Confused with simple diode circuits given I-V characteristics

Discussion in 'Homework Help' started by ModelIt, Apr 12, 2014.

  1. ModelIt

    Thread Starter New Member

    Apr 12, 2014
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    Hello, I am a bit confused with diode models and their application to a given circuit. I am currently working on these two problems and here's what I understand is the answer for them. I would appreciate if someone can double check me and correct me if it is wrong.
    problem 1)
    [​IMG]

    Since we are given current, and no DC voltage, I figure this is forward bias, thus V(in)=V(out) so graph would just be 2sint(wt) for Vout vs time.
    Or is it sine graph with peak of 1V, without the negative component of sine wave?


    problem 2)
    [​IMG]

    This problem a bit harder, I think that the diode is going to be reversed biased with I = 25mA, but not sure what happens to V(out)
    any suggestions will be appreciated, thx.
     
    Last edited: Apr 12, 2014
  2. shteii01

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    What is diode forward voltage?
     
  3. ModelIt

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    Apr 12, 2014
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    Would it be 0 for problem 1, and -10(breakdown voltage) for problem 2?
     
  4. shteii01

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    Looks like for Problem 1, forward voltage is 0 volts, like you say. So when Vin is 0 or more volts, you will have current though the resistor, and therefore you will have Vout. When Vin is less than 0 volts, that is where I am not sure, either you will have very small current going in reverse so there will be very small current through resistor and very small Vout, or the diode is simply OFF which means it is open circuit so there is no current through resistor and Vout is zero.
     
  5. ModelIt

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    Assuming it is about problem 1, by Vin<0, do you mean the part where sin wave is at negative region with peak of -2V?
    If so, would the current be constant for that of -I(0) = -1uA, as provided by the IV characteristics of a diode in problem 1?
     
  6. shteii01

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    Yes, that is what I mean. So for -2<=Vin<0: Vout=(-1 uA)(1000k)
     
  7. ModelIt

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    Apr 12, 2014
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    Thxm I tihnk I understand problem 1 fiarly well, so where 2=>Vin=>0 the Vout graph would just be a sine curve equal or above 0, correct?
     
  8. shteii01

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    If I understand correctly, in the interval 2=>Vin=>0, Vout will be the same as Vin.

    Basically you have half wave rectifier. Half the wave passes through the diode, the other half is blocked. The half that is zero and greater than zero passes through. The half that is less than zero generates some constant voltage so on the plot of Vout vs Time it will be just a horizontal line.

    Here is perhaps more detailed explanation: http://www.allaboutcircuits.com/vol_3/chpt_3/4.html
     
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  9. ModelIt

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    Apr 12, 2014
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    Yep, that makes sense, thx for link.
    Any suggestion on how to approach 2nd problem? I think it is reverse biased with breakdown voltage of -10V, so the (Vin-Vbr)/R1 = I = 25mA, but how do I proceed, and what is the significance breakdown voltage slope and given R2.
     
  10. shteii01

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    The slope is the resistance: http://en.wikipedia.org/wiki/File:Nonideal_diode_current-voltage_behavior.png
     
  11. shteii01

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  12. ModelIt

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  13. shteii01

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    The DC is just a level at which AC is moving up and down. So your Vin max is 15+2=17 volts, your Vin minimum is 15-2=13 volts.
     
  14. ModelIt

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    Apr 12, 2014
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    But what about the breakdown voltage and and DC Voltage being greater than it is as explained here http://www.allaboutcircuits.com/videos/54.html around 6:20 minute mark
     
  15. WBahn

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    Mar 31, 2012
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    Imagine putting the diode in a black box.

    Now imagine that I switch that black box with another black box that has the same I-V characteristic as the box containing the diode. Would you be able to tell the difference?

    Now imagine that you look inside that black box and what you see are three black boxes and a switch that senses the voltage and current at the terminals and connects one of the three black boxes to the main box's terminals depending on the voltage/current that it sees. Each of the smaller black boxes is responsible only for having the same I-V characteristic as one of the three line segments in the diode's characteristic over the same range of voltage/current as that line segment -- it doesn't matter what it looks like outside that range because the switch will disconnect it anyway.

