A Si 𝑝𝑛 step junction with NA = 1019 cm-3 (consider this to be nondegenerate),ND = 1015 cm-3, Ks=11.7, and 𝑛𝑖 = 1010 cm-3 is at room temperature (kT/q = 0.026 V).Using the depletion approximation and VA = 0, calculate: [20 Pts](a) Vbi; (b) xn, xp, and W (all in cm); (c) what percentage of W is on 𝑝-side? The 𝑛-side?(d) 𝑛𝑝 and 𝑝𝑛; (e) the electric field at x = 0Using the depletion approximation and VA = 0, draw:(f) the charge density and the electric field to the same x-axis scale. [15 Pts](g) the energy band diagram for the 𝑝𝑛 diode. Calculate EF-Ei on the quasi-neutral 𝑝- and 𝑛-sides and indicate these on the band diagram. [15 Pts]
Question
A Si 𝑝𝑛 step junction with NA = 1019 cm-3 (consider this to be nondegenerate),ND = 1015 cm-3, Ks=11.7, and 𝑛𝑖 = 1010 cm-3 is at room temperature (kT/q = 0.026 V).Using the depletion approximation and VA = 0, calculate: 20 Pts Vbi; (b) xn, xp, and W (all in cm); (c) what percentage of W is on 𝑝-side? The 𝑛-side?(d) 𝑛𝑝 and 𝑝𝑛; (e) the electric field at x = 0Using the depletion approximation and VA = 0, draw:(f) the charge density and the electric field to the same x-axis scale. 15 Pts the energy band diagram for the 𝑝𝑛 diode. Calculate EF-Ei on the quasi-neutral 𝑝- and 𝑛-sides and indicate these on the band diagram. [15 Pts]
Solution
This is a complex problem that involves several steps. Let's break it down:
(a) The built-in potential (Vbi) can be calculated using the formula:
Vbi = (kT/q) * ln((NA * ND) / ni^2)
Substituting the given values:
Vbi = 0.026 * ln((10^19 * 10^15) / (10^10)^2) Vbi = 0.026 * ln(10^24) Vbi = 0.026 * 55.26 Vbi = 1.44 V
(b) The depletion width (W) can be calculated using the formula:
W = sqrt((2 * Ks * ε0 * Vbi * (NA + ND)) / (q * NA * ND))
Substituting the given values:
W = sqrt((2 * 11.7 * 8.85 * 10^-14 * 1.44 * (10^19 + 10^15)) / (1.6 * 10^-19 * 10^19 * 10^15)) W = sqrt((2 * 11.7 * 8.85 * 10^-14 * 1.44 * 10^19) / (1.6 * 10^-19 * 10^19)) W = sqrt(2.33 * 10^-6) W = 1.53 * 10^-3 cm
The depletion region widths on the n-side (xn) and p-side (xp) can be calculated using the formulas:
xn = W * (NA / (NA + ND)) xp = W * (ND / (NA + ND))
Substituting the given values:
xn = 1.53 * 10^-3 * (10^19 / (10^19 + 10^15)) xn = 1.53 * 10^-3 * 0.9999 xn = 1.53 * 10^-3 cm
xp = 1.53 * 10^-3 * (10^15 / (10^19 + 10^15)) xp = 1.53 * 10^-3 * 0.0001 xp = 1.53 * 10^-7 cm
(c) The percentage of W on the p-side and n-side can be calculated as:
Percentage on p-side = (xp / W) * 100% Percentage on n-side = (xn / W) * 100%
Substituting the given values:
Percentage on p-side = (1.53 * 10^-7 / 1.53 * 10^-3) * 100% Percentage on p-side = 0.01%
Percentage on n-side = (1.53 * 10^-3 / 1.53 * 10^-3) * 100% Percentage on n-side = 99.99%
(d) The minority carrier concentrations on the p-side (np) and n-side (pn) can be calculated using the formulas:
np = ni^2 / NA pn = ni^2 / ND
Substituting the given values:
np = (10^10)^2 / 10^19 np = 10^1 cm^-3
pn = (10^10)^2 / 10^15 pn = 10^5 cm^-3
(e) The electric field at x = 0 (E0) can be calculated using the formula:
E0 = q * NA * xp / (Ks * ε0)
Substituting the given values:
E0 = 1.6 * 10^-19 * 10^19 * 1.53 * 10^-7 / (11.7 * 8.85 * 10^-14) E0 = 2.61 * 10^5 V/cm
(f) and (g) These parts of the question require graphical representation which cannot be provided in this text-based format. However, the charge density would be a step function with a jump at the junction (x = 0), and the electric field would be a linear function decreasing from its maximum value at x = 0. The energy band diagram would show the conduction and valence bands bending upwards on the n-side and downwards on the p-side, with the Fermi level constant throughout the device. The difference EF-Ei on the quasi-neutral p- and n-sides can be calculated using the formulas:
EF-Ei (p-side) = kT * ln(NA / ni) EF-Ei (n-side) = kT * ln(ND / ni)
Substituting the given values:
EF-Ei (p-side) = 0.026 * ln(10^19 / 10^10) EF-Ei (p-side) = 0.026 * 21.97 EF-Ei (p-side) = 0.57 V
EF-Ei (n-side) = 0.026 * ln(10^15 / 10^10) EF-Ei (n-side) = 0.026 * 11.51 EF-Ei (n-side) = 0.30 V
These values would be indicated on the energy band diagram.
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