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Solving for Zener Diode Voltage

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a problem where I need to calculate vo (i.e. output noise) and VO (i.e. DC output voltage) different resistances in a circuit with a Zener diode. The diode is reversed biased but I don't know the Vz. I have the IV curve but can't seem to wrap my head around how to calculate the load line for the Zener diode. Could someone give me some basic steps on calculating the load line and operating point, so I can calculate Vz?

    2. Relevant equations
    Unsure

    3. The attempt at a solution
    I have tried calculating the Thevenin voltage and resistance but still cannot get the load line correct as all the equations I have seen use the current through the diode to calculate the voltage and it doesn't make sense to me.

    I appreciate any help.
    Thank you.
     
  2. jcsd
  3. Apr 12, 2017 #2

    cnh1995

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    Welcome to PF!

    Please post the circuit diagram and problem statement.
    You need to show your attempt. It is mandatory here while posting in the HH forums.
     
  4. Apr 12, 2017 #3
    zener schematic.gif images_circuits_Zener-diode.gif

    VI=9.5V, RIN=1 kΩ, and vi=50.0mV.

    I tried using KVL at the top node above the diode, using VI + Id*Rin + Vd = 0. I set I to 0 and calculated that Vd was -9.5 V and then set V to 0 and got I was -9.5mA. I drew a load line and it appeared to cross the iv curve at v=-4.75 and i=-4.75mA but I am pretty sure that is wrong and not sure where to go from there.
     
  5. Apr 12, 2017 #4

    gneill

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    Were you given a value for the load resistor RL?
     
  6. Apr 12, 2017 #5
    Yes we have to solve for the Vout for a Load Resistance of 2K ohms and for a Load Resistor of 4K ohms. The want the Vo voltage and vo voltage at both.
     
  7. Apr 12, 2017 #6

    gneill

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    Okay. You've noted that the zener operates with a reverse bias. That's where the "Zener" breakdown occurs. That also means that the area of interest for the zener's characteristic curve is in the third quadrant, where both the component's voltage and current are negative.

    To make using the curve more convenient you can rotate it and use treat the current and voltage as positive values. So for the curve you've shown, where the breakdown occurs with a reverse bias of 5 V, the region of interest for the rotated curve will look something like this:

    upload_2017-4-12_21-30-15.png

    To tackle the DC operating point for the zener your idea of using a Thevenin equivalent is a good one. Start by removing the zener from the circuit and find the Thevenin equivalent for what remains. Start with the RL = 2 kΩ case. Can you place its load-line on the curve plot?
     
  8. Apr 12, 2017 #7
    First off, thank you for the informative reply. I've been driving myself crazy trying to figure this out.

    A couple of questions:
    1. Would the thevenin include the 2K ohms resistor? Or is it only the 1K ohm resistor?
    2. For the load line, would I use the thevenin voltage for zero current, or would I use a simple voltage divider circuit formula to get the voltage drop across the load resistor (2K ohms)?

    Thank you again.
     
  9. Apr 12, 2017 #8

    gneill

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    The Thevenin model would include the Rin and RL. The load line takes the open circuit voltage and short circuit current of the Thevenin equivalent as its end points. So work with the Thevenin equivalent.
     
  10. Apr 13, 2017 #9

    rude man

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    You can also model the zener as an ideal zener in series with a resistor ...
     
  11. Apr 13, 2017 #10
    Would you be so kind to draw me a simple schematic? Would you have the cathode of the ideal zener pointing up or down? What would the voltage and resistance be?

    I understand if you don't have the time but would be helpful.
     
  12. Apr 13, 2017 #11
    Does the Diode factor in to the Thevenin equivalent circuit?
     
  13. Apr 13, 2017 #12

    gneill

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    The ideal zener just provides the inflection points in the characteristic curve (the forward bias turn-on and reverse bias zener turn-on points). Look at the characteristic curve of your zener. Everything you need to know is there.

    Also note that as long as the zener is operating in "zener mode" you can model it as a voltage source in series with a resistor. If it might drop out of zener mode and turn off you can include an ideal diode in the path to accomplish that:

    upload_2017-4-13_11-40-5.png

    If there's a risk of the zener becoming forward biased, include another diode in parallel to handle that eventuality.

    No. That comes into play after you've reduced the source network to a Thevenin model (one voltage source, one resistor). You want to establish a load line for the source.
     
  14. Apr 13, 2017 #13
    So do I treat the diode as a 1 ohm resistor or remove it entirely? If I remove it, would it be an open circuit or short?
     
  15. Apr 13, 2017 #14

    gneill

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    Which diode are you referring to? The zener itself or the one in the equivalent model?
     
  16. Apr 13, 2017 #15
    I am wondering how to treat the diode to determine the thevenin voltage and resistance. So, I guess I would be referring to the zener diode in the original circuit?
     
  17. Apr 13, 2017 #16
    From my understanding of Thevenin/Norton equivalents, and I am still a novice on those, you need to calculate the Voc and Isc by removing the load. I am not sure what the load is in this case, since you said to include the 2K ohm resistor in the Thevenin and according to the schematic and problem that is the RL or load resistor.

    If I am mistaken here, please correct me.
     
  18. Apr 13, 2017 #17

    gneill

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    You remove the zener diode completely from the circuit. You want to determine the load-line of the power supply and resistor network, which is why you want its Thevenin equivalent. THEN you place that load line on the zener's characteristic curve to see where its operating point is.
     
  19. Apr 13, 2017 #18
    So to be clear, when I remove it, do I replace it with an open circuit or just remove the path completely so that it would be a simple circuit with 2 resistors in series.
     
  20. Apr 13, 2017 #19

    gneill

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    You want to find the operating point of the zener. As far as the zener is concerned, everything else, including the load resistor, is part of the power supply network. Remove only the zener and find the Thevenin equivalent of everything else.
     
  21. Apr 13, 2017 #20

    gneill

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    Seems to me that's the same thing. Just erase the zener from the diagram. Pretend it was never there.
     
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