PN Junction MCQ Quiz - Objective Question with Answer for PN Junction - Download Free PDF

Explanation:

Incorrect Statement Analysis for a Diode Operating in Forward Bias

In a forward-biased diode, the following processes occur that affect the depletion region and the potential barrier:

Correct Option:

Option 4: The reduction in the depletion region causes a heavy flow of minority carriers across the junction.

This statement is incorrect because, in a forward-biased diode, the reduction in the depletion region primarily facilitates the flow of majority carriers (electrons in the n-region and holes in the p-region) across the junction. The minority carriers (holes in the n-region and electrons in the p-region) do not play a significant role in the current flow in a forward-biased condition. The primary current flow in a forward-biased diode is due to the majority carriers overcoming the potential barrier and recombining at the junction.

Explanation of Correct Option:

When a diode is forward-biased, the external voltage applied across the diode reduces the potential barrier at the p-n junction. This reduction in the potential barrier is due to the narrowing of the depletion region, which occurs as the majority carriers are pushed towards the junction. The applied forward voltage causes the electrons in the n-region to move towards the p-region, and the holes in the p-region to move towards the n-region. As a result, the width of the depletion region decreases, allowing more majority carriers to cross the junction and recombine, leading to an increase in the current flow through the diode.

The key points to understand here are:

• The forward bias reduces the potential barrier, making it easier for majority carriers to cross the junction.
• The depletion region narrows, allowing for the increased flow of majority carriers.
• The current in a forward-biased diode is primarily due to the flow of majority carriers, not minority carriers.

Analysis of Other Options:

Option 1: The reduction in the width of the depletion region is due to the recombination of charge carriers and immobile ions near the junction.

This statement is correct. In forward bias, as the majority carriers move towards the junction, they recombine with the oppositely charged immobile ions (donors in the n-region and acceptors in the p-region), reducing the width of the depletion region.

Option 2: The reduction in the potential barrier occurs due to the narrowing of the depletion region.

This statement is correct. The applied forward voltage reduces the potential barrier at the junction by narrowing the depletion region, allowing more majority carriers to cross the junction.

Option 3: The reduction in the depletion region allows a majority carrier flow across the junction.

This statement is correct. The narrowing of the depletion region in forward bias facilitates the flow of majority carriers (electrons and holes) across the junction, leading to an increase in current.

In summary, the incorrect statement is option 4, as it incorrectly attributes the increased current flow in a forward-biased diode to the flow of minority carriers. The correct understanding is that the majority carriers are responsible for the increased current in a forward-biased diode.

Explanation:

A Zener diode has a breakdown voltage of Vz = 7 V at 300 K, with a temperature coefficient of 2.3 mV/°C. To find the new breakdown voltage Vz at 400 K, we need to consider the change in temperature and how it affects the breakdown voltage.

Calculation:

1. First, determine the temperature difference:

• The initial temperature (T1) = 300 K
• The final temperature (T2) = 400 K
• Temperature difference (ΔT) = T₂ - T₁ = 400 K - 300 K = 100 K

2. Convert the temperature difference from Kelvin to Celsius:

• Since the temperature coefficient is given in mV/°C, we need to use the equivalent Celsius temperature difference.
• Note: The temperature change is the same in Celsius and Kelvin, so ΔT = 100 °C.

3. Calculate the change in breakdown voltage using the temperature coefficient:

• Temperature coefficient = 2.3 mV/°C
• Change in breakdown voltage (ΔVz) = Temperature coefficient × Temperature difference
• ΔVz = 2.3 mV/°C × 100 °C = 230 mV

4. Convert the change in breakdown voltage to volts:

• ΔVz = 230 mV = 0.230 V

5. Determine the new breakdown voltage:

• Initial breakdown voltage (Vz) = 7 V
• New breakdown voltage (Vz_new) = Initial breakdown voltage + Change in breakdown voltage
• Vz_new = 7 V + 0.230 V = 7.23 V

Conclusion:

The new breakdown voltage at 400 K is 7.23 V, which corresponds to Option 2.

Explanation:

Typical Forward Voltage Across Silicon P-N Junction

Definition: The forward voltage of a silicon P-N junction refers to the voltage required to allow a current to pass through the junction in the forward bias condition. This voltage is crucial for the operation of silicon-based semiconductor devices such as diodes and transistors.

