Question 5.2: The junction described in Example 5–1 has a circular cross s...
The junction described in Example 5–1 has a circular cross section with a diameter of 10µm. Calculate x_{n_{0} }, x_{p_{0} } ,Q_{+} and ℰ_{0} for this junction at equilibrium (300 K). Sketch ℰ(x) and charge density to scale, as in Fig. 5–12.
A=\pi (5\times 10^{-4} )^{2} =7.85\times 10^{-7} cm^{2}
W= \left[\frac{2\epsilon V_{0} }{q} \left(\frac{1}{N_{a} }+\frac{1}{N_{d} } \right) \right] ^{{1}/{2}}
=\left[\frac{2\left(11.8\right) \left(8.85\times 10^{-14} \right)\left(0.796\right) }{1.6\times 10^{-19} } \left(10^{-18} +2\times 10^{-16} \right) \right] ^{{1}/{2}} =0.457\mu m
x_{n_{0} } =\frac{W}{1+{N_{d}}/{N_{a}}} =\frac{0.457}{1+5\times 10^{-3}} =0.455 \mu m
x_{p_{0} }=\frac{0.457}{1+200}=2.27\times 10^{-3}\mu m
Q_{+}=qAx_{n_{0} }N_{d}=qAx_{p_{0} }N_{a}=\left(1.6\times 10^{-19}\right) \left(7.85\times 10^{-7}\right) \left(2.27\times 10^{11}\right)
=2.85\times 10^{-14}C
ℰ_{0}=-\frac{q}{\epsilon } x_{n_{0} } N_{d}=-\frac{q}{\epsilon } x_{p_{0} } N_{a}=\frac{1.6\times 10^{-19} }{\left(11.8\right)\left(8.85\times 10^{-14}\right) } \left(2.27\times 10^{11}\right)
=-3.48\times 10^{4}{V}/{cm}
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