Why Does the Bottom Diode Short First in a Multi-Diode Circuit?

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Discussion Overview

The discussion revolves around the behavior of a multi-diode circuit, specifically why the bottom diode shorts first and the implications of this on the other diodes in the circuit. Participants explore the theoretical underpinnings of diode operation in relation to voltage levels and circuit behavior.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that the bottom diode shorts first, leading to the top two diodes becoming open circuited, proposing that this behavior is due to the inability to apply three different voltages to the same node.
  • Another participant recalls a previous discussion on a similar problem, indicating that when multiple diodes are tied to the same node, they can be treated as open circuits until one is shorted, which then reverse biases the others.
  • A different viewpoint describes the voltage at a node rising from zero, with the bottom diode conducting at 1 volt and preventing further voltage increase, while the other diodes remain reverse biased and unable to conduct.
  • This participant also raises assumptions about the power source's ability to conduct current in reverse and its lack of internal resistance, which may affect the circuit's behavior.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the diodes and the conditions under which they operate, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

Assumptions about the power source's characteristics, such as its ability to conduct current in reverse and its internal resistance, are not fully resolved. The discussion also highlights the complexity of analyzing circuits with multiple diodes and the conditions that affect their operation.

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I circled the given answer. The problem asks for I and V.

I guess the bottom diode shorts first and then that makes the top two diodes be open circuited. Is there any good explanation on why the circuit behaves like that? I guess it comes from the fact that three different voltages relative to ground cannot be applied to the same node? Therefore, the diode with the lowest voltage on its anode "wins" and the other diodes shut off due to the negative [tex]V_D[/tex]?
 
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vk6kro said:
We had that problem a couple of weeks ago. Maybe that poster is in your class?

https://www.physicsforums.com/showthread.php?t=376290

See the second post for a working link.

So when I have multiple diodes tied to the same node that would all be in forward bias if they were alone, I say they're all open circuits. Next, I choose the diode that when short circuited makes the others reverse bias?
 
If that helps... not sure it is right though.


I picture the voltage at V rising from zero when the power is applied. When it gets to 1 volt, the bottom diode starts to conduct and stops the voltage V getting any higher.

If the voltage tried to rise higher, there would be a voltage across a perfect diode which has no resistance. So, there would have to be infinite current flowing.
But infinite current isn't available because of the 1 K resistor.
So the voltage can't rise.

The other diodes are reverse biased so they can't conduct, so the voltage at V stays at 1 volt.

This assumes that the 1 volt power source can conduct current in reverse, not just deliver it, and also that it has no internal resistance.
 

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