Answer:
The magnitude of charged stored is [tex]Q=4.8*10^-^1^4C[/tex]
Explanation:
Junction built-in voltage [tex]V_{0}[/tex] = [tex]V_{T} ln(\frac{N_{A} N_{D} }{n_{1}^2 } )=25*10^-^3ln(\frac{10^7*10^1^6}{(1.5*10^1^0)^2} )=0.754V[/tex]
Width of depletion region is [tex]W_{dep} =x_{n}+ x_{p}=\sqrt{\frac{2E_{s} }{q}(\frac{1}{N_{A} } +\frac{1}{N_{P} })V_{0} }[/tex]
where [tex]E_{s}[/tex] is the electrical permitivity of silicon
[tex]E_{s}=11.7E_{0}=11.7*8.85*10^-^1^4F/cm=1.04*10^-^1^2F/cm\\ and\\q=1.6*10^-^1^9[/tex]
therefore
[tex]W_{dep}=\sqrt{\frac{2*1.04*10^-^1^2}{1.6*10^-^1^9} (\frac{1}{10^1^7}+\frac{1}{10^1^6} )(0.754) }=0.328*10^-^1^4cm=0.328[/tex]μ/m
[tex]x_{n}=W\frac{N_{A} }{N_{A}+N_{D} }=0.328\frac{10^1^7}{10^1^7+10^1^6} =0.298[/tex]μ/m
[tex]x_{P}=W\frac{N_{D} }{N_{A}+N_{D} }=0.328\frac{10^1^6}{10^1^7+10^1^6} =0.03[/tex]μ/m
Junction area=100 μ/m^2=[tex]100*10^-^8cm^2[/tex]
Magnitude of charge
[tex]|Q_{J}|=Q_{N}=Q_{p} =q\frac{N_{A}N_{D} }{N_{A}+N_{D} }AW_{dep} \\|Q_{J}|=1.6*10^-^1^9*\frac{10^1^7*10^1^6}{10^1^7+10^1^6} *100*10^-^8*0.328*10^-^4\\|Q_{J}|=4.8*10^-^1^4C[/tex]
