I'm playing with some transistor test data and having trouble understanding what is probably a very basic principle.

Below are my two graphs, plotted linearly and then with a logarithmic scale on the y axis. (Absolute vals for y taken.)

lin log

I understand that my linear values approach zero and log10(0) = -Inf, so the huge drop off in values as x -> 0 makes sense. But I haven't quite grokked converting back and forth yet.

I'm attempting to do some curve fitting to get the gradient in a pretty good linear region which can be used to calculate junction temperature etc. I'm plotting using semilogy, sighting a range to use as my linear region, then feeding that forward to perform the fitting. As the data stays in a protected structure, I want that range in terms of the non-log values (is it obvious my field is programming not maths..?).

How do I convert back? I can look at the graph and see that it is linear enough for my purposes between 10^-6 and 10^-1 but then how does that relate to my actual data?

log10(a) = b -> 10^b = a, yep, but using the range above:

10^(10^-6) = 1.0000023..

10^(10^-1) = 1.2589254..

And then I remembered I had taken the abs value so..

-10^(10^-6) = -0.9999..

-10^(10^-1) = -1.2589..

All of which are out of range of the original data.

I want to be able to say minY = someOperation(10^-6), maxY = someOperation(10^-2)

What do?

Thank you for reading, Marshall


Let's call the value on the horizontal $x$ (your Vd) and the value on the vertical $y$ (your ld)

From inspecting the semilog graph, you find that between two values $x_1$ and $x_2$, the decadic logarithm $z = \log_{10}(y)$ is approximately a linear function of $x$, with values $z_1 = \log_{10}(y_1)$ and $z_2 = \log_{10}(y_2)$ (in your example, $y_1 = 10^{-1}$ and $y_2 = 10^{-6}$ and therefore $z_1 = -1$ and $z_2 = -6$).

The linear equation for $z$ is $$ z = a x + b. $$ With a bit of algebra it follows that you can determine the coefficients $a$ and $b$ from the data about your two points via $$ a = \frac{z_2 - z_1}{x_2 - x_1} $$ and $$ b = z_1 - \frac{z_2 - z_1}{x_2 - x_1} x_1. $$

But what you are interested in is the equation for $y$: $$ y = 10^z = 10^{ax + b} $$ with $a$ and $b$ as computed above, or explicitly: $$ y = 10 ^ {\frac{z_2 - z_1}{x_2 - x_1} (x - x_1) + z_1}. $$

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  • $\begingroup$ Hi, thanks for the response. It's really good for me to step away from the code and have a mathematical explanation. I had remembered how to get the gradient from the elimination method (z_2 - z_1 = mx_2 - mx_1 etc.), and see that you've just subbed that back into the linear equation, but when the linear equation is z=ax+b, why is that z_1 and x_1 ? $\endgroup$ – El Loro Dec 2 '14 at 8:35
  • $\begingroup$ Ah, it doesn't matter which, because b is an element of both equations in the system of eqs that was developed to calculate the original value for a, namely z1 = mx1 + b & z2 = mx2 + b. $\endgroup$ – El Loro Dec 2 '14 at 8:55
  • $\begingroup$ Exactly. You're welcome. :-) $\endgroup$ – A. Donda Dec 2 '14 at 20:11

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