CUET UG Physics Booster Test 3-Junction Formation and Physics -

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CUET UG Physics Booster Test 3-Junction Formation and Physics

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3-Junction Formation and Physics

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QUESTION 1 OF 20

Regarding the conversion of a p-Si wafer to create a p-n junction

Answer Review

1 Regarding the conversion of a p-Si wafer to create a p-n junction

A Pentavalent impurities are uniformly distributed throughout the entire wafer volume.

B Only a targeted part of the p-Si wafer is doped with pentavalent impurities to form n-Si.

C The process creates a continuous atomic-level lattice with distinct p and n regions.

D The metallurgical junction is an abrupt physical boundary between two separate joined crystals.

💡 Short Explanation

Only a selected region is doped. A continuous crystal lattice is maintained. The p-n junction forms within the same wafer.

✅ Correct Answer Explanation

During fabrication, a small region of a p-type silicon wafer is doped with pentavalent impurities to form an n-region. This produces a continuous crystal structure containing both p-type and n-type regions.

❓ Why Other Options Are Incorrect

A: Doping is localized, not throughout the entire wafer.
D: A metallurgical junction is not formed by joining separate crystals.

🎯 Strategy

Used

Conceptual Understanding
Final Logic
Only statements B and C correctly describe wafer conversion.

🧠 Memory Trick

"One Crystal, Two Regions."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Formation of p-n Junction

2 If two separate slabs of p-type and n-type semiconductor are physically pressed together

A (B), (C), (D)

B (A), (C), (D)

C (A), (B), (C)

D (A), (B), (D)

💡 Short Explanation

Physical pressing does not create a proper p-n junction. Atomic-level continuity is absent. Charge flow is interrupted.

✅ Correct Answer Explanation

The surface roughness of semiconductor slabs is much larger than atomic spacing, preventing continuous atomic contact. Hence a proper p-n junction cannot be formed by pressing slabs together.

❓ Why Other Options Are Incorrect

Statement A: A functional p-n junction is not formed.

🎯 Strategy

Used

Statement Analysis
Final Logic
Only B, C and D are correct.

🧠 Memory Trick

"Pressed Slabs ≠ Junction."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Metallurgical Junction

3 Correct statements about the dynamic state during p-n junction formation

A (A), (B)

B (A), (C)

C (B), (D)

D (C), (D)

💡 Short Explanation

Initially diffusion dominates. Space-charge region grows during diffusion.

✅ Correct Answer Explanation

At the beginning, diffusion current is large while drift current is negligible. As diffusion continues, the depletion regions expand and an electric field develops.

❓ Why Other Options Are Incorrect

B: Diffusion is driven by concentration gradient, not electric field.
D: Holes leave behind ionized acceptors, which are negative ions.

🎯 Strategy

Used

Carrier Dynamics Analysis
Final Logic
Only A and C are correct.

🧠 Memory Trick

"Diffusion First, Drift Later."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diffusion Process in p-n Junction

4 Match List I with List II regarding current interplay

List I	List II
I. Diffusion current direction (conventional)	P. p to n
II. Drift current direction (conventional)	Q. Diffusion equals drift
III. Equilibrium condition	R. n to p
IV. Initial formation stage	S. Diffusion is large, drift is small

A I-P, II-R, III-Q, IV-S

B I-R, II-P, III-Q, IV-S

C I-P, II-R, III-S, IV-Q

D I-R, II-P, III-S, IV-Q

💡 Short Explanation

Diffusion current: p → n. Drift current: n → p. Equilibrium: currents balance.

✅ Correct Answer Explanation

Matching: I → P II → R III → Q IV → S

❓ Why Other Options Are Incorrect

They incorrectly interchange diffusion and drift current directions.

🎯 Strategy

Used

Matching Concepts
Final Logic
Only Option A correctly matches all terms.

🧠 Memory Trick

"Diffusion Follows Gradient, Drift Follows Field."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diffusion and Drift Currents

5 Incorrect statement regarding the electric field in the depletion layer

A It is established by the immobile ionized donors and acceptors.

B It causes holes on the n-side to drift to the p-side.

C It continues to increase indefinitely as the diffusion process proceeds endlessly.

D Its direction is from the positive space-charge region to the negative space-charge region.

💡 Short Explanation

Equilibrium eventually stops further growth. Electric field reaches a stable value.

✅ Correct Answer Explanation

The electric field increases only until equilibrium is achieved. At equilibrium, drift current balances diffusion current and further growth ceases.

❓ Why Other Options Are Incorrect

A, B, D: Correct descriptions of the depletion-layer electric field.

🎯 Strategy

Used

Conceptual Elimination
Final Logic
Electric field does not increase indefinitely.

