CHAPTER 14: SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
- Constants & Band Gaps: $\displaystyle {E_g(\text{Si}) = 1.1\text{ eV}}$, $\displaystyle {E_g(\text{Ge}) = 0.72\text{ eV}}$, $\displaystyle {e = 1.6 \times 10^{-19}\text{ C}}$, $\displaystyle {h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}}$, $\displaystyle {k_B = 1.38 \times 10^{-23}\text{ J/K}}$.
SECTION A (16 Marks)
Q.1. In an n-type semiconductor, which of the following statements is true?
Explanation: Pentavalent impurities (e.g., phosphorus, arsenic) donate free electrons, making electrons majority carriers (\\(\displaystyle {n_e \gg n_h}\\)) and holes minority carriers.
Q.2. At absolute zero temperature (\\(\displaystyle {0\text{ K}}\\)), a pure intrinsic semiconductor crystal behaves as:
Explanation: At $\displaystyle {0\text{ K}}$, thermal excitation is completely absent (\\(\displaystyle {k_B T = 0}\\)), so no covalent bonds are broken and the conduction band is completely empty.
Q.3. When a p-n junction diode is subjected to reverse biasing:
Explanation: The external reverse bias potential opposes majority carrier movement, pulling majority charge carriers away from the junction and expanding the region of uncompensated immobile ions.
Q.4. In an unbiased p-n junction, holes diffuse from the p-region to the n-region primarily because:
Explanation: Diffusion occurs naturally down the steep concentration gradient from high carrier density (p-side) to low carrier density (n-side).
Q.5. The forbidden energy band gap $\displaystyle {E_g}$ for a elemental germanium crystal at room temperature (\\(\displaystyle {300\text{ K}}\\)) is of the order of:
Explanation: Germanium has a forbidden energy gap of $\displaystyle {\approx 0.72\text{ eV}}$, whereas Silicon has $\displaystyle {E_g \approx 1.1\text{ eV}}$.
Q.6. Which impurity element must be doped into pure Silicon (\\(\text{Si}\\)) to obtain a p-type extrinsic semiconductor?
Explanation: Boron is a trivalent impurity (group 13 element) that creates acceptor hole energy levels just above the valence band.
Q.7. The potential barrier developed across a p-n junction diode in the absence of external biasing opposes the flow of:
Explanation: The built-in electric field $\displaystyle {\vec{E}_0}$ points from the n-side (+ve immobile donor ions) to the p-side (-ve immobile acceptor ions), opposing majority carrier diffusion.
Q.8. When a p-n junction diode is forward biased, the total current across the junction is primarily due to:
Explanation: Forward bias lowers the potential barrier height (\\(\displaystyle {V_0 - V}\\)), allowing vast numbers of majority carriers to diffuse across the junction.
Q.9. In a full-wave bridge or center-tapped rectifier operating from a $\displaystyle {50\text{ Hz}}$ AC mains supply, the fundamental frequency of the output ripple is:
Explanation: A full-wave rectifier conducts during both half-cycles of input AC, doubling the output pulse frequency (\\(\displaystyle {f_{\text{out}} = 2 f_{\text{in}} = 2 \times 50 = 100\text{ Hz}}\\)).
Q.10. The depletion layer of a p-n junction diode contains:
Q.11. Carbon (\\(\text{C}\\)), Silicon (\\(\text{Si}\\)), and Germanium (\\(\text{Ge}\\)) all belong to group 14 and have 4 valence electrons each. Their energy band gaps satisfy:
Explanation: Carbon (diamond) is an insulator (\\(\displaystyle {E_g \approx 5.4\text{ eV}}\\)), Silicon is a semiconductor (\\(\displaystyle {E_g \approx 1.1\text{ eV}}\\)), and Germanium has $\displaystyle {E_g \approx 0.72\text{ eV}}$.
Q.12. In the circuit diagram shown below, the ideal p-n junction diode $D_1$ has a potential of $-2\text{ V}$ at its p-side and is connected to ground (\\(0\text{ V}\\)) through a resistor at its n-side. The diode is:
Explanation: The anode (p-side) is at a lower potential (\\(\displaystyle {-2\text{ V}}\\)) than the cathode (n-side at $\displaystyle {0\text{ V}}$), placing the diode in reverse bias.
Directions for Q.13 to Q.16:
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is NOT the correct explanation of Assertion.
