CUET UG Physics Semiconductors p-n Junction Diode ForwardReverse Bias Rectifier Circuits

Must‑Know (15–20 detailed bullets)

• In a p-n junction diode, forward bias reduces the width of the depletion region; for example, applying +2V to p-side and 0V to n-side in a Si diode shrinks the barrier.
• Reverse bias increases the depletion width; e.g., connecting –5V to p-side and 0V to n-side widens the barrier and enhances electric field.
• The potential barrier for a silicon p-n junction is approximately 0.7 V at room temperature; for germanium, it is 0.3 V (verify from NCERT).
• In forward bias, current flows due to majority carriers: holes from p-side and electrons from n-side diffusing across the junction.
• In reverse bias, only a small leakage current flows due to minority carriers; this is typically in microamperes (µA) for Si diodes.
• Breakdown in reverse bias occurs at a specific voltage called the breakdown voltage; Zener diodes are designed to operate in this region.
• Diode current is given by the diode equation: ( I = I_0 \left( e^{\frac{eV}{kT}} - 1 \right) ), where ( I_0 ) is reverse saturation current.
• At room temperature (300 K), ( \frac{kT}{e} \approx 26 \, \text{mV} ), used in diode equation calculations (verify from NCERT).
• A half-wave rectifier uses one diode and conducts only during the positive half-cycle of AC input.
• The ripple factor for a half-wave rectifier is 1.21, indicating high pulsation in output DC.
• Full-wave rectifier uses two diodes (center-tapped transformer) or four diodes (bridge), conducting in both half-cycles.
• Ripple factor for full-wave rectifier is 0.48, significantly lower than half-wave.
• Efficiency of half-wave rectifier is about 40.6%; full-wave rectifier has efficiency of 81.2%.
• In a bridge rectifier, during positive half-cycle, diodes D1 and D3 conduct; during negative half-cycle, D2 and D4 conduct.
• No external voltage is required for diffusion current in p-n junction; it results from concentration gradient of charge carriers.
• Drift current in p-n junction is due to the electric field in the depletion region and balances diffusion current at equilibrium.
• The V-I characteristic curve of a diode shows exponential rise in current after knee voltage (~0.7 V for Si).
• In reverse bias, current remains nearly constant at ( -I_0 ) until breakdown, regardless of voltage increase.
• Rectifiers convert AC to pulsating DC; a filter (like capacitor) is needed to smooth the output.
• The peak inverse voltage (PIV) for a half-wave rectifier equals the peak AC voltage ( V_m ); for full-wave center-tapped, PIV = ( 2V_m ).

Difficulty Level

Intermediate — requires understanding of biasing effects, circuit behavior, and numerical parameters like PIV and ripple factor, but avoids advanced semiconductor physics.

Common CUET Traps (3 bullets)

• Trap: Assuming current flows in reverse bias due to majority carriers. Avoid: Only minority carriers contribute to small reverse saturation current; majority carriers are blocked.
• Trap: Confusing PIV values between half-wave and full-wave rectifiers. Avoid: Half-wave PIV = ( V_m ); center-tapped full-wave PIV = ( 2V_m ); bridge rectifier PIV = ( V_m ).
• Trap: Thinking rectifier efficiency exceeds 100% due to step-down transformers. Avoid: Efficiency is DC output power divided by AC input power; maximum theoretical is 81.2% for full-wave.

Practice MCQs (5 questions)

Q1. What is the typical barrier potential for a silicon p-n junction at room temperature?
A. 0.1 V
B. 0.3 V
C. 0.7 V
D. 1.0 V
Answer: C
Explanation: The barrier potential for silicon is approximately 0.7 V.
Why others fail: Option B (0.3 V) is correct for germanium, a common mix-up.

Q2. In a full-wave bridge rectifier, what is the peak inverse voltage (PIV) across each non-conducting diode if the peak AC voltage is ( V_m )?
A. ( V_m )
B. ( 2V_m )
C. ( V_m / 2 )
D. ( \sqrt{2} V_m )
Answer: A
Explanation: In bridge rectifier, PIV across each diode is ( V_m ).
Why others fail: Option B is PIV for center-tapped full-wave, often confused with bridge.

Q3. Which of the following best describes the current in a p-n junction under reverse bias?
A. Increases linearly with voltage
B. Remains constant at ( I_0 ) until breakdown
C. Decreases exponentially
D. Becomes zero immediately
Answer: B
Explanation: Reverse current is nearly constant at reverse saturation current ( I_0 ) until breakdown.
Why others fail: Option A seems intuitive but is false; reverse current does not increase linearly.

Q4. The ripple factor of a half-wave rectifier is:
A. 0.48
B. 1.0
C. 1.21
D. 2.2
Answer: C
Explanation: The ripple factor for half-wave rectifier is 1.21.
Why others fail: Option A (0.48) is for full-wave, a frequent confusion.

Q5. A p-n junction diode is forward biased. Which of the following occurs?
A. Depletion region widens and resistance increases
B. Depletion region narrows and resistance decreases
C. Barrier potential increases to 1.0 V
D. Only minority carriers contribute to current
Answer: B
Explanation: Forward bias reduces depletion width and lowers resistance, allowing current flow.
Why others fail: Option A describes reverse bias, a key conceptual trap.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ Si p-n junction barrier voltage = 0.7 V; Ge = 0.3 V.
• ⚠️ Forward bias: p connected to +ve, n to –ve terminal.
• ⚠️ Reverse bias: p to –ve, n to +ve — increases depletion width.
• ⚠️ Diode equation: ( I = I_0 (e^{eV/kT} - 1) ).
• ⚠️ ( kT/e ) at 300 K ≈ 26 mV — use in exponential calculations.
• ⚠️ Knee voltage ≈ barrier potential — current rises sharply beyond it.
• ⚠️ Reverse saturation current ( I_0 ) is in µA for Si, depends on temperature.
• ⚠️ No current in ideal diode below knee voltage in forward bias.
• ⚠️ Half-wave rectifier output frequency = input AC frequency.
• ⚠️ Full-wave rectifier output frequency = 2 × input frequency.
• ⚠️ Half-wave rectifier efficiency = 40.6%.
• ⚠️ Full-wave rectifier efficiency = 81.2%.
• ⚠️ Ripple factor: half-wave = 1.21, full-wave = 0.48.
• ⚠️ PIV: half-wave = ( V_m ), center-tapped full-wave = ( 2V_m ), bridge = ( V_m ).
• ⚠️ Bridge rectifier uses 4 diodes; no center tap needed.
• ⚠️ In bridge rectifier, two diodes conduct in series during each half-cycle.
• ⚠️ Diffusion current dominates in forward bias; drift in reverse.
• ⚠️ At equilibrium, diffusion current = drift current in p-n junction.
• ⚠️ Rectifiers produce pulsating DC — capacitor filter smoothens it.
• ⚠️ Mnemonic: “PIV Bridge = ( V_m )” — “Bridge is Best, PIV is Least”.