🧠 Memory Trick

"Equilibrium Stops Unlimited Growth."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Electric Field in Depletion Layer

6 The drift current is due to the motion of carriers from their ____ side to their ____ side across the junction, and is typically of the order of a few ____.

A majority, minority, mA

B minority, majority, µA

C majority, minority, µA

D minority, majority, mA

💡 Short Explanation

Drift current is caused by minority carriers. It is usually very small.

✅ Correct Answer Explanation

Minority carriers are swept across the junction by the electric field. The resulting drift current is generally of the order of microamperes.

❓ Why Other Options Are Incorrect

Majority carriers mainly contribute to diffusion current.
Drift current is not typically in mA range.

🎯 Strategy

Used

Current Mechanism Analysis
Final Logic
Drift current involves minority carriers and is very small.

🧠 Memory Trick

"Minority Moves, Microamp Flows."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Drift Current

7 If a region has an excess of x ionized donors and y ionized acceptors, the net charge of this region is

A −(x + y)e

B (x + y)e

C (x − y)e

D (y − x)e

💡 Short Explanation

Donors contribute positive charge. Acceptors contribute negative charge.

✅ Correct Answer Explanation

Net charge: \(Q=(+xe)+(-ye)Q=(x-y)e\)

❓ Why Other Options Are Incorrect

They do not correctly account for charge signs.

🎯 Strategy

Used

Charge Calculation
Final Logic
\(Q=(x-y)e\)

🧠 Memory Trick

"Positive Donors Minus Negative Acceptors."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Space-Charge Region

8 When holes diffuse from the p-side to the n-side, they leave behind ionized acceptors which

A are free to move and neutralize the n-side

B create a positive space-charge region on the p-side

C contribute directly to the diffusion current by flowing

D are bonded to surrounding atoms and thus remain immobile

💡 Short Explanation

Acceptor ions are fixed in the lattice. They cannot move.

✅ Correct Answer Explanation

Ionized acceptors remain bonded within the crystal lattice and therefore stay immobile while contributing to the negative space-charge region.

❓ Why Other Options Are Incorrect

A & C: Acceptor ions do not move.
B: They form a negative, not positive, space charge.

🎯 Strategy

Used

Ion Analysis
Final Logic
Ionized acceptors are fixed negative ions.

🧠 Memory Trick

"Acceptor Ions Stay Anchored."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Space-Charge Region

9 Incorrect statement concerning charge depletion

A The space-charge region is practically devoid of mobile charge carriers.

B The depletion process enhances the overall conductivity of the junction boundary.

C It is formed due to the initial movement of electrons and holes across the junction.

D The electric field strength in this region increases as diffusion continues initially.

💡 Short Explanation

Depletion region has very few mobile carriers. Conductivity decreases, not increases.

✅ Correct Answer Explanation

The depletion region is depleted of mobile charge carriers and therefore has high resistance and low conductivity.

❓ Why Other Options Are Incorrect

A, C, D: Correct descriptions of depletion-region formation.

🎯 Strategy

Used

Conceptual Elimination
Final Logic
Depletion reduces conductivity.

🧠 Memory Trick

"Depletion Means Fewer Carriers."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Depletion Region

10 If the depletion region thickness is 0.12 µm and an applied reverse bias increases this thickness by 25%, what will be the new thickness?

A 1.50×10⁻⁷ m

B 1.25×10⁻⁷ m

C 1.50×10⁻⁶ m

D 1.25×10⁻⁶ m

💡 Short Explanation

Increase = 25%

✅ Correct Answer Explanation

Initial thickness: \(0.12 \mu m=1.2\times {10}^{-7}m\) New thickness: \(1.2\times {10}^{-7}\times 1.25=1.5\times {10}^{-7}m\)

❓ Why Other Options Are Incorrect

They arise from incorrect percentage calculations or unit conversion errors.

🎯 Strategy

Used

Numerical Calculation
Final Logic
\(d_{new}=1.50\times {10}^{-7}m\)

🧠 Memory Trick

"Reverse Bias Widens Depletion Layer."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Depletion Layer Thickness

11 The barrier potential tends to prevent the movement of electrons from the ____ region into the ____ region, maintaining a condition of ____.

A p, n, disequilibrium

B n, p, equilibrium

C p, n, saturation

D n, p, depletion

💡 Short Explanation

Electrons naturally diffuse from n-side to p-side. Barrier potential opposes this diffusion. Equilibrium is established.

✅ Correct Answer Explanation

The built-in barrier potential develops due to charge separation near the junction and opposes the diffusion of electrons from the n-region to the p-region. This balance between diffusion and drift currents establishes equilibrium.

