Diode Lab vs Lab 0. You looked at the residuals of the fit, and they probably looked like random noise.

Transcription

1 Diode Lab vs Lab In Lab, the data was from a nearly perfect sine wave of large amplitude from a signal generator. The function you were fitting was a sine wave with an offset, an amplitude, a frequency, and a phase. You should have gotten a very good fit, with an amplitude and frequency that agreed very well with the signal generator setting, an offset near zero, and a phase that was different every time you pressed the Acquire button. You looked at the residuals of the fit, and they probably looked like random noise. Gnuplot s fit calculated the RMS (standard deviation) of the residuals. You drew error bars equal to that value, and probably found that the residuals were consistent with zero within the error about 2/3 of the time. The point was to learn how to use the data acquisition computers, and to use Gnuplot to fit some data. The situation was artificially simple, both to make it easy to get started, and also to show you what good agreement between data and a model looks like. This is the last time in the course where it will be that easy to fit data and get good agreement with the model. In the diode lab, the data will push the instruments to the limit, and the models won t be so perfect.

2 Lab looked trivially simple, but was not! Is there any systematic behavior in the residuals? YES, there is!

3 Basic Diode Lab Data You should have gotten data with the diode and a 1K resistor, at room temperature. The raw voltages don t look much like an exponential law 'diode.dat' u 2:1 '' u 4: e+6 You used Gnuplot to subtract the voltages at the two ends of the diode and divide by the resistor value to get the current through the diode. 1.6e-5 1.4e-5 1.2e-5 1e-5 8e-6 6e-6 4e-6 2e-6 'diode.dat' u ($1-$3):($3/R) -2e

4 Or did you actually see something like this?

5 Here is the source of the problem: the input from the function generator looks good but it is NOT

6 Here is the source of the problem: the input from the function generator looks good but it is NOT

7 If we introduce an appropriate RC filter, we can cut out the high frequency noise NO filter We still get two traces. Why is that? with filter

8 We still get two traces. Why is that? This is just a sampling issue, which we can solve by increasing the sampling frequency Low frequency High frequency

9 Log Plot and Line Fit Plotting the current vs the voltage does give you something that looks like an exponential law. But is it really? An exponential law will look like a line on a semi-log plot. So we made a log plot, and got a line..1 'diode.dat' u ($1-$3):($3/R) 1e-5 1e-6 1e-7 1e Blow up from volts to the max, and draw a line..1 1e-5 'diode.dat' u ($1-$3):($3/R) f(x) 1e-6 1e-7 1e-8 1e

10 Function Fit This looks pretty good, but it deviates from the line at the lower left. But we don t expect the line to be exact, the real function is expected to be I( V) = I [ e V V 1]. So we fit this form to the data and get this:.1 1e-5 'diode.dat' u ($1-$3):($3/R) I(x) 1e-6 1e-7 1e-8 1e-9 1e The real function makes the deviation look even worse! But the digitizer has a finite resolution. For your data, the minimum step size is 5 volts divided by 496 (12 bits), or about 1.2 millivolts. So we need to draw in the errorbars. An error of fixed size will look much bigger at the low end of the log plot.

11 Add Errorbars With the right errorbars, the disagreement isn t to terrible:.1 1e-5 'diode.dat' u ($1-$3):($3/R):(dv/R) I(x) 1e-6 1e-7 1e-8 1e-9 1e If we plot the ratio of the data to the fit, we can see that there s still a problem near zero voltage: 'diode.dat' u ($1-$3):(($3/R)/I($1-$3)):((dv/R)/I($1-$3)) It s only about one sigma, but it s systematic, not random.

