Electric Field in Space Charge Region MCQ Quiz - Objective Question with Answer for Electric Field in Space Charge Region - Download Free PDF

Last updated on Apr 1, 2025

Latest Electric Field in Space Charge Region MCQ Objective Questions

Electric Field in Space Charge Region Question 1:

The electric field profile in the depletion region of a PN junction in equilibrium is shown below. Which one of the following statement is TRUE ?

qImage67bf482a2b4e995d672d78b7

  1. The left side of junction is P-type and the right side is N-type.
  2. Both the N-type and P-type depletion regions are non-uniformly doped.
  3. If the P-type has a doping concentration of 1010 cm−3, then doping concentration in N-type region will be 1011 cm−3.
  4. None of the options

Answer (Detailed Solution Below)

Option 1 : The left side of junction is P-type and the right side is N-type.

Electric Field in Space Charge Region Question 1 Detailed Solution

Explanation:

The electric field profile in the depletion region of a PN junction in equilibrium is a critical aspect of understanding the behavior of semiconductor devices. Let's delve into the correct option and analyze the other options to provide a comprehensive solution.

Depletion Region of a PN Junction:

In a PN junction, the depletion region is formed when P-type and N-type semiconductors are joined together. The P-type semiconductor has an abundance of holes (positive charge carriers), while the N-type semiconductor has an abundance of electrons (negative charge carriers). When these two types are brought into contact, electrons from the N-side diffuse into the P-side and recombine with holes, and holes from the P-side diffuse into the N-side and recombine with electrons. This diffusion process creates a region devoid of free charge carriers, known as the depletion region.

The electric field in the depletion region is established due to the fixed ionized donor atoms (positive ions) on the N-side and acceptor atoms (negative ions) on the P-side. The electric field direction is from the N-side to the P-side, and it opposes the diffusion of charge carriers, thus establishing equilibrium.

Correct Option Analysis:

The correct option is:

Option 1: The left side of the junction is P-type and the right side is N-type.

This option is correct because the electric field direction in the depletion region of a PN junction at equilibrium is from the N-type to the P-type region. This indicates that the left side (where the electric field originates) is P-type, and the right side (where the electric field terminates) is N-type. The electric field profile typically shows a peak near the junction and decreases as we move away from the junction on both sides.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Both the N-type and P-type depletion regions are non-uniformly doped.

This option is incorrect. While it is possible to have non-uniform doping in some advanced semiconductor devices, a standard PN junction in equilibrium typically assumes uniform doping concentrations for simplicity. Non-uniform doping would complicate the electric field profile, but it does not inherently define the nature of the depletion region in equilibrium.

Option 3: If the P-type has a doping concentration of 1010 cm−3, then the doping concentration in the N-type region will be 1011 cm−3.

This option is also incorrect. The relationship between the doping concentrations of the P-type and N-type regions does not follow a fixed ratio like 1010 cm−3 to 1011 cm−3. The doping concentrations are typically chosen based on the desired electrical properties of the PN junction, and they can vary widely depending on the application. The given concentrations are arbitrary and do not reflect a universal rule.

Option 4: None of the options.

This option is incorrect because Option 1 is indeed correct. Therefore, stating that none of the options are correct would be inaccurate.

Conclusion:

Understanding the electric field profile in the depletion region of a PN junction is crucial for analyzing semiconductor devices. In equilibrium, the electric field direction is from the N-type region to the P-type region, confirming that the left side is P-type and the right side is N-type. This understanding helps in designing and optimizing various semiconductor components such as diodes, transistors, and integrated circuits. The analysis of other options further reinforces the correctness of Option 1, highlighting the importance of accurate doping concentration and uniformity in semiconductor physics.

Electric Field in Space Charge Region Question 2:

Consider an abrupt p-n junction act (T = 300 K) shown in the figure. The depletion region width Xn on the n-side of the junction is 0.2 μm and the permittivity of silicon (tsi)  is 1.044 × 10-12 F/cm.

