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Answer to Question #120804 in Electrical Engineering for Nana Safo Boama
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 Nd = 1016 and Na = 1017 and cj = 0.8pF, VR =5v
Solution
Carrier concentration of silicon at T= 300K is nt = 1.5*1010 CM-3
Potential barrier of pn junction is
Vbi = (kT/e) ln (NaNd/n2i) = VTln(NaNd/n2i)
= (0.026) ln (1016*1017/(1.5*1010)2)
= 0.757V
Junction capacitance
CJ = Cjo = (1+VR/Vbi)-1/2
where Cjo is junction
0.8*10-12 = (Cjo) ( 1 +5/0.757)-1/2
0.8*10-12 = (Cjo) (0.3626)
Cjo = 2.21pF
Therefore the zero biased junction capacitance at VR = 5V is 2.21pF
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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.
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.
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