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Semiconductor Electronics
Physics · Class 12 · Part 2NCERT pp. 323–3402 sets × 25 questions
Key facts and formulas for Semiconductor Electronics, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Resistivity ranges | Metal 10⁻² to 10⁻⁸ Ω m; semiconductor 10⁻⁵ to 10⁶ Ω m; insulator 10¹¹ to 10¹⁹ Ω m; ρ = 1/σ |
| Band gaps | C 5.4 eV, Si 1.1 eV, Ge 0.7 eV, Sn 0 eV; insulator E_g > 3 eV, semiconductor E_g < 3 eV |
| Intrinsic and doped | n_e = n_h = n_i; with doping n_e n_h = n_i²; donor level just below E_C, acceptor just above E_V |
| Dynamic resistance | r_d = ΔV / ΔI; forward barrier V₀ − V, reverse barrier V₀ + V |
| Photon wavelength | λ = hc / E with hc = 1240 eV nm |
| Transistor currents | I_E = I_B + I_C; α = I_C/I_E; β = α/(1 − α) = ΔI_C/ΔI_B |
| CE amplifier | A_v = β R_L / R_in; output is 180° out of phase with input |
| Zener regulator | I_series = (V_s − V_z)/R; I_Zener = I_series − I_load |
| Rectifier and filter | Full-wave ripple = 2 × input frequency; time constant τ = R_L C |
| Logic gates | NAND = NOT(A·B); NOR = NOT(A+B); De Morgan: NOT(A+B) = Ā·B̄; XOR = A·B̄ + Ā·B |
Three sample questions
Q1. Which statement about a vacuum diode is correct?
- Electrons flow only from the heated cathode to the anode
- Electrons flow equally in both directions
- Electrons flow from anode to cathode
- It needs no evacuated space
Hint and solution
Hint: Which electrode is heated, and which one then gives off electrons?
- Heated cathode → electrons → anode — A diode valve has a heated cathode and a cold anode. Only the cathode emits electrons.
- Anode cannot emit electrons — The anode stays cold, so no electrons come back. Current flows one way, like a valve.
- Space must be evacuated — Gas would block the electrons, so the tube is evacuated. That rules out option D.
- Answer: electrons flow from cathode to anode — So the correct statement is the first one. Option A.
Answer: A. Electrons are emitted by the heated cathode, so the tube conducts one way (a valve). The space must be evacuated.
Q2. Match the tube with its number of electrodes: (a) diode (b) triode (c) tetrode (d) pentode; (1) 3 (2) 5 (3) 2 (4) 4
- (a)-1, (b)-3, (c)-2, (d)-4
- (a)-3, (b)-4, (c)-1, (d)-2
- (a)-3, (b)-1, (c)-4, (d)-2
- (a)-4, (b)-1, (c)-3, (d)-2
Hint and solution
Hint: The prefix tells the count: di = 2, tri = 3, tetra = 4, penta = 5.
- di = 2, tri = 3, tetra = 4, penta = 5 — The name of each tube gives its electrode count.
- (a) diode → 2, i.e. (3) — A diode has cathode and anode, so (a) goes with (3).
- (b) triode → 3 (1); (c) tetrode → 4 (4); (d) pentode → 5 (2) — Similarly, triode matches (1), tetrode matches (4) and pentode matches (2).
- Match: (a)-3, (b)-1, (c)-4, (d)-2 — This pattern is the third option. Option C.
Answer: C. Diode 2, triode 3, tetrode 4, pentode 5.
Q3. A solid has resistivity 10⁻⁷ Ω m. According to the ranges in the chapter it is
- a metal
- a semiconductor only
- an insulator
- a semiconductor or an insulator
Hint and solution
Hint: Compare 10⁻⁷ Ω m with the metal range and the semiconductor range given in the chapter.
- Metal: ρ = 10⁻² to 10⁻⁸ Ω m — Metals have very low resistivity, from 10⁻² to 10⁻⁸ Ω m.
- Semiconductor: ρ = 10⁻⁵ to 10⁶ Ω m — Semiconductors lie higher, from 10⁻⁵ to 10⁶ Ω m.
- 10⁻⁷ is inside the first range, outside the second — The given value fits only the metal range.
- So the solid is a metal — The resistivity matches only the metal range, so the solid is a metal. Option A.
Answer: A. 10⁻⁷ lies in the metal range (10⁻² to 10⁻⁸) and outside the semiconductor range (10⁻⁵ to 10⁶).
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.