❓ Why Other Options Are Incorrect

A: Electrons do not naturally diffuse from p to n.
C: Saturation is not the equilibrium condition described.
D: Depletion is a region, not the resulting condition.

🎯 Strategy

Used

Conceptual Understanding

Application:

Identify the direction of electron diffusion and the role of barrier potential.

Final Logic:

Barrier potential opposes electron flow from n to p, maintaining equilibrium.

🧠 Memory Trick

"Barrier Blocks n → p Diffusion."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Barrier Potential

12 Equilibrium condition statements
1. Net current is zero
2. Drift current magnitude strictly equals diffusion current magnitude
3. Barrier potential reaches its maximum possible value to block all activity
4. The p-material is negative relative to the n-material

A (A), (B), (D)

B (A), (B), (C)

C (B), (C), (D)

D (A), (C), (D)

💡 Short Explanation

Diffusion and drift currents balance. Net current becomes zero. p-side is negative relative to n-side.

✅ Correct Answer Explanation

At equilibrium: Net current is zero. Drift current equals diffusion current in magnitude. The p-region is negative relative to the n-region due to ionized acceptors.

❓ Why Other Options Are Incorrect

Statement C: Barrier potential does not block all activity; drift and diffusion still occur.

🎯 Strategy

Used

Equilibrium Analysis

Application:

Evaluate the characteristics of equilibrium in a p-n junction.

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Equal Currents, Zero Net Flow."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Equilibrium in p-n Junction

13 If the reverse breakdown voltage magnitude is \(V_{br}\), and a user applies a reverse voltage \(V\) that exceeds this breakdown magnitude \(\left(V,\ V_{br}\right)\), the expression for the excess voltage causing sharp current increase is

A \(\left(V_{br},\ V\right)\)

B \(\left(V,\ V_{br}\right)\)

C \(-(V+V_{br})\)

D \(\left(V,\ V_{br}\right)\)

💡 Short Explanation

Breakdown begins when applied voltage exceeds breakdown voltage. Excess voltage equals applied voltage minus breakdown voltage.

✅ Correct Answer Explanation

\(Excess Voltage=V-V_{br}\) This excess reverse voltage is responsible for the rapid increase in reverse current after breakdown.

❓ Why Other Options Are Incorrect

A: Gives a negative value when \(V>V_{br}\).
C & D: Incorrect expressions.

🎯 Strategy

Used

Formula Application

Application:

Compute voltage beyond breakdown.

Final Logic:

\(V_{excess}=V-V_{br}\)

🧠 Memory Trick

"Beyond Breakdown = Applied − Breakdown."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Breakdown

14 When an external voltage is applied, the voltage drop occurs mostly across the depletion region because

A the p-side and n-side have very high resistance due to impurities

B the depletion region has an abundance of free charges

C the resistance of the depletion region is very high compared to the n-side and p-side

D the barrier potential intrinsically blocks voltage from reaching the metallic contacts

💡 Short Explanation

Depletion region lacks mobile carriers. High resistance causes maximum voltage drop.

✅ Correct Answer Explanation

Since the depletion region contains very few free charge carriers, it has a very high resistance compared to the p and n regions. Therefore, most of the applied voltage appears across it.

❓ Why Other Options Are Incorrect

A: p and n regions have relatively low resistance.
B: Depletion region lacks free charges.
D: Not the reason for voltage distribution.

🎯 Strategy

Used

Resistance Analysis

Application:

Determine where voltage drops in a diode.

Final Logic:

Higher resistance causes larger voltage drop.

🧠 Memory Trick

"High Resistance → High Voltage Drop."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Biasing of a p-n Junction

15 Under significant forward bias applied voltage
1. the barrier height is reduced significantly
2. the depletion layer width actively increases
3. only a negligible amount of carriers cross the junction
4. the current increases because more carriers have the required energy

A (A) and

B only

C only

D (B) and

💡 Short Explanation

Forward bias lowers barrier height. More carriers cross the junction. Current increases significantly.

✅ Correct Answer Explanation

A strong forward bias reduces both the barrier height and depletion width. As a result, many more carriers gain enough energy to cross the junction, increasing the current.

❓ Why Other Options Are Incorrect

B: Depletion width decreases, not increases.
C: Carrier crossing becomes substantial.

🎯 Strategy

Used

Forward-Bias Analysis

Application:

Examine the effect of forward voltage.

Final Logic:

A and D are correct.

🧠 Memory Trick

"Forward Bias → Lower Barrier, Higher Current."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Forward Bias

16 Correct statements about carrier dynamics in forward bias
1. The effective barrier height becomes \(V_{0}+V\)
2. If the applied voltage is small, the current is small
3. Carriers in the uppermost energy levels possess enough energy to cross
4. The effective barrier height is reduced

A (A), (B), (C)

B (B), (C), (D)

C (A), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Small forward bias gives small current. Barrier height decreases. High-energy carriers cross the junction.