(c) Assertion is correct, Reason is incorrect.
(d) Both Assertion and Reason are incorrect.
Q.13. Assertion (A): The electrical conductivity of an intrinsic semiconductor increases with an increase in temperature.
Reason (R): With a rise in temperature, thermal energy breaks covalent bonds, generating additional free electrons and holes in equal numbers.
Q.14. Assertion (A): Electrons have higher mobility than holes in a semiconductor.
Reason (R): Free electrons move in the conduction band, whereas holes move due to bound electron transfers in the valence band.
Q.15. Assertion (A): A p-n junction diode cannot be fabricated simply by pressing a p-type semiconductor block firmly against an n-type block.
Reason (R): Mechanical contact leaves surface roughness larger than interatomic spacing, preventing atomic-scale diffusion and depletion layer formation.
Q.16. Assertion (A): A half-wave rectifier converts alternating current into pulsating direct current with an efficiency of less than 50%.
Reason (R): A single diode conducts current during positive input half-cycles and blocks current during negative input half-cycles.
SECTION B (10 Marks)
Q.17. Distinguish clearly between an intrinsic semiconductor and an extrinsic semiconductor on the basis of charge carrier concentration and conductivity.
1. Intrinsic Semiconductor: Pure semiconductor without impurities. Thermal excitation generates equal electron and hole concentrations (\\(\displaystyle {n_e = n_h = n_i}\\)). Conductivity is very low at room temperature.
2. Extrinsic Semiconductor: Doped with specific impurity atoms. Carrier concentrations are highly unequal ($\displaystyle {n_e \gg n_h}$ in n-type, $\displaystyle {n_h \gg n_e}$ in p-type), obeying the mass action law $\displaystyle {n_e \cdot n_h = n_i^2}$. Conductivity is significantly higher.
Q.18. What is meant by the term 'doping' of a semiconductor? How does doping affect the forbidden energy band gap and electrical conductivity of a semiconductor?
Name the two dominant processes that occur during the formation of a p-n junction. Define depletion layer and potential barrier.
Doping: The deliberate addition of a desirable trace impurity (trivalent or pentavalent atoms, $\sim 1\text{ ppm}$) to a pure semiconductor to enhance electrical conductivity. It introduces discrete donor or acceptor energy levels very close to the conduction or valence bands, dramatically boosting carrier density without significantly altering the host crystal band gap.
OR Part:
1. Processes: (i) Diffusion (majority carriers move across concentration gradient) and (ii) Drift (minority carriers move under built-in electric field).
2. Depletion Layer: A thin region around the junction devoid of mobile charge carriers, containing only uncompensated fixed ions.
3. Potential Barrier: The built-in potential difference (\\(\displaystyle {V_0}\\)) across the depletion region that opposes further majority carrier diffusion.
Q.19. Draw neat, labeled energy band diagrams for an intrinsic semiconductor at $\displaystyle {T > 0\text{ K}}$, an n-type semiconductor, and a p-type semiconductor. Indicate the relative positions of donor and acceptor energy levels.
1. Intrinsic: CB contains few thermally excited electrons; VB contains equal number of holes.
2. N-type: Donor energy level $\displaystyle {E_D}$ lies just below the conduction band ($\displaystyle {\sim 0.01\text{ eV}}$ for Ge, $\displaystyle {\sim 0.05\text{ eV}}$ for Si).
3. P-type: Acceptor energy level $\displaystyle {E_A}$ lies just above the valence band ($\displaystyle {\sim 0.01\text{ eV}}$ for Ge, $\displaystyle {\sim 0.05\text{ eV}}$ for Si).
Q.20. Explain why the reverse current in a p-n junction diode reaches a saturation value at very small reverse voltages and remains nearly independent of applied voltage until breakdown.
Q.21. A semiconductor has an intrinsic carrier concentration $\displaystyle {n_i = 1.5 \times 10^{16}\text{ m}^{-3}}$. Upon doping with a pentavalent impurity, the electron concentration increases to $\displaystyle {n_e = 4.5 \times 10^{22}\text{ m}^{-3}}$. Calculate the minority hole concentration $\displaystyle {n_h}$ in the doped extrinsic semiconductor.
$\displaystyle {n_e \cdot n_h = n_i^2}$.
$\displaystyle {n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5.0 \times 10^9\text{ m}^{-3}}$.
