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the diode for a source potential that is a function of time, we need to repeat
the Newton-Raphson iteration for each of the different source voltage val-
ues at the different times. The sequence of the computation would proceed
as follows:

1. Generate the time array.
2. Generate the source potential for the different elements in the time
array.
3. For each time array entry, find the potential across the diode using
the Newton-Raphson method.
4. Obtain the current array from the potential array.
5. Plot the source potential and the current arrays as a function of the
time array.

Assuming that the source potential is given by:

V = V sin(2πft) (5.18)
s
0
and that f = 60 Hz, V = 5 V, kT = 0.025 V, R = 500 Ω, and the saturation current
0
I is 10 mA; the following script M-file finds the current in this circuit:
–6
s
Is=10^(-9);
R=500;
kT=1/40;
f=60;
V0=5;
t=linspace(0,2/f,600);
L=length(t);
K=200;
Vs=(V0*sin(2*pi*t*f))'*ones(1,K);
v=zeros(L,K);
i=zeros(L,K);

for k=1:K-1
v(:,k+1)=v(:,k)-(Is*(exp((1/kT)*v(:,k))-1)-...
(1/R)*(Vs(:,k)-v(:,k)))./...
((1/kT)*Is*exp((1/kT)*v(:,k))+1/R);
i(:,k+1)=(Vs(:,k+1)-v(:,k+1))/R;
end

plot(t,1000*i(:,K),'b',t,Vs(:,K),'g')

© 2001 by CRC Press LLC