Breakdown Voltage of a Junction MCQ Quiz - Objective Question with Answer for Breakdown Voltage of a Junction - Download Free PDF

Last updated on Apr 23, 2025

Latest Breakdown Voltage of a Junction MCQ Objective Questions

Breakdown Voltage of a Junction Question 1:

Consider avalanche breakdown in a silicon p+n junction. The n-region is uniformly doped with a donor density ND. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field Ecrit. Assume Ecrit to be independent of ND. If the built-in voltage of the p+n junction is much smaller than the breakdown voltage, VBR, the relationship between VBR and ND is given by

  1. VBR×ND constant

  2. ND×VBR constant
  3. ND×VBR constant
  4. ND/VBR constant

Answer (Detailed Solution Below)

Option 3 : ND×VBR constant

Breakdown Voltage of a Junction Question 1 Detailed Solution

Analysis:

In any type of PN junction;

Ecritical =Vbi+VRBW/2 

Ecritical =2[Vbi+VRB]W        ---(1)

Where,

Vbi: Built-in potential.

VRB: Reverse bias voltage.

W=2εq(1NA+1ND)(Vbi+VRB)         ---(2)

P+N junction is given so

NANDSo,1NA1ND          ---(3)

From equation (1) and (2)

[Vbi+VRB]=εEcrit22q[1ND+1NA] 

From equation (3),

VRB=εEcrit22q[1ND] 

 ∵  Vbi = 0 (Given)

VRB1ND 

VRB . ND  = constant

Breakdown Voltage of a Junction Question 2:

Compared to p-n junction with NA = ND = 1014/cm3, which one of the following statements is TRUE for a p-n junction with NA = ND=1020/cm3?

  1. Reverse breakdown voltage is lower and depletion capacitance is lower
  2. Reverse breakdown voltage is higher and depletion capacitance is lower
  3. Reverse breakdown voltage is lower and depletion capacitance is higher
  4. Reverse breakdown voltage is higher and depletion capacitance is higher

Answer (Detailed Solution Below)

Option 3 : Reverse breakdown voltage is lower and depletion capacitance is higher

Breakdown Voltage of a Junction Question 2 Detailed Solution

Concept:

Depletion capacitance.Cd=Aqϵs2VbiNA.NDNA+ND,

=Aqϵs2VbiN2  (Assuming NA = ND = N)

∴ Cd∝ N

And breakdown voltage decreases as doping increases.

So, the junction with NA = ND = 1020 cm-3 has, lower reverse breakdown voltage and higher depletion capacitance as compared to the one with NA = ND = 1014/cm3

Top Breakdown Voltage of a Junction MCQ Objective Questions

Compared to p-n junction with NA = ND = 1014/cm3, which one of the following statements is TRUE for a p-n junction with NA = ND=1020/cm3?

  1. Reverse breakdown voltage is lower and depletion capacitance is lower
  2. Reverse breakdown voltage is higher and depletion capacitance is lower
  3. Reverse breakdown voltage is lower and depletion capacitance is higher
  4. Reverse breakdown voltage is higher and depletion capacitance is higher

Answer (Detailed Solution Below)

Option 3 : Reverse breakdown voltage is lower and depletion capacitance is higher

Breakdown Voltage of a Junction Question 3 Detailed Solution

Download Solution PDF

Concept:

Depletion capacitance.Cd=Aqϵs2VbiNA.NDNA+ND,

=Aqϵs2VbiN2  (Assuming NA = ND = N)

∴ Cd∝ N

And breakdown voltage decreases as doping increases.

So, the junction with NA = ND = 1020 cm-3 has, lower reverse breakdown voltage and higher depletion capacitance as compared to the one with NA = ND = 1014/cm3

Consider avalanche breakdown in a silicon p+n junction. The n-region is uniformly doped with a donor density ND. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field Ecrit. Assume Ecrit to be independent of ND. If the built-in voltage of the p+n junction is much smaller than the breakdown voltage, VBR, the relationship between VBR and ND is given by

  1. VBR×ND constant

  2. ND×VBR constant
  3. ND×VBR constant
  4. ND/VBR constant

Answer (Detailed Solution Below)

Option 3 : ND×VBR constant

Breakdown Voltage of a Junction Question 4 Detailed Solution

Download Solution PDF

Analysis:

In any type of PN junction;

Ecritical =Vbi+VRBW/2 

Ecritical =2[Vbi+VRB]W        ---(1)

Where,

Vbi: Built-in potential.

