Questions & explanations
1. What is the primary function of a filter circuit used in a rectifier?
- To convert AC to DC
- To reduce the ripple in the rectified output
- To increase the peak voltage of the rectified output
- To change the frequency of the input AC
Answer: To reduce the ripple in the rectified output
The filter circuit in a rectifier is used to smooth out the pulsating DC by reducing the AC ripple component, resulting in a more steady DC output. The conversion of AC to DC is performed by the rectifier itself, not the filter. The filter does not significantly increase the peak voltage; its main purpose is ripple reduction. Changing the frequency of the input AC is not a function of the filter.
2. The depletion region in a p-n junction is formed due to:
- A) Drift of charge carriers under the influence of an electric field.
- B) Diffusion of majority carriers across the junction, leaving behind immobile ions.
- C) Recombination of electrons and holes at the junction.
- D) Thermal generation of electron-hole pairs at the junction.
Answer: B) Diffusion of majority carriers across the junction, leaving behind immobile ions.
When a p-type and n-type semiconductor are joined, majority carriers (holes from p-side, electrons from n-side) diffuse across the junction. As they leave, they expose fixed charged ions (negative on p-side, positive on n-side), creating a region devoid of mobile carriers called the depletion region. This sets up an internal electric field that opposes further diffusion.
3. Which of the following statements correctly distinguishes a semiconductor from a conductor and an insulator?
- A) Conductors have a large energy gap, semiconductors have a small gap, insulators have no gap.
- B) Conductors have no energy gap, semiconductors have a small gap, insulators have a large gap.
- C) All three have overlapping energy bands.
- D) The energy gap is the same for all three.
Answer: B) Conductors have no energy gap, semiconductors have a small gap, insulators have a large gap.
In conductors, the valence and conduction bands overlap (no energy gap), allowing free flow of electrons. Semiconductors have a small energy gap (about 1 eV) that can be overcome by thermal energy. Insulators have a large energy gap (>3 eV) that prevents electron flow under normal conditions. Thus, option B correctly describes the distinction.
4. Which of the following statements about a full‑wave rectifier is correct?
- A) It uses a single diode.
- B) It produces output only during one half of the input cycle.
- C) The ripple frequency of the output is twice the input frequency.
- D) Its efficiency is less than that of a half‑wave rectifier.
Answer: C) The ripple frequency of the output is twice the input frequency.
A full‑wave rectifier conducts during both halves of the input AC cycle, producing two output pulses per cycle. This makes the fundamental frequency of the ripple twice the input frequency. A full‑wave rectifier uses two or four diodes (not one), conducts during both halves, and its efficiency is higher than that of a half‑wave rectifier.
5. Which statement correctly describes the working of a Zener diode in reverse bias?
- It conducts only when forward biased like a normal diode.
- It breaks down at a specific reverse voltage (Zener voltage) and then conducts without damage if current is limited.
- It blocks all reverse voltages permanently.
- It emits light when reverse biased.
Answer: It breaks down at a specific reverse voltage (Zener voltage) and then conducts without damage if current is limited.
A Zener diode is specially designed to operate in the reverse breakdown region. At a well-defined reverse voltage called the Zener voltage, it begins conducting heavily, and as long as the current is limited by an external resistor, the diode is not damaged and maintains a nearly constant voltage across it.
6. For an LED to emit visible light, the band gap of the semiconductor material should be:
- Less than 0.1 eV.
- In the range of about 1.8 eV to 3.0 eV.
- Greater than 5 eV.
- Exactly 0.7 eV.
Answer: In the range of about 1.8 eV to 3.0 eV.
Visible light photons have energies roughly between 1.8 eV (red) and 3.0 eV (violet). In an LED, the photon energy equals the band gap (for efficient emission). Therefore the semiconductor band gap must lie in this range. Lower or higher energies correspond to infrared or ultraviolet, not visible light.
7. Which of the following is a donor impurity when added to silicon?
- A) Boron
- B) Aluminium
- C) Phosphorus
- D) Indium
Answer: C) Phosphorus
Donor impurities are pentavalent atoms (Group 15) that donate extra electrons. Phosphorus has five valence electrons; when added to silicon (tetravalent), it provides an extra free electron, forming an n-type semiconductor. Boron, aluminium, and indium are trivalent acceptors used for p-type doping.
8. In the V‑I characteristic curve of a p‑n junction diode, the reverse saturation current is primarily due to:
- A) Majority carriers
- B) Minority carriers
- C) Both majority and minority carriers equally
- D) Tunneling of electrons
Answer: B) Minority carriers
Under reverse bias, the majority carriers are repelled from the junction, so no current flows from them. The small reverse current (saturation current) is due to minority carriers (electrons in the p‑side and holes in the n‑side) that are thermally generated and drift across the junction.
9. In a half‑wave rectifier, the output waveform across the load resistor is:
- A) Pulsating DC for both halves of the input AC cycle.
- B) Pulsating DC for only one half of the input AC cycle.
- C) Continuous DC (constant voltage).
- D) The same as the input AC waveform.
Answer: B) Pulsating DC for only one half of the input AC cycle.
A half‑wave rectifier uses a single diode that conducts only during one half‑cycle (when forward biased). During the other half‑cycle, the diode is reverse biased and no current flows. Hence, the output is a pulsating DC that appears only during one alternation of the input AC.
10. When a p-n junction diode is reverse biased, the width of the depletion region:
- A) Increases
- B) Decreases
- C) Remains unchanged
- D) Becomes zero
Answer: A) Increases
In reverse bias, the external voltage adds to the built‑in potential, pulling majority carriers away from the junction. This exposes more immobile ions, widening the depletion region. In forward bias, the width decreases because carriers are pushed toward the junction.
11. In a Zener diode voltage regulator circuit, when the input voltage increases, what happens to the output voltage across the load?
- Increases proportionally.
- Decreases.
- Remains approximately constant.
- Becomes zero.
Answer: Remains approximately constant.
In a Zener voltage regulator, the Zener diode operates in its breakdown region, where the voltage across it remains nearly constant despite changes in input voltage or load current (within limits). Hence the output voltage across the load stays approximately constant.
12. Which of the following best describes the working of a solar cell?
- It converts electrical energy into light energy.
- It uses a Zener breakdown to generate voltage.
- It converts light energy into electrical energy using the photovoltaic effect.
- It works only in forward bias.
Answer: It converts light energy into electrical energy using the photovoltaic effect.
A solar cell is a p–n junction device that directly converts sunlight (photons) into electricity. Incident photons generate electron–hole pairs, which are separated by the built-in electric field, producing a voltage across the cell. This is the photovoltaic effect.