SECTION C (21 Marks)
Q.22. Explain the working principle of a p-n junction diode as a half-wave rectifier with the help of a neat circuit diagram. Draw the input AC voltage and output DC voltage waveforms.
1. Working Principle: Uses the unidirectional conduction property of a p-n junction diode (low resistance in forward bias, high resistance in reverse bias).
2. Positive Half-Cycle: Upper secondary terminal is positive relative to lower. Diode is forward biased, conducts current through load resistor $\displaystyle {R_L}$, producing output voltage.
3. Negative Half-Cycle: Upper secondary terminal is negative. Diode is reverse biased, blocks current flow, producing zero output voltage.
Q.23. Explain the working of a center-tapped full-wave rectifier circuit using two p-n junction diodes. Draw the input AC and output DC waveforms across the load resistor.
1. Setup: Uses a transformer with a center-tapped secondary winding and two diodes $\displaystyle {D_1}$ and $\displaystyle {D_2}$ connected to a common load resistor $\displaystyle {R_L}$.
2. Positive Half-Cycle: Terminal A is positive, B is negative relative to center tap. Diode $\displaystyle {D_1}$ is forward biased and conducts; $\displaystyle {D_2}$ is reverse biased.
3. Negative Half-Cycle: Terminal B is positive, A is negative. Diode $\displaystyle {D_2}$ is forward biased and conducts; $\displaystyle {D_1}$ is reverse biased.
4. Unidirectional Output: Current flows through $\displaystyle {R_L}$ in the exact same direction during both half-cycles, producing a continuous pulsating DC output.
Q.24. Describe how the V-I characteristics of a p-n junction diode are obtained experimentally in: (i) Forward bias, and (ii) Reverse bias. Draw typical V-I characteristic curves for a Silicon diode.
1. Forward Bias Characteristics: Current remains negligible until applied voltage exceeds the knee voltage ($\displaystyle {V_K \approx 0.7\text{ V}}$ for Si, $\displaystyle {0.3\text{ V}}$ for Ge). Beyond knee voltage, current increases exponentially with small voltage increments.
2. Reverse Bias Characteristics: Applied reverse voltage yields a tiny temperature-dependent reverse saturation current ($\displaystyle {I_s \sim \mu\text{A}}$ for Ge, $\displaystyle {\text{nA}}$ for Si). At critical reverse breakdown voltage (\\(\displaystyle {V_{BR}}\\)), reverse current increases abruptly due to avalanche multiplication.
Q.25. (a) State the role of a capacitor filter connected across the output load of a rectifier circuit.
(b) A p-n junction diode has a forward resistance of $\displaystyle {20\,\Omega}$. It is connected to a $\displaystyle {10\text{ V (rms)}}$ AC supply through a load resistor $\displaystyle {R_L = 980\,\Omega}$ in a half-wave rectifier configuration. Calculate: (i) Peak load current, and (ii) DC output load voltage.
(a) Capacitor Filter: A capacitor connected in parallel with load $\displaystyle {R_L}$ charges to peak supply voltage during conduction and discharges slowly through $\displaystyle {R_L}$ during non-conduction, smoothing out AC ripple voltage.
(b) Calculations:
Peak AC Voltage: $\displaystyle {V_m = \sqrt{2} \times V_{\text{rms}} = 1.414 \times 10\text{ V} = 14.14\text{ V}}$.
Total Forward Resistance: $\displaystyle {R_T = r_f + R_L = 20\,\Omega + 980\,\Omega = 1000\,\Omega}$.
(i) Peak Load Current: $\displaystyle {I_m = \frac{V_m}{R_T} = \frac{14.14\text{ V}}{1000\,\Omega} = 14.14\text{ mA}}$.
(ii) DC Load Current: $\displaystyle {I_{\text{dc}} = \frac{I_m}{\pi} = \frac{14.14\text{ mA}}{3.1416} \approx 4.50\text{ mA}}$.
DC Output Load Voltage: $\displaystyle {V_{\text{dc}} = I_{\text{dc}} \times R_L = (4.50 \times 10^{-3}\text{ A}) \times 980\,\Omega \approx 4.41\text{ V}}$.
Q.26. On the basis of energy band theory, distinguish clearly between conductors, semiconductors, and insulators. Draw neat energy band diagrams for each at room temperature.
