semiconductor formulas pyqs Sheet, NCERT Examples and PYQs
Complete revision page for formulas, diode equations, rectifiers, logic gates, NCERT Exercises 14.1 to 14.15, solved exam-style questions, assertion-reason and case studies.
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1. Complete Formula Sheet
Semiconductor Formulae
Diode Formulae
Rectifier Formulae
| Quantity | Half-Wave Rectifier | Full-Wave Rectifier |
|---|---|---|
| DC value | Idc = Im / π | Idc = 2Im / π |
| RMS value | Irms = Im / 2 | Irms = Im / √2 |
| Ripple factor | γ = 1.21 | γ = 0.482 |
| Maximum efficiency | η = 40.6% | η = 81.2% |
| Output frequency | fout = f | fout = 2f |
| PIV | PIV = Vm | Centre tap: 2Vm; Bridge: Vm |
2. Logic Gate Summary
| Gate | Boolean Expression | Output is 1 when | Note |
|---|---|---|---|
| OR | Y = A + B | At least one input is 1 | Inclusive OR |
| AND | Y = A.B | Both inputs are 1 | Product operation |
| NOT | Y = A̅ | Input is 0 | Inverter |
| NAND | Y = (A.B)̅ | Not both inputs are 1 | Universal gate |
| NOR | Y = (A + B)̅ | Both inputs are 0 | Universal gate |
| XOR | Y = A̅B + AB̅ | Inputs are different | Odd parity gate |
| XNOR | Y = AB + A̅B̅ | Inputs are same | Equivalence gate |
Universal Gates
NAND and NOR are universal gates because AND, OR and NOT operations can be built using only NAND gates or only NOR gates.
Truth Table Order
For two inputs, use the order 00, 01, 10, 11. This prevents mistakes in gate identification questions.
3. NCERT Exercise Source Diagrams
The following attached NCERT pages are embedded directly so that exercise diagrams 14.36, 14.37, 14.38, 14.39 and 14.40 retain their correct logic-gate symbols and connections.
4. NCERT Exercises with Solutions
14.1
Question: In an n-type silicon, which statement is true? (a) Electrons are majority carriers and trivalent atoms are dopants. (b) Electrons are minority carriers and pentavalent atoms are dopants. (c) Holes are minority carriers and pentavalent atoms are dopants. (d) Holes are majority carriers and trivalent atoms are dopants.
Solution: In n-type silicon, pentavalent impurity atoms donate extra electrons. Electrons become majority carriers and holes become minority carriers.
Final Answer: Correct option is (c).
Exam Tip: n-type means negative carriers, electrons, are majority carriers.
14.2
Question: Which statement of Exercise 14.1 is true for p-type semiconductors?
Solution: In p-type semiconductor, trivalent impurity creates holes. Holes are majority carriers and electrons are minority carriers.
Final Answer: Option (d) of 14.1 is true for p-type semiconductor.
Exam Tip: p-type means positive carriers, holes, are majority carriers.
14.3
Question: Carbon, silicon and germanium have four valence electrons each. These have band gaps (Eg)C, (Eg)Si and (Eg)Ge. Which statement is true?
Solution: Diamond has the largest band gap, silicon is intermediate, and germanium has the smallest band gap.
Final Answer: (Eg)C > (Eg)Si > (Eg)Ge, option (c).
Exam Tip: More insulating nature means larger band gap.
14.4
Question: In an unbiased p-n junction, holes diffuse from p-region to n-region because (a) free electrons in n-region attract them, (b) they move across the junction by potential difference, (c) hole concentration in p-region is more than n-region, (d) all the above.
Solution: Diffusion is due to concentration gradient. Holes move from higher hole concentration in p-region to lower hole concentration in n-region. Built-in field opposes diffusion after depletion layer forms.
Final Answer: Correct option is (c).
Exam Tip: Diffusion is due to concentration difference, not attraction or applied voltage.
14.5
Question: When a forward bias is applied to a p-n junction, it (a) raises potential barrier, (b) reduces majority carrier current to zero, (c) lowers potential barrier, (d) none.
Solution: Forward bias opposes the junction barrier and reduces depletion width. Majority carriers cross the junction easily.
Final Answer: Correct option is (c).
Exam Tip: Forward bias reduces barrier; reverse bias increases barrier.
14.6
Question: In half-wave rectification, what is output frequency if input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?
Solution: Half-wave rectifier gives one pulse per input cycle, so fout = f. Full-wave rectifier gives two pulses per input cycle, so fout = 2f.
