CUET UG Physics Booster Test 3-Diode Applications and Circuits -

🚀

CUET UG Physics Booster Test 3-Diode Applications and Circuits

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3-Diode Applications and Circuits

1 / 20

QUESTION 1 OF 20

Match List I (Physical Region/Component) with List II (Operational Characteristic)

List I	List II
1. p-side metallic contact	i. Acts as the conventional cathode connection
2. n-side metallic contact	ii. Contains immobile ion-cores devoid of charge carriers
3. Depletion region	iii. Acts as the conventional anode connection
4. p-n interface	iv. The metallurgical junction formed during wafer doping

Answer Review

1 Match List I (Physical Region/Component) with List II (Operational Characteristic)

List I	List II
1. p-side metallic contact	i. Acts as the conventional cathode connection
2. n-side metallic contact	ii. Contains immobile ion-cores devoid of charge carriers
3. Depletion region	iii. Acts as the conventional anode connection
4. p-n interface	iv. The metallurgical junction formed during wafer doping

A a–i, b–iii, c–ii, d–iv

B a–iii, b–i, c–ii, d–iv

C a–ii, b–iv, c–i, d–iii

D a–iv, b–ii, c–iii, d–i

💡 Short Explanation

p-side → anode n-side → cathode Depletion region contains immobile ions Interface forms the metallurgical junction

✅ Correct Answer Explanation

Matching: a → iii b → i c → ii d → iv

❓ Why Other Options Are Incorrect

They incorrectly assign anode/cathode roles or depletion region properties.

🎯 Strategy

Used

Component Identification

Final Logic:

Only Option B correctly matches all pairs.

🧠 Memory Trick

"P = Positive Side (Anode), N = Negative Side (Cathode)."

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

2 Which interpretation regarding the diode symbolic arrow is conceptually accurate?

A It traces the physical drift of electrons against the electric field.

B It points from the n-side to the p-side to indicate barrier potential polarity.

C It indicates the conventional direction of minority carrier injection from n to p.

D It indicates the conventional direction of current when the diode allows significant minority carrier injection under forward bias.

💡 Short Explanation

The arrow indicates conventional current direction.

✅ Correct Answer Explanation

Under forward bias, minority carrier injection becomes significant and current flows. The arrow represents this conventional current direction.

❓ Why Other Options Are Incorrect

A: Arrow does not represent electron drift.
B: Not used for barrier polarity.
C: Arrow indicates total conventional current, not specifically minority carrier injection.

🎯 Strategy

Used

Symbol Interpretation

Final Logic:

Arrow → conventional current direction.

🧠 Memory Trick

"Arrow Follows Conventional Current."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diode Symbol

3 Statements regarding current generation mechanisms
1. Forward current primarily results from the diffusion of injected minority carriers driven by a concentration gradient.
2. Reverse current primarily results from the drift of minority carriers swept across the junction by the electric field.
3. The magnitude of drift current under reverse bias is fundamentally limited by the applied voltage, not carrier concentration.
4. Drift current exists under forward bias but is negligible compared to the diffusion current.

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

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

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

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

💡 Short Explanation

Forward current → diffusion current. Reverse current → drift current. Reverse current depends on minority carriers.

✅ Correct Answer Explanation

Statements A, B, and D are correct. Reverse saturation current is limited by minority carrier concentration, not mainly by voltage.

❓ Why Other Options Are Incorrect

C: Incorrect because carrier concentration determines reverse saturation current.

🎯 Strategy

Used

Carrier Transport Analysis

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Forward = Diffusion, Reverse = Drift."

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

4 Correct statements about plotting diode characteristics
1. A rheostat or potentiometer is employed to systematically change the applied bias voltage.
2. Reversing the battery polarity demands switching the ammeter from mA to μA to capture the scale accurately.
3. The forward bias characteristic shows an exponential current rise in the first quadrant.
4. The reverse bias characteristic shows a sharply increasing current at 0.7 V.

