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Analyzing diode circuit with Constant Voltage Drop model

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data
    BMnUBqS.png

    2. Relevant equations
    Whenever assuming a diode is "on", replace that diode with a 0.7 voltage source.
    Whenever assuming a diode is "off", replace that diode with an open circuit.

    3. The attempt at a solution
    The problem is pretty straightforward but the format of the circuit is confusing to me so I'm not sure if I'm doing it correctly. Could someone please check my answers? Just concerned with part (a) for now:

    (a)
    3UGNRpN.jpg
     
  2. jcsd
  3. Mar 6, 2015 #2

    gneill

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    You might want to recheck your value for V. Note that the top of the first diode is at +0.7V, while the V terminal is just after the second diode which provides a 0.7V potential drop...
     
  4. Mar 6, 2015 #3
    Thanks, that's exactly what I wasn't sure about. So if the voltage at the node connecting the diodes is 0.7V, I would need to "step down" 0.7V to get to the V terminal right? So the voltage at V is 0 V?

    Also, aside from the wrong V value, is the method I used to solve for the currents correct?
     
  5. Mar 6, 2015 #4
    At the two cases v is equal to zero
     
  6. Mar 6, 2015 #5

    rude man

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    v1 = 0
    v2 is not 0.
     
  7. Mar 6, 2015 #6
    Yeah it looks like v2 in (b) can't be 0 with both the diodes on because then the current I would be negative, which isn't consistent with the Constant Voltage Drop Model.

    I'm having trouble figuring out what diode configuration to use though.

    With diode 1 (left) off and diode 2 (right) on, the voltage drop across diode 1 is still 0.7 V because of the shared node with diode 2, but the current would be 0 A which isn't consistent with the model.

    With diode 1 on and diode 2 off, the V terminal is at -5 V since there's no voltage drop across the 5k resistor due to there being no current, which means the voltage drop across diode 2 is 5.7 V since it's 0.7 V at the shared node with diode 1. Again, this isn't consistent with the model since no current flows through diode 2.

    With both diodes off, the voltage drop across diode 1 is 5 V and the drop across diode 2 is 10 V. This isn't consistent since no current is flowing.

    Where am I going wrong in my thought process? One of the configurations has to work.
     
  8. Mar 6, 2015 #7

    rude man

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    You mean because two diodes have the same anode voltage ("shared node"), their cathode voltages have to be the same too???
    I don't think so ...
     
  9. Mar 6, 2015 #8
    "With diode 1 (left) off and diode 2 (right) on, the voltage drop across diode 1 is still 0.7 V because of the shared node with diode 2, but the current would be 0 A which isn't consistent with the model."

    I mean with this configuration the anode voltage of diode 1 is 0.7 V but the cathode voltage is 0 V because of the ground, so the voltage drop is just 0.7 V?
     
  10. Mar 6, 2015 #9

    rude man

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    The anode voltage of diode 1 is not 0.7V. Diode 1 is off so how can its anode be 0.7V more positive than its cathode?
     
  11. Mar 6, 2015 #10
    I was under the impression that it'd be something like this:
    gpIS4.png
    Even though the diode's off I thought the anode's voltage would still be 0.7 V because of the other diode's anode, and its cathode would be 0 V because of the ground. If this isn't correct what would it be?
     
  12. Mar 6, 2015 #11

    gneill

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    With diode 1 out of the picture because it's assumed to be off (reverse biased), it's anode voltage cannot be presumed to be 0.7V. It will depend upon the potential drop from +5V that occurs across the 10k resistance. We can only presume that it will turn out to be less than 0.7V (and thus verify our assumption about D1 being off).

    The way to proceed is to determine the potential there by further analysis. Write a node equation for that node assuming that D1 is off and D1 on. Solve for the node voltage.
     
  13. Mar 6, 2015 #12
    Doing node voltage analysis at the anode:

    (Vn-5)/10k + (Vn-0.7)/5k = 0

    I get the voltage at the node, Vn = 2.133 V. This isn't less than 0.7 V though, did I do the analysis incorrectly?
     
  14. Mar 6, 2015 #13

    gneill

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    Nope. You've missed out the -5V source in the second term.
     
  15. Mar 6, 2015 #14
    Hm okay if I move the 0.7 V source below that 5k resistor, would I need to "step up" by 0.7 V from the -5 V source?
    So:

    (Vn-5)/10k + (Vn-(-5+0.7))/5k = 0

    (Vn-5)/10k + (Vn+4.3)/5k = 0

    Vn = -1.2 V
     
  16. Mar 6, 2015 #15

    gneill

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    Yup.
     
  17. Mar 7, 2015 #16

    rude man

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    Your picture is correct.
    So now you can ignore diode 1, right?
    So now, what's the voltage across the two resistors? The current? Then, what is the cathode voltage, considering the 5K resistor goes to -5V?
     
  18. Mar 7, 2015 #17
    Think I got it now. The current through the 2nd diode is 0.62 mA, and the cathode voltage of the 2nd diode is -1.2 - 0.7 = -1.9 V. (Solved for the anode voltage above).
     
  19. Mar 7, 2015 #18

    rude man

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    You got it 100%!
    You could also have gone
    i = 0.62 mA
    therefore. drop across the 5K is 0.62mA*5K = 3.1V
    but the bottom of the 5K is at -5V
    so the top must be at 3.1 - 5 = -1.9V.

    .
     
    Last edited: Mar 7, 2015
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