Question

A p-i-n junction has an intrinsic region thickness of 1-um (micrometer)

The p side has Na =10^15 and the n side has Nd=10^19.

a) Derive the electric field and potential using Poisson's equation.

b) Calculate the depletion region width (take into account the 1-um and the depletion region width for a regular pn junction)

Answer 1

(a)

From the above given data,depletion region will be formed in intrinsic region only. Hence the width of depletion region is

w=1um, and N_a=10¹⁵ and N_d=10¹⁹.

X_no+X_po=w=1*10^-6m and X_n0=X_po=0.5 um

and electron charge=q=1.6*10^-19 coulomb.

Electric field using Poission's equation is

E₀=-(q/$\varepsilon$)N_dx_n0

E₀=-(1.6*10^-19)(10¹⁹)(0.5*10^-6)/(8.854*10^-12)=90354.6V/m

Potential=V₀=-(1/2)(E₀)w=(1/2)(90354.6)(1*10^-6)=0.045V.

(b)

Depletion region width=W=[(2$\varepsilon$V₀/q)(N_a+N_d)/N_aN_d]^1/2

Substitute above values to get w

W=((2*8.854*10^-12)/(1.6*10^-19))(10¹⁵+10¹⁹)/10³⁴]^1/2

After solving above equation

Depletion region width=W=3.3*10^-4m