Question: Explain why a Zener diode is connected in reverse bias when used as a voltage regulator.

Solution: In reverse bias, the Zener diode enters breakdown at a fixed voltage V_Z. In this region, a large change in reverse current produces only a very small change in voltage. When the input voltage or load current changes, the Zener current adjusts and keeps the output voltage nearly equal to V_Z.

Final Answer: A Zener diode is reverse biased because its breakdown region provides nearly constant voltage V_Z.

Exam Tip: Write “reverse breakdown region” and “constant V_Z” together for full marks.

Question: A 9 V source is connected to a 5 V Zener diode through 200 Ω. If load current is 10 mA, find Zener current.

Solution: Current through series resistor is I_S = (V_in - V_Z) / R_S = (9 - 5) / 200 = 0.02 A = 20 mA. Since I_S = I_L + I_Z, I_Z = 20 mA - 10 mA = 10 mA.

Final Answer: I_Z = 10 mA.

Exam Tip: First find series current, then subtract load current.

Question: In the graph of V_O versus V_in for a Zener regulator, why does V_O become almost horizontal after V_in reaches a certain value?

Solution: Before breakdown, the output follows input approximately. Once the reverse voltage across the Zener reaches V_Z, the diode conducts in breakdown and clamps the load voltage. Further increase in V_in mainly increases current through R_S and the Zener diode, not V_O.

Final Answer: The horizontal part represents regulated output voltage V_O = V_Z.

Exam Tip: A flat V_O curve means regulation.

Question: In a Zener regulator, V_in and R_S are constant. If load resistance R_L decreases, what happens to I_L and I_Z?

Solution: Since V_O is approximately V_Z, load current I_L = V_Z / R_L. When R_L decreases, I_L increases. Series current I_S is nearly fixed because V_in, V_Z and R_S are fixed. Therefore I_Z = I_S - I_L decreases.

Final Answer: I_L increases and I_Z decreases.

Exam Tip: In regulator problems, use I_S = I_L + I_Z.

Question: State one reason why a resistor is connected in series with a Zener diode regulator.

Solution: In breakdown, Zener current can become large. The series resistor drops the excess voltage and limits current to a safe value. Without R_S, the Zener diode may be damaged due to excessive power dissipation.

Final Answer: R_S limits Zener current and prevents damage.

Exam Tip: Mention current limiting and power protection.

Question: Assertion: Zener diode can protect a circuit from overvoltage. Reason: It conducts strongly when reverse voltage exceeds V_Z.

Solution: The assertion is true because a Zener diode connected across a load can bypass excess current when voltage rises above V_Z. The reason is also true and explains the assertion.

Final Answer: Both assertion and reason are true, and reason is the correct explanation.

Exam Tip: In assertion-reason questions, test both statements separately first.

Question: Explain why an LED emits light when it is forward biased.

Solution: In forward bias, electrons from n-side and holes from p-side cross the junction and recombine. In direct band gap semiconductors, the recombination energy is released as photons. Therefore light is emitted.

Final Answer: LED emits light due to radiative recombination of electrons and holes in forward bias.

Exam Tip: Use the term “electron-hole recombination”.

Question: Which LED has larger band gap: red LED or blue LED? Explain.

Solution: Photon energy is E = hc / λ. Blue light has shorter wavelength than red light, so blue photons have higher energy. Since photon energy is approximately equal to band gap, blue LED has a larger band gap.

Final Answer: Blue LED has larger band gap.

Exam Tip: Shorter wavelength means higher photon energy.

Question: An LED emits light of wavelength 500 nm. Find photon energy in eV. Take hc = 1240 eV nm.

Solution: E = hc / λ = 1240 / 500 = 2.48 eV.

Final Answer: 2.48 eV.

Exam Tip: If wavelength is in nm, use hc = 1240 eV nm directly.

Question: The I-V curve of an LED shows a sharp rise after threshold voltage. Give the physical reason.

Solution: Below threshold, the junction barrier prevents significant carrier injection. After forward voltage becomes comparable to the barrier potential, large numbers of electrons and holes cross the junction and recombine. Hence current increases rapidly and light emission becomes appreciable.

Final Answer: Current rises sharply due to strong carrier injection after the junction barrier is overcome.

Exam Tip: Connect threshold voltage with carrier injection.

Question: A 6 V supply is used with an LED of forward drop 2 V and operating current 20 mA. Find suitable series resistance.

Solution: Voltage across resistor = 6 - 2 = 4 V. R = V / I = 4 / 0.02 = 200 Ω.

Final Answer: 200 Ω.

Exam Tip: The resistor carries the same current as the LED.

