How to best solve a system of equations like this one

Question

I need some help trying to solve a system of two equations with two unknowns.

Background: This is not homework, just hobby. I have some LEDs that I needed to identify, but the manufacturer datasheet was wrong. Diodes allow a current (I) to pass through them depending on the voltage (V) applied to the terminals. So it's a I(V) function. Giving a value for V, I can then measure the current, and repeating this again with a different V I have a system of two equations with two unknowns.

Diodes are usually modelled using the Shockley diode equation:

$I = Is\left (e^{\frac{Vd}{\eta Vt}} - 1 \right)$

Assuming the thermal voltage (Vt) is constant during the test, I need to solve a system of two equations:

$\begin{matrix} \\ a = x\left (e^{\frac{b}{y}} - 1 \right) \\ \\ c = x\left (e^{\frac{d}{y}} - 1 \right) \end{matrix}$

Where a, b, c and d are known, and x and y are the unknown values.

I solved it yesterday using software (shame on me). But today I tried it again with just paper and pencil. I've filled two and a half sheets to no avail. Tried solving x and failed, then tried again for y and this is the closer I could get with my rudiments of algebra:

$\left ( \frac{a}{x} + 1 \right )^{\frac{1}{b}}=\left ( \frac{c}{x} + 1 \right )^{\frac{1}{d}}$

And I'm not even sure it's correct, but for me it's a dead end.

How could I solve a system like this?

BTW, is there a good book on techniques to deal with this kind of stuff?

Thanks in advance.

Answer 1

I think that writing the ratio of the two equations makes the problem simpler : "x" is gone and you just have a function of "y". Plot the function (or compute it for a few values of "y") in order to locate even very rougthly where is the solution and how behaves the function. Then, just start Newton iterations. If you need more clarifications, just post.

Answer 2

I don't think there is a general solution for any $a, b, c, d$, but if $b$ is twice $d$ or the other way around, or $b = d$, you can do this: Divide one by the other and get

$ \frac a c = \frac{e^{b/y} - 1}{e^{d/y} - 1}$

which gives

$ e^{b/y} - \frac a c e^{d/y} + \frac a c - 1 = 0 $

which is a linear equation in $e^{b/y}$ if $b=d$ or a quadratic equation if $d = 2b$ or $b = 2d$. It may be analytically solvable for b/d being higher multiples of d/b, but it starts getting hard to solve then and you should probably use numerical methods like newtons method anyway.

Once solved for $e^{b/y} = k$, y is given by $\frac{b}{\ln{k}}$ and x by substitution of y into any of the two equations you began with.