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General Diode Question

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a basic circuit of a voltage source and two diodes in series:
    D1 is .7v @ 10mA
    D2 is .6v @ .1mA
    If I is 1mA, What is the voltage across the two diodes?

    2. Relevant equations
    VT~.026
    Vt= 60log (I(t)/Is
    I1 produces V1
    I2 produces V2
    so V2 = V1 + 60(mV)[log (I2/I1)]


    3. The attempt at a solution
    Just plain lost on how to start. I get that this needs to be turned into a ratio between the two votage/current values across the diodes, but where do I start?
     
  2. jcsd
  3. Jun 13, 2009 #2

    CEL

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    The inverse saturation current Is is different in each diode. Determine Is from the data you have at the currents 10mA and .1 mA and use it to find the voltages at 1mA.
     
  4. Jun 14, 2009 #3
    So solving for V2=V1 + 60mV[log(I2/I1)] I get D1=.64V and D2=.66v than using Kirchoff I can assume that the solution would be 1.3V? I know the whole thing is theoretical anyway because of the lack of resistance but is this the solution?
     
  5. Jun 14, 2009 #4

    CEL

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    I would use the expression [tex]I=I_se^{\frac{V}{V_T}}[/tex], using I =10mA and V=.7V, to obtain Is for D1. Then use I = .1mA and V = .6V, to obtain Is for D2.
    In reality, there should be a resistance limiting the current, but it is irrelevant, since we don't know the voltage of the source.
     
  6. Jun 14, 2009 #5

    Redbelly98

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    I don't think we can obtain Is1 or Is2 without knowing VT. But I agree, the diode equation is probably the key here.

    Note, they seem to be asking for the sum V1+V2. If so, it's not necessary to find V1 and V2 separately, we just need the sum.
     
  7. Jun 15, 2009 #6

    CEL

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    The OP already gave the value of VT: 26mV at 300K.
    I don't see how we can find the sum without knowing the indi8vidual values of V1 and V2. Remember that the two diodes have different characteristics.
     
  8. Jun 15, 2009 #7

    Redbelly98

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    Aha, I missed that. Thanks.

    Guess it's a moot point, because knowing VT we can find IS1 and IS2, as you said earlier, and then both V1 and V2 at 1 mA.

    So the key is using the 10 mA and 0.1 mA info to find IS1 and IS2, using the diode equation in post #4.

    p.s.
    Alternative, less straightforward method: one could also write out an expression for V1+V2, and find that it depends on the product i1·i2.
     
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