✅ Correct Answer Explanation

Forward bias reduces the effective barrier: \(V_{effective}=V_{0}-V\) Thus, carriers with sufficient energy can cross the junction, increasing current.

❓ Why Other Options Are Incorrect

Statement A: Correct relation is \(V_{0}-V\), not \(V_{0}+V\).

🎯 Strategy

Used

Barrier Potential Analysis

Application:

Determine the effect of forward bias.

Final Logic:

B, C, and D are correct.

🧠 Memory Trick

"Forward Bias Subtracts the Barrier."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Barrier Height Under Forward Bias

17 Match List I with List II regarding reverse bias physics

List I	List II
I. Effective barrier height	P. Widens
II. Depletion region width	Q. Decreases enormously
III. Diffusion current	R. (V₀ + V)
IV. Drift current	S. Is of the order of a few µA

A I-P, II-R, III-Q, IV-S

B I-R, II-P, III-Q, IV-S

C I-R, II-P, III-S, IV-Q

D I-Q, II-S, III-R, IV-P

💡 Short Explanation

Reverse bias increases barrier height. Depletion region widens. Diffusion current decreases sharply. Drift current remains small.

✅ Correct Answer Explanation

Correct matching: I → R II → P III → Q IV → S

❓ Why Other Options Are Incorrect

Incorrect pairing of reverse-bias effects.

🎯 Strategy

Used

Matching Concepts

Application:

Associate reverse-bias properties correctly.

Final Logic:

Only Option B gives the correct matching.

🧠 Memory Trick

"Reverse Bias: Barrier Up, Width Up."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Bias

18 Incorrect statement about reverse bias

A The direction of applied voltage is the same as the direction of barrier potential

B The effective barrier height suppresses the flow of majority carriers

C The reverse drift current is primarily dependent on the magnitude of the applied voltage

D The drift current is limited by the concentration of minority carriers on either side

💡 Short Explanation

Reverse current is mainly due to minority carriers. It remains nearly constant before breakdown.

✅ Correct Answer Explanation

The reverse drift current depends primarily on the concentration of minority carriers and remains almost independent of reverse voltage until breakdown occurs.

❓ Why Other Options Are Incorrect

A: Correct.
B: Correct.
D: Correct.

🎯 Strategy

Used

Reverse-Bias Current Analysis

Application:

Identify the incorrect statement.

Final Logic:

Reverse current is controlled by minority carriers, not applied voltage.

🧠 Memory Trick

"Before Breakdown, Reverse Current Stays Nearly Constant."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Saturation Current

19 If a diode operating in forward bias injects a minority hole diffusion current of 1.5 mA and a minority electron diffusion current of 3.5 mA, what is the total diode forward current?

A \(5.0\times {10}^{-3} A\)

B \(2.0\times {10}^{-3} A\)

C \(5.0\times {10}^{-6} A\)

D \(2.0\times {10}^{-6} A\)

💡 Short Explanation

Total current is the sum of electron and hole diffusion currents.

✅ Correct Answer Explanation

\(I=I_{h}+I_{e}I=1.5 mA+3.5 mAI=5.0 mAI=5.0\times {10}^{-3}A\)

❓ Why Other Options Are Incorrect

Obtained from incorrect addition or unit conversion.

🎯 Strategy

Used

Direct Calculation

Application:

Add the two diffusion current components.

Final Logic:

\(I=5.0\times {10}^{-3}A\)

🧠 Memory Trick

"Total Current = Electron Current + Hole Current."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Current in a Forward-Biased p-n Junction

20 The injected minority carriers at the junction boundary
1. have a significantly lower concentration compared to locations far from the junction
2. diffuse from the junction edge to the other end of their respective sides
3. give rise to a current whose magnitude is usually in mA
4. remain completely stationary due to the concentration gradient

A (A) and

B only

C (A) and

D only

💡 Short Explanation

Injected minority carriers have maximum concentration near the junction. They diffuse away from the junction. This produces diffusion current in the mA range.

✅ Correct Answer Explanation

After injection, minority carriers are most concentrated near the junction edge. They diffuse toward the opposite end of their respective regions, creating the forward diffusion current.

❓ Why Other Options Are Incorrect

A: Concentration near the junction is higher, not lower.
D: Carriers diffuse and are not stationary.

🎯 Strategy

Used

Carrier Injection Analysis

Application:

Analyze minority-carrier behavior under forward bias.

Final Logic:

B and C are correct.

🧠 Memory Trick

"Injected Carriers Spread Away from the Junction."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Minority Carrier Injection and Diffusion Current

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