Module VI: Semiconductor Physics - Question Solutions

Question 1: Fermi-Dirac Distribution Function

Question

Write Fermi-Dirac distribution function and explain the terms involved in it.

Fermi-Dirac Distribution Function

Mathematical Expression

f(E) = 1 / [1 + exp((E - E_F)/k_BT)]

Fermi-Dirac Distribution Function

Explanation of Terms

Symbol	Name	Description
f(E)	Fermi Function	Probability that an energy state E is occupied by an electron (ranges from 0 to 1)
E	Energy Level	The energy of the quantum state being considered (in Joules or eV)
E_F	Fermi Energy	The energy level at which the probability of occupation is exactly 0.5 (50%)
k_B	Boltzmann Constant	1.38 × 10⁻²³ J/K or 8.617 × 10⁻⁵ eV/K
T	Absolute Temperature	Temperature in Kelvin
k_BT	Thermal Energy	≈ 0.026 eV at room temperature (300K)

Physical Interpretation

Fermi-Dirac Distribution at Different Temperatures

Fermi-Dirac distribution showing step function at T=0K and smoothing at higher temperatures

Key Properties:

At E = E_F: f(E_F) = 1/[1 + e⁰] = 1/2 = 0.5 (always 50% probability)
At T = 0K: f(E) = 1 for E < E_F and f(E) = 0 for E > E_F (step function)
At T > 0K: The distribution smooths out; some electrons above E_F, some empty states below E_F
For E >> E_F: f(E) ≈ exp[-(E - E_F)/k_BT] (Maxwell-Boltzmann limit)
For E << E_F: f(E) ≈ 1 (states are almost certainly filled)

Significance

Describes the statistical distribution of electrons in energy states at thermal equilibrium
Essential for understanding electrical conductivity in metals and semiconductors
Accounts for Pauli exclusion principle (fermions cannot occupy same state)
Used to calculate carrier concentrations in semiconductors

Question 2: Fermi Level in Intrinsic Semiconductor

Question

Show that for an intrinsic semiconductor, fermi level lies midway of conduction band and valence band. (OR) Derive an expression for Fermi level for an intrinsic semiconductor.

Proof

Energy Band Diagram of Intrinsic Semiconductor

Mathematical Derivation

Step 1: Electron concentration in conduction band:

n = N_C exp[-(E_C - E_F)/k_BT]

where N_C is the effective density of states in conduction band:

N_C = 2(2πm_e*k_BT/h²)^(3/2)

Step 2: Hole concentration in valence band:

p = N_V exp[-(E_F - E_V)/k_BT]

where N_V is the effective density of states in valence band:

N_V = 2(2πm_h*k_BT/h²)^(3/2)

Step 3: For an intrinsic semiconductor, the number of electrons equals the number of holes:

n = p = n_i

Step 4: Equating the two expressions:

N_C exp[-(E_C - E_F)/k_BT] = N_V exp[-(E_F - E_V)/k_BT]

Step 5: Taking natural logarithm of both sides:

ln(N_C) - (E_C - E_F)/k_BT = ln(N_V) - (E_F - E_V)/k_BT

Step 6: Rearranging:

(E_F - E_V) - (E_C - E_F) = k_BT × ln(N_C/N_V)

2E_F - E_V - E_C = k_BT × ln(N_C/N_V)

Step 7: Solving for E_F:

E_F = (E_C + E_V)/2 + (k_BT/2) × ln(N_V/N_C)

Step 8: Since N_C/N_V = (m_e*/m_h*)^(3/2):

E_F = (E_C + E_V)/2 + (3k_BT/4) × ln(m_h*/m_e*)

Final Result

E_F = (E_C + E_V)/2 + (3k_BT/4) × ln(m_h*/m_e*)

Special Case: If m_e* = m_h* (equal effective masses), then ln(1) = 0:

E_F = (E_C + E_V)/2

Fermi level lies exactly at the middle of the band gap

Key Points

For most semiconductors, m_e* ≈ m_h*, so E_F lies very close to the middle
The second term is temperature-dependent and usually small at room temperature
At T = 0K, Fermi level is exactly at the midpoint
This result applies only to intrinsic (undoped) semiconductors

Question 3: Mobility of Electrons - Numerical

Question

Explain mobility of electrons. The resistivity of Cu is 1.72 × 10⁻⁸ ohm-m. Calculate the mobility of electrons in Cu given that number of electrons per unit volume is 10.41 × 10²⁸ m⁻³.

Definition of Mobility

Electron Mobility (μ):

Mobility is defined as the drift velocity acquired by an electron per unit electric field. It represents how quickly an electron can move through a material when subjected to an electric field.

