Question #120804

1.Asilicon pn-junction at T= 300K is doped at Nd =10^16 and Na=10^17. the junction capacitance is to be cj=0.8pf. when a reverse bias voltage of VR=5v is applied . find the zero-biased junction.

Expert's answer

Given silicon pn-junction at T= 300K, doped at *N*_{d }= 10^{16} and *N*_{a} = 10^{17} and cj = 0.8pF, *V*_{R} =5v

Solution

Carrier concentration of silicon at T= 300K is *n*_{t} = 1.5*10^{10} CM^{-3}

Potential barrier of pn junction is

*V*_{bi} = (kT/*e*) ln (*N*_{a}*N*_{d}/*n*^{2}_{i}) = *V*_{T}ln(*N*_{a}N_{d}/*n*^{2}_{i})

= (0.026) ln (10^{16}*10^{17}/(1.5*10^{10})^{2})

= 0.757V

Junction capacitance

*C*_{J}* = C*_{jo} = (1+V_{R}/V_{bi})^{-1/2}

where C_{jo }is junction

0.8*10^{-12} = (C_{jo}) ( 1 +5/0.757)^{-1/2}

0.8*10^{-12} = (C_{jo}) (0.3626)

C_{jo} = 2.21pF

Therefore the zero biased junction capacitance at *V*_{R} = 5V is 2.21pF

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Assignment Expert11.06.20, 18:42Dear Michael Yeboah, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Michael Yeboah07.06.20, 17:501.Asilicon pn-junction at T= 300K is doped at Nd =10^16 and Na=10^17. the junction capacitance is to be cj=0.8pf. when a reverse bias voltage of VR=5v is applied . find the zero-biased junction.

Michael Yeboah07.06.20, 17:491.Asilicon pn-junction at T= 300K is doped at Nd =10^16 and Na=10^17. the junction capacitance is to be cj=0.8pf. when a reverse bias voltage of VR=5v is applied . find the zero-biased junction.

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