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Assumptions with diodes

  1. Feb 6, 2017 #1
    1. The problem statement, all variables and given/known data
    I am working on this problem, where I need to find I and V
    upload_2017-2-6_18-42-8.png



    2. Relevant equations


    3. 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
     
  2. jcsd
  3. Feb 6, 2017 #2

    gneill

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    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?
     
  4. Feb 6, 2017 #3
    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
     
  5. Feb 6, 2017 #4

    gneill

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    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?
     
  6. Feb 6, 2017 #5
    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
     
  7. Feb 6, 2017 #6

    gneill

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    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?
     
  8. Feb 6, 2017 #7
    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
     
  9. Feb 6, 2017 #8

    gneill

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    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.
     
  10. Feb 6, 2017 #9

    cnh1995

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    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.
     
  11. Feb 6, 2017 #10
    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
     
  12. Feb 6, 2017 #11
    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
     
  13. Feb 6, 2017 #12

    cnh1995

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    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.
     
  14. Feb 6, 2017 #13

    gneill

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    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.
     
  15. Feb 6, 2017 #14
    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
     
  16. Feb 6, 2017 #15

    cnh1995

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    What if the circuit had more than two diodes? That way would be time consuming.

    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.
     
  17. Feb 6, 2017 #16

    gneill

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    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.
     
  18. Feb 6, 2017 #17
    It's fine, i think it is better to understand the whole process
     
  19. Feb 6, 2017 #18
    OK, I see what you mean.
     
  20. Feb 6, 2017 #19
    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 ?
     
  21. Feb 6, 2017 #20
    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 aproach 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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