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I PN junction current

  1. Jun 12, 2016 #1
    Hi. I'm trying to understand how the current in a PN junction depends on various parameters. I have found this formula in Ashcroft and Mermin. It looks like the current will increase if we decrease the density of donors and/or acceptors, Na and Nd. Is this correct?

    YIQyokv.jpg
     
    Last edited: Jun 12, 2016
  2. jcsd
  3. Jun 12, 2016 #2

    mfb

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    There is still ##n_i^2##.
     
  4. Jun 13, 2016 #3
    ##n_i^2## is the intrinsic carrier concentration, which doesn't depend on ##N_a## or ##N_d##, so how could that affect the result?
     
  5. Jun 13, 2016 #4

    mfb

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    Ah wait, you have the doping in the denominators, not the actual number density of electrons/holes. Hmm, then it looks strange.
     
  6. Jun 13, 2016 #5

    phyzguy

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    The saturation current is caused by the thermal generation of carriers in the depletion region, which are then swept out of the depletion region by the built-in field. As the doping concentration(Na and Nd) goes down, the depletion region gets larger, so there is a larger volume of depleted semiconductor where generation takes place. I think the equation is correct.
     
  7. Jun 13, 2016 #6
    Actually this particular derivation in Ashcroft and Mermin doesn't include any generation in the depletion region, but anyway you're right that the formula does make sense when thinking about a reverse biased diode. If we let ##p_{no}## denote the equilibrium density of holes on the N side, and since ##n_i^2/N_d=p_{no}##, we can write for the hole current under reverse bias

    [tex]
    J_p=en_i^2\frac{D_p}{L_pN_d}=e\frac{D_p}{L_d}p_{no}
    [/tex]

    So the reverse hole current is proportional to the density of holes on the N side, which makes sense.

    When the diode is forward biased I would like the hole current to depend on the density of holes on the P side. This is of course possible. If ##V_{bi}## is the built-in potential, ##V_T## the thermal voltage, and ##p_{po}## the density of holes on the P side, we have

    [tex]
    p_{no}=p_{po}e^{-V_{bi}/V_T}
    [/tex]

    Which leads to a hole current that is proportional to ##p_{po}=N_a## and decreases as the built-in potential increases.

    [tex]
    J_p=e\frac{D_p}{L_d}N_ae^{-V_{bi}/V_T}(e^{V/V_T}-1)
    [/tex]

    So the answer to my initial question must be that the current increases when you decrease ##N_a## because this decreases the built-in potential

    [tex]
    V_{bi}=V_T\log{\frac{N_aN_d}{n_i^2}}
    [/tex]

    WELL, still puzzling that a low carrier density gives the highest current....
     
    Last edited: Jun 13, 2016
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