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Determine VPS, quiescent value, and resistor value

  1. Sep 20, 2013 #1
    1. The problem statement, all variables and given/known data
    1) What is the VPS voltage source value in Volts

    a) 1 b) 2 c)2.5 d)4 e) 5

    2) Determine the Resistor value

    a)1.75k b) 2 k e) 2.5k d)2.285 k e) 5k

    3) What is the ID quiescent value approximately in mA’s?

    a) 2 mA b) 1.7 mA c) 1.5 mA d) 1 mA e) 0.6 mA



    2. Relevant equations
    V=IR
    Id=(Vps/R)-(Vd/R)


    3. The attempt at a solution
    I get Id=1.7 mA for part 3. I just found that by using the graph it looked liked 1.7 so thats what I assume is the answer. I think R can be found using the graph and ohms law Vd=5 V and I= 2 mA which gives R=2.5kΩ. What im wondering is why do I use the farthest values here? Why dont I use the Vd and Id values are where the lines cross? Vps=RId+Vd=2.5(1.7)+.75=5v for part 1 this can be found using KVL because I had already found R, Id, and Vd.

    I uploaded the entire quiz and a picture so that you could see the graph
     

    Attached Files:

    Last edited: Sep 20, 2013
  2. jcsd
  3. Sep 20, 2013 #2

    gneill

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    Staff: Mentor

    The black line is called a Load Line. It depends only upon the source voltage and its series resistance. It is essentially independent of the load attached. In this case the "load" is the diode, which has its own characteristic curve (blue). Where the two intersect is the operating point of the circuit. It's a graphic way to "solve" for the operating conditions of a circuit containing nonlinear elements (like a diode).
     
  4. Sep 20, 2013 #3

    CWatters

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    You don't have to use the furthest values.

    Ohms law V=IR looks like the standard equation for a straight line y=mx+c where c=0 and the slope of the line m = R.

    So you could calculate the slope of the line using any two points to get R.

    This has real life applications because you cannot always measure the short circuit current. Imagine you wanted to measure the internal resistance of a 12V car battery. There is no way you could measure the short circuit current to get a data point at V=0. Trying to short circuit a 12V car battery is dangerous, the current could be 100's of amps and the battery terminals might melt and throw hot lead at you or worse. However you could load up the battery with some known safe currents and measure the corresponding voltage. Then plot the graph, calculate the slope, and get the internal resistance that way.
     
  5. Sep 21, 2013 #4
    Ok so then that would make my logic correct by the way thanks to both of you for conceptual info?
     
  6. Sep 21, 2013 #5
    R=V/I if I put in .75/1.75=.44 k ohms hmmmm what did I do wrong
     
  7. Sep 21, 2013 #6
    Hey on question 4 what equations should I use I can't find an equation the equates Reverse bias voltage and reverse bias current
     
  8. Sep 21, 2013 #7
    Ah ha so if the load line depends on the source and series resistance then Vd=Vps right?
     
  9. Sep 21, 2013 #8

    gneill

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    Staff: Mentor

    That would give the resistance of the diode at the operating point: the potential across the diode divided by the diode current.

    To find R you want the slope of the load line at that operating point. But since the load line is a straight line, you can take the rise/run for the line using any convenient pair of points that lie on it. Its end points are very convenient indeed...
     
  10. Sep 21, 2013 #9

    gneill

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    Staff: Mentor

    No, Vd is potential across the diode, which is Vp minus the drop across R. It's a series circuit, so you need to count all the potential drops.
     
  11. Sep 21, 2013 #10
    Vps=2.5Id+Vd ?
     
  12. Sep 21, 2013 #11

    gneill

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    Is that "2.5" meant to be the 2.5kΩ resistance of R? If so, the equation looks fine.
     
  13. Sep 21, 2013 #12
    Well by the graph I know Id=1.7 then Vps=(2.5)(1.7) + Vd
     
  14. Sep 21, 2013 #13
    Yes the R is 2.5
     
  15. Sep 21, 2013 #14

    gneill

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    Sure. But I don't understand why you're agonizing over Vps. Didn't you already decide that Vps is 5V from the load line endpoint? It will not change.
     
  16. Sep 21, 2013 #15
    Oh so that's all I have to do on a problem like this is just say that 5V is Vps
     
  17. Sep 21, 2013 #16

    gneill

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    Yes, since you can read it directly off the given load line.
     
  18. Sep 22, 2013 #17

    CWatters

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    Yes.

    In most real life problems you would know Vps and use that to construct the load line.

    This problem is slightly different in that they are asking you to work out Vps from a load line someone has already constructed.
     
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