0259 ln 0731v 10 ni 15 10 which is exactly

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Unformatted text preview: × 1016 ⎞ EFi , p − EF = kBT ln ⎜ 0 ⎟ = 0.0259 ln ⎜ = 0.365 eV ⎝ 1.5 × 1010 ⎟ ⎠ ⎝ ni ⎠ . From the figure, we find that Vbi = 0.731 V. (c) From (7.10), we have ⎡⎛ 2 × 1016 ⎞ 2 ⎤ ⎛ Na Nd ⎞ Vbi = kBT ln ⎜ 2 ⎟ = 0.0259 ln ⎢⎜ = 0.731V 10 ⎟ ⎥ ⎝ ni ⎠ ⎢⎝ 1.5 × 10 ⎠ ⎥ ⎣ ⎦ which is exactly the amount from the figure (but (7.10) is just described as the product of the two equations from part (a)). (d) We now use (7.28), (7.29), and (7.16 ) to give 1/2 x p = xn = 0.154 µ m E peak eN d xn 1.6 × 10 −19 ⋅ 2 × 10 22 ⋅ 0.154 × 10 −6 =− =− = 47.6 kV / cm εs 11.7 ⋅ 8.854 × 10 −12 Consider a uniformly doped GaAs pn junction with doping concentrations of Na = 2 x 1015 cm- 3 and Nd = 4 x 1016 cm- 3. Plot the built- in potential barrier Vbi versus temperature over the ra...
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