Reverse Breakdown Voltage of a 1N4002 Diode

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Discussion Overview

The discussion revolves around the reverse breakdown voltage of the 1N4002 diode, focusing on the interpretation of the I-V characteristics and the implications of circuit configuration. Participants explore the conditions under which breakdown occurs and the significance of voltage measurements in relation to the diode's behavior.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether the reverse breakdown voltage can be 0V, suggesting that the current direction might need to be reconsidered.
  • Others propose that the breakdown voltage is more realistically around -100V, aligning with typical characteristics of diodes.
  • A participant mentions that the I-V curve indicates reverse breakdown occurs close to 0V, typically specified at a reverse current of 50µA.
  • Concerns are raised regarding the configuration of the circuit and the placement of the ammeter, which may affect the interpretation of the breakdown voltage.
  • There is a suggestion that the current measurement setup might lead to confusion about the polarity and the actual behavior of the diode.
  • One participant confirms that adjusting the measurement setup yields a more typical I-V characteristic for the diode.

Areas of Agreement / Disagreement

Participants express differing views on the reverse breakdown voltage, with some suggesting it is near 0V while others argue for a value closer to -100V. The discussion remains unresolved regarding the exact breakdown voltage and the implications of circuit configuration.

Contextual Notes

There are limitations in the assumptions made about the circuit configuration and the definitions of current direction, which may affect the interpretation of the breakdown voltage. The discussion does not resolve these uncertainties.

user101
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http://img151.imageshack.us/img151/1953/51953732lu5.png

In this case, is the reverse breakdown voltage at V = 0V since there is negative voltage?
 
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seems rather bizarre 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).
 
http://img99.imageshack.us/img99/4204/partacircuitwh4.jpg

That's my circuit config.

Instructions:

"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."
 
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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:

http://img150.imageshack.us/img150/5097/picture2mh8.jpg
 
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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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