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pn junction from 5 Ω cm resistivity

Short answer

Find doping from resistivity: Na = 1/(5×1.6e-19×450) = 2.77×10^15 cm^-3, Nd = 1/(5×1.6e-19×1350) = 9.25×10^14 cm^-3. V0 = 0.02586 ln(NaNd/ni²) = 0.601 V. Total W = 1.07 μm (Wn = 0.80 μm, Wp = 0.27 μm). Maximum field |E0| = 1.13×10^4 V/cm.

A pn junction semiconductor has resistivity of 5Ω cm. If mobility of holes is 450 cm²/Vs, and electron mobility is three times the mobility of holes at room temperature, find i) Built in potential ii) Depletion width that lies in n-region and p-region respectively iii) Built in electric field at x=0.

Assumptions: silicon at 300 K; both the p-side and the n-side have resistivity 5 Ω\Omega cm; ni=1.45×1010n_i = 1.45\times10^{10} cm−3^{-3}, εr=11.9\varepsilon_r = 11.9, kT/e=0.02586kT/e = 0.02586 V; abrupt junction, full ionisation.

Given: μh=450\mu_h = 450 cm2^2/Vs, μe=3μh=1350\mu_e = 3\mu_h = 1350 cm2^2/Vs.

Doping levels from resistivity

Majority carriers dominate, so ρ=1/(eNμ)\rho = 1/(eN\mu):

Na=1ρeμh=15×1.602×10−19×450=2.77×1015 cm−3Nd=1ρeμe=15×1.602×10−19×1350=9.25×1014 cm−3\begin{aligned} N_a &= \frac{1}{\rho e\mu_h} = \frac{1}{5 \times 1.602\times10^{-19} \times 450} = 2.77\times10^{15}\ \text{cm}^{-3} \\ N_d &= \frac{1}{\rho e\mu_e} = \frac{1}{5 \times 1.602\times10^{-19} \times 1350} = 9.25\times10^{14}\ \text{cm}^{-3} \end{aligned}

(i) Built-in potential

V0=kTeln⁡NaNdni2=0.02586ln⁡2.77×1015×9.25×1014(1.45×1010)2=0.02586×23.22=0.601 V\begin{aligned} V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} = 0.02586 \ln\frac{2.77\times10^{15} \times 9.25\times10^{14}}{(1.45\times10^{10})^2} \\ &= 0.02586 \times 23.22 = 0.601\ \text{V} \end{aligned}

(ii) Depletion widths

With ε=11.9×8.854×10−14=1.054×10−12\varepsilon = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12} F/cm:

W=2ε(Na+Nd)V0eNaNd=2×1.054×10−12×3.70×1015×0.6011.602×10−19×2.77×1015×9.25×1014=1.067×10−4 cm=1.067 μmWn=NaNa+NdW=0.75×1.067=0.800 μmWp=NdNa+NdW=0.25×1.067=0.267 μm\begin{aligned} W &= \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}} \\ &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times 3.70\times10^{15} \times 0.601}{1.602\times10^{-19} \times 2.77\times10^{15} \times 9.25\times10^{14}}} \\ &= 1.067\times10^{-4}\ \text{cm} = 1.067\ \mu\text{m} \\ W_n &= \frac{N_a}{N_a+N_d}W = 0.75 \times 1.067 = 0.800\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W = 0.25 \times 1.067 = 0.267\ \mu\text{m} \end{aligned}

(iii) Built-in field at x=0x = 0

∣E0∣=eNdWnε=1.602×10−19×9.25×1014×8.00×10−51.054×10−12=1.13×104 V/cm\begin{aligned} |E_0| &= \frac{eN_dW_n}{\varepsilon} = \frac{1.602\times10^{-19} \times 9.25\times10^{14} \times 8.00\times10^{-5}}{1.054\times10^{-12}} \\ &= 1.13\times10^{4}\ \text{V/cm} \end{aligned}

It is directed from n to p. Check: 12∣E0∣W=0.5×1.126×104×1.067×10−4=0.601\frac{1}{2}|E_0|W = 0.5 \times 1.126\times10^4 \times 1.067\times10^{-4} = 0.601 V.

Answer: V0=0.601V_0 = 0.601 V; Wn=0.800W_n = 0.800 μ\mum, Wp=0.267W_p = 0.267 μ\mum (W=1.07W = 1.07 μ\mum); ∣E0∣=1.13×104|E_0| = 1.13\times10^4 V/cm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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