Answer :
Answer:
The resistance is 24.9 Ω
Explanation:
The resistivity is equal to:
[tex]R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm[/tex]
The area is:
A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²
[tex]w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })[/tex]
If NA is greater, then, the term 1/NA can be neglected, thus the equation:
[tex]w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })[/tex]
Where
V = 0.44 V
E = 11.68*8.85x10¹⁴ f/cm
[tex]V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V[/tex]
[tex]w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um[/tex]
The length is:
L = 10 - 0.335 = 9.665 um
The resistance is:
[tex]Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm[/tex]