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Homework Help: How to measure maximum current in diode config?

  1. Sep 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the peak current in the circuit config:

    V1 = 15v amplitude, 1khz freq.
    R = 1.5k
    Vd = 0.7v

    Draw the transfer characteristic.

    http://img341.imageshack.us/img341/9761/picture2ge7.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution

    Transfer Characteristic

    I only understand the transfer characteristic if the diode is flipped. When the diode is flipped, The source has to supply more than 0.7V to turn the diode on, then after there will be a constant voltage drop of .7v across the diode as you keep increasing the voltage.

    However, in my case, I'm guessing that the voltage Vs would only turn the diode on when it is decreasing, so the cathode part of the diode is not greater than Vo?

    Peak Current
    KVL: -V1 - Vdiode + Vr = 0
    Vr / R = (V1 + Vdiode) / R = Ir

    so peak current = Ir = (V1 + Vdiode) / R

    Is my derivation correct? Thankss!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 21, 2007 #2


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    Staff: Mentor

    The voltage source shown is a sine wave AC source, so its output at it's "+" terminal is a 15Vpp or 30Vpp (depending on what the problem statement actually means by 15V amplitude) sine wave about ground, when the "-" terminal is grounded. So it doesn't matter which way the diode is pointing, the AC waveform ensures that it will turn on for half of each AC cycle.

    So first you need to clarify what is meant by "V1 = 15V amplitude". Does that mean that v(t) = V1 sin(wt), in which case the total amplitude is 30Vpp, or does it mean that the total amplitude is 15Vpp, in which case v(t) = 7.5V sin(wt)?

    Then, you should draw what the sine wave output at the + terminal would look like if the diode were not there, and then think about what part of the waveform gets chopped off when the diode is connected.

    Does that help?
  4. Sep 21, 2007 #3
    It's peak amplitude is 15V, so the Vpp would be 30V.

    For my waveform of Vout (or Vr, w/e you want to call it), I have the diode cutting out the positive wave cycles from the AC source, leaving only the negative sinusoids. My sinusoids are in quad IV. This is for the graph of Vo. Thanks for the clarification, but I think my question is asking the rectifier transfer characteristic. I think they mean Vout (y axis) vs V1 (x axis).

    If you have the diode in the reversed direction for example, the graph of Vout vs V1 would equal 0 up until the turn on voltage Vf, then slowly increasing with a slope of R / (R+rd) if you're using the battery plus resistance model. I can visualize that without a problem, it's just now with this diode reversed, I'm totally confused with Vout vs V1.
  5. Sep 22, 2007 #4
    Can anyone confirm the peak current derivation I have in the first post:

  6. Sep 22, 2007 #5


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    Staff: Mentor

    I'm not tracking what you are saying (sorry), but try picturing it this way. Start the AC voltage source out at zero, and let it go through the positive half-cycle. You are correct that the diode is an open circuit for this half-cycle, so the Vout is zero (pulled to ground by the resistor). Then, as the AC voltage source swings negative, as it swings down through -0.7V, it snaps on the diode. The output voltage then follows the AC voltage source's waveform, but it stays 0.7V behind, right? It does this through the full negative sine waveform, and then the whole process repeats through the next positive cycle (well, almost -- see the Quiz Question below). So if all of that is true, then what is the small error in your equations in the previous post?

    Quiz Question -- If you draw the first few cycles of the Vout waveform, what is different between the first positive cycle that I described, versus the second and subsequent positive cycles?

    EDIT -- for the Quiz Question, assume that there is some parasitic capacitance in parallel with the resistor, so that the RC time constant is about half of the period of the AC voltage source...
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