At the junction, the approximate absolute value of the peak electric field (KV/cm) is ______

F1 S.B 10.7.20 Pallavi D10

Answer (Detailed Solution Below) 30 - 32

Electric Field in Space Charge Region Question 2 Detailed Solution

Concept:

From charge neutrality equation:

xn ND = xp NA

\({x_p} = \frac{{{x_n}{N_D}\;}}{{{N_A}}}\) 

For NA ND:

\(\frac{{{N_D}}}{{{N_A}}} \ll 1\) 

xp xn

Width of junction is given by:

x = xn + xp

xn xp

x = xn         

Also, the peak electric field is given by Poisson’s equation as:

\({E_{peak}} = \frac{V}{E}{x_n}{N_D}\) 

Calculation:

Given:

xn = 0.2 μm = 0.2 × 10-4 cm

ND = 1016/cm3

ϵ = 1.044 × 10-12 F/cm

\({E_{peak}} = \frac{{1.6\; \times \;{{10}^{ - 19}}}}{{1.044\; \times\; {{10}^{ - 12}}}} \times 0.2 \times {10^{ - 4}} \times {10^{16}}\) 

Epeak = 30.65 KV/cm

Electric Field in Space Charge Region Question 3:

The electric field profile in the depletion region of a p-n junction in equilibrium is shown in the figure. Which one of the following statements is NOT TRUE?

F1 R.D Madhu 23.10.19 D7

  1. The left side of the junction is n-type and the right side is p-type
  2. Both the n-type and p-type depletion regions are non-uniformly doped
  3. The potential difference across the depletion region is 700 mV
  4. If the p-type region has a doping concentration of 1015 cm-3, then the doping concentration in the n-type region will be 1016 cm-3

Answer (Detailed Solution Below)

Option 3 : The potential difference across the depletion region is 700 mV

Electric Field in Space Charge Region Question 3 Detailed Solution

Concept:

The potential difference across the junction called the built-in voltage is given as:

\({V_0} = - \frac{1}{2}{E_{max}}.W\)

V0 → Built-in voltage

W → Depletion width

Emax → Maximum electric field at the junction.

Putting on the respective values we get:

\({V_0} = - \frac{1}{2} \times {10^4} \times 1 \cdot 1 \times {10^{ - 4}}\)

|V0| = 0.55 Volts.

So, Option-(3) which states that the potential difference across the depletion region is 700 mV is NOT TRUE.

F1 S.B Madhu 15.11.19 D 6

Option 1Since the electric field is positive towards the right side of the axis, the left side of the junction must have positive immobile ions and the right side must have negative immobile ions. This indicates that the left side of the junction must be n-type and the right side must be p-type. ∴ The statement in Option 1 is correct.

Option 2Since the length of the depletion region is unequal on the n-side and p-side, the depletion regions will be non-uniformly doped. ∴ The statement in Option 2 is correct.

Alternate:

V0 = - ∫ E⋅ dl i.e the Voltage is the area under the Electric field & distance characteristics.

So the Area of the above-given triangle is \( = \frac{1}{2} \times {E_{max}} \times W\)

Electric Field in Space Charge Region Question 4:

A uniformly doped p+n junction at T = 300 K is designed such that, at a reverse bias of VR = 10 V, the maximum electric field is limited to \(\left| {{E_{max}}} \right| = 2.5 \times {10^5}V/cm.\) The maximum doping concentration in the n-region is _________ × 1016 cm-3

(Up to 2-decimal places). (Take Vbi ≪ VR)

r = 11.7, ϵ0 = 8.854 × 10-14, e = 1.6 × 10-19C)

Answer (Detailed Solution Below) 1.5 - 2.5

Electric Field in Space Charge Region Question 4 Detailed Solution

Concept:

The maximum Electric field is given by

\(\left| {{E_{max}}} \right| = \frac{{q{N_d}{x_n}}}{\epsilon}\) 

Where xn ≅ W   (∵ of p+n junction the depletion region will extend to the lightly doped side)

