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Si crystal resistance with As then B doping

Short answer

Arsenic gives Nd=5×10¹³ cm⁻³ → n≈Nd, σ=e n μe≈1.08×10⁻² S cm⁻¹, R=1/σ≈92.6 Ω. Boron gives Na=5×10¹⁶ cm⁻³; compensation yields p≈Na−Nd=4.995×10¹⁶ cm⁻³, σ≈e p μh≈3.6 S cm⁻¹, R≈0.278 Ω.

Find the resistance of 1 cm³ silicon crystal doped with arsenic, the doping density is such that every Arsenic atom sites every 10⁹ silicon atoms. Atomic concentration of silicon is 5×10²² cm⁻³, ni = 1×10¹⁰ cm⁻³, μₑ = 1350 cm²V⁻¹s⁻¹ and μₕ = 450 cm²V⁻¹s⁻¹. Find the resistance if the above silicon sample is further doped with Boron, the doping density is such that every Boron atom sites every 10⁶ silicon atoms.

Take the sample as a 1 cm cube (L = 1 cm, A = 1 cm²), so R=ρL/AR = \rho L/A is numerically equal to ρ\rho in Ω cm.

Given: Si atomic concentration 5×1022 cm−35\times10^{22}\ \text{cm}^{-3}, ni=1010 cm−3n_i = 10^{10}\ \text{cm}^{-3}, μe=1350\mu_e = 1350, μh=450 cm2V−1s−1\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}, e=1.6×10−19e = 1.6\times10^{-19} C.

Part 1: Arsenic doped (1 As per 10910^9 Si)

Nd=5×1022109=5×1013 cm−3N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3} n≈Nd=5×1013,p=ni2n=10205×1013=2×106 cm−3n \approx N_d = 5\times10^{13},\qquad p = \frac{n_i^2}{n} = \frac{10^{20}}{5\times10^{13}} = 2\times10^{6}\ \text{cm}^{-3} σ=e(nμe+pμh)=1.6×10−19(5×1013×1350+2×106×450)=1.08×10−2 S cm−1R=1σ=92.6 Ω\begin{aligned} \sigma &= e(n\mu_e + p\mu_h)\\ &= 1.6\times10^{-19}(5\times10^{13}\times1350 + 2\times10^{6}\times450)\\ &= 1.08\times10^{-2}\ \text{S cm}^{-1}\\ R &= \frac{1}{\sigma} = 92.6\ \Omega \end{aligned}

Part 2: Further doped with boron (1 B per 10610^6 Si)

Na=5×1022106=5×1016 cm−3N_a = \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}

Since Na≫NdN_a \gg N_d, the sample becomes p-type:

p=Na−Nd=5×1016−5×1013=4.995×1016 cm−3p = N_a - N_d = 5\times10^{16} - 5\times10^{13} = 4.995\times10^{16}\ \text{cm}^{-3} n=ni2p=2.0×103 cm−3n = \frac{n_i^2}{p} = 2.0\times10^{3}\ \text{cm}^{-3} σ≈epμh=1.6×10−19×4.995×1016×450=3.596 S cm−1R=13.596=0.278 Ω\begin{aligned} \sigma &\approx ep\mu_h = 1.6\times10^{-19}\times4.995\times10^{16}\times450\\ &= 3.596\ \text{S cm}^{-1}\\ R &= \frac{1}{3.596} = 0.278\ \Omega \end{aligned}
SampleMajority carrier (cm⁻³)R (1 cm cube)
Pure Si (for reference)n=p=1010n = p = 10^{10}3.47×105 Ω3.47\times10^{5}\ \Omega
As dopedn=5×1013n = 5\times10^{13}92.6 Ω
As + B dopedp=4.995×1016p = 4.995\times10^{16}0.278 Ω

Answer: with arsenic only, R≈92.6 ΩR \approx 92.6\ \Omega; after boron doping, R≈0.278 ΩR \approx 0.278\ \Omega (the sample is now p-type).

✓ Answer checked by Claude Opus 5.5 (high effort)

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