Question

3 If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3 , find the junction built-in voltage. Assume ni = 1.5 × 1010/cm3 . Also, find the width of the depletion region (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 100 μm2 .

Answer 1

Answer:

The magnitude of charged stored is $Q=4.8*10^-^1^4C$

Explanation:

Junction built-in voltage $V_{0}$ = $V_{T} ln(\frac{N_{A} N_{D} }{n_{1}^2 } )=25*10^-^3ln(\frac{10^7*10^1^6}{(1.5*10^1^0)^2} )=0.754V$

Width of depletion region is $W_{dep} =x_{n}+ x_{p}=\sqrt{\frac{2E_{s} }{q}(\frac{1}{N_{A} } +\frac{1}{N_{P} })V_{0} }$

where $E_{s}$ is the electrical permitivity of silicon

$E_{s}=11.7E_{0}=11.7*8.85*10^-^1^4F/cm=1.04*10^-^1^2F/cm\\ and\\q=1.6*10^-^1^9$

therefore

$W_{dep}=\sqrt{\frac{2*1.04*10^-^1^2}{1.6*10^-^1^9} (\frac{1}{10^1^7}+\frac{1}{10^1^6} )(0.754) }=0.328*10^-^1^4cm=0.328$μ/m

$x_{n}=W\frac{N_{A} }{N_{A}+N_{D} }=0.328\frac{10^1^7}{10^1^7+10^1^6} =0.298$μ/m

$x_{P}=W\frac{N_{D} }{N_{A}+N_{D} }=0.328\frac{10^1^6}{10^1^7+10^1^6} =0.03$μ/m

Junction area=100 μ/m^2=$100*10^-^8cm^2$

Magnitude of charge

$|Q_{J}|=Q_{N}=Q_{p} =q\frac{N_{A}N_{D} }{N_{A}+N_{D} }AW_{dep} \\|Q_{J}|=1.6*10^-^1^9*\frac{10^1^7*10^1^6}{10^1^7+10^1^6} *100*10^-^8*0.328*10^-^4\\|Q_{J}|=4.8*10^-^1^4C$