Working Principle: In a P-N junction, when a forward bias voltage is applied (positive to the P-type and negative to the N-type), the potential barrier at the junction decreases, allowing charge carriers (electrons and holes) to move across the junction. This movement of charge carriers constitutes a current. The voltage at which this current starts to significantly flow is known as the forward voltage.

Typical Value: For a silicon P-N junction, the typical forward voltage is approximately 0.7 volts. This value is a characteristic of silicon material and is essential for the proper functioning of silicon-based electronic components.

Correct Option Analysis:

The correct option is:

Option 4: 0.7 V

This option correctly identifies the typical forward voltage of a silicon P-N junction. At approximately 0.7 volts, the potential barrier is sufficiently reduced to allow a significant current to flow through the junction, enabling the device to conduct in the forward bias condition.

Additional Information

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

Option 1: 1.2 V

This option overestimates the forward voltage of a silicon P-N junction. While 1.2 V might be relevant for other types of semiconductor materials or specific high-power applications, it is not typical for standard silicon P-N junctions.

Option 2: 2.3 V

This value is significantly higher than the typical forward voltage for a silicon P-N junction. A forward voltage of 2.3 V is more characteristic of light-emitting diodes (LEDs) made from materials like gallium arsenide or gallium nitride, not silicon.

Option 3: 0.3 V

This option underestimates the forward voltage of a silicon P-N junction. A forward voltage of 0.3 V is typical for germanium P-N junctions, not silicon. Germanium-based devices have a lower potential barrier compared to silicon.

Conclusion:

Understanding the typical forward voltage of a silicon P-N junction is essential for designing and analyzing electronic circuits that use silicon-based components. The correct forward voltage of approximately 0.7 V ensures that the device operates efficiently and reliably in its intended applications. Other options provided either overestimate or underestimate this value, highlighting the importance of knowing the specific characteristics of the semiconductor material in use.

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.

Ans.(3)

Sol.

At equilibrium in a p-n junction, the forward diffusion current in the depletion region is balanced with a reverse drift current, so that the net current is zero.

In a pn junction, if the doping concentration increases, the recombination of electrons and holes increases, thereby increases the voltage across the barrier.

\(V = \frac{{KT}}{q}{\rm{ln}}\left( {\frac{{{N_a}{N_d}}}{{n_i^2}}} \right)\)

Explanation:

Peak inverse voltage:

The peak inverse voltage rating of a diode may be defined as the maximum value of the reverse voltage that a PN-junction (or diode) can withstand without damaging (or destruction).

When the voltage applied to a diode is more than PIV, it is likely to result in breakdown at the junction.

Important Points

PIV for different rectifiers is shown below:

• Half Wave rectifier: Vm
• Full Wave centre tap rectifier: 2Vm
• Full Wave Bridge rectifier: Vm​

A diode can have two different capacitances namely Junction capacitance and diffusion capacitance.

Junction Capacitance:

It is dominant when the diode is reverse biased and is the result of the charge stored in the Depletion layer.

The junction capacitance is given by the formula:

\({C_j} = \frac{{A\epsilon}}{W}\)   ---(1)

W = Width of the depletion region calculated as:

\(W= \sqrt {\frac{{2\epsilon_s(V_{bi}-V_a)(N_a+N_d)}}{N_aN_d}} \)   ---(2)

V_a = Applied Voltage

N_a and N_d are the acceptor and donor doping concentrations respectively.

Substituting the above in Equation (1), we get:

\(C=\frac{A\epsilon}{ \sqrt {\frac{{2\epsilon_s(V_{bi}-V_a)(N_a+N_d)}}{N_aN_d}} }\)

\(C=\sqrt {\frac{{{A^2\epsilon_s}{N_A}{N_D}}}{{2\left( {{V_{bi}} - {V_a}} \right)\left( {{N_A} + {N_D}} \right)}}} \)

∴ The junction capacitance of a p-n junction depends on both the doping concentration and the applied voltage.

Diffusion Capacitance:

It is dominant when the diode is forward and is the result of storage charges when forward biased.

The diffusion capacitance is given by the formula:

\({C_d} = \frac{{{I_{DQ}}\times\tau }}{{η {V_T}}}\)

IDQ = Quiescent current of the diode.

τ = Minority carrier lifetime

VT = Thermal voltage

η = Constant = 2 for silicon and 1 for germanium

• The main difference between the conventional rectifier and bridge rectifier is that it produces almost double the Peak inverse voltage as a full-wave center-tapped transformer rectifier using the same secondary voltage.
• The advantage of using this circuit is that no center-tapped transformer is required.
• In center-tapped rectifier each diode uses only one-half of the transformer secondary voltage, so the DC output is comparatively small, also it is difficult to locate the center-tap on the secondary winding of the transformer and the diodes used must have high Peak-inverse voltage.