Final Answer: Half-wave: 50 Hz. Full-wave: 100 Hz.
Exam Tip: Full-wave rectifier doubles ripple frequency.
14.7
Question: A p-n photodiode is fabricated from a semiconductor with band gap 2.8 eV. Can it detect wavelength 6000 nm?
Solution: Photon energy E = 1240 / λ(in nm) eV = 1240 / 6000 = 0.207 eV. This is less than band gap 2.8 eV, so electrons cannot be excited across the gap.
Final Answer: No, it cannot detect 6000 nm radiation.
Exam Tip: Detection needs hν ≥ Eg.
14.8
Question: Silicon atoms per m3 are 5 × 1028. It is doped with 5 × 1022 arsenic atoms per m3 and 5 × 1020 indium atoms per m3. Given ni = 1.5 × 1016 m-3. Find electrons and holes and decide n-type or p-type.
Solution: Arsenic is donor and indium is acceptor. Net donor concentration ND - NA = 5 × 1022 - 5 × 1020 = 4.95 × 1022 m-3. Thus n ≈ 4.95 × 1022 m-3. From mass action law, p = ni2 / n = (1.5 × 1016)2 / (4.95 × 1022) ≈ 4.55 × 109 m-3.
Final Answer: n ≈ 4.95 × 1022 m-3, p ≈ 4.55 × 109 m-3; material is n-type.
Exam Tip: Compare donor and acceptor densities first.
14.9
Question: In an intrinsic semiconductor, Eg = 1.2 eV. Hole mobility is much smaller and independent of temperature. Find ratio of conductivity at 600 K and 300 K, assuming ni = n0 exp[-Eg / (2kBT)].
Solution: Conductivity is proportional to ni when mobility is taken nearly constant. Ratio σ600 / σ300 = exp[-Eg/(2kB × 600) + Eg/(2kB × 300)]. With kB = 8.6 × 10-5 eV/K, exponent = Eg/(2kB) × (1/300 - 1/600) = 1.2/(2 × 8.6 × 10-5) × 1/600 ≈ 11.63.
Final Answer: σ600 / σ300 ≈ exp(11.63) ≈ 1.1 × 105.
Exam Tip: Use eV with kB in eV/K.
14.10
Question: For a p-n diode I = I0 exp(eV/2kBT - 1). Given I0 = 5 × 10-12 A and T = 300 K: (a) forward current at 0.6 V, (b) increase if voltage increases to 0.7 V, (c) dynamic resistance, (d) current if reverse bias changes from 1 V to 2 V.
Solution: At 300 K, 2kBT/e ≈ 0.052 V. For V = 0.6 V, exponent = 0.6/0.052 ≈ 11.63, so I ≈ 5 × 10-12 × exp(11.63) ≈ 5.6 × 10-7 A. For V = 0.7 V, exponent = 13.57, I ≈ 3.9 × 10-6 A. Increase ≈ 3.3 × 10-6 A. Dynamic resistance near this change rd ≈ ΔV/ΔI = 0.1/(3.3 × 10-6) ≈ 3.0 × 104 Ω. In reverse bias, current is approximately -I0 and nearly unchanged.
Final Answer: (a) 5.6 × 10-7 A, (b) increase ≈ 3.3 × 10-6 A, (c) rd ≈ 3.0 × 104 Ω, (d) reverse current remains nearly 5 × 10-12 A in magnitude.
Exam Tip: Reverse saturation current is almost independent of reverse voltage before breakdown.
14.11
Question: Two circuits of Fig. 14.36 use NOR gates. Show circuit (a) acts as OR gate and circuit (b) acts as AND gate.
Solution: In (a), first NOR gives (A + B)̅. The second NOR is used as NOT, so Y = [(A + B)̅]̅ = A + B, an OR gate. In (b), A and B are first inverted using NOR as NOT, giving A̅ and B̅. These go into a NOR gate: Y = (A̅ + B̅)̅ = A.B by De Morgan theorem.
Final Answer: Circuit (a) is OR; circuit (b) is AND.
Exam Tip: NOR with tied inputs behaves as NOT.
14.12
Question: Write the truth table for a NAND gate connected as given in Fig. 14.37. Identify the exact logic operation.
Solution: The two inputs of NAND are connected together to A. Therefore Y = (A.A)̅ = A̅.
| A | Y |
|---|---|
| 0 | 1 |
| 1 | 0 |
Final Answer: It acts as a NOT gate.