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

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

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

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

💡 Short Explanation

Forward current is measured in mA. Reverse current is measured in μA. Forward characteristic is exponential.

✅ Correct Answer Explanation

A rheostat controls voltage, forward current rises exponentially, and reverse current is usually very small, requiring a microammeter.

❓ Why Other Options Are Incorrect

D: Reverse current does not rise sharply at 0.7 V; breakdown occurs at much larger reverse voltages.

🎯 Strategy

Used

Experimental Analysis

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"Forward → mA, Reverse → μA."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Experimental V-I Characteristics

5 Incorrect statement about threshold voltage and barrier height

A The value of threshold voltage is dependent on the semiconductor material used.

B Below the threshold voltage, only a small number of uppermost-energy carriers can cross the junction.

C Increasing the applied forward voltage significantly reduces the barrier height, facilitating the exponential current increase.

D The threshold voltage strictly equals 0 V for both silicon and germanium due to ideal diode characteristics.

💡 Short Explanation

Real diodes have non-zero threshold voltages.

✅ Correct Answer Explanation

Typical threshold voltages are approximately 0.2 V for Ge and 0.7 V for Si.

❓ Why Other Options Are Incorrect

B, C, A: Correct descriptions.

🎯 Strategy

Used

Fact Verification

Final Logic:

Threshold voltage is not zero.

🧠 Memory Trick

"Ge = 0.2 V, Si = 0.7 V."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Threshold Voltage

6 Before the cut-in voltage is reached, the effective potential barrier height obstructing majority carriers under forward bias V is represented by:

A (V₀ + V)

B (V₀ − V)

C (V − V₀)

D V₀ / V

💡 Short Explanation

Forward bias opposes the built-in potential.

✅ Correct Answer Explanation

\(Barrier Height=V_{0}-V\) Forward bias reduces the barrier.

❓ Why Other Options Are Incorrect

They do not represent the forward-bias barrier relationship.

🎯 Strategy

Used

Formula Recall

Final Logic:

\(V_{effective}=V_{0}-V\)

🧠 Memory Trick

"Forward Bias Reduces Barrier."

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

7 Under equilibrium, the reverse saturation current reflects a balance heavily favoring ________ current and its magnitude is typically proportional to ________.

A Diffusion, 10⁻³ A

B Drift, 10³ A

C Drift, 10⁻⁶ A

D Diffusion, 10⁰ A

💡 Short Explanation

Reverse saturation current is a drift current. Typical magnitude is microampere.

✅ Correct Answer Explanation

Minority carriers drift across the junction due to the electric field, producing a reverse current of order μA.

❓ Why Other Options Are Incorrect

They use incorrect current mechanisms or unrealistic magnitudes.

🎯 Strategy

Used

Current Mechanism Recall

Final Logic:

Reverse saturation current is drift current in μA range.

🧠 Memory Trick

"Reverse → Drift → μA."

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

8 Statements regarding the mechanism at breakdown voltage
1. The critical reverse bias voltage forces a sharp, sudden increase in reverse current.
2. The phenomenon suppresses the flow of electrons from n to p completely.
3. The device power dissipation spikes rapidly, threatening destruction through overheating.
4. General-purpose diodes are purposefully rated to operate safely within this zone.

A (A), (C)

B (B), (C)

C (C), (D)

D (A), (D)

💡 Short Explanation

Breakdown causes a sudden increase in current. Excessive heating may destroy the diode.

✅ Correct Answer Explanation

At breakdown, reverse current rises sharply and can damage the junction if not externally limited.

❓ Why Other Options Are Incorrect

B: Current increases rather than being suppressed.
D: Ordinary diodes are not intended for breakdown operation.

🎯 Strategy

Used

Breakdown Analysis

Final Logic:

A and C are correct.

🧠 Memory Trick

"Breakdown = Huge Current + Heat."

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

9 Because the V-I characteristic of a diode is non-linear, the dynamic resistance:

A is mathematically uniform regardless of the operating bias voltage.

B is fundamentally defined as the ratio of a small change in voltage to a small change in current around an operating point.