VRB: Reverse bias voltage.

W=2εq(1NA+1ND)(Vbi+VRB)         ---(2)

P+N junction is given so

NANDSo,1NA1ND          ---(3)

From equation (1) and (2)

[Vbi+VRB]=εEcrit22q[1ND+1NA] 

From equation (3),

VRB=εEcrit22q[1ND] 

 ∵  Vbi = 0 (Given)

VRB1ND 

VRB . ND  = constant

Breakdown Voltage of a Junction Question 5:

Compared to p-n junction with NA = ND = 1014/cm3, which one of the following statements is TRUE for a p-n junction with NA = ND=1020/cm3?

  1. Reverse breakdown voltage is lower and depletion capacitance is lower
  2. Reverse breakdown voltage is higher and depletion capacitance is lower
  3. Reverse breakdown voltage is lower and depletion capacitance is higher
  4. Reverse breakdown voltage is higher and depletion capacitance is higher

Answer (Detailed Solution Below)

Option 3 : Reverse breakdown voltage is lower and depletion capacitance is higher

Breakdown Voltage of a Junction Question 5 Detailed Solution

Concept:

Depletion capacitance.Cd=Aqϵs2VbiNA.NDNA+ND,

=Aqϵs2VbiN2  (Assuming NA = ND = N)

∴ Cd∝ N

And breakdown voltage decreases as doping increases.

So, the junction with NA = ND = 1020 cm-3 has, lower reverse breakdown voltage and higher depletion capacitance as compared to the one with NA = ND = 1014/cm3

Breakdown Voltage of a Junction Question 6:

Consider avalanche breakdown in a silicon p+n junction. The n-region is uniformly doped with a donor density ND. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field Ecrit. Assume Ecrit to be independent of ND. If the built-in voltage of the p+n junction is much smaller than the breakdown voltage, VBR, the relationship between VBR and ND is given by

  1. VBR×ND constant

  2. ND×VBR constant
  3. ND×VBR constant
  4. ND/VBR constant

Answer (Detailed Solution Below)

Option 3 : ND×VBR constant

Breakdown Voltage of a Junction Question 6 Detailed Solution

Analysis:

In any type of PN junction;

Ecritical =Vbi+VRBW/2 

Ecritical =2[Vbi+VRB]W        ---(1)

Where,

Vbi: Built-in potential.

VRB: Reverse bias voltage.

W=2εq(1NA+1ND)(Vbi+VRB)         ---(2)

P+N junction is given so

NANDSo,1NA1ND          ---(3)

From equation (1) and (2)

[Vbi+VRB]=εEcrit22q[1ND+1NA] 

From equation (3),

VRB=εEcrit22q[1ND] 

 ∵  Vbi = 0 (Given)

VRB1ND 

VRB . ND  = constant

Breakdown Voltage of a Junction Question 7:

For a symmetric silicon p-n junction the doping on p-side and n-side are N = Nd = 1017 cm-3. If the peak electric field at breakdown is 5 × 105 V/cm. The reverse breakdown voltage in this junction is ______ V.

(Take ϵr = 11.8, KT = 26 mV, ni = 1.5 x 1010 cm -3)

Answer (Detailed Solution Below) 15 - 16

Breakdown Voltage of a Junction Question 7 Detailed Solution

The maximum electric field is given by:

E0=qϵNd×xpo

E0=qϵNd×W2

W=(2ϵE0qNd)

19 D4

Contact potential in terms of the maximum electric field is

V0=12E0W

Vbr+V0=12×(5×105)(2ϵE0qNd)

Given:

ϵ = ϵr ϵ0 = 11.8 × 8.85 × 10-14 F/cm

E0 = 5 × 105 V/cm

Nd = 1017 cm-3

Vbr+V0=11.8×8.85×1014×(5×105)21.6×1019×1017

= 16.32 V

V0=KTqln(NaNdni2)

V= 0.817 V

Vbr = 15.502 V

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