0
$\begingroup$

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

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

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: $$ y = 10 ^ {\frac{z_2 - z_1}{x_2 - x_1} (x - x_1) + z_1}. $$

$\endgroup$
3
  • $\begingroup$ 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, 2014 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, 2014 at 8:55
  • $\begingroup$ Exactly. You're welcome. :-) $\endgroup$
    – A. Donda
    Dec 2, 2014 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.