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

[xcvxcvvc]

http://img38.imageshack.us/img38/6127/88280233.jpg
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]?

 

[vk6kro]

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.

 

[xcvxcvvc]

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?

 

[vk6kro]

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.

 

[DeeNos]

I would like to provide a response to the ideal model of diode problem presented in the given image.

The circuit shown in the image consists of three diodes with different voltage levels applied to them. The problem asks for the current and voltage values in the circuit. Based on the given information, it can be observed that the bottom diode has the lowest voltage applied to its anode, while the top two diodes have higher voltages applied to their anodes.

The behavior of diodes in a circuit can be explained by their characteristic curve, which is a plot of the diode current versus the applied voltage. In an ideal diode, the current is zero when the voltage is below a certain threshold, known as the forward voltage drop. Once the voltage exceeds this threshold, the current increases rapidly.

In this circuit, the bottom diode has the lowest voltage applied to its anode, which is below the forward voltage drop. This means that the diode is in the off state and no current flows through it. As a result, the voltage drop across this diode is zero.

On the other hand, the top two diodes have higher voltages applied to their anodes, which are above the forward voltage drop. This means that these diodes are in the on state and current flows through them. The voltage drop across these diodes will be equal to the forward voltage drop.

The reason why the bottom diode shorts first and the top two diodes become open circuited is due to the concept of voltage division. When multiple voltages are applied to the same node, the voltage will be divided among the components based on their impedance. In this case, the bottom diode has the lowest impedance, causing it to have the lowest voltage drop and therefore, shorting the circuit. This also explains why the other two diodes are open circuited, as the voltage drop across them is now zero.

In summary, the behavior of diodes in a circuit can be understood by their characteristic curve and the concept of voltage division. The ideal model of diodes provides a simplified understanding of their behavior, but in practical circuits, other factors such as temperature and manufacturing variations may affect their performance.