12 Fit Residuals Now let s check the residuals, the difference between the data and the fit. This emphasizes the high-voltage end, where the current is large. 1e-6 5e-7 'diode.dat' u ($1-$3):($3/R-I($1-$3)) -5e-7-1e-6-1.5e-6-2e There s lots of scatter there! And the errorbars calculated from the digitizer step size aren t big enough to explain it: 1e-6 5e-7 'diode.dat' u ($1-$3):($3/R-I($1-$3)):(dv/R) -5e-7-1e-6-1.5e-6-2e

13 Horizontal Errors But, we have only considered the errors in the current (vertical) measurements. There is also an error in the voltage (horizontal) measurements. Since the exponential is so steep, even a small horizontal error can make a point appear to have a large vertical discrepancy. If we plot the data with the right horizontal errorbars as well as vertical errorbars (which are invisibly small in this region!), the agreement with the function is pretty good: 1.8e-5 1.6e-5 1.4e-5 1.2e-5 1e-5 8e-6 6e-6 4e-6 2e-6 'diode.dat' ev 1 u ($1-$3):($3/R):(1.41*dv):(dv/R) I(x) Gnuplot, like most fitting programs, can t cope with horizontal errors. But there is a trick that is a reasonable approximation to doing the right thing: add an additional vertical error equal to the horizontal error times the slope of the function. The lab procedure spells out how to do this.

14 Residuals with Equivalent Vertical Errors Plot the residuals, but using the effective error combining both vertical and horizontal contributions: 1e-6 5e-7-5e-7-1e-6-1.5e-6 'diode.dat' ev 5 u ($1-$3):($3/R-I($1-$3)):(ierr($1-$3)) -2e The residuals are now consistent with zero within the (combined vertical and horizontal) error. Plotting the residuals from all the data points, we get 1e-6 5e-7 'diode.dat' ev 5 u ($1-$3):($3/R-I($1-$3)):(ierr($1-$3)) -5e-7-1e-6-1.5e-6-2e

15 Pulls Plot The basic voltage measurement error is the same for all the points, but since the exponential slope varys so much, the error bars of different points are very different in size. The plot autoscales to the biggest errorbars, which actually contain the least information. In cases like that, the best thing to plot is the residual divided by the error, which is often called the pull for short. The errorbar is then the error divided by the error, or 1. for all points. 3 2 'diode.dat' ev 5 u ($1-$3):(($3/R-I($1-$3))/(ierr($1-$3))):(1) This shows that the data is consistent with the fit, over the full range, within the (combined) errorbars. If anything, the errorbars look too large...

16 Least Squares Fits If we don t have any idea about the errors of the measurements going into a fit, we have to treat each point equally in the fit. Usually the fit tries to minimize the sum of the squares of the difference between the function and the data. This is called a least squares fit. All we can do to check the quality of the fit is to look at the residuals. And we don t really know what the errors on the fit output are. If the residuals look random and the scatter is uniform, then it s sensible to conclude that the measurement errors are about the same on all the data points. We can then use the scatter of the measurements to infer the measurement errors. This allows us to estimate the error on the fit output. Gnuplot does a least squares fit if you don t give a third column. It also uses the scatter (whether or not the residuals look uniform and random!) to calculate errors.

17 Chi-Square Fits If we do have an external error estimate for each point going into a fit, we should give more weight to the points with smaller errors. The quantity minimized is the sum of the squares of the [residuals divided by errors] (i.e., sum of squares of pulls). χ 2 = meas [ y meas y pred ( parameters) ] 2 σ 2 The usual symbol for this quantity is the Greek letter chi, so it is called a chi-square fit. If the model used actually does fit the the data within the measurement errors used, then the square of the residual is on the average equal to the square of the error. So the average value of each term in the sum is 1., and the chi-square sum is expected to be roughly the number of data points.

