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Fermi level shift with As then B

Short answer

EF – EFi = kT ln(n/ni). With 10¹⁵ As/cm³, n = 10¹⁵ cm⁻³, so EF – EFi ≈ 0.288 eV above intrinsic level. After adding 2×10¹⁶ B/cm³, p = Na – Nd = 1.9×10¹⁶ cm⁻³, EFi – EF = kT ln(p/ni) ≈ 0.364 eV, so EF lies 0.364 eV below EFi.

An n-type Si wafer has been doped uniformly with 10¹⁵ Arsenic atoms per cm³. Calculate the position of the Fermi energy level with respect to the intrinsic Fermi energy level EFi at 27 °C. If this sample is further doped with 2×10¹⁶ Boron atoms per cm³, where will the Fermi level be shifted?

For non-degenerate Si,

EF−EFi=kTln⁡nni,EFi−EF=kTln⁡pniE_F - E_{Fi} = kT\ln\frac{n}{n_i}, \qquad E_{Fi} - E_F = kT\ln\frac{p}{n_i}

Assume: T=300T = 300 K (27∘27^\circC), kT=0.02586kT = 0.02586 eV, ni=1.45×1010n_i = 1.45\times10^{10} cm−3^{-3}, full ionisation.

With 101510^{15} As cm−3^{-3}

n=Nd=1015 cm−3EF−EFi=0.02586ln⁡10151.45×1010=0.02586×11.14=0.288 eV\begin{aligned} n &= N_d = 10^{15}\ \text{cm}^{-3} \\ E_F - E_{Fi} &= 0.02586\ln\frac{10^{15}}{1.45\times10^{10}} = 0.02586\times11.14 \\ &= 0.288\ \text{eV} \end{aligned}

EFE_F is 0.288 eV above EFiE_{Fi}.

After adding 2×10162\times10^{16} B cm−3^{-3}

Na>NdN_a > N_d, so the sample is now p-type:

p=Na−Nd=2×1016−1015=1.9×1016 cm−3EFi−EF=0.02586ln⁡1.9×10161.45×1010=0.02586×14.09=0.364 eV\begin{aligned} p &= N_a - N_d = 2\times10^{16} - 10^{15} = 1.9\times10^{16}\ \text{cm}^{-3} \\ E_{Fi} - E_F &= 0.02586\ln\frac{1.9\times10^{16}}{1.45\times10^{10}} = 0.02586\times14.09 \\ &= 0.364\ \text{eV} \end{aligned}

EFE_F is 0.364 eV below EFiE_{Fi}, so it has shifted down by 0.288+0.364=0.6520.288 + 0.364 = 0.652 eV.

 Ec ---------------------------
     EF (As only) .. +0.288 eV
 Ei - - - - - - - - - - - - - -
     EF (As + B)  .. -0.364 eV
 Ev ---------------------------

Answer: EF−EFi=0.288E_F - E_{Fi} = 0.288 eV (n-type); after boron, EF−EFi=−0.364E_F - E_{Fi} = -0.364 eV (p-type).

✓ Answer checked by Claude Opus 5.5 (high effort)

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