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Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations of 1.5 x 10¹⁶ m⁻³. Doping by indium increases nₕ to 4.5 x 10²² m⁻³. The doped semiconductor is of
Options
(a) n-type with electron concentration nₑ = 5 x 10²² m⁻³
(b) p-type with electron concentration nₑ = 2.5 x 10¹⁰ m⁻³
(c) n-type with electron concentration nₑ = 2.5 x 10²³ m⁻³
(d) p-type with electron concentration nₑ = 5 x 10⁹ m⁻³
Correct Answer:
p-type with electron concentration nₑ = 5 x 10⁹ m⁻³
Explanation:
nᵢ² = nₑ nₕ
(1.5 x 10¹⁶)² = nₑ (4.5 x 10²²)
⇒ nₑ = 0.5 x 10¹⁰ or nₑ = 5 x 10⁹
Given nₕ = 4.5 x 10²² ⇒ nₕ >> nₑ
semiconductor is p-type and
nₑ = 5 x 10⁹ m⁻³.
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Topics: Electronic Devices
(124)
Subject: Physics
(2479)
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