1. Conductors: Valence and conduction bands overlap (\\(\displaystyle {E_g = 0}\\)), or conduction band is partially filled, yielding extremely high free electron availability.
2. Insulators: Large forbidden energy gap ($\displaystyle {E_g > 3\text{ eV}}$, e.g., diamond $\displaystyle {E_g \approx 5.4\text{ eV}}$). Conduction band is completely empty at room temperature.
3. Semiconductors: Small forbidden energy gap ($\displaystyle {E_g < 3\text{ eV}}$, e.g., Si $\displaystyle {1.1\text{ eV}}$, Ge $\displaystyle {0.72\text{ eV}}$). Thermal energy excites some electrons across $\displaystyle {E_g}$ at room temperature.
Q.27. (a) Why does the resistivity of a pure semiconductor decrease with an increase in temperature, whereas that of a metallic conductor increases?
(b) Calculate the maximum wavelength of electromagnetic radiation required to create an electron-hole pair in a Silicon crystal (\\(\displaystyle {E_g = 1.1\text{ eV}}\\)).
(a) In metallic conductors, carrier density $\displaystyle {n}$ is fixed and high; temperature rise increases lattice vibrations, reducing relaxation time $\displaystyle {\tau}$ (\\(\displaystyle {\rho \propto 1/\tau}\\)). In semiconductors, carrier density $\displaystyle {n}$ increases exponentially with temperature ($\displaystyle {n \propto e^{-E_g / (2 k_B T)}}$), overriding the slight decrease in $\displaystyle {\tau}$ and causing resistivity to fall.
(b) Wavelength Calculation:
$\displaystyle {\lambda_{\max} = \frac{hc}{E_g} = \frac{1240\text{ eV}\cdot\text{nm}}{1.1\text{ eV}} \approx 1127\text{ nm} = 1.127\,\mu\text{m}}$ (Infrared region).
Q.28. Explain the mechanism of conduction in p-type and n-type semiconductors using covalent bond models. Show that extrinsic semiconductors remain electrically neutral as a whole despite possessing majority charge carriers.
1. N-type Conduction: Pentavalent donor atom (e.g., As) replaces a Si atom. Four electrons form covalent bonds; fifth electron is weakly bound (\\(\displaystyle {\sim 0.05\text{ eV}}\\)) and easily donated to the conduction band at room temperature.
2. P-type Conduction: Trivalent acceptor atom (e.g., B) replaces a Si atom. Three electrons form covalent bonds, leaving one bond incomplete (a hole). Valence electrons jump into this hole, causing hole movement.
3. Electrical Neutrality: Extrinsic semiconductors are electrically neutral because every parent impurity atom added is electrically neutral ($\text{protons} = \text{electrons}$ in every atom).
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Intrinsic and Extrinsic Semiconductor Physics.
A pure semiconductor like Germanium or Silicon free of impurities is called an intrinsic semiconductor. At room temperature, intrinsic charge carriers (electrons and holes) exist in small, equal numbers (\\(\displaystyle {n_e = n_h = n_i}\\)), resulting in low conductivity. Doping with pentavalent or trivalent impurity atoms forms n-type or p-type extrinsic semiconductors with dramatically higher conductivity.
(i) At absolute zero temperature (\\(\displaystyle {0\text{ K}}\\)), pure Silicon acts as:
(a) Non-metal (b) Metal (c) Perfect Insulator (d) Superconductor
(ii) Which impurity atom must be doped into pure Silicon to produce a p-type semiconductor?
(a) Phosphorus (b) Boron (c) Arsenic (d) Antimony
(iii) In an n-type semiconductor, the majority charge carriers are:
(a) Holes (b) Free electrons (c) Positive ions (d) Protons
(iv) The product of electron concentration $\displaystyle {n_e}$ and hole concentration $\displaystyle {n_h}$ in thermal equilibrium satisfies:
(a) $\displaystyle {n_e \cdot n_h = n_i}$ (b) $\displaystyle {n_e \cdot n_h = n_i^2}$ (c) $\displaystyle {n_e / n_h = n_i^2}$ (d) $\displaystyle {n_e + n_h = n_i}$
Q.30. Case Study 2: P-N Junction Physics and Rectifier Applications.
A p-n junction is formed by bringing p-type and n-type semiconductor regions into atomic contact. A thin depletion layer devoid of mobile charge carriers forms at the junction, creating a built-in potential barrier. The junction offers very low resistance when forward biased and extremely high resistance when reverse biased, enabling rectifier action.
