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Basic Diode Analysis

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data

    First question,
    In the picture attached (One diode and Resistor), when you are calculating V, is the V drop across the diode or the resistor?

    Second Question,
    In the second picture (circuit with current source) Why is the nodal analysis equation,
    1A=V/100ohms + (V-0.7)/100ohms

    Forward voltage drop = 0.7, V is voltage at the node in the middle.

    The part I don't get is the (V-0.7)/100ohms, why are we taking the voltage across the diode then dividing by the resistor? Shouldn't it be voltage across both the diode and the resistor?

    This forward voltage drop stuff is really confusing me..
     

    Attached Files:

  2. jcsd
  3. Apr 5, 2014 #2

    gneill

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    Staff: Mentor

    Neither, really. It's the potential at the terminal with respect to the circuit's reference node. In this case the reference node is implied and not shown explicitly. Note that there are two voltage sources, +5 V and - 5 V, which are indicated. These have an implied common reference point. If you want you can sketch in the voltage sources and the common reference to make it clear:

    attachment.php?attachmentid=68355&stc=1&d=1396742587.gif

    For nodal analysis the idea is to treat a series branch as a whole and find the net potential difference across the resistive part and divide by that resistance to determine the current. The diode drop, if the diode is forward biased, can be treated as a voltages source of 0.7 V. That makes the voltage drop across the resistor V - 0.7.
     

    Attached Files:

  4. Apr 6, 2014 #3
    So the resistor does not have to be in the middle of potential difference you are trying to measure? and the equation really is V-0.7-0/100ohms ?
     
  5. Apr 6, 2014 #4

    gneill

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    Staff: Mentor

    You can rearrange the order of the components in a series circuit (a branch) any way at all and won't change the result. After all, the potential changes are summed, and it doesn't matter in what order a sum is performed.

    Yes, that really is the equation :smile:
     
  6. Apr 6, 2014 #5
    I never had this kind of problem before so this was confusing. So just to make sure, lets say for this circuit,

    10s946b.jpg

    The equation for current is I = (10-0.7-0)/(2.7Kohms), assuming the diode is 0.7volts forward biased.
     
  7. Apr 6, 2014 #6

    gneill

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    Staff: Mentor

    Yup!
     
  8. Apr 6, 2014 #7
    thnx that clears it up :)
     
  9. Apr 6, 2014 #8
    Another question about the same circuit, what if the diode was reverse biased? I know the current will be zero since there won't be any flow of current, but what about voltage across the diode? Will it be 0.7 volts or zero too?
     
  10. Apr 6, 2014 #9

    gneill

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    Staff: Mentor

    A reverse biased diode is, essentially, equivalent to an open circuit (or at least a very, very, high resistance). As such, essentially all the available potential drop will occur across it. So for example, if in our diagram the diode's polarity was reversed, the full 10V would present across it.
     
  11. Apr 6, 2014 #10

    So in terms of equations,
    10-IR-Vd = 0
    Vd = 10volts

    or
    since the polarity switched
    10-IR+Vd = 0
    Vd= -10
     
  12. Apr 6, 2014 #11

    gneill

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    Staff: Mentor

    Sure, if you choose Vd to be defined as the cathode to anode potential for the diode.
     
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