Question

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.

Answer 1

Given silicon pn-junction at T= 300K, doped at N_d = 10¹⁶ and N_a = 10¹⁷ and cj = 0.8pF, V_R =5v

Solution

Carrier concentration of silicon at T= 300K is n_t = 1.5*10¹⁰ CM^-3

Potential barrier of pn junction is

V_bi = (kT/e) ln (N_aN_d/n²_i) = V_Tln(N_aN_d/n²_i)

= (0.026) ln (10¹⁶*10¹⁷/(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