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

Concept:

From charge neutrality equation:

x_n N_D = x_p N_A

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

For N_A ≫ N_D:

\(\frac{{{N_D}}}{{{N_A}}} \ll 1\) 

∴ x_p ≪ x_n

Width of junction is given by:

x = x_n + x_p

∵ x_n ≫ x_p

x = x_n         

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

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

Calculation:

Given:

x_n = 0.2 μm = 0.2 × 10^-4 cm

N_D = 10¹⁶/cm³

ϵ = 1.044 × 10^-12 F/cm

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

E_peak = 30.65 KV/cm

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

E_max → 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.

Option 1: Since 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 2: Since 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 \( = \frac{1}{2} \times {E_{max}} \times W\)

Concept:

The maximum Electric field is given by

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

Where x_n ≅ 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 x_n in |E_max| 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

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;

Application:

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

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:

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.

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.

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

E_max → 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.

Option 1: Since 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 2: Since 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 \( = \frac{1}{2} \times {E_{max}} \times W\)

Concept:

From charge neutrality equation:

x_n N_D = x_p N_A

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

For N_A ≫ N_D:

\(\frac{{{N_D}}}{{{N_A}}} \ll 1\) 

∴ x_p ≪ x_n

Width of junction is given by:

x = x_n + x_p

∵ x_n ≫ x_p

x = x_n         

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

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

Calculation:

Given:

x_n = 0.2 μm = 0.2 × 10^-4 cm

N_D = 10¹⁶/cm³

ϵ = 1.044 × 10^-12 F/cm

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

E_peak = 30.65 KV/cm

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;

Application:

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

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:

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.

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.

From Charge neutrality equation:

xpoNA= xnoND

Since Xpo< X_no

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

For \(0 < x < {x_{no}}\)

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

Similarly, for \(- {x_{po}} < x < 0\)

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

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

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

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.

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

E_max → 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.

Option 1: Since 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 2: Since 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 \( = \frac{1}{2} \times {E_{max}} \times W\)