Solving $ 20I - 0.1092 \times \ln{\left(\frac{I}{0.002}\right)} = 7 $

Question

I have come across the first usage of a logarithm since college in solving a circuit equation, and I cannot remember for the life of me how to solve log problems. What I have thus far is:

$$ 20I - 0.1092 \times \ln{\left(\frac{I}{0.002}\right)} = 7 $$

I think I need to raise both sides to e, but hoping someone could give me a basic refresher on how to do this.

Answer 1

David, just follow along. I'm going to use variable names for your constants, though, per a fine comment below this answer.

So please find for the following development, these variables: $R=20\:\Omega$, $\eta\;V_T=0.1092\:\text{V}$, $I_0=2\:\text{mA}$, and $V_\text{CC}=7\:\text{V}$. $\eta$ is the emission co-efficient and represents the non-ideality as charges pass through a depletion region. $V_T=\frac{k\;T}{q}$ and is the statistical thermal voltage. $R$ is a resistor in series with the diode. And $I_0$ is a reference current against which $I$ is compared for these purposes. Finally, $V_\text{CC}$ is the applied voltage to the series circuit.

That said, I think the OP got a sign wrong in the equation provided. But I'm taking it as given rather than attempt to correct it. It really should be the KVL equation, $V_\text{CC} - R\;I - \eta\;V_T \cdot \ln{\left(\frac{I}{I_0}\right)} = 0\:\text{V}$. (See Note below.)

(For those not versed in the units used in electronics for the OP's question, please forgive my use of variables with units here. They do apply, though. So I'm keeping them. Dimensional analysis still applies, of course.)

$$\begin{align*} R\;I - \eta\;V_T \cdot \ln{\left(\frac{I}{I_0}\right)} &= V_\text{CC}\\\\ \frac{R\;I}{\eta\;V_T}-\frac{V_\text{CC}}{\eta\;V_T} &= \ln{\left(\frac{I}{I_0}\right)}\\\\ e^{^{\frac{R\;I}{\eta\;V_T}-\frac{V_\text{CC}}{\eta\;V_T}}} &= \frac{I}{I_0}\\\\ 1 &= \frac{I}{I_0}\cdot e^{^{-\frac{R\,I}{\eta\,V_T}+\frac{V_\text{CC}}{\eta\,V_T}}}\\\\ e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}} &= \frac{I}{I_0}\cdot e^{^{-\frac{R\,I}{\eta\,V_T}}}\\\\ \frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}} &= \frac{R\,I}{\eta\,V_T}\cdot e^{^{-\frac{R\,I}{\eta\,V_T}}}\\\\ -\frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}} &= -\frac{R\,I}{\eta\,V_T}\cdot e^{^{-\frac{R\,I}{\eta\,V_T}}}\\\\&\text{set }u=-\frac{R\,I}{\eta\,V_T}\\\\&\therefore\\\\ u\,e^u&=-\frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}}\\\\ u&=\operatorname{LambertW}\left(-\frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}}\right)\\\\ -\frac{R\,I}{\eta\,V_T}&=\operatorname{LambertW}\left(-\frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}}\right)\\\\ I&=-\frac{\eta\,V_T}{R}\cdot\operatorname{LambertW}\left(-\frac{R\,I_0}{\eta\,V_T}\cdot e^{^{-\frac{V_\text{CC}}{\eta\,V_T}}}\right) \end{align*}$$

There are two obvious branches of LambertW that produce a real number result for $I$. These are $I=2.89494375302447\times 10^{-31}$ and $I=0.378629016951324$.

[Note: Had the correct KVL equation been applied (it wasn't, above), the answer would instead be the main branch's $I=0.322251273110783$.]

Answer 2

I will do this assuming the constants are exact, but note that $20,7,.002$ each have only one significant figure if they are measured numbers so you should look closely at error propagation. You cannot use the simple linear model we usually do because $0.002$ could be $\pm 25\%$. I will work to four figures. First simplify $$20I - 0.1092 \times \ln{\left(\frac{I}{0.002}\right)} = 7\\ 20I-0.1092\ln(I)+0.1092\ln(0.002)=7\\ 20I-0.1092\ln(I)=7.678$$ Now logs are slowly varying. If we ignore the $\ln(I)$ term we have $I=0.3839$ so it screams for fixed point iteration $$I=\frac 1{20}(7.678+0.1092\ln(I))$$ We start with $I=\frac{6.321}{20}=0.3839$ and iterate to convergence. In three iterations we have $I=0.3786$.

Answer 3

Working with whole numbers, the equation write $$20 x-\frac{273 }{2500}\log (500 x)=7$$ Let $500 x=e^t$ to make $$\frac{1}{25}e^t-\frac{273 }{2500}t=7$$ and, as said in comments, the solution is given in terms of Lambert function. $$t=-\frac{2500}{39}-W_{-1}\left(-\frac{100}{273 }e^{-2500/39}\right)\approx 5.24341$$ Back to $x$, this gives $x \approx 0.378629$.

If you do not want to use Lambert function, only numerical methods would give the result. Neglecting the logarithmic term, let us start with $x_0=\frac 7 {20}$ and use Newton method which will give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.3500000000 \\ 1 & 0.3786466187 \\ 2 & 0.3786290170 \end{array} \right)$$

Answer 4

I changed the variable $I$ to $x$, then asked WolframAlpha to solve your equation, it gave two solutions:

$$x_1 \approx 2.89494 \times {10}^{-31}$$ and $$x_2 \approx 0.378629.$$

I agree with the others in saying that the problem does not admit an analytic solution. Only a numerical approximation is possible.