18 Gnuplot Chi-Square Fits Gnuplot does chi-square fits if you give it a third column containing measured or calculated errors on points. It reports the degrees of freedom which is number of data points minus number of fit parameters, and the reduced chisquare which is the chi-square divided by degrees of freedom. Here s the results for the above fit, using combined vertical and horizontal errors degrees of freedom (ndf) : 998 rms of residuals (stdfit) = sqrt(wssr/ndf) : variance of residuals (reduced chisquare) = WSSR/ndf : This is actually too good, which probably means that the errors used in the fit are larger than they should be. Remember that the pulls-plot showed that nearly all the pulls were zero within the errors, more than the expected 2/3. For comparison, here s the results using only the vertical error from digitizer quantization: degrees of freedom (ndf) : 998 rms of residuals (stdfit) = sqrt(wssr/ndf) : variance of residuals (reduced chisquare) = WSSR/ndf : This is an awful chi-square! Remember that the residuals plot with only vertical errors had huge scatter for large voltage, so this should not be surprising.

19 Chi-Square Fits and Parameter Errors If reduced chisquare is much greater than 1, then something is wrong. Either the model function cannot reproduce the shape of the data (which could be a problem with the model, or a problem with the data), or the program didn t converge properly, or the errors assigned to the data are too small. So play with the model function to see if you can make it show all the features of the data, check convergence (fit function vs data, lambda-value), plot residuals and pulls to see where the problem is. If reduced chi-square is much less than 1, almost always the errors assigned to the data are too large. That s what happened in the above example. Remember Gnuplot scales parameter errors by the square root of the reduced chi-square. If the reduced chi-square is close to 1 like it s supposed to be, this doesn t change things much. If it s much less than 1, probably you used errors that were too big, so the standard fit parameter errors were probably too big, and the re-scaled errors are probably about right. If it s much greater than 1, Gnuplot s errors have been inflated. But you shouldn t really trust anything in such cases.

20 Improving Data with Different Resistors The function predicts negative current at negative voltage, and our data shows positive current at negative voltage. But, if we extrapolate the fit, the negative current is very small. We wouldn t really be able to see it, because the digitizer step size is too coarse. On a linear scale: 5e-8 4e-8 'diode.dat' u ($1-$3):($3/R) I(x) 3e-8 2e-8 1e-8-1e The data you took at the end, with no voltage going into the circuit, allows you to measure the offset of the digitizer, which should help explain this disagreement. We can t reduce the digitizer step size, but we can increase the voltage signal that we get for a given current, but using a larger resistor value. Using 1 million Ohms instead of 1K would increase the signal by a factor of 1, making it more easily visible over the digitizer. This will make the negative-voltage range measureable, but will limit the range at positive voltage. So we will need to combine measurements with different resistors.

21 Combining Data Files Gnuplot can t fit multiple data files. So to fit all of the data, you will need to combine the files together into one. We need to tell Gnuplot which resistor value to use for which data points. But Gnuplot only thinks about one line of the data file at a time, so we have to get the resistor value into each line of the file. That s what the AddCols program is for. You can find it on the course web page. Alternatively, and possibly more convenientely, you can use either Gnumeric or Excel. When you fit the combined data, it s OK to leave in all the points which look far from the line, as long as you tell Gnuplot what the errors on each point are. This will vary from file to file because the resistor values are different, and you may also have used different digitizer inputs which have different quantization errors for different files. You have to make sure that Gnuplot has enough information on each line of the file to calculate the right error. That error should include the contribution of the horizontal error times the function slope.

22 Fitting Combined Data Your data will probably cover more than 6 orders of magnitude. Having a model that works over a factor of a million is pretty impressive! You should probably exclude the highest-voltage points from the fit of the combined data file. They are probably rather inconsistent with the simple exponential model. It s interesting to plot the measured current divided by fitted function, with errorbars. If you excluded the right range, this should be flat over the fit range with 2/3 of the points with errorbars crossing 1.. Above the fit range, it will systematically deviate from the simple exponential model. The final lab session will be on the temperature dependence of the current-voltage relation. Both V and I depend on temperature rather strongly. You will dip diodes in boiling water, ice water, dry-ice and alcohol, and liquid nitrogen. You will have to decide what resistor value(s) to use, and what voltage ranges to use.