Question

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less than the p-type doping, what is the resistance in ohms outside the depletion region on the n side of the junction. Use three significant digits and fixed point notation. The diode is square with an edge length of 60 microns, with a bias of 0.44V. Assume p and n mobilities are 500 & 1500 cm^2/(V*s) respectively. Correct Answer: 24.9

Answer 1

The resistance is 24.9 Ω

Step-by-step explanation:

The resistivity is equal to:

`R=(1)/(N_(o)*u*V ) =(1)/(4.48x10^(15)*1500*106x10^(-19) ) =0.93ohm*cm`

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

`w=\sqrt{(2E(V_(o)-V) )/(p)((1)/(N_(A) )+(1)/(N_(D) ))`

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

`w=\sqrt{(2E(V_(o)-V) )/(p)((1)/(N_(D) ))`

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

`V_(o) =(KT)/(p) ln((N_(A)*N_(D))/(n_(i)^(2) ) , if n_(i)=1.5x10^(10)cm^(-3) \\V_(o)=0.02585ln((4.48x10^(18)*4.48x10^(15) )/((1.5x10^(10))^(2) ) )=0.83V`

`w=\sqrt{(2*11.68*8.85x10^(-14)*(0.83-0.44) )/(1.6x10^(-19)*4.48x10^(15) ) } =3.35x10^(-5) cm=0.335um`

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

`Re=(pL)/(A) =(0.93*9.665x10^(-4) )/(0.36x10^(-4) ) =24.9ohm`