Why is it not possible for D2 to be off in this diode problem?

In summary: I don't understand really is how to know that it won't be possible and that the assumption is incorrect. My professor talks about if the voltage on top of the open circuit (if D2 is off) than...If the voltage on top of the open circuit (if D2 is off) than D1(left) will be off and D2(right) will be on. If the voltage on top of the open circuit (if D2 is on) than D1(left) will be on and D2(right) will be off.
  • #1
Cocoleia
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4

Homework Statement


I am working on this problem, where I need to find I and V
upload_2017-2-6_18-42-8.png

Homework Equations

The Attempt at a Solution


I know the right answer will be when D1(left) is off and D2(right) is on, we went over the solution in class. I wanted to do the 4 assumptions just to try to understand. I am not able to figure out what happens when D2 is off, and why it's not possible. Can someone please explain? Thanks
 
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  • #2
Cocoleia said:
I am not able to figure out what happens when D2 is off, and why it's not possible. Can someone please explain?
Make the assumption that it is off and determine the potentials at the nodes of the circuit. What would be the potential at the node where D2 connects?
 
  • #3
gneill said:
Make the assumption that it is off and determine the potentials at the nodes of the circuit. What would be the potential at the node where D2 connects?
Well the top of the open circuit would be at 0 since it would be like the ground. And the bottom of the open circuit would be -5V ? If I have the right logic
 
  • #4
The circuit has a ground and that should be used as the reference node. Write a node equation at the top of D2. By the way, are the diodes to be taken as ideal diodes or will they have a forward bias voltage?
 
  • #5
gneill said:
The circuit has a ground and that should be used as the reference node. Write a node equation at the top of D2. By the way, are the diodes to be taken as ideal diodes or will they have a forward bias voltage?
These ones were ideal.

For the top of D2 you have I1 = I, i guess, since the current in an open circuit is 0 A
 
  • #6
Cocoleia said:
These ones were ideal.

For the top of D2 you have I1 = I, i guess, since the current in an open circuit is 0 A
Sorry, I don't see any I1 on the diagram. Yes, the current in an open circuit is 0 A. What open circuit are you referring to? The output terminals?
 
  • #7
gneill said:
Sorry, I don't see any I1 on the diagram. Yes, the current in an open circuit is 0 A. What open circuit are you referring to? The output terminals?
By I1 I mean the current going through the 10k, sorry. I thought when the diode is off you replace it with an open circuit ? That is what I was referring to
 
  • #8
Cocoleia said:
By I1 I mean the current going through the 10k, sorry. I thought when the diode is off you replace it with an open circuit ? That is what I was referring to
Okay. It's best to define everything so there's no confusion.

With D1 removed there will be only one path for current to follow in the circuit. A straightforward way to determine a particular potential along that path is to write a node equation for that location. You could go the route of first determining the current, then finding the potential drops across the components and doing a "KVL walk" to the location from a known potential, too. It's your choice.
 
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  • #9
Cocoleia said:
By I1 I mean the current going through the 10k, sorry. I thought when the diode is off you replace it with an open circuit ? That is what I was referring to
How do you know it is open circuited?
Cocoleia said:
These ones were ideal.
So you can replace them by short circuits and see how currents flow in them. That should give you an idea which diode is on and which one is off.
 
  • #10
cnh1995 said:
How do you know it is open circuited?

So you can replace them by short circuits and see how currents flow in them. That should give you an idea which diode is on and which one is off.
I said it was open circuited because I am asking the specific case where I assume D2 is off. When it is off we replace with an open circuit, when it is on with a short circuit ? Or have I completely misunderstood
 
  • #11
Cocoleia said:
I said it was open circuited because I am asking the specific case where I assume D2 is off. When it is off we replace with an open circuit, when it is on with a short circuit ? Or have I completely misunderstood
cnh1995 said:
How do you know it is open circuited?

So you can replace them by short circuits and see how currents flow in them. That should give you an idea which diode is on and which one is off.
What I don't understand really is how to know that it won't be possible and that the assumption is incorrect. My professor talks about if the voltage on top of the open circuit (if D2 is off) than the voltage on the bottom... I don't get it
 
  • #12
Cocoleia said:
When it is off we replace with an open circuit, when it is on with a short circuit ?
Yes, you replace it with an open circuit but it would be better if you didn't "assume" anything at first.

Just write KVL and KCL equations for the given circuit with two diodes. What decides whether an ideal diode is on or off is the direction of current through it. Just by looking at the circuit, you can't assume a particular diode to be off unless it is reverse biased.
If D2 is off, D1 is on and if D1 is off, D2 is on. You need to understand why D2 is on even when D1 is present in the circuit and why D1 is off. So you should analyse the circuit with two diodes, without assuming anything.
 
  • #13
Cocoleia said:
What I don't understand really is how to know that it won't be possible and that the assumption is incorrect. My professor talks about if the voltage on top of the open circuit (if D2 is off) than the voltage on the bottom... I don't get it
Diodes can only conduct current if they are forward biased. If you remove a diode from a circuit and there is a potential difference between the nodes where it was connected, and if that potential difference is in the right direction and great enough to forward bias the diode, then the diode will conduct when it is in the circuit.
 