μ = v_d/E

where v_d = drift velocity (m/s), E = electric field (V/m)

Units: m²/V·s or cm²/V·s

Physical Significance:

Higher mobility means electrons move more freely through the material
Mobility depends on: temperature, impurity concentration, crystal structure
Related to mean free time (τ) between collisions: μ = eτ/m*
Conductivity is directly proportional to mobility: σ = neμ

Numerical Solution

Given Data:

Resistivity of Cu, ρ = 1.72 × 10⁻⁸ Ω·m
Electron concentration, n = 10.41 × 10²⁸ m⁻³
Charge of electron, e = 1.6 × 10⁻¹⁹ C

Formula:

The relationship between conductivity, carrier concentration, and mobility:

σ = neμ

Since σ = 1/ρ:

1/ρ = neμ

μ = 1/(neρ)

Calculation:

μ = 1/(n × e × ρ)

μ = 1/(10.41 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.72 × 10⁻⁸)

μ = 1/(10.41 × 1.6 × 1.72 × 10²⁸⁻¹⁹⁻⁸)

μ = 1/(28.65 × 10¹)

μ = 1/286.5

μ = 3.49 × 10⁻³ m²/V·s

Mobility of electrons in Cu, μ = 3.49 × 10⁻³ m²/V·s = 34.9 cm²/V·s

Typical Mobility Values

Material	Electron Mobility (cm²/V·s)
Copper	~35
Silicon (intrinsic)	~1350
Germanium	~3900
GaAs	~8500

Question 4: Effect of Indium Doping on Silicon

Question

Predict the effect on the electrical properties of silicon at room temperature if every millionth silicon atom is replaced by an atom of indium.
Given: Concentration of silicon atoms = 5 × 10²⁸ /m³
Intrinsic conductivity of silicon = 4.4 × 10⁻⁴ /Ω-m
Mobility of holes = 0.048 m²/V·sec

Understanding the Problem

Indium Doping:

Indium (In) is a Group III element with 3 valence electrons. When it replaces a silicon atom (Group IV, 4 valence electrons), it creates an acceptor level and produces a p-type semiconductor. Each indium atom creates one hole.

Given Data:

Concentration of Si atoms, N_Si = 5 × 10²⁸ m⁻³
Doping ratio = 1 in 10⁶ (every millionth atom)
Intrinsic conductivity, σᵢ = 4.4 × 10⁻⁴ Ω⁻¹m⁻¹
Mobility of holes, μₕ = 0.048 m²/V·s
Charge of electron, e = 1.6 × 10⁻¹⁹ C

Step 1: Calculate acceptor (Indium) concentration

N_A = N_Si/10⁶

N_A = (5 × 10²⁸)/10⁶

N_A = 5 × 10²² m⁻³

Step 2: Determine hole concentration

At room temperature, all acceptor atoms are ionized, so:

p ≈ N_A = 5 × 10²² m⁻³

(This is much greater than intrinsic carrier concentration n_i ≈ 10¹⁶ m⁻³)

Step 3: Calculate conductivity due to holes (extrinsic)

For a p-type semiconductor, conductivity is dominated by holes:

σ = p × e × μₕ

σ = 5 × 10²² × 1.6 × 10⁻¹⁹ × 0.048

σ = 5 × 1.6 × 0.048 × 10²²⁻¹⁹

σ = 0.384 × 10³

σ = 384 Ω⁻¹m⁻¹

Step 4: Calculate resistivity

ρ = 1/σ = 1/384

ρ = 2.6 × 10⁻³ Ω·m

Step 5: Compare with intrinsic silicon

Intrinsic resistivity:

ρᵢ = 1/σᵢ = 1/(4.4 × 10⁻⁴) = 2273 Ω·m

Ratio of conductivities:

σ_doped/σᵢ = 384/(4.4 × 10⁻⁴) = 8.7 × 10⁵

Results Summary

Property	Intrinsic Si	Doped Si (with In)
Type	Intrinsic	p-type
Conductivity	4.4 × 10⁻⁴ Ω⁻¹m⁻¹	384 Ω⁻¹m⁻¹
Resistivity	2273 Ω·m	2.6 × 10⁻³ Ω·m

The conductivity increases by a factor of approximately 8.7 × 10⁵ (nearly a million times).
The doped silicon becomes a p-type semiconductor with σ = 384 Ω⁻¹m⁻¹

Effects of Indium Doping

Silicon becomes p-type semiconductor (majority carriers are holes)
Conductivity increases dramatically (~10⁶ times)
Fermi level shifts towards the valence band
Hole concentration >> electron concentration
Temperature dependence of conductivity changes