Applications of Semiconductor Devices
NIOS · Class 12 · Physics
Flashcards for Applications of Semiconductor Devices — NIOS Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Get startedWhat is rectification? Why is it needed in everyday electronics?
Answer
Rectification is the process of converting an alternating current (AC) signal into a direct current (DC) signal. Why needed: • Most household appliances (phones, laptops, TVs) require DC to operate.
Explain how a p-n junction diode acts as a half-wave rectifier. What happens during positive and negative half cycles?
Answer
Half-Wave Rectifier Working: • AC input is applied to the diode through a step-down transformer. Positive Half Cycle (0 to T/2): → Diode is FORWARD BIASED. → Current flows through load resistance R_…
A half-wave rectifier has a peak input voltage of 15.7 V. Calculate the DC output voltage.
Answer
Given: V_m (peak voltage) = 15.7 V Formula for half-wave rectifier: V_dc = V_m / π Step 1: Substitute values V_dc = 15.7 / π V_dc = 15.7 / 3.14 Step 2: Calculate V_dc = 5 V Answer: V_dc = 5 V Not…
State the formula for DC output voltage and DC current in a half-wave rectifier.
Answer
Key Formulas – Half-Wave Rectifier: 1. DC Voltage: V_dc = V_m / π 2. DC Current: I_dc = V_dc / R_L = V_m / (π × R_L) Where: • V_m = Peak (maximum) AC voltage (in Volts) • R_L = Load resistanc…
What is Peak Inverse Voltage (PIV) of a diode? Why is it important in selecting a diode for a rectifier?
Answer
Peak Inverse Voltage (PIV): • It is the MAXIMUM reverse voltage that a diode can withstand without breaking down. In a half-wave rectifier: • During the negative half cycle, the entire peak AC voltag…
How does a full-wave rectifier work? What is the role of a centre-tapped transformer?
Answer
Full-Wave Rectifier Working: Components: 2 diodes (D1 and D2), centre-tapped step-down transformer, load R_L. Centre-tapped transformer: • Has two identical secondary windings in series with a centr…
A full-wave rectifier has a peak input voltage of 31.4 V and load resistance of 500 Ω. Find the DC output voltage and current.
Answer
Given: V_m = 31.4 V R_L = 500 Ω Formula for full-wave rectifier: V_dc = 2 × V_m / π I_dc = V_dc / R_L Step 1: Calculate V_dc V_dc = 2 × 31.4 / π V_dc = 62.8 / 3.14 V_dc = 20 V Step 2: Calculate I_d…
State the formulas for DC voltage and current in a full-wave rectifier. How do they differ from a half-wave rectifier?
Answer
Full-Wave Rectifier Formulas: 1. DC Voltage: V_dc = 2V_m / π 2. DC Current: I_dc = 2V_m / (π × R_L) Comparison with Half-Wave: | Parameter | Half-Wave | Full-Wave | |-----------|-----------|…
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