Question #120890

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 *nt* = 1.5*10^{10} cm^{-3}

Potential barrier of pn junction is

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

= (0.026) ln (10^{16}*10^{17}/(1.5*1010)^{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

Learn more about our help with Assignments: Electrical Engineering

## Comments

## Leave a comment