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Diffusion current

  1. Feb 12, 2015 #1
    Hello,

    The diffusion current in a semiconductor is caused by difference in concentration of holes and electrons. My question is it related to Coulomb's law (electric force due to electric field) ?

    Thank you
     
  2. jcsd
  3. Feb 12, 2015 #2
  4. Feb 12, 2015 #3
    I have already read it but my question is because of what if there is difference of concentration of charged carriers there is diffusion current? is it related to Coulomb's law about electric force between 2 charged particules?
     
  5. Feb 12, 2015 #4
    What do you think what kind of force generates the electric field?
     
  6. Feb 12, 2015 #5
    I mean that the electric charges will generate electric field thus attraction of 2 opposite charges like what is stated in Coulombs law
     
  7. Feb 12, 2015 #6
    Aha, if you want to distinguish between cause and effect, read first about Fick's diffusion law.
    See also on p.15 about diffusion current densities in semiconductors in this pdf.
    Hope, it helps.
     
  8. Feb 12, 2015 #7
    haha yes but why here coulombs law does not apply here
     
  9. Feb 12, 2015 #8
    It does apply after equillibrium state in semiconductor is established and the potential barrier formed.
     
  10. Feb 12, 2015 #9
    No Im talking about diffusion current why it is not explained with Coulombs law
     
  11. Feb 12, 2015 #10
    Becouse classical phenomenon of diffusion is not caused by electrostatic repulsion among particles, but concentration changes.
    However, it is important to remember diffusion current is essential for explanation of properties of semiconductors P-N junctions (diodes, tranzistors etc.) And there is a famous Einstein's relation between diffusion constant D and electrical mobility of carriers μq:
    D/μq=k⋅T/q
     
  12. Feb 12, 2015 #11
    ah ok thanks a lot
     
  13. Feb 13, 2015 #12
    Hello again,

    In a diode, there is a resistance also in the p and n region, and electric field is required to move the charges in forward bias conditions. My question is when there is diffusion current is the electric field helping the diffusion current because of the resistance? thus we can say that there is also drift current?

    Thank you
     
  14. Feb 13, 2015 #13
    It is not good to think of a semiconductor resistances as key point here. In p-n junction, n-type region has a majority electrons carrier concentration, and p-type region has a majority holes carrier concentration. Electrons diffuse from n side to p side, and the holes diffuse from p side to n side. If the electrons and holes were not charged, diffusion would continue until concentrations of holes and electrons on both sides were the same. Since they are charged, charge separation in material creates E-field. This E-field opposes diffusion and at some point there is balance where diffusion process stops. This results in a built in potential due to E-field generated and so called depletion layer. Now, in a forward biased state the net effect is such that when external voltage is applied, it reduces the "barrier voltage": when it exceeds the barrier voltage the current flow through the p-n junction quickly increases. Example of this depedence is Shockley's ideal diode equation:
    7cbe19104958cabb4dd28cd46ad0384a.png
    Of course, there will be also a drift current. But remember drift and diffusion components are just helpful tool in the analysis, to distinguish between causes and effects. In reality there is just one current through the diode. For example, another indirect effect of the external voltage to diffusion current can be as follows. The rate at which diffusion occurs depends on the average velocity of carriers and distance between scattering events in the material. Raising the temperature increases the termal velocity of carriers, and diffusion occurs faster at higher temperatures in semiconductors. Higher the external voltage to forward biased diode, higher the current through it and diode acquires higher operating temperature. Therefore, diffusion current also increases.
     
  15. Feb 13, 2015 #14
    Great. But why when we calculate the loss of power due to resistance of the diode we take into consideration the total current in the diode? why cant we just take the drift current caused by the applied electric field?
     
    Last edited: Feb 13, 2015
  16. Feb 13, 2015 #15
    Becouse, in reality it is the "total" current that flows and counts (so loss is PD=VD⋅I). I made that clear:
    regards
     
  17. Feb 14, 2015 #16
    Just to make things clear. Here in the photo they represented a real diode by 4 resistances and an ideal diode. I know it is a representation but when we represent the p and n region by 2 resistors so there is a drift current. But inside this 2 regions there is drift and diffusion current. So you mean by same current is that the overall current consisting of drift and diffusion current is all the same inside the diode right? (because it is a serie circuit)
     

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  18. