emitter voltage is

For the circuit shown in fig., let the value of βR =0.5 and βF = 50. The saturation current is 10-16 AQ. The current I1 is

The leakage current of a transistor are ICBO = 5μA and ICEO = 0.4 mA, and IB =30 μAQ. The value of β is

The leakage current of a transistor are ICBO = 5μA and ICEO = 0.4 mA, and IB =30 μAQ. The value of IC is

For a BJT, IC = 5 mA, IB = 50 μA and ICBO = 0.5μA.Q. The value of β is

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A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is N_E = 10⁸ cm^-3, N_B= 5.2 x 10¹⁶ cm^-3 and N_C = 10¹⁵ cm^-3 .For V_EB = 0.6 V, the p_B at x =0 is

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x = x_B is

An npn bipolar transistor having uniform doping of N_E= 10¹⁸ cm^-3 N_B= 10¹⁶ cm^-3 and N_c = 6 x 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = - 2 V and V_BC = 0.6 V. The geometry of transistor is shown in fig
Q. The minority carrier concentration at x" = 0 is

An pnp bipolar transistor has uniform doping of N_E = 6 x 10¹⁷cm^-3, N_B= 2 x 10¹⁶ cm^-3 and N_C= 5 x 10¹⁴cm^-3. The transistor is operating is inverse-active mode. The maximum V_CB voltage, so that the low injection condition applies, is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The α is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The β is

The following currents are measured in a uniformly doped npn bipolar transistor:
I_nE = 1.20 mA, I_pE = 0.10 mA, I_nC = 1.18 mA
I_R = 0.20 mA, I_G = 1 μA, I_pC0 = 1 μA
Q.
The γ is

A silicon npn bipolar transistor has doping concentration of N_E = 2 x 10¹⁸cm^-3, N_B =10¹⁷cm^-3 and N_C = 15 x 10¹⁶ cm^-3. The area is 10^-3 cm² and neutral base width is 1 μm. The transistor is biased in the active region at V_BE = 0.5 V. The collector current is
(D_B = 20 cm²/s)

A uniformly doped npn bipolar transistor has following parameters:
N_E = 10¹⁸ cm^-3 N_B = 5 x 10¹⁶ cm^-3,
N_c = 2 x 10¹⁹ cm^-3,
D_E = 8 cm² /s , D_B = 15 cm² /s , D_c = 14 cm² /s
x_E = 0.8 μm, x_B = 0.7 μm
The emitter injection efficiency γ is

In bipolar transistor biased in the forward-active region the base current is I_B = 50 μA. and the collector currents is I_C= 27 μA. The α is

For the transistor in fig., I_S = 10¹⁵ A, β_F = 100, β_R = 1. The current I_CBO is

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A
Q. The base-emitter voltage is

For the circuit shown in fig., let the value of β_R=0.5 and β_F = 50. The saturation current is 10^-16A
Q. The current I₁ is

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA
Q. The value of β is

The leakage current of a transistor are I_CBO = 5μA and I_CEO = 0.4 mA, and I_B =30 μA
Q. The value of I_C is

For a BJT, I_C = 5 mA, I_B = 50 μA and I_CBO = 0.5μA.
Q. The value of β is