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Circuit analysis with diodes and multiple current/voltage sources

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data

    This isn't a homework question, just one I can't get the correct answer to. The circuit diagram is attached, and I need to find the voltage at V.

    Circuit.jpg

    3. The attempt at a solution

    Assume the voltage drop across each diode is 0.6V. Let V_5 be the voltage drop across the current source. Using the voltage polarities and current directions I assumed in the diagram, I get the four equations below.

    i_1 = 0.005 A. (1)

    By KCL, i_1 = i_2 + i_3. (2)

    Applying KVL to the left inner loop and then Ohm's law gives: -2000i_2 + V_5 = 0.6V. (3)

    Applying KVL to the outside loop and then Ohm's law gives: -1000i_3 + V_5 = -9.4V. (4)

    When I solve these 4 equations I get:
    i_1 = 0.005 A
    i_2 = -0.00167 A
    i_3 = 0.00667 A
    V_5 = -2.73 V

    The problem is that i_2 is negative, which means that current is flowing backwards through diode D_1.

    What have I done wrong?

    Many thanks in advance.

    Note: I know that the voltage at V (relative to ground) is the voltage across resistor R_1, which is i_3*1000 (V). But I suspect that my value for i_3 is wrong because my value for i_2 is wrong.
     
    Last edited: Jul 23, 2013
  2. jcsd
  3. Jul 23, 2013 #2

    berkeman

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    Staff: Mentor

    I'm not quite understanding your KVL equations. What's V_5? Sorry if I'm missing the obvious.

    I would probably just stick with the KCL method for the whole problem. Write the KCL equations for the two nodes on either side of the current source, and try solving it that way. And in the end, check to see if the answer is reasonable (like, is D1 conducting or not).
     
  4. Jul 23, 2013 #3
    Thanks for your reply. At the top of my solution, I've written that V_5 is the voltage drop across the current source and that I've assumed a 0.6V drop across the diodes.

    Here is extra working for my KVL equations:

    For left loop: -(voltage drop across R_2) - (voltage drop across D_1) + (voltage drop across current source) = 0

    *By Ohm's law, the voltage drop across R_2 is i_2*R_2, which is 2000i_2.
    *The voltage drop across the diode I assumed to be 0.6V.

    Then the equation becomes -2000i_2 - 0.6 + V_5 = 0, or -2000i_2 + V_5 = 0.6.

    For outside loop: -(voltage drop across R_1) + 10V - (voltage drop across D_2) + V_5 = 0.

    *By Ohm's law, the voltage drop across R_1 is i_3*R_1, which is 1000i_3.
    *The voltage drop across the diode I assumed to be 0.6V.

    Then the equation becomes -1000i_3 + 10 - 0.6 + V_5 = 0, or -1000i_3 + V_5 = -9.4.
     
    Last edited: Jul 23, 2013
  5. Jul 23, 2013 #4

    gneill

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    Simplifying hint: If you (temporarily) removed the center branch from the circuit, can you calculate the potential difference that would appear between its former connection points (it's a simple loop that remains)? Would that potential difference forward bias the removed branch's diode or reverse bias its diode?
     
  6. Jul 24, 2013 #5
    Thanks for your reply.

    If I remove the centre branch, then the current throughout the circuit is just i = 5mA. The voltages (relative to ground) at the former connection points are:

    * V_A = voltage across resistor R_1 = i*R_1 = 5mA*1k ohm = 5V.
    * V_B = 10V - voltage across diode D_2 = 10V - 0.6V = 9.4V.

    V_A - V_B is negative, which means current would flow from B to A and therefore reverse bias the removed branch's diode.

    Does this mean my initial calculations are in fact correct? I'm still confused.
     
  7. Jul 24, 2013 #6

    gneill

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    Well, since you've just shown that the central branch conducts no current, any equations you write that assume current there won't apply... that branch is "dead" and can be removed.

    There is only the one outer loop that is conducting current, and you've just solved it for the potential you're looking for :wink:
     
  8. Jul 24, 2013 #7
    Thanks, I see what you mean. I found this example that shows how to work out whether a diode is conducting, but I'm not sure about their reasoning. Are they saying that "if you assume a diode is off and you then get a positive voltage across it, then the diode is actually on", and "if you assume a diode is on and then get a negative voltage across it, then the diode is actually off"?

    I'm confused about why they mention the direction of the current iD1 in part c, but don't mention the direction of the current iD2 in part b.

    Please help me understand how to determine the state of a diode.

    Diode state.PNG
     
  9. Jul 24, 2013 #8

    gneill

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    That's basically it. You make assumptions and then test them. If you guessed correctly then all is well and you can proceed. If you guessed wrong, correct your assumptions and again proceed.
    They are just mentioning some pertinent facts about each circuit which confirm or invalidate the assumptions made. In (b) it is assumed that D2 is conducting, which determines the direction of the current (diodes conduct current in one direction only). Looking at the circuit you can see that the 3V voltage source does indeed want to push current in that direction, so that particular assumption is confirmed. However, their text draws your attention to the more important fact that the potential across the points where D1 was "removed" would in fact cause D1 to conduct, which contradicts the assumption that D1 is off. A contradiction forces you to reject the model.
    You look at the circuit and make an educated guess about which diodes are conducting and which are not. Test the assumptions by checking the resulting current directions through the conducting diodes and the potentials across the "open" diode connections. If all assumptions are confirmed then you're good to go. If any assumptions are disproved, alter your assumptions accordingly and re-test.
     
  10. Jul 24, 2013 #9
    Thanks so much. One last question sorry: If I get a contradiction to one diode but not the other, which is what happened in (b), then this doesn't necessarily mean that I have one diode state correct does it? It just means that my combination of on/off diode states was wrong and to try another combination?

    So in (b), although D2 being 'on' is satisfied, D1 being 'off' is not satisfied and hence we conclude that our combination of states for D1 and D2 is wrong (but we can't conclude that D2 has the correct state. And in fact, in part (c), we find that D2 is actually OFF).
     
  11. Jul 24, 2013 #10

    gneill

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    Yes, that's the idea.
     
  12. Jul 24, 2013 #11
    Great, thanks.
     
  13. Jul 24, 2013 #12
    Do you know how they got the diode voltages, 7V and -3V, in my previous attachment?

    Thanks again.
     
  14. Jul 24, 2013 #13

    gneill

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    Sure. Determine the individual potentials at the open terminals from some common reference point. The potential across the gap is the difference in those potentials.
     
  15. Jul 24, 2013 #14
    I'm confused about how to do that because no current flows through the 10V battery and the 4k resistor when diode D1 is off.
     
  16. Jul 24, 2013 #15

    gneill

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    If no current flows through a resistor, what's the potential difference across the resistor?
     
  17. Jul 24, 2013 #16
    Zero I suppose. I guess that means the voltage (relative to the -ve battery terminals) at the positive terminal of the off diode is 10V. The other voltage is 3V. So their difference is 7V.

    Are the negative terminals of each battery both at ground? If the 3V battery isn't grounded, then the second voltage isn't 3V.
     
  18. Jul 24, 2013 #17

    gneill

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    Yup.
    You determine the potentials with respect to a common node. The bottom wire connects both battery's negative terminals, so it makes a convenient reference node which you can designate as "ground".
     
  19. Jul 24, 2013 #18
    Thanks for all your help!
     
  20. Jul 24, 2013 #19

    gneill

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    You're very welcome.
     
  21. Jul 24, 2013 #20
    Cheers.
     
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