Question 4.4: In a very long p- type Si bar with cross- sectional area = 0...
In a very long p- type Si bar with cross-sectional area=0.5 cm^{2} andN_{a} =10^{17} cm^{-3}, we inject holes such that the steady state excess hole concentration is 5\times 10^{16} cm^{-3} at x = 0. What is the steady state separation between F_{p} and E_{c} at x=1000\mathring{A}? What is the hole current there? How much is the excess stored hole charge? Assume that \mu_{p} =500{cm^{2}}/{V_{-s} } and \pmb{\tau}_{p} =10^{-10} s
D_{p} =\frac{KT}{q} \mu_{p} =0.0259\times 500=12.95{cm}/{s}
L_{p}=\sqrt{D_{p}\tau_{p} } =\sqrt{12.95\times 10^{-10} } =3.6\times 10^{-5} cm
p=p_{0}+\Delta pe^{-\frac{x}{L_{p} } } =10^{17}+5\times 10^{16}e^{-\frac{10^{-5} }{3.6\times 10^{-5} } }
=1.379 \times 10^{17} =n_{i}e^{(E_{i}-F_{b} )/KT} =(1.5\times 10^{10}cm^{-3} )e^{{(E_{i}-F_{b} )/KT}}
E_{i}-F_{b}=(\ln \frac{1.379\times 10^{17} }{1.5\times 10^{10} } )\cdot 0.0259=0.415eV
E_{c}-F_{b}=1.1/2eV+0.415eV=0.965 eV
We can calculate the hole current from Eq. (4–40)
J_{p}(x)=-qD_{p} \frac{dp}{dx} =-qD_{p}\frac{d\delta p}{dx}= q\frac{D_{p}}{L_{p}}\Delta pe^{-x/L_{p}} =q\frac{D_{p}}{L_{p}}\delta p(x) (4–40)
I_{p} =-qAD_{p} \frac{dp}{dx} =qA\frac{D_{p}}{L_{p}} (\Delta p)x^{-\frac{x}{L_{p} } }
1.6\times 10^{-19} \times 0.5\times \frac{12.95}{3.6\times 10^{-5}} \times 5\times 10^{16}e^{-\frac{10^{-5} }{3.6\times 10^{-5} } }
=1.09\times 10^{3}A
Q{p} =qA(\Delta p)L_{p}
=1.6\times 10^{-19} (0.5)(5\times 10^{16} )(3.6\times 10^{-5})
=1.44\times 10^{-7}C
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