# Nonuniform Doping of Semiconductor

Figure below is a part of the energy band diagram of a P-type semiconductor bar under equilibrium conditions (i.e., $E_F$ is constant). The valence band edge is sloped because doping is nonuniform along the bar. Assume that $E_V$ rises with a slope of $∆/L$.

(a) Write an expression for the electric field inside this semiconductor bar.

(b) Within the Boltzmann approximation, what is the electron concentration n(x) along the bar? Assume that $n(x=0)= n_0$. Express your answer in terms of $n_0$, $∆$ and $L$

The built-in potential inside the semiconductor varies with distance x from origin as

$V(x)= -1/e*(E_c (x)-E_F )=(1/e) (E_F-E_c (x) )=(∆*x)/e L+Const$

Where $E_c (x)>E_F$ is the energy of the conduction band and the value of the constant is determined from initial conditions

$V(0)=Const$ at $x=0$

The electric field along the semiconductor is (Greek Ε)

$Ε(x)=-d V/(d x )=-Δ/e L$

which is a constant value along all the length of semiconductor.

The Fermi-Dirac distribution function is

$f(E)= 1/(1+e^{(E-E_F)/KT})$

Which becomes the Boltzmann distribution for $(E-E_F )>3KT$ :

$f_B (E)=e^{-(E-E_F)/KT}$

Thus the Boltzmann approximation for electron concentration is

$n=N_c*e^{-((E c-E_F)/KT) }$

And in our case for

$E_c (x)-E_F=(Δ*x)/L+ C$ we have $n(x)=N_c*e^{-(Δ*frac{x}{L}+ C)/KT}$

$n(0)=N_c*e^{-C/KT}$ therefore $n(x)=n(0) e^{-(Δ*x)/(L K T)}$

The total current in the semiconductor is null (equilibrium) (Greek E is electric field below)

$J=J_{diff}-J_{cond}=e D*(d n/d x)-en(x)*μ*Ε=0$

$(d n(x)/d x)=-n(x)*Δ/L K T$

$D*n(x)*Δ/L K T=n(x)*μ*Δ/e L$

$D/μ=KT/e$ q.e.d