## Zener Diode Characteristic Curve

 I'm try to model the zener diode characteristic curve for a range of just negative of the avalanche region to just positive of the Vpn region. This should be a really simple circuit and a simple DC Sweep. I've had no issue doing the same with the regular diode, but for some reason, I'm getting something that appears to be exactly backwards. Several of my students have worked with me and none of us have come up with an answer. Would you please check my circuit? I've tried switching the orientation of the voltage source, changing the sweep limits, changing the zener choice from the list available, etc. Nothing resembles the "curve" that I know the zener is supposed to show. Thanks as usual for your help, Carol Strong UAH by strongc October 08, 2020 Hi @strongc, the circuit appears to be private. Can you make it public so we can take a look? by mrobbins October 08, 2020 Now Public, sorry! by strongc October 09, 2020 You have to plot the current through the diode versus the voltage across the diode. Your actual plot is against the voltage of the source, not the diode. by vanderghast October 09, 2020

 Take note of the axis (which one is horizontal and which one is vertical). Haven't found the way to exchange them. I added a ground to specify where is the reference (a voltage is always a difference about TWO points. When a single point is given, that implies that the second point is the referenced ground) in order to simplify the graph. Or, if you prefer, the voltage across the diode is the difference of potential between, first, the current source and the diode, and, second, the diode and the resistor. When the latter is zero, that simplify the equation to be used to get the voltage ACROSS the diode. +1 vote by vanderghast October 09, 2020 I like this concept. Here's my simplification: Resistor in series with an ideal current source is basically irrelevant, so I've removed R1. Then I flipped around the elements a bit, and extended the DC Sweep to include both positive and negative currents: Take a look at this simulation. I think you'll find that the two "plateau" areas on the I-V curve are where you expect: one around +0.7V, and one around -5.1V (for this particular Zener part number). by mrobbins October 09, 2020

 Thanks to both of you. I'll look closely at your ideas this weekend and I might suggest a description of both options to my students without actually giving them a circuit. By now they should be able to put together a circuit and play from someone's description. I'm excited to try this!! C. Strong UAH +1 vote by strongc October 09, 2020

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