# p+n step junction width

Problem 5:

A silicon p+n step junction diode has a doping of $Na= 10^{17} cm^{-3}$ and $Nd=10^{15} cm^{-3}$.

(a) What is the built-in potential.

(b) What are $x_n$, $x_p$, $W$, and the electric field at $x = 0$.

(c) If the applied voltage is $V_a= 0.4V$, calculate the new values of $x_n$, $x_p$, $W$,  and the electric field at $x=0$.

(d) If the applied voltage is $V_a= -3V$  calculate new values of $x_n$, $x_p$, $W$, and the electric field at $x=0$.

(e) From the results in (d), determine the percent change in $x_n$ and $W$ from the $V_a= 0 V$ case.

a)

$V_{barrier} = K*T/e *ln(N_a*N_d/ni^2)$

(http://en.wikipedia.org/wiki/P%E2%80%93n_junction#Electrostatics)

$n_i(silicon) =1.5*10^{10} cm^{-3}$

(http://www-inst.eecs.berkeley.edu/~ee105/fa05/handouts/discussions/Discussion1.pdf)

$K =1.38*10^{-23}$

$T =300 K$

$V_{barrier} =0.69397 V$

b)

$W=\sqrt {2*\epsilon/e * (N_a+N_d)/(N_a*N_d)*V_b}$

$\epsilon(silicon) =11.7*\epsilon_0 =1.04*10^{-12} F/cm$

$W =9.5456*10^{-5} cm =0.9546 microns$

$E = Vb/W =0.694/0.9546*10^{-6} =7.27*10^5 V/m$

$x_n*N_d = x_p*N_a$

$x_n+x_p =W$

$x_n/x_p= N_a/N_d$

$x_p(N_a/N_d +1) =W$

$x_p =W/(N_a/N_d +1) =0.9546*10^{-6} /(100+1) =0.00945 *10^{-6} m =9.45 nm$

$x_n =W – x_p =(0.9546-0.00945)*10^{-6} =0.9451 *10^{-6} m = 0.9451 microns$

c)

$W_1=\sqrt {2*\epsilon/e * (N_a+N_d)/(N_a*N_d)*(V_b-V_a)}$

$V_a =+0.4 V$

$W =0.6213*10^{-6} m$

$x_p =W/(N_a/N_d +1) = 0.6213*10^{-6}/101 =0.00615 *10^{-6} m =6.15 nm$

$x_n = W-x_p = (0.6213 -0.00615) =0.615*10^{-6} m =0.615 microns$

$E = (V_b-V_a)/W =4.73*10^5 V/m$

d)

$W_2=\sqrt {2*\epsilon/e * (N_a+N_d)/(N_a*N_d)*(V_b-V_a)}$

$V_a =-3 V$

$W = 2.2*10^{-4} cm =2.2 microns$

$x_p = W/(N_a/n_d +1) = 2.2/101 =0.0218 microns$

$x_n =(22-0.218) =2.18 microns$

$E = (V_b-V_a)/W =(0.694+3)/2.2*10^{-6} =1.68*10^6 V/m$

e)

$(W_2-W_0)/W_0 = (2.2-0.9451)/0.9451 =1.3277 =132.8 %$

$(x_n2-x_n0)/x_n0 = (2.18-0.615)/0.615 =2.545 =254.5%$