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Our teacher in class instructed us to experiment with a resistor/diode circuit in LTspice. The goal is to display the I-V characteristic of the resistor, as well as that of the diode, and test them with different resistor values in order to get different operating points, but I'm struggling to achieve that outcome.

Here's what's provided on his slide that we should base our simulation on: enter image description here

And here's what I'm supposed to get : enter image description here

Since there are several operations in the LTspice file, I believe I might need to make some adjustments (perhaps using a DC source instead of AC?) to achieve the desired outcome. Could you assist me in obtaining the expected result?

I'm also unsure about what to do with all the other operations. He covered them during the lecture, but it was so quick that I couldn't grasp it.

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  • \$\begingroup\$ You are being asked to cause LTspice to create an I-V chart that looks exactly like the nice picture there? \$\endgroup\$
    – periblepsis
    Commented Sep 30, 2023 at 21:45
  • \$\begingroup\$ Well something similar with the straight line (resistor characteristic) and the exponential (the diode characteristic) \$\endgroup\$
    – c.leblanc
    Commented Sep 30, 2023 at 21:47
  • \$\begingroup\$ LTspice doesn't support two different x-axis choices. For the resistor, you'd like to have Vdd-V(Vdd,Vd) as the x-axis. But for the diode you want V(Vd). Not sure how you get those two pasted on top of each other. Not at the same time, anyway. \$\endgroup\$
    – periblepsis
    Commented Sep 30, 2023 at 21:52
  • \$\begingroup\$ Also, note that in the given image the value for Vd is about 720 mV and the value for Vdd is 10 V. But the picture shows a Vd that is halfway to Vdd. But that's not even close to reality. So the x-axis as shown isn't in proper perspective. A linear chart just won't look like that. \$\endgroup\$
    – periblepsis
    Commented Sep 30, 2023 at 22:19
  • \$\begingroup\$ numbers are not important I just gave the picture to give an idea of what kind of graph we need to get .. I don't really get what he wanted us to do with the simulation and all those operations he showed us :( \$\endgroup\$
    – c.leblanc
    Commented Sep 30, 2023 at 22:29

3 Answers 3

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LTSpice allows you to use an expression when plotting a measured value. The right type of curves can be achieved by doing a DC sweep and calculating the resistor current.

The sweep is performed on the diode and resistor separately within the same circuit.

In the attached plot the resistor loadline is plotted by subtracting it from 10mA.

The scaling of the plot needs to be adjusted manually as 10V across a 1n914 causes about 16A to flow and the normal autoscaling will shrink the interesting part of the plot so it is too small to see.

If the scaling is zoomed in even further the value of the intersection can be seen.

The diode I have chosen gives slightly different intersection than your example.

Schematic enter image description here

Plot with 10mA by 10V scaling

enter image description here

Plot with 10mA by 1V scaling

enter image description here

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Since there are several operations in the LTspice file, I believe I might need to make some adjustments (perhaps using a DC source instead of AC?) to achieve the desired outcome.

  1. Only the DC analysis operation is relevant to this problem. The AC and transient analyses don't help you with this problem. It's not clear if your instructor included them just to confuse you or because you'll use them for some other part of the problem set you haven't shared with us.

  2. The source shown in your circuit is set up to have a transient output, not an AC output. Even though the transient is a sine wave, this won't cause the source to produce any output in an AC simulation. (This doesn't matter to you because you want to do a DC simulation, but I wanted to mention this potential gotcha in case you have additional problems that do require AC simulation)

  3. The DC analysis takes over the DC setting of the source, so you won't have to make any changes to make any changes to the source to make the analysis work.

  4. The DC analysis shown here seems to be trying to control a source named V2, but there isn't any V2 in the circuit. So you will have to adjust the analysis operation to have it control a source that is actually present in your circuit.

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I found the answer in a video of an old lecture, the point was to plug a second voltage source parallel to the diode and apply the DC sweep on that one. Also using different value of the resistor with '.step PARAM' makes it possible to see different characteristic lines with different operating point.

enter image description here

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