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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.
1
2020-06-10T07:49:39-0400

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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Assignment Expert
11.06.20, 18:42

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Michael Yeboah
07.06.20, 17:50

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.

Michael Yeboah
07.06.20, 17:49

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.