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  • #14
cnh1995 said:
Yes, you replace it with an open circuit but it would be better if you didn't "assume" anything at first.

Just write KVL and KCL equations for the given circuit with two diodes. What decides whether an ideal diode is on or off is the direction of current through it. Just by looking at the circuit, you can't assume a particular diode to be off unless it is reverse biased.
Ok, our professor had said that you can just try all four assumptions (D1on/D2on, D1off/D2off, D1on/D2off, and D1off/D2on) and then it will be clear that only one gives possible answers for your current and voltage and those will be the final answers to the problem
 
  • #15
Cocoleia said:
Ok, our professor had said that you can just try all four assumptions (D1on/D2on, D1off/D2off, D1on/D2off, and D1off/D2on) and then it will be clear that only one gives possible answers for your current and voltage and those will be the final answers to the problem
What if the circuit had more than two diodes? That way would be time consuming.

cnh1995 said:
You need to understand why D2 is on even when D1 is present in the circuit and why D1 is off. So you should analyse the circuit with two diodes, without assuming anything.
When D2 is open, D1 will conduct but that doesn't prove D2 is on and D1 is off. Similarly, when D1 is off, D2 conducts but that doesn't prove anything. You need to take both the diodes into account at the same time.
 
  • #16
I think we're getting off the track of Cocoleia's stated desire to examine one particular case. We should address that before offering to analyze the circuit from scratch.
 
  • #17
gneill said:
I think we're getting off the track of Cocoleia's stated desire to examine one particular case. We should address that before offering to analyze the circuit from scratch.
It's fine, i think it is better to understand the whole process
 
  • #18
cnh1995 said:
When D2 is open, D1 will conduct but that doesn't prove D2 is on and D1 is off. Similarly, when D1 is off, D2 conducts but that doesn't prove anything. You need to take both the diodes into account at the same time.

OK, I see what you mean.
 
  • #19
cnh1995 said:
When D2 is open, D1 will conduct but that doesn't prove D2 is on and D1 is off. Similarly, when D1 is off, D2 conducts but that doesn't prove anything. You need to take both the diodes into account at the same time.
So let's say I had this example:
upload_2017-2-6_20-4-18.png


I would automatically say that they couldn't both be on, since it would not give a logical answer for V. I would also say they couldn't both be off, since the voltages are both bigger than -3V, so there would be some kind of current (I don't know if this is a good assumption)
Now I am stuck between D1 on and D2 off, or D2 on and D1 off.
I would say that D2 needs to be on since 2V is bigger so it will have the current? But I don't know if this is true. That's just my logic. Does any of it make sense? Or it's still a bad strategy to take assumptions ?
 
  • #20
Cocoleia said:
I said it was open circuited because I am asking the specific case where I assume D2 is off. When it is off we replace with an open circuit, when it is on with a short circuit ? Or have I completely misunderstood
You can do that. Assume that d2 is open and d1 is on. But if d1 is on (0volt throw it) then d2 should be on that causes d1 to be off, i think that could only be a transient condition.

In other words what you did is a good approach one way will be stable an the other won't. Is like thinking what would hapen if you put a open switch before d2 and you close it.
 
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  • #21
gneill said:
Diodes can only conduct current if they are forward biased. If you remove a diode from a circuit and there is a potential difference between the nodes where it was connected, and if that potential difference is in the right direction and great enough to forward bias the diode, then the diode will conduct when it is in the circuit.
To the OP - I think the above is the best approach. It is simple, direct, and will get you to your answer.

Just think about that, if the ideal diode was off in the circuit, removing it does not change anything. And if you remove it, and there is no forward bias in that circuit, it must be off. If there is forward bias, it is on.

Go from there.
 
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  • #22
Cocoleia said:
I would say that D2 needs to be on since 2V is bigger so it will have the current? But I don't know if this is true. That's just my logic. Does any of it make sense?
Yes. I agree with your logic.
 
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1. What is the purpose of making assumptions with diodes?

Making assumptions with diodes allows us to simplify complex circuits and analyze them more easily. It also helps us understand the behavior of diodes in different circuit configurations.

2. What are some common assumptions made with diodes?

Some common assumptions include assuming ideal diode behavior (i.e. zero voltage drop when forward biased and infinite resistance when reverse biased), neglecting the reverse saturation current, and assuming a constant voltage drop for all diodes in a circuit.

3. What are the limitations of making assumptions with diodes?

Making assumptions with diodes can lead to inaccurate results, especially in circuits with high frequencies or high currents. It also does not account for variations in diode characteristics and temperature effects.

4. How do you determine if assumptions can be made with diodes in a circuit?

Assumptions can be made with diodes in a circuit if the circuit is operating at low frequencies and currents, and if the diodes are not experiencing extreme conditions such as high temperatures or large voltage variations.

5. Can assumptions be made with all types of diodes?

Assumptions can be made with most common types of diodes, such as silicon and germanium diodes. However, it is important to be aware of any unique characteristics of specialized diodes, such as Zener diodes or Schottky diodes, that may require different assumptions to be made.

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