    For the vertical line segment, you need a circuit that allows any current to flow while producing no voltage across it. What circuit element has that property?

    For the horizontal line segment, you need a circuit that allows any voltage to appear across it while allowing no current to flow. What circuit element has that property?

    For the sloped line segment, you need a circuit that produces -10V across it when no current is flowing and that sees a change in voltage of 250mV for every 1mA that the current changes by. What circuit has that property?

    Once you've identified the equivalent circuits for all three regions all you need to do is then identify when you original circuit is in each region and replace the diode with the appropriate circuit before analyzing it.
     
  16. ModelIt

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    Apr 12, 2014
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    Current flow with no voltage would be an short circuit(just wire).
    No current with voltage potential is open circuit.
    And the sloped line is resistive circuit, provided with R2=250 ohms.

    Now to identification of when circuit in which region, since V(in) has a constant 15 volt of DC, by the analysis, it cant be short circuit. So it can be either open circuit or resistive(breakdown region), but I am not sure when it would be which
     
  17. WBahn

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    But if you use a circuit that consists of just a resistor, then you will have a block box that has 0V across it when there is no current flowing through it. Your black box needs to have -10V across it when there is no current flowing.

    Apples and oranges. You are talking about one component of the voltage across Vin. What matters is the total voltage across the diode. Two different voltages.

    You know that Vin will vary between +13V and +17V. You know that there will be some current flowing in the resistor. Call this current (flowing from left to right) Is.

    Now sketch a plot showing Is as a function of Vout. Since it will depend on Vin, draw several different lines, say at 0.5V intervals, giving you a total of 9 lines. After you plot the first couple you should see the pattern and be able to plot the rest very easily.

    Now, plot the diode characteristic on this same set of axes. Keep in mind the polarity differences between Vout and Vd and between Is and Id.

    Since both lines (the line for a given Vin and the line for the diode) are for the same voltage and same current, the actual voltage and current has to be where the lines intersect.

    This is known as a load-line analysis.
     
  18. ModelIt

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    Apr 12, 2014
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    Is I(s) = V(in)/1k or I(s) = V(in)/R(2)
    If I(s)= V(in)/1k, I(s) vs V(out) would just be pts (13V,13mA) to (17V,17mA). So in essence my Vout would be equal to V(in), but I think this true if I am given ideal model and not the specific one that is provided. If use the actual model provided for this problem, as you mention the polarity my I(s) vs V(out) will be in breakdown region so would it be just straight line with slope of 1 for V(out)?
     
  19. WBahn

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    Neither. Vin is NOT the voltage across R1 or R2.

    Forget about the diode for this first part. You are trying to establish the relationship between Vout and Is for various values of Vs (i.e., Vin). Use the following diagram.

    [​IMG]

    Q1) For a given value of Vs (say 13V), what the voltage Vout will be if Is is zero? Plot that point.

    Q2) For that same value of Vs, what will the current Is be if Vout is zero? Plot that point.

    Now pick a few other values of Vout and/or Is that are between these limits and plot them. Do you notice anything about this collection of points?

    Now do this for Vs=17V.

    Now do it for Vs = {13.5V, 14V, 14.5V, 15V, 15.5V, 16V, 16.5V}. If you have seen the pattern that is at work, you should be able to do these in just a couple minutes.

    Just do that much and post your graph. Once we get this part right, we'll worry about the diode.
     
  20. ModelIt

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    Apr 12, 2014
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    Ok, so if Vs=13V and Is=0mA, I know Vout = Vs-IsR1=Vs
    Now, for 2nd part if Vout is zero, Is=Vs/R1

    And if I use condition for Q1 as Is=1ma, i would get a point of 12V
    so my graph for Vout vs Is would just be essentially y=x, as for w/e conditions i will get pts that are (13V,13mA),(13.5v,13.5mA) etc.
     
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