Answer:
a) 222/cm^3
b) The Ec with respect to Ef and Ev = -0.107ev
c) Resistance, R = 467 ohms
Explanation:
(a) From the velocity and the applied electric field, we can calculate the mobility of holes:
υdp = µpε, µp = υdp/ε = 2×10^5/1000
= 200cm2/V.s
From a), we find Nd is equal to 4.5×1017/cm3
. Hence,
n = Nd = 4.5×10^17/cm3
, and p = ni^2/n
= ni^2/Nd
= 10^20 / 4.5×10^17 = 222/cm3
.
Clearly, the minority carrier is hole.
(b) The Fermi level with respect to Ec is
Ef = Ec - kTln(Nd/Nc) = Ec - 0.107 eV.
(c) R = ρL/A. Using Equation, we first calculate the resistivity of the sample:
σ = q(µn n + µp p) ≈ qµn n = 1.6×10^-19 × 400 × 4.5×10^17 = 28.8/Ω-cm, and
ρ= σ
-1 = 0.035 Ω-cm.
Therefore, R = (0.035) × 20µm / (10µm× 1.5µm) = 467 Ω.
