Semiconductor Analysis PDF

# Sheet 1(Basics of Semiconductor) 1. **(a)** Determine the concentration of free electrons and holes in a sample of germanium at 300°K which has a concentration of donor atoms equal to 2x10¹⁴ atoms/cm³ and a concentration of acceptor atoms equal to 3x10¹⁴ atoms/ c...

# Sheet 1(Basics of Semiconductor) 1. **(a)** Determine the concentration of free electrons and holes in a sample of germanium at 300°K which has a concentration of donor atoms equal to 2x10¹⁴ atoms/cm³ and a concentration of acceptor atoms equal to 3x10¹⁴ atoms/ cm³. Is this p- or n-type germanium? In other words, is the conductivity due primarily to holes or to electrons? (n_i for Ge at 300°K=2.5x10¹³ atoms/cm³) - **(b)** Repeat part (a) for equal donor and acceptor concentrations of 10¹⁵ atoms/cm³. Is this p- or n-type germanium? - **(c)** Repeat part (a) for donor concentration of 10¹⁶ atoms/cm³ an acceptor concentration of 10¹⁴ atoms/cm³. ## Solution **(a)** - ND + p = NA + n - P + (ND - NA) - n = 0 - n p = n_i² - p = -(ND-NA) ± √(ND - NA)² + 4n_i² / 2 - Choose the ‘+’ sign since p > 0 and substituting n_i =2.5x10¹³ atoms/cm³ one obtains - p = -1 x 10¹⁴ + √10²⁸ + 2.5 x 10²⁷ / 2 - p = 1.06 x 10¹⁴ holes/cm³ - P + (ND - NA ) – (n_i²/p) = 0 - Multiplying both sides by p - P² + (ND - NA )p – n_i² = 0 - n = P + (ND - N_i ) =1.06 x 10¹⁴ – 1 х 10¹⁴ - n =0.06 x 10¹⁴ electrons/cm³ - Therefore, the sample is p-type **(b)** - ND = NA - ∴ n² = p² = n_i² - P = n - n = p = n_i = 2.5 x 10¹³ / cm³ - The sample is an intrinsic semiconductor **(c)** - Since ND >> NA - n ≈ N_D =10¹⁶ electrons/cm³ - p = n_i² / n = 6.25x10²⁶/10¹⁶ - p = 6.25x10¹⁰ holes/cm³ - The sample is n-type # 2) Calculate the resistance of a bar 10 µm long, 3 µm wide, and 1 µm thick, made from the following material : - (a) intrinsic silicon. - (b) n-doped silicon with ND=10¹⁶ /cm³ - (c) p-doped silicon with Na=10¹⁸ /cm³ - (d) Extrinsic silicon with Na=3x10¹⁴ /cm³, Np=2x10¹⁴/cm³ - (e) aluminum with resistivity of 2.8 μΩ.cm. (for doped silicon, assume μ_n=2.5 μ_p=1200 cm²/V.s) (for intrinsic silicon μ_n=1350 cm²/V.s and μ_p=480 cm²/V.s, n_i =1.5 x 10¹⁰ /cm³) (Remember : R=pL/A) ## Solution # 3) A sample of germanium is doped to the extent of 2x10¹¹donor atoms/ cm³ and 3x10¹⁴ acceptors atoms/ cm³ at the temperature of the sample the resistivity of intrinsic germanium is 60 ohms.cm. Assume that the value of the mobility of holes and electrons μ_n=1800 cm²/V.s and μ_p=3800 cm²/V.s. if the applied electric filed intensity is 2v/cm, find the total conduction current density ## Example **(a)** A sample of germanium is doped to the extent of 3x10¹⁴ donor atoms/cm³ and 2x10¹⁴ acceptor atoms/cm³. At the temperature of the sample the resistivity of pure (intrinsic) germanium is 60 Ω-cm. Assume that the value of the mobility of holes and electrons is approximately the same as at 300°K (μ_n=1800cm²/V.s and μ_p=1800cm²/V.s). If the applied electric field intensity is 2V/cm, find the total conduction current density. ## Solution The current density J that results from an electric field is obtained by: - J = q (nμ_n + pμ_p) ε =δε A\cm² - To find n and p, we first find n_i - 1/p= 1/δq(nμ_n + pμ_p) - For intrinsic germanium, p = n = n_i - ρ_i = qn_i(μ_n + μ_p) - n_i = ρ_i/ q(μ_n + μ_p) - n_i = 1.6×10^-19×60×(3800+1800) - n_i =1.86×10¹³ cm^-3 - From mass action law and neutral equation - ND +p = Na+n - n+ (NA - ND ) -p = 0 - n p = n_i² - n+ (NA - ND ) – (n_i² /n) = 0 - Multiplying both sides by n - n² + (NA - ND )n – n_i² = 0 - n = -(NA-ND) + √(N_i - N₂)²+ 4n_i² / 2 - n= 1×10¹⁴ + √10²⁸ +4×(1.86×10¹³)² / 2 - n=1.0335×10¹⁴ electrons / cm³ - p = n_i² /n - p = n + (N_i - ND) - p = 1.0335 x 10¹⁴ – 1 x 10¹⁴ - p = 0.0335 x 10¹⁴ holes/cm³ - J =q(nμ_n + pμ_p)ε - J = 1.6×10^-19 (1.0335×10¹⁴ × 3800 +0.0335×10¹⁴ ×1800)×2 - J = 0.1276 A/cm² # 4) In a germanium sample, ND = 10¹⁴/cm³, NA = 7 × 10¹³/ cm³. It is kept at a low temperature, where pure Ge would have resistivity of 60 Ω-cm. Find the current density that would result by applying an electric field as 2 V/cm to this sample. - (a) 0. 522 A/cm² - (b) 1. 022 A/cm² - (c) 0.0522 A/cm² - (d) 0.0452 A/cm² ## Solution: Given that ND = 10¹⁴, NA = 7 × 10¹³, p = 60. Then conductivity is given by σ = [(μ_n + μ_p)n_i]q - σ = 5600 × n_i; ×1.6×10^-19 Therefore, intrinsic concentration is - n_i = 1/ 60(5600×1.6×10^-19) = 1.86×10¹³ Thus, the number of electrons is given by - n = √(ND - NA/2)² + n_i² - n = √(3×10¹³/2)² + (1.86×10¹³)² - n = √(3×10¹³ + 10³ x 2.38) = 3.88×10¹³ Then number of holes is given by - p = −1.5×10¹³ + 2.38×10¹³ = 0.88 × 10¹³ Thus the resistivity on application of voltage is - σ = (ημ_n + μ_pp) × q - σ = (3.88×10¹³ × 3600 + 1800 × 0.88 × 10¹³) 1.6×10^-18 - σ = 0.0261 S/m The current density on application of voltage is given by - J =σE - J = 0.0261×2 - J = 0.0522 A/cm² Ans. (c) # 5) Determine the forward current of a Ge diode operating at room temperature with a forward voltage of 100 mV across it with reverse saturation current 20 μΑ. - (a) 1.936 mA - (b) 0.0486 mA - (c) 0.639 mA - (d) 0.936 mA ## Solution: The forward current is given by - I_F = I₀ [e^V₀/nV_T − 1] Substituting given values and using η = 1 for Ge and V_T at room temperature 300 K as ≈26 mV, we have - I_F = 20×10^-6 e^{100 x 10^-3/ 26 ×10^-3} -1 - I_F = 20×10^-6[47.46 -1] - I_F = 0.92 mA Ans. (d)