## Diode Parameter Extraction & Plotting

After blowing up a MAX4372 high-side current amp, I thought a Schottky diode would do the trick as a voltage limiter for the delicate current-sensing inputs. The comments to that posting showed I might be close, but that I hadn’t figured it right.

The first step is finding out how the diode behaves.

Clip a multimeter (set to the 2 V range) across the diode, clip another multimeter (set to maybe 200 mA in series with the diode, then connect a bench power supply through a 100 kΩ (more or less) resistor to limit the diode current across everything.

Twiddle the power supply knob, record voltage and current pairs, type them into a spreadsheet (say, OpenOffice, but Excel is probably similar). That gives you a table & plot that looks like this:

 mV uA 20.1 0.4 40.1 1.4 60.0 3.5 80.6 8.2 100.2 18.0 125.4 48.9 150.5 131.3 175.0 343.0 183.8 481.0

Nothing surprising there: the current has an exponential relation to the forward voltage.

An exponential relation cries out for a semilog plot, so add a third column figuring the natural log (a.ka. ln or log-to-the-base-e) of the current values in the second column. The equation is just

`=LN(B3)`

copied down the column as needed.

That gives you another table & plot, thusly:

 mV uA ln(current) 20.1 0.4 -0.92 40.1 1.4 0.34 60.0 3.5 1.25 80.6 8.2 2.10 100.2 18.0 2.89 125.4 48.9 3.89 150.5 131.3 4.88 175.0 343.0 5.84 183.8 481.0 6.18

The trick is to add a regression line to the data, which you do by selecting the data series, other-clicking, selecting “Add Regression Line”, selecting the regression line, other-clicking, selecting “Show Equation”, then futzing around until the equation shows enough decimal places. Also extend the X & Y axes so you can see the Y-axis intercept on the left and the current at the MAX4273’s Absolute Max rating of 300 mV on the right.

I threw out the first measurement point, as it didn’t quite fit the rest of the data. My measurement accuracy isn’t all that great below a microamp, sooo that seemed justified. Check the raw data and see for yourself.

The regression equation is, comfortingly, ln(current) = 0.040 * voltage – 1.187.

The slope of 0.040 = kT/q, which says the temperature of my basement laboratory is 464 K, a tad warmer than the actual 286 K. Feeding the actual temperature in, the slope should be 0.046.

What that really means is that the ideality factor n = 1.62. We usually forget about that little Fudge Factor, but here it is in action: 0.040 = nkT/q.

The Y-axis intercept is -1.187, which means:

saturation current = exp(-1.187) = 0.3 uA = 300 nA.

Not a number you’ll get from the datasheet, of course.

For more on all that, consult the Wikipedia diode entry.

It’s worth mentioning that the slope depends linearly on the temperature. The exponent causes the far end of that nice line to whip the current around something nasty.

Anyhow, with numbers in hand, it’s back to the schematic… and a bit of SPICE simulation that uses a canned diode model.