Problems about diodes

Discussion in 'Homework Help' started by PsySc0rpi0n, Mar 8, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hi...

    I'm trying to solve some diodes problems and I need some help!

    Teacher says:

    For a given diode with:

    Is = 2nA
    T=293K
    Si ⇒ η=2

    and is asking us to draw the characteristic curve Vd(Id).
    What I need to do is for a few Vd values, calculate Id with Shockley equation, right?
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    Yes, but it might be a good idea to know what the result ought to look like. Do you know what to expect?
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    If I know what you mean, yes I know what to expect... A curve that is almost zero up to around 0.6 or 0.7V of Vd and after this, the curve will rise exponentially...
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    While you are doing this, you might ask yourself how much you have to increase or decrease the voltage by in order to change the current by an order of magnitude (i.e., a factor of ten, also know as a decade change in current).
     
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Another question is about
    I haven't done that math but I have fifteen like 15 values of Vd from 0V upt to 0.85V and plotted the curve in Excel and the curve is just like what I expected.
    But If I understood you, you're asking me what range of Vd values I have to consider to make the current change my a factor of 10, is that it?
     
  6. WBahn

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    Mar 31, 2012
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    Yes. If you have Vd=0.50V you get a particular current. How much do you have to increase that by to get 10x that current. Then ask the same question but starting with a voltage of Vd=0.65V.
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ah, I think what you want me to see... That around 0.67V and further, the current will increase exponentially, right? And before that, the current is very small, around a few \displaystyle{\mu}As.
     
  8. WBahn

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    Nope. That's not what I'm trying to get you to see.

    Take a look at your plot and try to answer that question. For each data point that you have, look at your graph and see how much you would need to increase the voltage by to increase the current by a factor of ten.

    Did you take your data at evenly spaced voltages? If so, then what do you see if you plot, for each voltage data point, the ratio of the current at the next higher voltage data point to the current at the present voltage data point?
     
  9. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I have plotter these values in excel and got what is in attached picture.
     
  10. WBahn

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    So look at your data and your plot and consider the following:

    Q1) What is the current at Vd=0.7V?
    Q2) What is 10x that current?
    Q3) What value of Vd would produce that voltage?
    Q4) How much of an increase in Vd does that represent?

    Now answer the same questions but start at a different voltage:

    Q5) What is the current at Vd=0.63V?
    Q6) What is 10x that current?
    Q7) What value of Vd would produce that voltage?
    Q8) How much of an increase in Vd does that represent?

    What do you notice about the answers to Q4 and Q8?
     
  11. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I'm now trying to solve some other exercises about "load line" and anothe
    That \displaystyle{\Delta V = \~11mV} for both situations!
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I need to move forward...

    I'm working now with the "Load Line" of a diode and I don't know how to say the other technical term which translated as is, will be something like "rest function point" but probably you can't even figure out what I'm trying to talk about. Anyway, what I'm look for is the point in the diode characteristic curve that matches the load line when Vd = 0V and Id = 0A...

    You know what I mean?


    Edited;

    Well, I guess I found it... Quiescent Point, Q-Point or DC operating point... Right?
     
    Last edited: Mar 9, 2015
  13. WBahn

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    Mar 31, 2012
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    You got the right conclusion, but I think you meant to say that it is ~110mA.

    The point is that, regardless of the voltage or the current in the diode, increasing the voltage by a fixed amount will result in a fixed percentage increase in the current.

    See if you can derive that result mathematically.
     
  14. WBahn

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    Mar 31, 2012
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    Right -- that's the name you are looking for.

    But in general the load line will intersect the diode characteristic at some point other than Vd=0V and Id=0A. The whole idea of load line analysis is that you have two circuits that are connected together at two points, known as the "port". If we look at one circuit in isolation we can plot a set of voltage and current values for the port that are consistent with that circuit and we can plot those on a graph. We can do the same for the other circuit and plot them on the same graph. When the two circuits are connected, they must each operate at the same value for voltage and current at the port, and that point is where the two curves intersect.
     
  15. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Well, but we work with Load Line and Quiescent Point, with only one circuit...

    And that's what I'm working at now!

    We have an exercise that asks to find the Q-Point for a circuit like the one attached.

    Then he says that the diode has the following data:
    \eta = 2; I_{s} = 2nA; T = 293K

    So, what I did was to find the net equation for the circuit and calculate Id for Vd = 0 and calculate Vd for Id = 0. This allows me to draw the Load Line.
    ThenI've used Shockley's equation and the circuit's net equation to find Vd and Id for the given conditions which resulted in Q-Point (0.74V; 4.63mA). I don't know if it's correct!
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    Need to attach the circuit.
     
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