Newton Raphson method issues with differentiation

Question

I am trying to solve for the roots of the follwing equation using the newton raphson method $$f\left(x\right)\ =-I+I_{ph}-I_s\times\left(e^\frac{q\times(V_c+I\times R_c)}{n\times k\times T}-1\right)$$

The solution that I am trying to find is: $$I_{n+1}=\ I_n-\frac{I_n-I_{ph}+I_s\times\left(e^\frac{q\times(V_c+I_n\times R_c)}{n\times k\times T}-1\right)}{\frac{{q\times I}_s\times{R_c\times\left(e^\frac{q\times(V_c+I_n\ timesR_c)}{n\times k\times T}-1\right)}_\ }{n\times k\times T}+1}$$

However when i reproduce the method myself I arrive to: $$I_{n+1}=\ I_n-\frac{I_n-I_{ph}+I_s\times\left(e^\frac{q\times(V_c+I_n\times R_c)}{n\times k\times T}-1\right)}{\frac{{q\times I}_s\times{R_c\times\left(e^\frac{q\times(V_c+I_n\times R_c)}{n\times k\times T}\right)}_\ }{n\times k\times T}+1}$$

I have used matlab to find the derivative and do the calculations, I dont know where the -1 term in the denominator has gone. The Method does not work without it.

Answer 1

No idea how the first variant could be correct as Newton method, it could be a copy-paste error.

That it works is because the fixed-point iteration scheme $$x_+=x-h(x)f(x)$$ always has fixed points at the roots of $f$, and is rather flexible in $h$ if you only want to get convergence, that is, contractivity around a specific root $x_*$ of $f$. Then $$|h(x_*)f'(x_*)-1|<1$$ is a sufficient condition to have the iteration contractive in some neighborhood of $x_*$. In general, the smaller the left side, the larger the region and the faster the convergence. The extra $-1$ in the first formula could count as such an insignificant perturbation of $h(x)=f'(x)^{-1}$.

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To get quadratic convergence like in the Newton method, you need that $$h(x_*)=f'(x_*)^{-1}$$ at the root $x_*$. The easiest way to guarantee this is to have $h(x)=f'(x)^{-1}$ everywhere. Note however that any modification of $f$ leads to a different Newton iteration, for instance modifying $f$ by a factor $g$ gives $$ x_+ = x - \frac{f(x)g(x)}{f'(x)g(x)+f(x)g'(x)}=x-\frac{f(x)}{f'(x)+f(x)\frac{g'(x)}{g(x)}}. $$ An interesting factor is $g(x)=e^{i\epsilon x}$ giving $x_+=x-\frac{f(x)}{f'(x)+iϵf(x)}$ if you want to find complex roots.

Answer 2

$$f_{(V,I)} = 0 = I_{ph} - I_0\left( exp(\frac{V+IR_s}{V_t}) - 1\right) - I$$

You can simplify the problem by evaluating the function around $V_{sh}$ rather than $V$ or $I$. The $V_{sh}$ is defined as $V_{sh} = V+IR_s$. Then rewrite the equation using $V_{sh}$, such that:

$$f_{(V,I)} = 0 = I_{ph} - I_0\left( exp(\frac{V_{sh}}{V_t}) - 1\right) - \frac{V_{sh}-V}{R_s}$$.

Take partial derivative $\frac{\partial f_{(V_{sh}, V)}}{ \partial V_{sh}}$:

$$\frac{\partial I}{ \partial V} = - \frac{I_0}{V_t} exp(\frac{V_{sh}}{V_t}) - \frac{1}{R_s}$$

Solve iteratively for $I$, given some initial value for $V_{sh}$, remember that $V_{sh}=V$ at $V_{oc}$, which is a good initial guess, since you also know $I=0$ at $V_{oc}$.

$$V_{sh(i+1)} = V_{sh(i)} - \frac{I_{ph} - I_0\left( exp(\frac{V_{sh}}{V_t}) - 1\right) - \frac{V_{sh}-V}{R_s}}{ - \frac{I_0}{V_t} exp(\frac{V_{sh}}{V_t}) - \frac{1}{R_s} }$$

Upon convergence, the current is then:

$$I = \frac{V_{sh}-V}{R_s}$$

You can get the whole I-V characteristic this way, even continue with Bishops reverse-bias model, and/or second diode, $R_{sh}$ resistance, etc.

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P.S. The $V_t$ gathers all the terms for temperature, diode factor and the other constants.