Exam Tip: NAND with joined inputs is an inverter.
14.13
Question: Two circuits in Fig. 14.38 consist of NAND gates. Identify the logic operation carried out by the two circuits.
Solution: In (a), first NAND gives (A.B)̅. The second NAND has tied inputs, so it inverts this output: Y = [(A.B)̅]̅ = A.B. In (b), first two NAND gates act as NOT gates producing A̅ and B̅. Final NAND gives Y = (A̅.B̅)̅ = A + B.
Final Answer: Circuit (a) is AND; circuit (b) is OR.
Exam Tip: NAND logic uses De Morgan theorem repeatedly.
14.14
Question: Write the truth table for Fig. 14.39 consisting of NOR gates and identify the logic operation.
Solution: First NOR gives P = (A + B)̅. The second NOR has both inputs connected to P, so Y = (P + P)̅ = P̅ = A + B.
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Final Answer: The circuit performs OR operation.
Exam Tip: NOR followed by NOR-inverter becomes OR.
14.15
Question: Write the truth table for Fig. 14.40 consisting of NOR gates only. Identify the logic operation performed by the two circuits.
Solution: In (a), one NOR gate has both inputs tied to A, so Y = (A + A)̅ = A̅; it is NOT. In (b), first two NOR gates produce A̅ and B̅. The last NOR gives Y = (A̅ + B̅)̅ = A.B.
Truth table for (a):
| A | Y |
|---|---|
| 0 | 1 |
| 1 | 0 |
Truth table for (b):
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Final Answer: Circuit (a) is NOT; circuit (b) is AND.
Exam Tip: NOR can build NOT and AND using De Morgan theorem.
5. CBSE, NEET, JEE, IB, IGCSE and A-Level Practice
PN Junction Diode
Question: Explain why current increases rapidly in forward bias but remains very small in reverse bias.
Solution: Forward bias lowers the barrier potential and narrows depletion region, so majority carriers cross the junction easily. Reverse bias increases barrier and depletion width, so only minority carrier current flows.
Final Answer: Forward current is due to majority carriers; reverse current is small minority carrier current.
Exam Tip: Mention barrier potential and depletion width.
Photodiode
Question: Why is a photodiode generally operated in reverse bias?
Solution: Reverse bias widens depletion region and makes the current due to light-generated minority carriers easy to detect. The photocurrent is approximately proportional to light intensity.
Final Answer: Reverse bias improves sensitivity to incident light.
Exam Tip: Photodiode current increases with light intensity.
Rectifier Frequency
Question: A 60 Hz AC input is applied to a full-wave rectifier. Find ripple frequency.
Solution: In full-wave rectification both half cycles appear in output. Thus ripple frequency = 2f = 120 Hz.
Final Answer: 120 Hz.
Exam Tip: Half-wave gives f; full-wave gives 2f.
Zener Regulator
Question: A Zener diode of 6 V regulates a load. If supply varies above breakdown and series resistor is suitable, what is Vout?
Solution: In reverse breakdown, Zener voltage remains nearly constant. The load is connected parallel to Zener, hence output equals Zener voltage.
Final Answer: Vout ≈ 6 V.
Exam Tip: Zener is used in reverse breakdown region.
Solar Cell
Question: State the energy conversion in a solar cell and name the effect involved.
Solution: Incident photons create electron-hole pairs near the junction. The junction field separates charges and produces emf.
Final Answer: Light energy is converted to electrical energy by photovoltaic effect.
Exam Tip: Photovoltaic effect is not the same as photoconductive effect.
LED
Question: Why does an LED emit light only when forward biased?
Solution: Forward bias allows electrons and holes to recombine at the junction. Energy released during recombination appears as photons.
Final Answer: LED emits light due to radiative recombination in forward bias.
Exam Tip: Colour depends on band gap.
Doping
Question: Explain how adding phosphorus to silicon changes its conductivity.
Solution: Phosphorus is pentavalent and contributes one extra electron per impurity atom. Electron concentration increases, so conductivity increases.
Final Answer: It forms n-type silicon with electrons as majority carriers.
Exam Tip: Pentavalent dopants are donors.
NAND as Universal Gate
Question: Show how a NOT gate is made using a NAND gate.
Solution: Connect both inputs of NAND gate to the same input A. Output Y = (A.A)̅ = A̅.
Final Answer: Tied-input NAND gate works as NOT gate.
Exam Tip: Same trick works for NOR gate too.
