# How to convert an I V curve for a diode into a straight line

Discussion in 'Electronic Basics' started by J. Payne, Nov 4, 2003.

1. ### J. PayneGuest

Having don some measuring of the I V characteristics of some diodes, I
am stuck with the problem of converting them into a staight line, so
that I can get some useful information out of them, and also to check
that the effect that is caused is caused for the reason I expect.
Now, before you go an say "plot V against ln(I)", I have tried that,
and, getting another curve, am now fairly sure that this is due to an
ohmic resistance (I have checked this on spice, and the approximate
shapes look similar).
Now, my question is, how do I show that this effect is due to the
additional resistance and not anything else - ie - is there a function
I can perform on the data to make it linear?
Cheers

2. ### Jacobe HazzardGuest

I would think your best bet would be to produce an equation that closely
fits your measured value (all sorts of software can do this fitting for
you). Then coming up with an inverse function is easy, and your only task is
to account for all the constants in your equation (or come up with an
equation for which you can).

3. ### Kevin AylwardGuest

There is an equation for the supply diode resister combination. I have
it derived and shown at the end of the page at the following link
http://www.anasoft.co.uk/EE/widlarlambert/widlarlambert.html

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

4. ### SteveGuest

Draw a "line of best fit"

5. ### J. PayneGuest

Cheers, thats the kind of thing I was looking for, however, its a
little far from the scope of what I'm doing (this is A-level Physics).
However, I had another idea today and am wondering whether it would
work. As the I V curve converges towards becomming a staight lin for
larger values of V, which is basically the resistor's characteristic
curve, I could use the gradient of this to find the ohmic resistance.
Then I calculate the voltage drop across this (V=IR) and subtract that
from the drop I measured. This will give me the voltage across the
perfect diode within. Then my new curve (V across ideal diode part vs
I) should follow the standard exponential pattern. - Would this work?

And steve, what exactly do you mean? My line is a curve, not straight,
and drawing an arbitrary line through it will achieve nothing - the
purpose of such lines is to smooth out random errors in data.

6. ### Roger JohanssonGuest

Use a log/lin diagram and you will get a straight line.

7. ### John PopelishGuest

I think you need to do a curve fit for a function that has one term is
proportional to the log of current, and one term that is proportional
to the current, and solve for the coefficients that best fit your
data. The definition of 'best' is also something you need to think
about. The log curve is the ideal diode characteristic, and the
proportional term is the resistive drop.