(i) In an unbiased p-n junction, the depletion layer consists of:
(a) Free mobile electrons only (b) Mobile holes only (c) Immobile ions (d) Neutral atoms
(ii) When a p-n junction is reverse biased, the depletion layer width:
(a) Decreases (b) Increases (c) Remains unchanged (d) Shrinks to zero
(iii) A semiconductor p-n junction diode is widely used as a:
(a) Oscillator (b) Amplifier (c) Rectifier (d) Modulator
(iv) In a full-wave center-tapped rectifier operating with $\displaystyle {50\text{ Hz}}$ input AC, the ripple output frequency is:
(a) $\displaystyle {25\text{ Hz}}$ (b) $\displaystyle {50\text{ Hz}}$ (c) $\displaystyle {100\text{ Hz}}$ (d) $\displaystyle {200\text{ Hz}}$
SECTION E (15 Marks)
Q.31. (a) Explain with circuit diagrams the experimental setup used to plot the V-I characteristics of a p-n junction diode in forward bias and reverse bias. Sketch typical V-I curves and define dynamic resistance.
(b) The V-I characteristic of a Silicon diode shows that at a forward voltage of $\displaystyle {0.75\text{ V}}$, the forward current is $\displaystyle {10\text{ mA}}$, and at $\displaystyle {0.80\text{ V}}$, the forward current rises to $\displaystyle {30\text{ mA}}$. Calculate the dynamic forward resistance of the diode in this region.
(a) Dynamic Resistance Definition: The ratio of a small change in voltage $\displaystyle {\Delta V}$ across the junction to the corresponding change in current $\displaystyle {\Delta I}$: $\displaystyle {r_d = \frac{\Delta V}{\Delta I}}$.
(b) Calculation:
$\displaystyle {\Delta V_F = 0.80\text{ V} - 0.75\text{ V} = 0.05\text{ V}}$.
$\displaystyle {\Delta I_F = 30\text{ mA} - 10\text{ mA} = 20\text{ mA} = 20 \times 10^{-3}\text{ A}}$.
$\displaystyle {r_d = \frac{\Delta V_F}{\Delta I_F} = \frac{0.05\text{ V}}{20 \times 10^{-3}\text{ A}} = \frac{50}{20} = 2.5\,\Omega}$.
Q.32. (a) Explain the processes of diffusion and drift in the formation of a p-n junction. How do these processes lead to the establishment of a depletion layer and a potential barrier?
(b) Explain why a p-n junction diode offers very low resistance in forward bias and extremely high resistance in reverse bias.
(a) With the help of a neat circuit diagram, explain the working of a full-wave bridge rectifier using four p-n junction diodes. Draw the input AC and output DC waveforms.
(b) Explain how a smoothing capacitor filter converts pulsating DC output into steady DC voltage.
(a) Junction Formation:
1. Diffusion: High concentration gradients cause majority holes from p-side to diffuse into n-side and majority electrons from n-side to diffuse into p-side.
2. Depletion Region: Recombination near the boundary leaves behind uncompensated positive donor ions on the n-side and negative acceptor ions on the p-side, forming a space charge region devoid of mobile carriers.
3. Drift: The space charge builds an internal electric field $\displaystyle {\vec{E}_0}$ from n to p that drives minority carriers in opposition to diffusion. Equilibrium is reached when $\displaystyle {I_{\text{diffusion}} = I_{\text{drift}}}$.
OR Part (a) Bridge Rectifier: Uses 4 diodes in a bridge arrangement. Diodes $\displaystyle {D_1, D_3}$ conduct during positive half-cycles; $\displaystyle {D_2, D_4}$ conduct during negative half-cycles. Current flows through load $\displaystyle {R_L}$ in the exact same direction during both half-cycles.
Q.33. (a) What is an intrinsic semiconductor? Derive the expression for electrical conductivity $\displaystyle {\sigma}$ of a semiconductor in terms of charge carrier concentrations (\\(\displaystyle {n_e, n_h}\\)) and mobilities (\\(\displaystyle {\mu_e, \mu_h}\\)).