C decreases exponentially as the reverse bias voltage approaches breakdown.

D is calculated exclusively using the static V/I ratio.

💡 Short Explanation

\(r_{d}=\frac{\Delta V}{\Delta I}\)

✅ Correct Answer Explanation

Dynamic resistance is determined locally around an operating point because the V-I curve is non-linear.

❓ Why Other Options Are Incorrect

A: It changes with operating point.
C: Not generally true.
D: Static resistance and dynamic resistance are different.

🎯 Strategy

Used

Definition Recall

Final Logic:

\(r_{d}=\frac{\Delta V}{\Delta I}\)

🧠 Memory Trick

"Dynamic = Small Change Resistance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dynamic Resistance

10 Referring to the data calculated in Example 14.4, what is the approximate ratio of the reverse-bias dynamic resistance (at −10 V) to the forward-bias dynamic resistance (at 15 mA)?

A 1.0 × 10⁻⁶

B 1.0 × 10⁶

C 1.0 × 10³

D 1.0 × 10⁻³

💡 Short Explanation

Reverse dynamic resistance is enormously larger than forward dynamic resistance.

✅ Correct Answer Explanation

From NCERT Example 14.4: \(r_{reverse}\approx {10}^{7}\Omega r_{forward}\approx 10\Omega \frac{r_{reverse}}{r_{forward}}=\frac{{10}^{7}}{10}={10}^{6}\)

❓ Why Other Options Are Incorrect

They do not match the calculated resistance ratio.

🎯 Strategy

Used

Numerical Ratio Calculation
Final Logic
\(\frac{r_{reverse}}{r_{forward}}={10}^{6}\)

🧠 Memory Trick

"Reverse Resistance is Millions of Times Larger."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dynamic Resistance and V-I Characteristics

11 Question: Correct statements on the physical limits of unidirectional flow
1. Forward biasing drops the effective barrier to (V₀ - V), facilitating majority carrier flow.
2. Reverse biasing raises the effective barrier to (V₀ + V), suppressing majority carrier flow.
3. The reverse saturation drift current remains non-zero due to thermally generated minority carriers.
4. A perfect unidirectional diode has zero resistance in reverse bias.

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

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

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

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

💡 Short Explanation

Forward bias lowers the barrier potential. Reverse bias increases the barrier potential. Minority carriers cause reverse saturation current.

✅ Correct Answer Explanation

Forward bias reduces the effective barrier height from \(V_{0}\)to \(\left(V_{0},\ V\right)\), allowing majority carriers to cross the junction more easily. Reverse bias increases the barrier to \(\left(V_{0},\ V\right)\), preventing majority carrier flow. However, thermally generated minority carriers continue to drift across the junction, producing a small reverse saturation current. Therefore statements A, B and C are correct.

❓ Why Other Options Are Incorrect

Option D: Reverse resistance is extremely high, not zero.
Options containing D become incorrect.

🎯 Strategy

Used

Elimination

Application: Identify the incorrect statement and eliminate options containing it.

Final Logic: D is false, therefore only option A remains.

🧠 Memory Trick

Forward ↓ Barrier, Reverse ↑ Barrier

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Rectification and Unidirectional Conduction

12 Question: Match List I (Signal Stage) with List II (Waveform Profile)

List I	List II
1. Input signal at transformer primary	a. Continuous half sinusoids
2. Output across RL in half-wave rectifier	b. Alternating full sinusoids
3. Output across RL in full-wave rectifier	c. Steady DC with exponential discharge ripples
4. Output across capacitor filter	d. Spaced half sinusoids with zero-voltage gaps

A a-ii, b-iv, c-i, d-iii

B a-i, b-ii, c-iii, d-iv

C a-iii, b-i, c-iv, d-ii

D a-ii, b-iii, c-i, d-iv

💡 Short Explanation

Input is AC sinusoidal. Half-wave gives separated pulses. Full-wave gives continuous pulses. Capacitor filtering produces nearly DC output.

