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Si with 10¹⁵ Sb: carriers, resistance, Fermi level

Short answer

Initial n-type: n = 10^15 cm^-3, p = 2.1×10^5 cm^-3; for 1 cm cube, R = 1/(e N_d μ_e) = 4.63 Ω, Fermi level 0.288 eV above intrinsic. Adding 10^22 cm^-3 boron makes p-type, R ≈ 1.25×10^-5 Ω.

Sample of silicon wafer is doped with 10¹⁵ Antimony atoms/cm³. Find the carrier concentrations, its resistance and the shift in Fermi level from its intrinsic Fermi level at 27°C. If this sample is further doped with 10²² Boron atoms/cm³, what will be the change in its resistance. [Graph attached with the paper: log-log plot of electron drift mobility μₑ (cm² V⁻¹ s⁻¹, 50 to 10⁴) versus temperature T (100 to 1000 K) for n-type Si with curves for Nd = 10¹⁴, 10¹⁶, 10¹⁷, 10¹⁸, 10¹⁹ cm⁻³; inset shows ln(μₑ) vs ln(T) rising as T^(3/2) for impurity scattering and falling as T^(−3/2) for lattice scattering.]

Antimony is a donor. The sample size is not given, so the resistance is found for a 1 cm × 1 cm × 1 cm cube (R in Ω = ρ in Ω cm).

Data used: ni=1.45×1010 cm−3n_i = 1.45\times10^{10}\ \text{cm}^{-3}, kT=0.02585kT = 0.02585 eV at 300 K, μh=450 cm2V−1s−1\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}. From the attached graph, at 300 K and Nd=1015N_d = 10^{15} (between the 101410^{14} and 101610^{16} curves) μe≈1350 cm2V−1s−1\mu_e \approx 1350\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}.

Carrier concentrations

n≈Nd=1015 cm−3,p=ni2n=(1.45×1010)21015=2.1×105 cm−3n \approx N_d = 10^{15}\ \text{cm}^{-3},\qquad p = \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{10^{15}} = 2.1\times10^{5}\ \text{cm}^{-3}

Resistance

σ=e(nμe+pμh)≈eNdμe=1.6×10−19×1015×1350=0.216 S cm−1ρ=4.63 Ω cm  ⇒  R=4.63 Ω (1 cm cube)\begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx eN_d\mu_e\\ &= 1.6\times10^{-19}\times10^{15}\times1350 = 0.216\ \text{S cm}^{-1}\\ \rho &= 4.63\ \Omega\,\text{cm} \;\Rightarrow\; R = 4.63\ \Omega\ (\text{1 cm cube}) \end{aligned}

Fermi level shift

EFn−EFi=kTln⁡Ndni=0.02585ln⁡10151.45×1010=0.288 eV above EFi\begin{aligned} E_{Fn}-E_{Fi} &= kT\ln\frac{N_d}{n_i} = 0.02585\ln\frac{10^{15}}{1.45\times10^{10}}\\ &= 0.288\ \text{eV above } E_{Fi} \end{aligned}

Further doped with 102210^{22} boron cm⁻³

Now Na≫NdN_a \gg N_d, so the sample becomes strongly p-type:

p=Na−Nd≈1022 cm−3,n=ni2/p≈0.02 cm−3p = N_a - N_d \approx 10^{22}\ \text{cm}^{-3},\qquad n = n_i^2/p \approx 0.02\ \text{cm}^{-3}

At such heavy doping impurity scattering dominates; the hole mobility falls to about its minimum value, taken as μh≈50 cm2V−1s−1\mu_h \approx 50\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} (assumed; the graph gives only electron mobility).

σ=epμh=1.6×10−19×1022×50=8×104 S cm−1ρ=1.25×10−5 Ω cm  ⇒  R=1.25×10−5 Ω\begin{aligned} \sigma &= ep\mu_h = 1.6\times10^{-19}\times10^{22}\times50 = 8\times10^{4}\ \text{S cm}^{-1}\\ \rho &= 1.25\times10^{-5}\ \Omega\,\text{cm} \;\Rightarrow\; R = 1.25\times10^{-5}\ \Omega \end{aligned}

Answer: n=1015n = 10^{15}, p=2.1×105 cm−3p = 2.1\times10^5\ \text{cm}^{-3}; R≈4.63 ΩR \approx 4.63\ \Omega (1 cm cube); EFE_F is 0.288 eV above EFiE_{Fi}. After boron doping, R≈1.25×10−5 ΩR \approx 1.25\times10^{-5}\ \Omega, so the resistance falls by a factor of about 3.7×1053.7\times10^{5}.

Note: 102210^{22} cm⁻³ is 20% of the Si atom density; such a sample is degenerate and almost metallic, so this result is only an estimate.

✓ Answer checked by Claude Opus 5.5 (high effort)

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