Question

A si pn junction ( Na= 1016 cm-4 and Nd = 4 × 10 16 cm-3) is biased with Va= -3v . Calculate the built-in potential and depletion layer width ?

Answer 1

$Given silicon pn-junction at T= 300K, doped at Nd = 1016 and Na = 1017 and cj = 0.8pF, VR=5v Solution Carrier concentration of silicon at T= 300K is nt = 1.5*1010 cm-3 Potential barrier of pn junction is Vbi = (kT/e) ln (NaNd/n2i) = VTln(NaNd/n2i) = (0.026) ln (1016*1017/(1.5*1010)2) = 0.757V Junction capacitance Cj= Cjo (1+VR/Vbi)-1/2 where Cjo is junction 0.8*10-12 = (Cjo) ( 1 +5/0.757)-1/2 0.8*10-12 = (Cjo) (0.3626) Cjo = 2.21pF Therefore the zero biased junction capacitance at VR = 5V is 2.21pF$