Project 3 Iterative solution

Project 3
At the end of this exercise you will be able to:
• Solve a problem using an iterative solution
• Extract a parameter model for a device or system to match given data
• Solve using a system of equations, and find the root solution
Diodes solutions require iterative and nonlinear solvers. Let’s first consider the
most basic diode problem. We can model a diode with this equation
 (Eq.1). The complication comes when you place this diode in a
circuit. Consider the case where the diode is in series with a resistor R. The
current voltage relationship is, of course, Ohm’s Law, V=IR. Note that I is on both
sides of the equation below, suggesting several ways to solve for it. We will solve
for I using the diode equation above and nodal analysis using the circuit below.
Figure 1
These plots show several different resistances, but your answer will look
different. (Notice that the y-axis is a log scale!)
We will write some code in class to get you started with each of these problems.
There are two parts to this project. Both parts should be contained in the SAME
python script called project3.py. The two parts of the problem will have
functions in common and it is best if both parts use the common functions
rather than implementing the same function twice.
Problem 1: Determine the current, I, thorough the circuit in Figure 1 by doing a
nodal analysis for the circuit and using the current through the diode defined by
the diode equation. Then solve using optimize.fsolve to find the voltage across
1
qV
nkT s I Ieæ ö = ç ÷ - è ø
( )
1
V IR q
nkT s I Ie - æ ö = - ç ÷ è ø
0.4 0.6 0.8 1 1.2 1.4 1.6 1 .10 7
1 .10 6
1 .10 5
1 .10 4
1 .
10 3
0.01
0.1
1
10
100
Diode Current as a functionof diode resistance
Diode Voltage (V)
Diode Current (A)
the diode. Use these parameters for the diode: Is=1e-9, n=1.7, R=11k ohms, T =
350 and applied voltage V from 0.1 to 2.5 V. Analyze with steps of 0.1 V.
Create one plot with two curves: (a) log(Diode Current) vs Source Voltage; and (b)
log(Diode Current) vs Diode Voltage.
Your method is to write the equation using nodal analysis for the node connecting
the resistor and the diode, and then write the current equation for the diode in
terms of the voltage across it. This should be something like
, where Err should be zero when the value for Vd, the
voltage across the diode, is correctly chosen.
Problem 2: We are going to repeat the diode problem above for a diode where
the parameters are not known. The function for this diode is shown below.
def DiodeI(Vd,A,phi,n,T):
 k = 1.380648e-23
 q = 1.6021766208e-19
 Vt = n*k*T/q
 Is = A*T*T*np.exp(-phi*q/(k*T))
 return Is*(np.exp(Vd/Vt)-1)
Where phi is the barrier height, T is the temperature, R is the lead resistance, A is
the cross-sectional area of the diode, n is the ideality, and Vd is the voltage across
the diode.
a) Assume this diode is placed in the circuit in Figure 1 where the lead
resistance R is included. Using the methods you developed in problem 1,
create a function that returns the currents through this diode for supplied
voltage Vs (where Vs is an array of voltages). The returned current should
also be an array. In this function you will solve for the current at each
supplied voltage as you did in problem 1 and return the array of currents.
b) You are supplied a data set named DiodeIV.txt. The objective is to find the
three missing diode parameters. This is a diode which may have uncommon
( ) d
d
V V Err Idiode V
R R = - +
material parameter values. You will be given A, and T, of the measured data
and you must determine the other parameters n, phi, and R.
Use scipy optimize leastsq, which will require you optimize one parameter at a
time, iteratively, until the solution is achieved. (It may be possible to optimize for
all three parameters at once. However, problems like this often require the
parameters to be optimized one at a time, so we’ll take that approach.)
Ø from scipy import optimize
Ø and then use optimize.leastsq( …
Read the DiodeIV.txt. This file contains two columns of data. The first column is
the source voltage (that is, the voltage across the diode and resistor) and the
second column is the diode current. The Area is 1e-8 and the temperature is 375
Kelvin. Try an initial value of Phi of 0.8, an initial Ideality of 1.5, and an initial
Resistor value of 10000 ohms. Your objective is to find the actual values for Phi,
Ideality, and the Resistor.
To do this you will have to solve for the voltage across the diode, then calculate
the diode current for a set of parameters. Then optimize the values of these
parameters to give the same currents as a function of source voltage as those in
the file.
You will have to write a function that calculates the residual error. Consider
returning absolute or normalized error from the residual function which helps fit
the low amplitude parts of the curve as well as the high bias parts. It looks
something like this:
Absolute Error = (ynew-ylast)
Normalized Error = (ynew-ylast)/(ynew+ylast+1e-15)
You may also want to find the errors in logs of currents rather than actual
currents.
When you use optimize.leastsq, the first parameter will be the residual function
and the second will be the guess of the parameter you are trying to find. This is
then the first parameter in that residual function. Since you are optimizing three 
parameters you may want to write identical residual functions with n, phi, and R
as the first parameter to facilitate this process.
Then you could optimize them something like this:
while (errtolerance and iteration<niter):
phi = optimize.leastsq(residualp, phi, all the other parameters including n and R)
n = optimize.leastsq(residualn, n, all the other parameters including n and R)
R = optimize.leastsq(residualR, R, all the other parameters including n and R)
You may get warnings during the first few iterations. This is due to the inability to
drive the residual to 0 when the parameters are still way off. Such warnings are
normal and should not be a concern.
Err calculation is up to you. One choice is the difference between the parameters
in each pass. When the parameters stop changing you might want to stop. You
probably want to limit the number of times through the loop to prevent an
infinite loop in the case where your code doesn’t converge.
Another way to determine if you should stop is based on the residual values
passed back by the last leastsq call in the loop. The residual will be an array of
values. You could take the absolute value of each entry, sum the absolute values,
and divide by the number of values. This will give you the average size of the
errors at each data point. When this value gets small enough, you can stop. (This
is how my solution works…)
At the bottom of the while loop described above (or whatever loop you choose to
use), print the iteration number and the values of the 3 parameters so that we
can track the progress following each iteration. When you have found your
values, plot the log(Diode Current) vs source voltage for the data in the file and
the predictions from your model on the same plot. You should get an excellent fit
for the top part of the curve. Do not be surprised if the section below the knee
doesn’t match well.
Note that leastsq returns an array, and the first element in the array is the array
of optimized values. Therefore, when optimizing a single value, you want to
invoke it like this:
r_val_op = optimize.leastsq(opt_r,r_val,args=(…))
r_val = r_val_opt[0][0]