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Piece-wise linear diode model

  1. May 2, 2009 #1
    I am NOT trying to solve this as a homework problem, I'm just trying to dig out basic concepts that I'm missing when using this battery+resistance diode model.


    I'm having trouble seeing why, when solving for current, KCL equation does NOT work out as:

    1) I= (0-Va)/rd
    2) I=(Va-(-5))/R

    where Va is the voltage at node a, rd is the diode resistance, and Vd0 is the diode voltage drop.

    I understand that the resistors are in series, and there is only one current, so the correct Kirchoff's equation should be:

    I=(0-Vd0)/(rd+R)

    But each resistor is responsible for a (different) voltage drop. The position of the resistors, even if they are in series, is specific. We can see this by measuring voltages before and after a resistor. Hopefully somebody can provide a fullproof explanation for this.
     

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  3. May 2, 2009 #2
    Here's an example directly from the book which very clearly shows the question that I have:

    They give us the equation for the current, since there is only one current, and the resistors are in series.

    My question is, why is the equation for I NOT:

    1) I=(Vdd-Vd0)/R
    2) I=(Vd0-0)/rd

    Each resistor is responsible for a voltage drop, meaning that it is important where each resistor is located. Where am I going wrong?
     

    Attached Files:

  4. May 2, 2009 #3

    Redbelly98

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    VDD - VD0 gives the sum of the voltage drops across R and rD, hence the given equation for I.

    No, because (Vdd-Vd0) is not the voltage across R.

    No, because Vd0 is not the voltage drop across rD
     
  5. May 2, 2009 #4
    Why not? I'm just following from one voltage source to another, isn't that how KCL works?
     
  6. May 3, 2009 #5

    Redbelly98

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    It's KVL that we want to use here ... KCL just tells us that the current is the same through each element in the loop.

    According to KVL:

    VDD - ID*R - VD0 - ID*rD = 0​

    You can rearrange that to get

    VDD - VD0 = ID*R + ID*rD

    Note the right-hand-side has the voltages across both resistors.
     
  7. May 3, 2009 #6
    Thanks for that, made me realize something I didn't recall:

    Consider the entire voltage across the diode, first; then, KVL becomes:

    I=(-vd-(-5))/R

    from node a to Vss. Now, realizing that for this diode model, the voltage across the diode, (variable) vd:

    vd=Vd0+I*rd

    where Vd0 is the intrinsic threshold voltage (constant).

    Substituting in the original equation for I and playing around with some algebra leads us to:

    I=(-Vd0+5)/(R+rd)

    which I'm hoping is the right answer. Thanks for the refresher.
     
  8. May 3, 2009 #7

    Redbelly98

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    Looks good, assuming -V = -5V in your first circuit.
     
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