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Load- Line Analysis of Diode

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Determining the best resistance for my simple circuit consisting of a diode, battery, and resistor. Doing this using Load- Line Analysis

    2. Relevant equations
    ID = VD - V / R - Given by KVL


    image014.png (I can disregard n-pretend )
    3. The attempt at a solution

    IMG_0817.jpg

    Here I have attempted using a 1KΩ resistor, obviously too small because I have a voltage drop of about 8.2V across the resistor so then there should be a drop of approx. 1.8 V across the diode. However using the diode equation this gives me way too large of a current.
     
  2. jcsd
  3. Sep 21, 2016 #2

    Charles Link

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    I don't know that your sketch of ## I=I_0 (exp^{eV/(n k_b T)}-1) ## is real accurate. Otherwise your load line looks close to correct. The ## I_0 =1 E-12 ## is only an estimate. Did you compute the exponential with ## e=1.602 E-19 ##, ## k_b=1.381 E-23 ##, and ## T=300 ## with ## V=.7 ##? And n=2? The diode equation is only an estimate, but I think experimentally you are likely to get close to I=9 mA with ## R=1 k \Omega ##. I think the ## V_{on}=.7 \, Volts ## for a silicon diode is a pretty good number.
     
  4. Sep 21, 2016 #3
    Dear Charles, yes I did compute it with all of those values, however the computation I perform with a 1KΩ gives me a current that is too high for my diode, thus I know that I need a resistor with higher resistance. I will attempt to use the online DESMO software and see what that will give me. Should be a tad bit more accurate with the graph. From my work though is it accurate the steps I took to determine the load line? I am simply asking because this is what I THINK is correct, but this is also my first type encountering such a load - line problem.

    Thanks!
     
  5. Sep 21, 2016 #4

    Charles Link

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    You did it correctly. The load line is very useful also in cases where the DC voltage varies for constant resistance. In that case the x-intercept (Voltage of the voltage source) will vary with the entire load line moving in the horizontal direction. Every so often I have encountered problems that use the load line method to solve them. ## \\ ## For your diode current, I do think if you use ## V_{Diode}=.7 ## and compute current ## I=(V_0-.7)/R ## it will give you a fairly accurate answer without the load line method. I googled the Schockley diode equation, and they did not give a precise value for ## I_0 ## or n. I think you can get considerable variation in these parameters from one diode to another. ## \\ ## Just an additional item about the load line method: You begin with a graph of your working component, in this case ## I=I(V_{Diode}) ## i.e. of ## I ## vs. ## V_{Diode} ##. In a series circuit(It only works for the series circuit) you have ## V_0=I R +V_{diode} ##. This equation can be written (solving for ## I=I_{Diode} ) ## as ## I=-(1/R) V_{Diode}+(V_0/R) ## which is a straight line (with ## I=y ## and ## V_{Diode}=x ##) that has slope ## m=-1/R ## and the x-intercept is ## V_0 ##. (Usually you use ## \ ## ## y=mx+b ##, and ## b ## is the y-intercept), but here the x-intercept is important. The intersection of the load line with the graph of ## I=I(V_{Diode}) ## is the operating point.
     
    Last edited: Sep 21, 2016
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