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Reverse Saturation Current in P-N Junction

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data

    An abrupt Si p-n junction of cross sectional area 10-4cm-2 has the following properties:

    p side:

    number densities of acceptor impurity atoms:
    NA = 1017cm-3

    minority carrier lifetime:
    τn = 0.1 us

    hole mobility:
    μp = 200cm2/Vs

    electron mobility:
    μn = 700cm2/Vs

    n side:

    number densities of donor impurity atoms:
    NA = 1015cm-3

    minority carrier lifetime:
    τn = 10 us

    hole mobility:
    μp = 1300cm2/Vs

    electron mobility:
    μn = 450cm2/Vs

    (I've assumed that μp is hole mobility and μn is electron mobility although I'm not 100% on this)

    What is the reverse saturation current Is?

    2. Relevant equations

    Reverse saturation current:

    Is = eA[(Dhpno / Lh) + (Denpo / Le)]

    Dh = kTμh / e
    De = kTμe / e

    Lh = √(τhDh)
    Le = √(τeDh)

    pno = ni2 / NA
    npo = ni2 / ND

    3. The attempt at a solution

    With these equations I should be able to find the reverse saturation current. However, I'm not entirely sure on which values of mobility μ I should be using.

    It seems like I can find the reverse saturation current in either the p side or the n side. I thought I'd work it out for both sides and hope that would be equal but I came a little unstuck when finding Lh and Le as it seems I need two values for τ on each side.

    In a PN junction, the reverse saturation current is due to the diffusive flow of minority electrons from the p-side to the n-side and the minority holes from the n-side to the p-side.

    I feel like I'm lacking some understanding, I was wondering if anyone could help clear this up for me?
  2. jcsd
  3. Apr 5, 2015 #2


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    Science Advisor

    Do you understand that the subscript h (hole) is the same as the subscript p (positive) and the subscript e (electron) is the same as the subscript n (negative). Given that, it seems like you have everything you need. You are given the mobilities, form which you can calculate the diffusion coefficients (the D's) and you are give the lifetimes, from which you can calculate the Diffusion lengths (the L's). Where are you stuck?

    Conceptually, the reverse saturation current is the sum of minority holes on the N-side flowing to the P-side and minority electrons on the P-side flowing to N-side, that's why there are two terms.

    Edit: I see your question now. You are asking which mobility to use, since you have the electron and hole mobility on both sides. The idea is that on the P-side of the junction, there are a large number of holes, and very few electrons. Those electrons which do exist on the P-side can diffuse into the depletion region and get swept to the N-side, contributing to the reverse current. So, on the P-side, it is the electron diffusion coefficient that matters, so you want to use the electron mobility on the P-side to calculate the electron diffusion coefficient. Similarly, you want to use the hole mobility on the N-side to calculate the hole diffusion coefficient on the N-side. Does this make sense?
    Last edited: Apr 5, 2015
  4. Apr 6, 2015 #3
    Yes, I think so.

    Does this mean I can calculate the reverse saturation current from both the p side and the n side?

    In the equation for reverse saturation current (Is) does the two bracketed terms find each of those 2 currents, electrons from p side across to n side and vice versa, which are then summed to find the total current?
  5. Apr 6, 2015 #4


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    Yes. The (Dhpno / Lh) term is holes diffusing from the N-side, and the other term is electrons diffusing from the P-side.
  6. Apr 6, 2015 #5
    Great, I think I've worked through it, I'll add my working:

    Dhpno / Lh = (kTμh / e) * (ni2/ND) / Lh
    Where μh = μp from n side = 450cm2/Vs

    Dhpno / Lh = 1.08E8

    Denpo / Le = (kTμe / e) * (ni2/NA) / Le
    Where μe = μn from p side = 700cm2/Vs

    Denpo / Le = 1.35E7

    Is = eA[Dhpno / Lh + Denpo / Le] = 1.94E-15 A

    Does that look right?
  7. Apr 6, 2015 #6


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    It looks right to me. I think in your original post you have mixed up the mobilities on the n-side - the hole mobility should be the smaller value. But you used the correct one in your calculation, and I got the same final answer as you did. Make sure and carry the units along - you have written Dh*pno / Lh = 1.08E8, but there are no units, so many graders would take off points.
  8. Apr 6, 2015 #7
    Yeah I did write them down the wrong way round.
    Thanks for your help. I take your point about the units, I normally do it but I just whipped through it quickly. I'll add them in before I write it up!
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