\({x_n} \cong \sqrt {\frac{{2\epsilon\left( {{V_{bi}} + {V_R}} \right)}}{{q{N_d}}}} \cong \sqrt {\frac{{2\epsilon{V_R}}}{{q{N_d}}}} \) 

Calculation:

\(\left| {{E_{max}}} \right| = \frac{{q{N_d}{x_n}}}{\epsilon},\;with\;{x_n} = \sqrt {\frac{{2\epsilon{V_R}}}{{q{N_d}}}} \) 

Putting xn in |Emax| equation, we get,

\(\Rightarrow \left| {{E_{max}}} \right| = \frac{{q{N_d}}}{\epsilon}\sqrt {\frac{{2\epsilon{V_R}}}{{q{N_d}}}} \) 

\(\Rightarrow {\left| {{E_{max}}} \right|^2\epsilon}{^2}q{N_d} = {q^2}N_d^22\epsilon{V_R}\) 

\({N_d} = \frac{{{{\left| {{E_{max}}} \right|}^2\epsilon}}}{{2q{V_R}}}\) 

Putting values, we get,

\({N_d} = \frac{{\left( {11.7} \right)\left( {8.854 \times {{10}^{ - 14}}} \right){{\left( {2.5 \times {{10}^5}} \right)}^2}}}{{2\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {10} \right)}}\) 

= 2.023 × 1016cm-3

Electric Field in Space Charge Region Question 5:

An abrupt p-n junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the p-n junction?

  1. 2nd Review of GATE 2017 Questions PART 2.docx 7
  2. 2nd Review of GATE 2017 Questions PART 2.docx 8
  3. 2nd Review of GATE 2017 Questions PART 2.docx 9
  4. 2nd Review of GATE 2017 Questions PART 2.docx 10

Answer (Detailed Solution Below)

Option 1 : 2nd Review of GATE 2017 Questions PART 2.docx 7

Electric Field in Space Charge Region Question 5 Detailed Solution

Concept:

When two ‘p’ type and ‘n’ type semiconductors come into contact to form a p-n junction diode as shown in the figure, there is a diffusion of majority carries which takes place, leaving behind uncompensated ions behind, as shown;

2nd Review of GATE 2017 Questions PART 2.docx 11

Application:

The figure for charge density within the transition region is as shown below:

2nd Review of GATE 2017 Questions PART 2.docx 12

According to Poisson’s equation, the electric field is given as

\(\frac{{d{\text{E}}\left( x \right)}}{{dx}} = \frac{q}{\epsilon}\left( {N_d^ + - N_a^ - } \right)\)

This shows that E(x) will vary linearly with distance.

So, the electric field distribution is as shown:

2nd Review of GATE 2017 Questions PART 2.docx 13

Top Electric Field in Space Charge Region MCQ Objective Questions

Consider the charge profile shown in the figure. The resultant potential distribution is best described by

Gate EC 2016 paper 3 Images-Q55

  1. Gate EC 2016 paper 3 Images-Q55.1
  2. Gate EC 2016 paper 3 Images-Q55.2
  3. Gate EC 2016 paper 3 Images-Q55.3
  4. Gate EC 2016 paper 3 Images-Q55.4

Answer (Detailed Solution Below)

Option 4 : Gate EC 2016 paper 3 Images-Q55.4

Electric Field in Space Charge Region Question 6 Detailed Solution

Download Solution PDF

From Poisson’s equation

\( \nabla^2 {\rm{V}} = \frac{{ - {\rm{\rho }}\left( {\rm{x}} \right)}}{\epsilon}\)

For one dimensional charge density

\(\frac{{{{\rm{d}}^2}{\rm{V}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \frac{{ - {\rm{\rho }}\left( {\rm{x}} \right){\rm{\;}}}}{\epsilon}\)

for x < 0, \({\rm{\rho }}\left( {\rm{x}} \right) = - {{\rm{\rho }}_2}\)

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{V}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \frac{{{{\rm{\rho }}_2}}}{\epsilon}{\rm{\;}}\)

Solving this we get

\({{\rm{V}}_ - }\left( {\rm{x}} \right) = \frac{{{{\rm{\rho }}_2}}}{\epsilon}{{\rm{x}}^2} + {{\rm{c}}_1}{\rm{x}} + {{\rm{c}}_2}\)      \({\rm{x}} < 0\) 

where C1 and C2 are arbitrary constants.