✅ Correct Answer Explanation

Input AC consists of alternating full sinusoids. Half-wave rectification produces spaced half sinusoids. Full-wave rectification converts both half cycles into continuous half sinusoids. A capacitor filter smooths the output into DC with small ripple.

❓ Why Other Options Are Incorrect

Incorrect waveform-stage matching.
Capacitor filter output is not AC.
Half-wave and full-wave outputs are interchanged.

🎯 Strategy

Used

Option Grouping

Application: Match each waveform with its physical stage.

Final Logic: Only option A correctly matches all stages.

🧠 Memory Trick

AC → Half-wave → Full-wave → Filtered DC

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Rectification of AC

13 Question: Incorrect statement about series load operation in a half-wave rectifier
1. The series load receives current only during alternating half-cycles of the input source.
2. The reverse saturation current during the non-conducting half-cycle is considered mathematically negligible.
3. The reverse breakdown voltage of the diode must be securely below the peak AC voltage to ensure safe functioning.
4. The output voltage waveform tracks the input waveform solely when the diode's anode is positive relative to its cathode.

A (A)

B (B)

C (C)

D (D)

💡 Short Explanation

Breakdown voltage must be greater than peak reverse voltage. Otherwise the diode may get damaged.

✅ Correct Answer Explanation

For safe operation: \(V_{BR}>V_{peak}\) Therefore breakdown voltage should be above the peak reverse voltage. Statement C incorrectly says it should be below.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Concept Verification

Application: Apply diode safety condition.

Final Logic: Breakdown voltage must exceed peak reverse voltage.

🧠 Memory Trick

Breakdown > Peak Voltage

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Half-Wave Rectifier

14 Question: While the output of a half-wave rectifier is unidirectional, it mathematically represents:
1. a constant integration of the input AC frequency.
2. purely DC power without any AC frequency components.
3. pulses of the shape of half sinusoids.
4. an exact mirror of the reverse bias drift current.

A (A)

B (B)

C (C)

D (D)

💡 Short Explanation

Only one half-cycle is transmitted. Output remains pulsating.

✅ Correct Answer Explanation

A half-wave rectifier allows only one half-cycle of the AC input. The resulting output consists of half-sinusoidal pulses in one direction.

❓ Why Other Options Are Incorrect

A: Not an integration process.
B: Output is not pure DC.
D: Unrelated to rectification.

🎯 Strategy

Used

Concept Identification

Application: Recognize the waveform produced by half-wave rectification.

Final Logic: Half-wave rectifier outputs half-sinusoidal pulses.

🧠 Memory Trick

Half-Wave = Half Sine

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Half-Wave Rectifier Output

15 Question: In the centre-tapped secondary configuration, if the upper terminal A voltage phase angle is positive, the lower terminal B phase angle is _________, enforcing a _________ state upon diode D2.

A Positive, Forward-biased

B Negative, Reverse-biased

C Negative, Forward-biased

D Positive, Reverse-biased

💡 Short Explanation

Ends of secondary are 180° out of phase. One diode conducts while the other remains reverse biased.

✅ Correct Answer Explanation

When A is positive, B becomes negative. Hence D1 conducts and D2 remains reverse biased.

❓ Why Other Options Are Incorrect

They violate transformer phase relationships.

🎯 Strategy

Used

Elimination

Application: Apply phase opposition principle.

Final Logic: Positive at A means negative at B.

🧠 Memory Trick

Opposite Ends → Opposite Phases

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Full-Wave Rectifier

16 Question: Though each diode in a full-wave dual diode configuration rectifies strictly for half the total cycle...

A the alternating conduction states seamlessly combine at the common junction to supply uninterrupted continuous output current.

B they completely nullify each other's output across the load resistor.

C they amplify the total secondary voltage by a factor of two.

D they block the secondary transformer's current altogether.

💡 Short Explanation

In a full-wave rectifier, D₁ conducts during one half-cycle and D₂ conducts during the other half-cycle, ensuring current through the load in the same direction during both halves.