6. Assertion-Reason Questions
AR 1: Doping
Assertion: Doping increases conductivity of a semiconductor. Reason: Doping increases the number of charge carriers.
Answer: Both are true and Reason correctly explains Assertion.
AR 2: Depletion Region
Assertion: Depletion region contains no mobile charge carriers. Reason: Electrons and holes recombine near the junction leaving immobile ions.
Answer: Both are true and Reason correctly explains Assertion.
AR 3: Reverse Bias
Assertion: Reverse current is very small before breakdown. Reason: Reverse current is mainly due to minority carriers.
Answer: Both are true and Reason correctly explains Assertion.
AR 4: Zener Diode
Assertion: Zener diode is used as voltage regulator. Reason: In breakdown region its voltage remains nearly constant for large current change.
Answer: Both are true and Reason correctly explains Assertion.
AR 5: Logic Gates
Assertion: NAND gate is called a universal gate. Reason: All basic gates can be made using NAND gates only.
Answer: Both are true and Reason correctly explains Assertion.
7. Case Studies
Case Study 1: PN Junction Diode
A p-n junction is formed by joining p-type and n-type semiconductors. Diffusion creates a depletion layer and barrier potential. Forward bias reduces barrier while reverse bias increases it.
Questions: 1. Why is depletion region formed? 2. What happens in forward bias? 3. Which carriers dominate forward current? 4. What is reverse saturation current?
Answers: 1. Recombination near junction. 2. Barrier decreases. 3. Majority carriers. 4. Small current due to minority carriers.
Case Study 2: Rectifier
A rectifier converts AC into pulsating DC. A half-wave rectifier uses one half cycle, while a full-wave rectifier uses both half cycles.
Questions: 1. Which rectifier has higher efficiency? 2. What is output frequency of full-wave rectifier for 50 Hz input? 3. Why is filter used? 4. What does PIV mean?
Answers: 1. Full-wave. 2. 100 Hz. 3. To reduce ripple. 4. Peak inverse voltage.
Case Study 3: Zener Regulator
A Zener diode is connected in reverse bias across load resistance. Series resistance limits current.
Questions: 1. Which bias is used? 2. What is regulated output? 3. Why series resistor is needed? 4. What happens when input rises?
Answers: 1. Reverse bias. 2. VZ. 3. To limit current. 4. Extra voltage drops across series resistor while output remains nearly constant.
Case Study 4: Photodiode
A reverse biased photodiode produces photocurrent when illuminated. Current increases with intensity.
Questions: 1. Which bias is used? 2. What creates photocurrent? 3. How does intensity affect current? 4. Name one application.
Answers: 1. Reverse bias. 2. Light-generated electron-hole pairs. 3. Current increases. 4. Optical communication or light sensor.
Case Study 5: Solar Cell
A solar cell works without external bias and uses photovoltaic effect to generate emf.
Questions: 1. What is Voc? 2. What is Isc? 3. What energy conversion occurs? 4. Which effect is involved?
Answers: 1. Open-circuit voltage. 2. Short-circuit current. 3. Light to electrical energy. 4. Photovoltaic effect.
Case Study 6: Logic Gates
Logic gates perform Boolean operations on binary inputs 0 and 1. NAND and NOR gates are universal.
Questions: 1. Output of AND for 1,1? 2. Output of OR for 0,1? 3. NAND of 1,1? 4. NOR of 0,0?
Answers: 1. 1. 2. 1. 3. 0. 4. 1.
Case Study 7: Semiconductor Doping
Adding controlled impurity to pure semiconductor changes carrier concentration. Pentavalent dopants form n-type; trivalent dopants form p-type.
Questions: 1. Name majority carriers in n-type. 2. Name majority carriers in p-type. 3. What is donor impurity? 4. State mass action law.
Answers: 1. Electrons. 2. Holes. 3. Pentavalent impurity. 4. n p = ni2.
8. Quick Revision Notes
Most Important Definitions
- Intrinsic semiconductor: pure semiconductor with n = p.
- Extrinsic semiconductor: doped semiconductor.
- Depletion region: region near junction without mobile carriers.
- Zener breakdown: sharp reverse breakdown used for voltage regulation.
Most Important Formulas
- σ = n e μe + p e μh
- n p = ni2
- I = I0[exp(eV/kBT) - 1]
- Full-wave fout = 2f
Common Mistakes
- Confusing diffusion with drift.
- Writing p-type majority carriers as electrons.
- Forgetting full-wave output frequency doubles.
- Drawing NAND/NOR bubbles incorrectly.