An abrupt silicon - junction has its side doped to

PHYSICS · Multi-part · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Semiconductor physics and devices · Semiconductor physics and devices

Solid State Physics, Ch Junctions and devices, finished

Problem

An abrupt silicon pp-nn junction has its pp side doped to NA=3.0×1016 cm−3N_A=3.0\times10^{16}\,\mathrm{cm^{-3}}, known from the growth. The nn-side doping NDN_D is unknown but of the same order, so neither side's share of the depletion width can be neglected. Measured at T=300 KT=300\,\mathrm{K}, the depletion capacitance per unit area CC at two reverse biases VV (with V<0V<0 for reverse bias) gives: V=−1.0 VV=-1.0\,\mathrm{V}: 1/C2=2.11×1015 cm4 F−21/C^2 = 2.11\times10^{15}\,\mathrm{cm^4\,F^{-2}}. V=−3.0 VV=-3.0\,\mathrm{V}: 1/C2=4.52×1015 cm4 F−21/C^2 = 4.52\times10^{15}\,\mathrm{cm^4\,F^{-2}}. Use εs=11.7\varepsilon_s=11.7 for silicon, ε0=8.854×10−14 F cm−1\varepsilon_0=8.854\times10^{-14}\,\mathrm{F\,cm^{-1}}, e=1.602×10−19 Ce=1.602\times10^{-19}\,\mathrm{C}, ni=1.0×1010 cm−3n_i=1.0\times10^{10}\,\mathrm{cm^{-3}} and kBT/e=25.85 mVk_BT/e=25.85\,\mathrm{mV}. ((a)) Show that 1/C2=2(Vbi−V)eεsε0Neff1/C^2 = \dfrac{2(V_{bi}-V)}{e\varepsilon_s\varepsilon_0 N_{\rm eff}} with Neff=NAND/(NA+ND)N_{\rm eff}=N_AN_D/(N_A+N_D), use the two data points to extract NeffN_{\rm eff}, and from it find NDN_D. ((b)) Extract the built-in voltage VbiV_{bi} from the data, and compare it with (kBT/e)ln⁡(NAND/ni2)(k_BT/e)\ln(N_AN_D/n_i^2) for the dopings you now have.

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