[proFit-list] about implicit function fitting

[pro Fit Support]

Dear Cüneyt

This message got caught in my spam filter, so I am late to reply.

Your question is a good one, and the reply might be of general interest.

There are basically two ways to fit an implicit function, one being intuitive and the other being less intuitive but more interesting (and probably faster if you have a slow function and a large number of data points).




The intuitive approach is to use a root solver algorithm to solve the diode equation, i.e. to calculate the root of

I - Is*exp(q*(V-I*Rs)/(n*k*T)-1) - (V-I*Rs)/Rp

i.e. to find the value of I where this equation returns zero.

To do so, I recommend using Python for defining the function and one of Python’s root finders (e.g. in SciPy) to find the root. The function basically looks like this:

————

## function DiodeModelFunction

## input 1e-12, active, 'Is'
## input 0, active, 'Rs'
## input 100, active, 'Rp'
## input 300, active, 'T'
## input 1, active, 'n'

import math
from scipy import optimize

k = 1.380649e-23
q =  1.602176634e-19 


def f(I, V, Is, Rs, Rp, T, n):
    global k, q
    return I - Is * math.exp(q *(V-I*Rs)/(n*k*T)-1) - (V-I*Rs)/Rp


def DiodeModelFunction(x, Is, Rs, Rp, T, n):
    global k, q
    Imax = 2 * (Is * math.exp(q*x/(n*k*T) - 1) + x / Rp)   ## estimate for an upper limit of I
    sol = optimize.root_scalar(f, bracket = [-1, Imax], args = (x, Is, Rs, Rp, T, n), method = 'brentq')
    return sol.root

————

(Not sure if I used a good guess for Imax: It should be an estimate for the current that is higher than the true current. You might also use a high enough constant value there.)

Note: This only works if you have SciPy installed with your Python. If you don’t have that, you can use Pascal scripting and the Analyze(…) function, but that’s slightly trickier. Let me know if you need help with that.




The non-intuitive approach is to define a function that should return zero:

————

## function ImplicitDiodeFunction

## input 0, inactive, 'I'
## input 1e-12, active, 'Is'
## input 0, active, 'Rs'
## input 100, active, 'Rp'
## input 300, active, 'T'
## input 1, active, 'n'

k = 1.380649e-23
q =  1.602176634e-19 


import math

def ImplicitDiodeFunction(x, I, Is, Rs, Rp, T, n):
    global k, q
    return I - Is * math.exp(q *(x-I*Rs)/(n*k*T)-1) - (x-I*Rs)/Rp

————

Then, prepare a data window that has three columns, the first one with the x (voltage) values, the second one with the values of the current, and the third one all zeroes.

Now you can fit the ImplicitDiodeFunction to this dataset using the first and second column for x and I and the third one (the zeroes) as the y-value. This also allows you to fit the parameters. You can’t plot this function, though.


Best

Kurt
QuantumSoft




> On 8 Sep 2021, at 20:21, Dr. M. Cuneyt Haciismailoglu <mcuneyt at uludag.edu.tr> wrote:
> 
> Hi, 
> Has anybody tried to fit an implicit function? My function is diode equation:
> I=Is*(exp(q*(V-I*Rs)/(n*k*T)-1)+(V-I*Rs)/Rp
> where q, k and T are known constants, V is independent variable, I is dependent variable and Is, n, Rs and Rp are parameters to be estimated.
> Thanks in advance
> 
> cüneyt
> 
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