Question: Give two advantages of LED over an incandescent lamp.

Solution: LED consumes less power and has longer life. It also switches fast, produces less heat and is mechanically more robust.

Final Answer: Low power consumption and long life.

Exam Tip: For application questions, write simple practical advantages.

Question: Why is a photodiode operated in reverse bias?

Solution: In reverse bias, the depletion region becomes wider. Incident light creates electron-hole pairs in or near this region. The electric field quickly separates the carriers, producing reverse photocurrent proportional to light intensity.

Final Answer: Photodiode is reverse biased to obtain fast and sensitive photocurrent response.

Exam Tip: Write “wide depletion region” and “photocurrent proportional to intensity”.

Question: If the intensity of light incident on a photodiode is increased, what happens to reverse current?

Solution: Higher light intensity means more photons strike the junction per second. More electron-hole pairs are produced, so reverse photocurrent increases.

Final Answer: Reverse current increases with light intensity.

Exam Tip: Photodiode current is a measure of illumination.

Question: A photodiode produces 8 µA current at intensity I. Assuming linear response, find current at intensity 2.5I.

Solution: Photocurrent is proportional to light intensity. Therefore I_photo' = 2.5 × 8 µA = 20 µA.

Final Answer: 20 µA.

Exam Tip: Check whether the question says response is linear.

Question: In reverse-bias characteristic curves of a photodiode, why are curves for higher illumination lower on the current axis?

Solution: Reverse current is conventionally plotted in the negative current direction. Higher illumination produces larger reverse photocurrent, so the curve shifts farther in the negative current direction.

Final Answer: Higher illumination gives larger reverse current, plotted as more negative current.

Exam Tip: Direction convention matters in photodiode graphs.

Question: Explain one use of a photodiode in an optical fibre communication system.

Solution: At the receiver end, light pulses from the optical fibre fall on a reverse biased photodiode. The photodiode converts changes in light intensity into corresponding electrical current pulses, which are processed as the signal.

Final Answer: It acts as a light detector and converts optical pulses into electrical pulses.

Exam Tip: Use “receiver end” for communication applications.

Question: Name the diode used for detecting light and state its biasing condition.

Solution: The diode used for detecting light is a photodiode. It is normally operated in reverse bias so that small changes in light intensity produce measurable current changes.

Final Answer: Photodiode; reverse bias.

Exam Tip: Do not confuse photodiode with LED; LED emits light, photodiode detects light.

Question: What is photovoltaic effect? How does it help in working of a solar cell?

Solution: Photovoltaic effect is the generation of emf when light falls on a suitable semiconductor junction. In a solar cell, photons create electron-hole pairs. The junction electric field separates them and produces a potential difference across the terminals.

Final Answer: Photovoltaic effect converts light energy directly into electrical energy in a solar cell.

Exam Tip: Always mention electron-hole pair generation and separation.

Question: A solar cell works without external bias. Why?

Solution: Incident light itself generates charge carriers. The built-in electric field of the p-n junction separates these carriers and produces emf. Therefore no external battery is required.

Final Answer: Light energy and the junction field produce emf without external bias.

Exam Tip: Solar cell is not a reverse-biased photodiode in normal operation.

Question: Define open circuit voltage V_oc and short circuit current I_sc for a solar cell.

Solution: V_oc is the terminal voltage when no external current is drawn from the solar cell. I_sc is the current when the output terminals are shorted, so terminal voltage is zero.

Final Answer: V_oc: voltage at I = 0. I_sc: current at V = 0.

Exam Tip: On the I-V graph, V_oc is voltage-axis intercept and I_sc is current-axis intercept.

Question: A solar cell receives 1.5 W of light energy and delivers 0.45 W electrical power. Find efficiency.

Solution: Efficiency η = Output power / Input power = 0.45 / 1.5 = 0.30. In percentage, η = 30%.

Final Answer: 30%.

Exam Tip: Convert efficiency into percentage only after taking the ratio.

Question: Why does the I-V curve of an illuminated solar cell lie in the power-producing region?

Solution: Under illumination, the solar cell supplies current to an external load while maintaining terminal voltage. This means electrical power is delivered by the cell rather than absorbed by it, so the curve lies in the generator or power-producing region.

Final Answer: Because the illuminated cell delivers power to the load.

Exam Tip: Sign convention may vary; focus on power delivered to external circuit.

Question: A solar cell gives 0.6 V and 0.25 A to a load. Find output power.

Solution: P = VI = 0.6 × 0.25 = 0.15 W.

Final Answer: 0.15 W.

Exam Tip: For solar cell output, use load voltage and load current.