Thus, V(x) is an upward parabola.

Similarly \({{\rm{V}}_ + }\left( {\rm{x}} \right) = \frac{{ - {{\rm{\rho }}_1}}}{\epsilon}{{\rm{x}}^2} + {{\rm{c}}_3}{\rm{x}} + {{\rm{c}}_4}\) is a downward parabola for x > 0

At x = 0, V-(x) = V+(x) and V(x) will be constant for x < b and x > a and will have no discontinuity. Thus, we see option D is correct answer.

In a p+n junction diode under reverse bias, the magnitude of electric field is maximum at

  1. The edge of the depletion region of the p – side

  2. The edge of the depletion region on the n – side 

  3. The p+n junction

  4. The centre of the depletion region on the n – side

Answer (Detailed Solution Below)

Option 3 :

The p+n junction

Electric Field in Space Charge Region Question 7 Detailed Solution

Download Solution PDF

The depletion region width in a region is inversely proportional to the doping density of that region

Since doping in p side is higher, the depletion region width of p side will be less

The maximum/minimum value of the electric field always occurs at the junction interface.

The variation of electric field in depletion region is shown in the figure

So option (c) is correct.
pn jnction

The electric field profile in the depletion region of a p-n junction in equilibrium is shown in the figure. Which one of the following statements is NOT TRUE?

F1 R.D Madhu 23.10.19 D7

  1. The left side of the junction is n-type and the right side is p-type
  2. Both the n-type and p-type depletion regions are non-uniformly doped
  3. The potential difference across the depletion region is 700 mV
  4. If the p-type region has a doping concentration of 1015 cm-3, then the doping concentration in the n-type region will be 1016 cm-3

Answer (Detailed Solution Below)

Option 3 : The potential difference across the depletion region is 700 mV

Electric Field in Space Charge Region Question 8 Detailed Solution

Download Solution PDF

Concept:

The potential difference across the junction called the built-in voltage is given as:

\({V_0} = - \frac{1}{2}{E_{max}}.W\)

V0 → Built-in voltage

W → Depletion width

Emax → Maximum electric field at the junction.

Putting on the respective values we get:

\({V_0} = - \frac{1}{2} \times {10^4} \times 1 \cdot 1 \times {10^{ - 4}}\)

|V0| = 0.55 Volts.

So, Option-(3) which states that the potential difference across the depletion region is 700 mV is NOT TRUE.

F1 S.B Madhu 15.11.19 D 6

Option 1Since the electric field is positive towards the right side of the axis, the left side of the junction must have positive immobile ions and the right side must have negative immobile ions. This indicates that the left side of the junction must be n-type and the right side must be p-type. ∴ The statement in Option 1 is correct.

Option 2Since the length of the depletion region is unequal on the n-side and p-side, the depletion regions will be non-uniformly doped. ∴ The statement in Option 2 is correct.

Alternate:

V0 = - ∫ E⋅ dl i.e the Voltage is the area under the Electric field & distance characteristics.

So the Area of the above-given triangle is \( = \frac{1}{2} \times {E_{max}} \times W\)

Consider an abrupt p-n junction act (T = 300 K) shown in the figure. The depletion region width Xn on the n-side of the junction is 0.2 μm and the permittivity of silicon (tsi)  is 1.044 × 10-12 F/cm.

