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pn junction 10¹⁸/10¹⁶: built-in potential and widths

Short answer

V0 = (kT/e) ln(Na Nd / ni^2) = 0.814 V. Total depletion width W = sqrt[2ε(Na+Nd)V0/(e Na Nd)] = 0.329 μm, with Wn = 0.326 μm, Wp = 3.26 nm.

A heavily doped p side with acceptor concentration of 10¹⁸ cm⁻³ is connected to n-side with donor concentration of 10¹⁶ cm⁻³. Calculate the built-in potential and depletion width in n-side and p-side and overall depletion width of pn junction. The intrinsic concentration is 1.45×10¹⁰ cm⁻³ and temperature is 300 K.

Formulas (abrupt junction, depletion approximation):

V0=kTeln⁡NaNdni2,W=2ε(Na+Nd)V0eNaNd,Wn=NaWNa+Nd,Wp=NdWNa+NdV_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}, \quad W = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}, \quad W_n = \frac{N_a W}{N_a+N_d}, \quad W_p = \frac{N_d W}{N_a+N_d}

Constants: kT/e=0.02586kT/e = 0.02586 V at 300 K, e=1.602×10−19e = 1.602\times10^{-19} C, ε=εrε0=11.9×8.854×10−14=1.054×10−12\varepsilon = \varepsilon_r\varepsilon_0 = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12} F/cm (silicon assumed).

Given: Na=1018N_a = 10^{18} cm−3^{-3}, Nd=1016N_d = 10^{16} cm−3^{-3}, ni=1.45×1010n_i = 1.45\times10^{10} cm−3^{-3}, T=300T = 300 K.

Built-in potential

NaNdni2=1018×1016(1.45×1010)2=4.76×1013V0=0.02586×ln⁡(4.76×1013)=0.02586×31.49=0.814 V\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{18}\times10^{16}}{(1.45\times10^{10})^2} = 4.76\times10^{13} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{13}) = 0.02586 \times 31.49 \\ &= 0.814\ \text{V} \end{aligned}

Total depletion width

W=2×1.054×10−12×(1018+1016)×0.8141.602×10−19×1018×1016=3.29×10−5 cm=0.329 μm\begin{aligned} W &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{18}+10^{16}) \times 0.814}{1.602\times10^{-19} \times 10^{18} \times 10^{16}}} \\ &= 3.29\times10^{-5}\ \text{cm} = 0.329\ \mu\text{m} \end{aligned}

Width on each side

Wn=NaNa+NdW=10181.01×1018×0.329=0.326 μmWp=NdNa+NdW=10161.01×1018×0.329=3.26×10−3 μm=3.26 nm\begin{aligned} W_n &= \frac{N_a}{N_a+N_d}W = \frac{10^{18}}{1.01\times10^{18}} \times 0.329 = 0.326\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W = \frac{10^{16}}{1.01\times10^{18}} \times 0.329 = 3.26\times10^{-3}\ \mu\text{m} = 3.26\ \text{nm} \end{aligned}

Check: NaWp=1018×3.26×10−7=3.26×1011N_aW_p = 10^{18}\times3.26\times10^{-7} = 3.26\times10^{11} cm−2^{-2} =NdWn= N_dW_n. Since this is a p+^+n junction, almost all the depletion region lies in the lightly doped n-side. (The peak field is E0=eNdWn/ε=4.95×104E_0 = eN_dW_n/\varepsilon = 4.95\times10^4 V/cm.)

Answer: V0=0.814V_0 = 0.814 V; Wn=0.326W_n = 0.326 μ\mum; Wp=3.26W_p = 3.26 nm; W=0.329W = 0.329 μ\mum.

✓ Answer checked by Claude Opus 5.5 (high effort)

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