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CBSE · Class 12 · Physics

Semiconductor Electronics

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
Silicon has an intrinsic carrier concentration of 1 × 1016 m−3. It is doped with 9 × 1020 donor atoms per m3. The hole concentration is:
  1. 1.11 × 1011 m−3
  2. 9 × 1020 m−3
  3. 1 × 1016 m−3
  4. 1.11 × 10−5 m−3
2.
Doping pure germanium with indium makes it:
  1. n-type, with electrons as majority carriers
  2. an insulator
  3. p-type, with holes as majority carriers
  4. p-type, with electrons as majority carriers
3.
The input to a full-wave rectifier is 1000 Hz AC. The frequency of the output ripple is:
  1. 0 Hz
  2. 1000 Hz
  3. 2000 Hz
  4. 500 Hz
4.
A 2 V battery drives current through a forward-biased ideal diode and an 800 Ω resistor in series. The current is:
  1. 2.5 mA
  2. 0 mA
  3. 0.25 mA
  4. 3.375 mA
5.
In an intrinsic semiconductor, the number of free electrons is:
  1. equal to the number of holes
  2. zero
  3. more than the number of holes
  4. less than the number of holes
6.
A solid whose band gap is 0.2 eV is:
  1. a superconductor
  2. an insulator
  3. a conductor
  4. a semiconductor
7.
The p side of a junction diode is at −4 V and the n side at 3 V. The diode is:
  1. reverse biased
  2. broken down
  3. forward biased
  4. unbiased
8.
Assertion (A): In silicon doped with arsenic, the majority carriers are electrons.
Reason (R): Pentavalent impurities make p-type semiconductors.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Pure silicon has ni = 1 × 1016 m−3. It is doped with 6 × 1023 pentavalent atoms per m3. Find the electron and hole concentrations, and name the type.
10.
Silicon has 5 × 1028 atoms per m3. It is doped with 1 atom of arsenic for every 108 silicon atoms, and ni = 1 × 1016 m−3. Find the number of donor atoms per m3 and the hole concentration.
11.
A 14 V battery is connected to two branches in parallel: one has an ideal diode (forward biased) and a 950 Ω resistor; the other has an ideal diode that is reverse biased and a 900 Ω resistor. Find the current drawn from the battery.
12.
For a silicon diode, the current is 20 mA at 0.61 V and 46 mA at 0.77 V. Find its dynamic resistance in this range.
13.
A 100 Hz AC voltage of peak value 28 V is fed to (a) a half-wave rectifier, (b) a full-wave rectifier, each with ideal diodes. Find the output ripple frequency and the peak output voltage in each case.
14.
(a) A solid has a band gap of 9 eV. What kind of material is it? (b) Silicon is doped with aluminium. What type does it become, and what are its majority carriers?
15.
A silicon diode (0.7 V drop when conducting) and an 850 Ω resistor are in series with a 17 V battery. Find the current when the diode is forward biased and when it is reverse biased (ignore leakage).
16.
In a circuit, the p side of a diode is at 6 V and the n side at −10 V. Is the diode forward or reverse biased, and will an ideal diode conduct?
Section C — Word problems4 × 4 = 16 marks
17.
Silicon with ni = 5 × 1015 m−3 is doped with 3 × 1020 pentavalent atoms per m3. Find the majority and minority carrier concentrations, name the type, and say what happens to the band diagram.
18.
Two silicon diodes (0.7 V drop each) are joined in series, both forward biased, with resistors of 200 Ω and 1000 Ω, across a 19 V battery. Find the current and the voltage across each resistor. What changes if one diode is turned round?
19.
A full-wave rectifier with ideal diodes is fed a 60 Hz supply of peak 20 V and drives a 300 Ω load. Explain how it works, and find the ripple frequency and the peak load current.
20.
Diode A has its p side at 2 V and n side at −2 V; diode B has its p side at −4 V and n side at 5 V. State the bias of each, which one conducts, and explain what happens to the depletion layer in each case.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Physics

Answer key

Semiconductor Electronics · AW-12P-SEM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 1.11 × 1011 m−3nₑnₕ = ni2; nₑ ≈ N_D
2(c) p-type, with holes as majority carriersindium is trivalent
3(c) 2000 Hzboth halves of each cycle give a pulse
4(a) 2.5 mAI = 2/800
5(a) equal to the number of holes—
6(d) a semiconductorsmall gap → semiconductor; large → insulator; overlap → conductor
7(a) reverse biasedforward bias: p side at the higher potential
8(c) A is true, but R is false.A is true: arsenic is pentavalent and gives a spare electron. R is false.
9nₑ ≈ 6 × 1023 m−3; nₕ = 1.67 × 108 m−3; n-typenₕ = ni2/nₑ
105 × 1020 m−3; 2 × 1011 m−3N_D = 5 × 1028 ÷ 108; nₕ = ni2/N_D
1114.7 mAthe reverse-biased branch carries no current
126.15 Ωr = ΔV/ΔI
13(a) 100 Hz, 28 V; (b) 200 Hz, 28 Videal diodes pass the full peak
14(a) an insulator; (b) p-type, holes—
1519.2 mA; 0 mAI = (17 − 0.7)/850
16forward biased; it conductsconduction needs the p side higher
17electrons ≈ 3 × 1020 m−3; holes = 8.33 × 1010 m−3; n-type; a donor level appears just below the conduction bandminority = ni2/majority
1814.7 mA; 2.93 V and 14.7 V; with one diode reversed, no current flowsI = (19 − 1.4)/(200 + 1000)
19the two diodes conduct on alternate half-cycles, so current flows through the load the same way in both; 120 Hz; 66.7 mAI0 = V0/R
20A: forward biased; B: reverse biased; A conducts; forward bias narrows the depletion layer, reverse bias widens it—