At the junction, the approximate absolute value of the peak electric field (KV/cm) is ______

F1 S.B 10.7.20 Pallavi D10

Answer (Detailed Solution Below) 30 - 32

Electric Field in Space Charge Region Question 9 Detailed Solution

Download Solution PDF

Concept:

From charge neutrality equation:

xn ND = xp NA

\({x_p} = \frac{{{x_n}{N_D}\;}}{{{N_A}}}\) 

For NA ND:

\(\frac{{{N_D}}}{{{N_A}}} \ll 1\) 

xp xn

Width of junction is given by:

x = xn + xp

xn xp

x = xn         

Also, the peak electric field is given by Poisson’s equation as:

\({E_{peak}} = \frac{V}{E}{x_n}{N_D}\) 

Calculation:

Given:

xn = 0.2 μm = 0.2 × 10-4 cm

ND = 1016/cm3

ϵ = 1.044 × 10-12 F/cm

\({E_{peak}} = \frac{{1.6\; \times \;{{10}^{ - 19}}}}{{1.044\; \times\; {{10}^{ - 12}}}} \times 0.2 \times {10^{ - 4}} \times {10^{16}}\) 

Epeak = 30.65 KV/cm

An abrupt p-n junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the p-n junction?

  1. 2nd Review of GATE 2017 Questions PART 2.docx 7
  2. 2nd Review of GATE 2017 Questions PART 2.docx 8
  3. 2nd Review of GATE 2017 Questions PART 2.docx 9
  4. 2nd Review of GATE 2017 Questions PART 2.docx 10

Answer (Detailed Solution Below)

Option 1 : 2nd Review of GATE 2017 Questions PART 2.docx 7

Electric Field in Space Charge Region Question 10 Detailed Solution

Download Solution PDF

Concept:

When two ‘p’ type and ‘n’ type semiconductors come into contact to form a p-n junction diode as shown in the figure, there is a diffusion of majority carries which takes place, leaving behind uncompensated ions behind, as shown;

2nd Review of GATE 2017 Questions PART 2.docx 11

Application:

The figure for charge density within the transition region is as shown below:

2nd Review of GATE 2017 Questions PART 2.docx 12

According to Poisson’s equation, the electric field is given as

\(\frac{{d{\text{E}}\left( x \right)}}{{dx}} = \frac{q}{\epsilon}\left( {N_d^ + - N_a^ - } \right)\)

This shows that E(x) will vary linearly with distance.

So, the electric field distribution is as shown:

2nd Review of GATE 2017 Questions PART 2.docx 13

The electric field profile in the depletion region of a PN junction in equilibrium is shown below. Which one of the following statement is TRUE ?

qImage67bf482a2b4e995d672d78b7

  1. The left side of junction is P-type and the right side is N-type.
  2. Both the N-type and P-type depletion regions are non-uniformly doped.
  3. If the P-type has a doping concentration of 1010 cm−3, then doping concentration in N-type region will be 1011 cm−3.
  4. None of the options

Answer (Detailed Solution Below)

Option 1 : The left side of junction is P-type and the right side is N-type.

Electric Field in Space Charge Region Question 11 Detailed Solution

Download Solution PDF

Explanation:

The electric field profile in the depletion region of a PN junction in equilibrium is a critical aspect of understanding the behavior of semiconductor devices. Let's delve into the correct option and analyze the other options to provide a comprehensive solution.

Depletion Region of a PN Junction:

In a PN junction, the depletion region is formed when P-type and N-type semiconductors are joined together. The P-type semiconductor has an abundance of holes (positive charge carriers), while the N-type semiconductor has an abundance of electrons (negative charge carriers). When these two types are brought into contact, electrons from the N-side diffuse into the P-side and recombine with holes, and holes from the P-side diffuse into the N-side and recombine with electrons. This diffusion process creates a region devoid of free charge carriers, known as the depletion region.

The electric field in the depletion region is established due to the fixed ionized donor atoms (positive ions) on the N-side and acceptor atoms (negative ions) on the P-side. The electric field direction is from the N-side to the P-side, and it opposes the diffusion of charge carriers, thus establishing equilibrium.