(b) An intrinsic Germanium crystal at $\displaystyle {300\text{ K}}$ has $\displaystyle {n_i = 2.5 \times 10^{19}\text{ m}^{-3}}$. Electron and hole mobilities are $\displaystyle {\mu_e = 0.38\text{ m}^2\text{V}^{-1}\text{s}^{-1}}$ and $\displaystyle {\mu_h = 0.18\text{ m}^2\text{V}^{-1}\text{s}^{-1}}$. Calculate the intrinsic conductivity and resistivity of the Germanium crystal.
(a) Total current density in a semiconductor is the sum of electron and hole current densities:
$\displaystyle {J = J_e + J_h = n_e e v_e + n_h e v_h}$.
Since drift velocity $\displaystyle {v = \mu E}$, $\displaystyle {J = e (n_e \mu_e + n_h \mu_h) E}$.
Electrical Conductivity: $\displaystyle {\sigma = \frac{J}{E} = e (n_e \mu_e + n_h \mu_h)}$.
For intrinsic semiconductor (\\(\displaystyle {n_e = n_h = n_i}\\)): $\displaystyle {\sigma_i = n_i e (\mu_e + \mu_h)}$.
(b) Calculation:
$\displaystyle {\sigma_i = (2.5 \times 10^{19}\text{ m}^{-3}) \times (1.6 \times 10^{-19}\text{ C}) \times (0.38 + 0.18)\text{ m}^2\text{V}^{-1}\text{s}^{-1}}$.
$\displaystyle {\sigma_i = 4.0 \times 0.56 = 2.24\,\Omega^{-1}\text{m}^{-1} \text{ (or S/m)}}$.
Intrinsic Resistivity: $\displaystyle {\rho_i = \frac{1}{\sigma_i} = \frac{1}{2.24} \approx 0.446\,\Omega\cdot\text{m}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. If the electrical conductivity of a semiconductor is caused solely by thermal breakdown of covalent bonds, the material is called:
2. The electrical conductivity of pure Silicon or Germanium can be increased by:
3. When pure Germanium is doped with Arsenic (\\(\text{As}\\)), the resulting material becomes:
Explanation: Arsenic is a pentavalent donor impurity that provides excess free conduction electrons.
4. The resistivity of semiconductors at room temperature typically lies in the range:
5. In an insulator, the forbidden energy band gap between the valence band and conduction band is of the order of:
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Ohm's law is not strictly obeyed by semiconductor devices like p-n junction diodes.
Reason (R): The V-I characteristic of a p-n junction diode is non-linear and depends on the direction of applied voltage.
2. Assertion (A): The temperature coefficient of resistance is negative for semiconductors.
Reason (R): Raising the temperature breaks covalent bonds, liberating large numbers of charge carriers and increasing conductivity.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Resistivity and Temperature Behavior of Materials
Materials are broadly classified as conductors, semiconductors, and insulators based on resistivity. Metals have low resistivity (\\(\displaystyle {10^{-8}\,\Omega\cdot\text{m}}\\)). Insulators have resistivity $\displaystyle {10^{18}}$ times higher. Semiconductors exhibit a characteristic decrease in resistivity with increasing temperature, and their resistivity can be modified via doping.
1. When heated, the resistivity of a semiconductor decreases from $\displaystyle {2.5\,\Omega\cdot\text{m}}$ to $\displaystyle {0.5\,\Omega\cdot\text{m}}$. The percentage decrease in resistivity is:
(a) 20% (b) 40% (c) 60% (d) 80%
2. The temperature coefficient of resistance is negative for:
(a) Copper (b) Gold (c) Carbon (d) Silver
3. The product of resistivity (\\(\displaystyle {\rho}\\)) and electrical conductivity (\\(\displaystyle {\sigma}\\)) for any conductor depends on:
(a) Area of cross-section (b) Temperature (c) Length (d) None of these (always equals 1)
1. (d) 80% (\\(\displaystyle {\frac{2.5 - 0.5}{2.5} \times 100\% = \frac{2.0}{2.5} \times 100\% = 80\%}\\)).
2. (c) Carbon (Carbon is a metalloid/semiconductor with negative temperature coefficient).
3. (d) None of these ($\displaystyle {\rho \cdot \sigma = \rho \cdot (1/\rho) = 1}$, a constant).
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
CONCEPTUAL SHORT QUESTIONS
1. Why does the width of the depletion layer decrease during forward bias and increase during reverse bias?
2. Why are photodiodes preferably operated under reverse bias condition for light intensity detection?
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