In this case, is the reverse breakdown voltage at V = 0V since there is negative voltage?
seems rather bizzare that you can have 0V as breakdown voltage. perhaps the current is defined in the other direction, and so you should really flip the y-axis... and breakdown would be at -100V.. that's how I see it. look at the circuit itself (if there is one) and see what situations induce this graph
This has to be the oddest "diode" I've seen!
But according to the I-V curve, yes, reverse breakdown happens very close to 0V (typically you spec the voltage at a reverse current of 50muA).
That's my circuit config.
"Use Analysis/Setup to establish a DC linear sweep using default model parameters for -110V < V < + 10V in increments of 0.01V. Plot the I-V characteristic of the diode by using the Trace setup to add the diode current to the plot output. Be sure that the voltage axis is the voltage across only the diode, and not the diode plus resistor, which is usually the default voltage. From the plot, determine the reverse bias breakdown voltage."
according to your circuit diagram (and assuming stanard usage of symbols), when V1 is positive, the diode is FORWARD biased. so non-sensible to talk about reverse breakdown at all... you sure the ammeter is plugged in with the right polarities? -100V breakdown voltage make more sense... and more like a real diode too.
Hmm, you're right, I think the current marker is supposed to be BEFORE the diode. I get a more normal looking diode IV characteristic:
yes, perhaps by putting the meter AFTER the diode, the computer takes it as measuring the current into that terminal and not out of and hence leading to the switch in polarity.
Yup, this is right. Phew!
The 1N4002 is a pretty standard 100V rectifier (breakdown near -100V).
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