Correct Option Analysis:

The correct option is:

Option 1: The left side of the junction is P-type and the right side is N-type.

This option is correct because the electric field direction in the depletion region of a PN junction at equilibrium is from the N-type to the P-type region. This indicates that the left side (where the electric field originates) is P-type, and the right side (where the electric field terminates) is N-type. The electric field profile typically shows a peak near the junction and decreases as we move away from the junction on both sides.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Both the N-type and P-type depletion regions are non-uniformly doped.

This option is incorrect. While it is possible to have non-uniform doping in some advanced semiconductor devices, a standard PN junction in equilibrium typically assumes uniform doping concentrations for simplicity. Non-uniform doping would complicate the electric field profile, but it does not inherently define the nature of the depletion region in equilibrium.

Option 3: If the P-type has a doping concentration of 1010 cm−3, then the doping concentration in the N-type region will be 1011 cm−3.

This option is also incorrect. The relationship between the doping concentrations of the P-type and N-type regions does not follow a fixed ratio like 1010 cm−3 to 1011 cm−3. The doping concentrations are typically chosen based on the desired electrical properties of the PN junction, and they can vary widely depending on the application. The given concentrations are arbitrary and do not reflect a universal rule.

Option 4: None of the options.

This option is incorrect because Option 1 is indeed correct. Therefore, stating that none of the options are correct would be inaccurate.

Conclusion:

Understanding the electric field profile in the depletion region of a PN junction is crucial for analyzing semiconductor devices. In equilibrium, the electric field direction is from the N-type region to the P-type region, confirming that the left side is P-type and the right side is N-type. This understanding helps in designing and optimizing various semiconductor components such as diodes, transistors, and integrated circuits. The analysis of other options further reinforces the correctness of Option 1, highlighting the importance of accurate doping concentration and uniformity in semiconductor physics.

Electric Field in Space Charge Region Question 12:

Consider the charge profile shown in the figure. The resultant potential distribution is best described by

Gate EC 2016 paper 3 Images-Q55

  1. Gate EC 2016 paper 3 Images-Q55.1
  2. Gate EC 2016 paper 3 Images-Q55.2
  3. Gate EC 2016 paper 3 Images-Q55.3
  4. Gate EC 2016 paper 3 Images-Q55.4

Answer (Detailed Solution Below)

Option 4 : Gate EC 2016 paper 3 Images-Q55.4

Electric Field in Space Charge Region Question 12 Detailed Solution

From Poisson’s equation

\( \nabla^2 {\rm{V}} = \frac{{ - {\rm{\rho }}\left( {\rm{x}} \right)}}{\epsilon}\)

For one dimensional charge density

\(\frac{{{{\rm{d}}^2}{\rm{V}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \frac{{ - {\rm{\rho }}\left( {\rm{x}} \right){\rm{\;}}}}{\epsilon}\)

for x < 0, \({\rm{\rho }}\left( {\rm{x}} \right) = - {{\rm{\rho }}_2}\)

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{V}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \frac{{{{\rm{\rho }}_2}}}{\epsilon}{\rm{\;}}\)

Solving this we get

\({{\rm{V}}_ - }\left( {\rm{x}} \right) = \frac{{{{\rm{\rho }}_2}}}{\epsilon}{{\rm{x}}^2} + {{\rm{c}}_1}{\rm{x}} + {{\rm{c}}_2}\)      \({\rm{x}} < 0\) 

where C1 and C2 are arbitrary constants.

Thus, V(x) is an upward parabola.

Similarly \({{\rm{V}}_ + }\left( {\rm{x}} \right) = \frac{{ - {{\rm{\rho }}_1}}}{\epsilon}{{\rm{x}}^2} + {{\rm{c}}_3}{\rm{x}} + {{\rm{c}}_4}\) is a downward parabola for x > 0

At x = 0, V-(x) = V+(x) and V(x) will be constant for x < b and x > a and will have no discontinuity. Thus, we see option D is correct answer.

