Answer:
a.
i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2 × 10¹¹ cm⁻³
ii. electron concentration, n₀ = 1.33 × 10¹¹ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³
b.
i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2.205 × 10⁻³ cm⁻³
ii. electron concentration, n₀ = 1.47 × 10⁻³ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³
c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.
Explanation:
a. For Germanium, intrinsic concentration n₁ = 2 × 10¹³ cm⁻³.
i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,
n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂
n₀ = 1/2[(2 × 10¹⁵ cm⁻³ - 0) +√[(2 × 10¹⁵ cm⁻³ - 0)² + 4(2 × 10¹³ cm⁻³)²] ]
n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]
n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4.0016 × 10³⁰ cm⁻⁶] ]
n₀ = 1/2[(2 × 10¹⁵ cm⁻³ + 2.0004 × 10¹⁵ cm⁻³ ]
n₀ = 1/2[4.0004 × 10¹⁵ cm⁻³ ]
n₀ = 2.0002 × 10¹⁵ cm⁻³ ≅ 2 × 10¹⁵ cm⁻³
The hole concentration p₀ is gotten from
n₀p₀ = n₁²
p₀ = n₁²/n₀ = (2 × 10¹³ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/2 × 10¹⁵ cm⁻³
p₀ = 2 × 10¹¹ cm⁻³
ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,
p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁
p₀ = 1/2[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³) +√[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³)² + 4(2 × 10¹³ cm⁻³)²] ]
p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]
p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9.0016 × 10³⁰ cm⁻⁶] ]
p₀ = 1/2[(3 × 10¹⁵ cm⁻³ + 3.0003 × 10¹⁵ cm⁻³ ]
p₀ = 1/2[6.0003 × 10¹⁵ cm⁻³ ]
p₀ = 3.00015 × 10¹⁵ cm⁻³ ≅ 3 × 10¹⁵ cm⁻³
Te electron concentration n₀ is gotten from
n₀p₀ = n₁²
n₀ = n₁²/p₀ = (2 × 10¹³ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/3 × 10¹⁵ cm⁻³
n₀ = 1.33 × 10¹¹ cm⁻³
b. For GaAs, intrinsic concentration n₁ = 2 × 10⁶ cm⁻³.
i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,
n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ] since N₁ > N₂ and N₁ - N₂ = 2 × 10¹⁵ cm⁻³ >> n₁ = 2 × 10⁶ cm⁻³
n₀ = (N₁ - N₂) = 2 × 10¹⁵ cm⁻³ - 0 = 2 × 10¹⁵ cm⁻³
The hole concentration p₀ is gotten from
n₀p₀ = n₁²
p₀ = n₁²/n₀ = (2.1 × 10⁶ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/2 × 10¹⁵ cm⁻³
p₀ = 2.205 × 10⁻³ cm⁻³
ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 10¹⁶ cm⁻³,
p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ] since N₂ > N₁ and N₂ - N₁ = 10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³ = 3 × 10¹⁵ cm⁻³ >> n₁ = 2.1 × 10⁶ cm⁻³
p₀ ≅ N₂ - N₁ = 3 × 10¹⁵ cm⁻³
The electron concentration n₀ is gotten from
n₀p₀ = n₁²
n₀ = n₁²/p₀ = (2.1 × 10⁶ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/3 × 10¹⁵ cm⁻³
n₀ = 1.47 × 10⁻³ cm⁻³
c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.