✅ Correct Answer Explanation

The two diodes conduct alternately. Their outputs combine across the load resistor to produce a continuous unidirectional current, resulting in full-wave rectification.

❓ Why Other Options Are Incorrect

(B): Outputs add constructively; they do not cancel.
(C): Rectification does not amplify voltage.
(D): The transformer current is not blocked.

🎯 Strategy

Used

Concept-Based Elimination

🧠 Memory Trick

"D₁ then D₂ → Continuous DC"

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Full-Wave Rectifier

17 Question: Capacitor dynamic response statements
1. When the rectified pulse rises, the capacitor charges rapidly toward the peak AC voltage.
2. When the rectified pulse drops, the diode blocks backward flow, forcing the capacitor to discharge exclusively through RL.
3. The resulting output voltage traces the peak of the pulse, then decays exponentially until the next pulse recharges it.
4. The capacitor charges through the load RL and discharges through the diode.

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

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

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

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

💡 Short Explanation

A capacitor charges quickly to the peak value and discharges slowly through the load between successive peaks.

✅ Correct Answer Explanation

Statements A, B and C correctly describe capacitor-filter operation. Statement D is incorrect because the capacitor charges through the diode and discharges through the load resistor RL.

❓ Why Other Options Are Incorrect

D is incorrect.
Any option containing D becomes invalid.

🎯 Strategy

Used

Statement Verification

🧠 Memory Trick

Charge through diode → Discharge through load

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Capacitor Filter

18 Question: Correct statements regarding series inductance filtering
1. An inductor opposes variations in current, naturally smoothing the pulsating output.
2. It is placed in series with the load resistance RL.
3. It effectively filters out the high-frequency AC ripple components from the rectified wave.
4. It functions by short-circuiting the DC component directly to ground.

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

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

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

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

💡 Short Explanation

An inductor opposes changes in current and smooths ripple when connected in series with the load.

✅ Correct Answer Explanation

Statements A, B and C are correct. Inductors oppose rapid current variations and reduce AC ripple. Statement D is incorrect because an inductor does not short-circuit DC to ground.

❓ Why Other Options Are Incorrect

D is false.
Options containing D are eliminated.

🎯 Strategy

Used

Elimination Method

🧠 Memory Trick

Inductor blocks ripple, passes DC

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Inductor Filter

19 Question: If an electronic designer wishes to analyze the exponential voltage decay across the load of a filtered rectifier, the critical rate-of-fall parameter they must calculate is:

A RL / C

B C / RL

C C RL

D 1 / (C RL)

💡 Short Explanation

The capacitor discharge rate depends on the RC time constant.

✅ Correct Answer Explanation

The time constant is \(\tau =R_{L}C\) A larger value of \(R_{L}C\) produces a slower voltage decay and less ripple.

❓ Why Other Options Are Incorrect

A: Incorrect expression.
B: Incorrect expression.
D: Reciprocal of the time constant.

🎯 Strategy

Used

Formula Recall

🧠 Memory Trick

Time Constant = R × C

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Filter Circuits and Time Constant

20 Question: What is the typical order of magnitude of the reverse saturation current observed inside the junction when evaluating stable diode leakage?

A 1.0 × 10⁰ A

B 1.0 × 10⁻³ A

C 1.0 × 10⁻⁶ A

D 1.0 × 10³ A

💡 Short Explanation

Reverse saturation current is extremely small and generally lies in the microampere range.

✅ Correct Answer Explanation

The reverse saturation current arises from minority carriers and is typically of the order of \({10}^{-6} A\) for ordinary p-n junction diodes.

❓ Why Other Options Are Incorrect

A: Too large.
B: Milliampere range is much larger than leakage current.
D: Physically unrealistic.

🎯 Strategy

Used

Order-of-Magnitude Estimation

🧠 Memory Trick

Reverse Leakage → Microampere (μA)

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

CUET UG Physics Booster Test 3-Diode Applications and Circuits

Test Complete!

Answer Review