Between the peak point and the valley point of the tunnel diode, there is a negative resistance region.

Tunnel diode:

• It is a highly doped PN Junction diode, used for low voltage high-frequency switching applications.
• It works on the tunneling principle.
• When compared to a normal p-n junction diode, the tunnel diode has a narrow depletion width.
• In normal forward-biased operation, it exhibits the “Negative resistance region” as shown:

• The negative differential resistance in their operation, allows them to be used as oscillators, amplifiers, and switching circuits.
• Their low capacitance allows them to function at microwave frequencies.
• Tunnel diodes are not good rectifiers, as they have relatively high leakage current when reverse biased.

Concept:

The dynamic resistance can be defined from the I-V characteristic of a diode in forward bias. It is defined as the ratio of a small change to voltage to a small change in current, i.e.

\({r_d} = \left( {\frac{{{\rm{\Delta }}V}}{{{\rm{\Delta }}I}}} \right) = \frac{{\eta {V_T}}}{I}\)

VT = Thermal voltage

I = Bias current

\(V_T = \frac{kT}{q}\)

V_T ∝ T

∴ The dynamic resistance of the diode is directly proportional to the temperature. 

The dynamic resistance is given by the inverse of the slope of i-v characteristics as shown: 

Zener Breakdown:

• The breakdown that occurs in reverse bias conditions in a narrow junction diode is known as Zener breakdown.
• In a heavily doped PN junction, under high reverse voltage (around 5 V), the electrons tunnel through the depletion region due to extremely small depletion region width.
• Even though the electric field is very high the electrons are able to tunnel through due to very thin or narrow depletion regions.
• This causes high reverse current to flow even in reverse bias. The junction is not destroyed can be used as a normal PN diode in forward bias.

Avalanche Breakdown:

• In thinly doped PN junction diodes Zener effect is not possible, but under very high reverse voltage (< 8-10 V) the high electric fields (lower than Zener diode) cause the electrons to accelerate.
• The accelerated electrons break free more electrons (called impact ionization) this newly free electron, in turn, create more charge carriers causing a kind of avalanche effect.
• This results in a huge reverse current, but the junction is destroyed to this effect.

Important difference between Avalanche and Zener breakdown:

Tunnel Diode:

• Symbol:

• It is a highly doped PN Junction diode, used for low voltage high-frequency switching applications.
• It works on the tunneling principle.
• It has a high charge carrier velocity.
• When compared to a normal p-n junction diode, the tunnel diode has a narrow depletion width.
• In normal forward-biased operation, it exhibits the “Negative resistance region” as shown:

• The negative differential resistance in their operation, allows them to be used as oscillators, amplifiers, and switching circuits.
• Their low capacitance allows them to function at microwave frequencies.
• Microwave frequencies range between 109 Hz (1 GHz) to 1000 GHz
• Tunnel diodes are not good rectifiers, as they have relatively high leakage current when reverse biased.

Reverse saturation current:

• The reverse saturation current of the diode increases with an increase in the temperature.
• The rise is 7% /°C for both germanium and silicon and approximately doubles for every 10°C rise in temperature.

Mathematically if the reverse saturation current is I₀₁ at a temperature T₁ and I₀₂ at temperature T₂, then: 

I₀₂ = I₀₁ 2^{(T₂-T₁)/10}

The reverse saturation current is given by:

\({I_o} = A.q.n_i^2\left[ {\frac{{{D_p}}}{{{L_P}{N_D}}} + \frac{{{D_n}}}{{{L_n}{N_A}}}} \right]\)

\({I_o} \propto n_i^2\;\)

Also,

\(n_i^2=N_cN_ve^{-{\frac{E_g}{KT}}}\)

\(n_i^2 \propto {e^{ - \frac{{{E_g}}}{{{KT}}}}}\;\)

The diode for which Eg/KT will be more will have less ni and subsequently will have less reverse saturation current.

• In forward bias, the resistance of a PN junction is of the order of Ω, approximately 100 Ω
• Reverse bias PN junction has a resistance of the order of MΩ 

• The ideal diode will acts as a short circuit in forward bias and open circuit in reverse bias condition.
• Hence the forward resistance of the ideal diode is zero and reverse resistance of the ideal diode is infinity.