Electric Field in Space Charge Region Question 13:

The graph of the electric field in the transition region of an abrupt p - n junction is as given below

F2 S.B 20.5.20 pallavi D1

Let Na, Nd be doping concentration in P and n side of P - N junction then

  1. Na > Nd
  2. Na < Nd
  3. Na = Nd
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Na > Nd

Electric Field in Space Charge Region Question 13 Detailed Solution

From Charge neutrality equation:

xpoNA= xnoND

Since Xpo< Xno

Na > Nd

Alternate Approach:

From the diagram of the electric field in the given figure slope of the electric field \(\frac{{\partial E}}{{\partial x}}\) is more on the left side than the right side

Depletion

For \(0 < x < {x_{no}}\)

\(\frac{{\partial E}}{{\partial x}} = \frac{q}{\varepsilon }{N_d}\)

Similarly, for \(- {x_{po}} < x < 0\)

\(\frac{{\partial E}}{{\partial x}} = \frac{{ - q}}{\varepsilon }{N_a}\)

\({\left| {\frac{{\partial E}}{{\partial x}}} \right|_{{x_{po}}}} > {\left| {\frac{{\partial E}}{{\partial x}}} \right|_{{x_{no}}}}\)

\(\frac{q}{\varepsilon }{N_a} > \frac{q}{\varepsilon }{N_d} \Rightarrow {N_a} > {N_d}\)

Electric Field in Space Charge Region Question 14:

In a p+n junction diode under reverse bias, the magnitude of electric field is maximum at

  1. The edge of the depletion region of the p – side

  2. The edge of the depletion region on the n – side 

  3. The p+n junction

  4. The centre of the depletion region on the n – side

Answer (Detailed Solution Below)

Option 3 :

The p+n junction

Electric Field in Space Charge Region Question 14 Detailed Solution

The depletion region width in a region is inversely proportional to the doping density of that region

Since doping in p side is higher, the depletion region width of p side will be less

The maximum/minimum value of the electric field always occurs at the junction interface.

The variation of electric field in depletion region is shown in the figure

So option (c) is correct.
pn jnction

Electric Field in Space Charge Region Question 15:

The electric field profile in the depletion region of a p-n junction in equilibrium is shown in the figure. Which one of the following statements is NOT TRUE?

F1 R.D Madhu 23.10.19 D7

  1. The left side of the junction is n-type and the right side is p-type
  2. Both the n-type and p-type depletion regions are non-uniformly doped
  3. The potential difference across the depletion region is 700 mV
  4. If the p-type region has a doping concentration of 1015 cm-3, then the doping concentration in the n-type region will be 1016 cm-3

Answer (Detailed Solution Below)

Option 3 : The potential difference across the depletion region is 700 mV

Electric Field in Space Charge Region Question 15 Detailed Solution

Concept:

The potential difference across the junction called the built-in voltage is given as:

\({V_0} = - \frac{1}{2}{E_{max}}.W\)

V0 → Built-in voltage

W → Depletion width

Emax → Maximum electric field at the junction.

Putting on the respective values we get:

\({V_0} = - \frac{1}{2} \times {10^4} \times 1 \cdot 1 \times {10^{ - 4}}\)

|V0| = 0.55 Volts.

So, Option-(3) which states that the potential difference across the depletion region is 700 mV is NOT TRUE.

F1 S.B Madhu 15.11.19 D 6

Option 1Since the electric field is positive towards the right side of the axis, the left side of the junction must have positive immobile ions and the right side must have negative immobile ions. This indicates that the left side of the junction must be n-type and the right side must be p-type. ∴ The statement in Option 1 is correct.

Option 2Since the length of the depletion region is unequal on the n-side and p-side, the depletion regions will be non-uniformly doped. ∴ The statement in Option 2 is correct.

Alternate:

V0 = - ∫ E⋅ dl i.e the Voltage is the area under the Electric field & distance characteristics.

So the Area of the above-given triangle is \( = \frac{1}{2} \times {E_{max}} \times W\)

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