Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
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Find the longest wavelength of light that can create an electron–hole pair in silicon (E_g = 1.1 eV) and in germanium (0.7 eV).
Given: E_g(Si) = 1.1 eV; E_g(Ge) = 0.7 eV
Use: λ = hc/E_g = 1240 eV nm / E_g
Si: λ = 1240/1.1 = 1127 nm
Ge: λ = 1240/0.7 = 1771 nm
Answer: 1130 nm; 1770 nm (both infrared)
A solid has a band gap of 5.4 eV. Is it a conductor, semiconductor or insulator? What is the cut-off wavelength to excite an electron across the gap?
Given: E_g = 5.4 eV
Use: Compare with 3 eV
E_g > 3 eV so it is an insulator (diamond)
λ = 1240/5.4 = 230 nm (ultraviolet)
Answer: Insulator; 230 nm
In doped silicon nₑ = 4.5 × 10²² m⁻³ and nᵢ = 1.5 × 10¹⁶ m⁻³. Find the hole concentration.
Given: nₑ = 4.5 × 10²²; nᵢ = 1.5 × 10¹⁶
Use: nₑ nₕ = nᵢ²
nₕ = (1.5 × 10¹⁶)² / 4.5 × 10²² = 2.25 × 10³² / 4.5 × 10²² = 5 × 10⁹ m⁻³
Answer: 5 × 10⁹ m⁻³
Silicon is doped with arsenic and germanium with boron. Name the type of each, and the majority carriers.
Given: As is pentavalent; B is trivalent
Use: Pentavalent gives n-type; trivalent gives p-type
Si + As: n-type, majority carriers electrons
Ge + B: p-type, majority carriers holes
Answer: n-type (electrons); p-type (holes)
In an n-type sample nₑ = 10²² m⁻³ and nᵢ = 10¹⁶ m⁻³. By what factor do electrons exceed holes?
Given: nₑ = 10²²; nᵢ = 10¹⁶
Use: nₕ = nᵢ²/nₑ
nₕ = 10³² / 10²² = 10¹⁰ m⁻³
nₑ/nₕ = 10²² / 10¹⁰ = 10¹²
Answer: 10¹²
A junction has a built-in potential of 0.7 V. Find the effective barrier for a forward bias of 0.5 V and a reverse bias of 5 V.
Given: V₀ = 0.7 V
Use: V₀ − V (forward); V₀ + V (reverse)
Forward: 0.7 − 0.5 = 0.2 V
Reverse: 0.7 + 5 = 5.7 V
Answer: 0.2 V; 5.7 V
An ideal diode and a 1 kΩ resistor are connected to a 5 V battery in forward bias. Find the current. Repeat with a silicon diode with a 0.7 V drop.
Given: V = 5 V; R = 1 kΩ
Use: I = (V − V_D)/R
Ideal: I = 5/1000 = 5 mA
Silicon: I = (5 − 0.7)/1000 = 4.3 mA
Answer: 5 mA; 4.3 mA
The diode above is reversed. What is the current (ideal diode)? What voltage appears across the diode?
Given: Reverse bias; ideal
Use: Ideal diode: infinite reverse resistance
I = 0
The full 5 V appears across the diode
Answer: 0 A; 5 V
The forward current of a diode changes from 10 mA to 20 mA when the voltage changes from 0.7 V to 0.8 V. Find the dynamic resistance.
Given: ΔV = 0.1 V; ΔI = 10 mA
Use: r_d = ΔV/ΔI
r_d = 0.1 / 0.01 = 10 Ω
Answer: 10 Ω
The input frequency to a rectifier is 50 Hz. Find the output frequency for a half-wave and for a full-wave rectifier.
Given: f = 50 Hz
Use: Half-wave: f; full-wave: 2f
Half-wave: 50 Hz
Full-wave: 100 Hz
Answer: 50 Hz; 100 Hz
A transformer gives 12 V rms to a half-wave rectifier with a silicon diode (0.7 V drop). Find the peak output voltage.
Given: V_rms = 12 V
Use: V_peak = √2 V_rms − 0.7
V_peak(input) = 1.414 × 12 = 17.0 V
Peak output = 17.0 − 0.7 = 16.3 V
Answer: about 16.3 V
How many diodes are needed for (a) a half-wave rectifier, (b) a centre-tap full-wave rectifier, (c) a bridge rectifier?
Given: —
Use: From the circuits
(a) 1
(b) 2
(c) 4
Answer: 1; 2; 4
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In n-type silicon the dopants are pentavalent atoms (P, As, Sb), which give extra free electrons. So electrons are the majority carriers and holes are the minority carriers.
(a) is false (the dopants are not trivalent), (b) is false (electrons are the majority), (d) is false. The true statement is (c).
In p-type semiconductors the dopants are trivalent atoms (B, Al, In), which create holes. Holes are the majority carriers and electrons the minority carriers.
Hence statement (d) is true.
The band gaps are about 5.4 eV for carbon (diamond), 1.1 eV for silicon and 0.7 eV for germanium. So (Eg)C > (Eg)Si > (Eg)Ge: statement (c) is true.
Diffusion is the movement of carriers from a region of higher concentration to a region of lower concentration. The hole concentration is far higher in the p-region than in the n-region, so holes diffuse into the n-region. The potential difference (barrier) in fact opposes this movement. Answer: (c).
In forward bias the applied voltage V opposes the built-in potential V₀, so the effective barrier becomes V₀ − V and the depletion region narrows. The majority carrier current increases. Answer: (c).
Half-wave: the diode conducts in one half of every cycle, giving one output pulse per cycle, so the output frequency is 50 Hz.
Full-wave: both half cycles produce output pulses, so there are two pulses per cycle and the output frequency is 100 Hz.