I started today with yet another variant of the “do-now” problem:

You have sensor whose resistance varies from 1kΩ to 4kΩ with the property it measures and a 6v power supply. Set the parameters for the following circuit to get an output voltage that varies from 1v (at 1kΩ) to 2v (at 4kΩ).

I did not give them the solution, but promised them a shortcut solution at the end of class.

My co-instructor then proceeded to develop voltage sources, current sources, load and source resistances, load lines (though he did not use that name nor prove that it was a line), Thévenin circuits, and Norton circuits. He showed how Thévenin and Norton circuits were indistinguishable from the outside, but ran out of time before getting to the Thévenin equivalent of a voltage divider. His presentation was in a bit different order than I would have done it—I would have introduced the non-ideal voltage source first, plotted its load line, and then talked about the resistance of the ideal voltage source. I’ll have to think about whether I now like his order better or not.

The shortcut solution that I was going to give the students relied on that last step. I did suggest that the students look at splitting the circuit into a voltage divider and the load resistance RC, and come up with a Thévenin equivalent for the voltage divider. Since we had already determined that a 3V source with a 2kΩ resistor would convert the load resistance into the desired voltages, all we need to do is to come up with a voltage divider that changes the 6v source into a 3v source with a 2kΩ resistance. Setting RA=RB=4kΩ does that.

I think that it is interesting to compare this circuit with the circuit using series resistors from Wednesday’s lecture (using a 6v power supply for each), using a gnuplot script. Although both designs meet the specs for the voltage at the endpoints, they behave differently in between.

The following gnuplot script illustrates the differences:

parallel(r1,r2)=r1*r2/(r1+r2) s_circuit(r,ra,rb) = Vin*(r+rb)/(r+ra+rb) sp_circuit(r,ra,rb) =Vin*parallel(r,rb)/(ra+parallel(r,rb)) Vin=6 set title "Calibration curve for resistance->voltage conversions" set xlabel "Resistance RC (ohms)" set ylabel "Output voltage (V)" set xrange [1000:4000] set key top left unset label unset arrow plot s_circuit(x,10000,1000) title "series RA=10kohm, RB=1kohm",\ sp_circuit(x,4000,4000) title "series-parallel RA=4kohm, RB=4kohm"

The series-parallel conversion might be more appropriate if we need more sensitivity at the low end, and the all-series one if we need more uniform sensitivity. That decision depends on how the resistance is related to the phenomenon we are measuring.

The series-parallel conversion circuit always draws more current from the 6v source than the all-series one, with 1.25mA at the low end and 1mA at the high end, instead of 0.5mA at the low end and 0.4mA at the high end.

I’m trying to decide whether to spend any more class time on this problem, or simply hand out this solution. We will certainly revisit Thévenin equivalents, since they are such a useful tool.

There is no class Monday, and we have a lot of “stuff” to cover before the students can do next Thursday’s lab, so I need to see how much I can pack into the next lecture without losing the students. At a minimum, they’ll need

- complex numbers: . I’ll probably have to do this one by fiat, rather than any sort of elegant math, and there won’t be time in this lecture to do a nice development of phasors—that will have to wait until after the lab.
- differential equation for capacitors: , derived from (which they got reminded of earlier this week).
- current for a capacitor with a sinusoidal voltage gives
- complex impedance (voltage over current)
- behavior of RC circuits (just looking at amplitude, not phase)
- gnuplot commands for setting up impedance functions for capacitors and series and parallel circuits (which can be done with a handout, rather than lecture time).

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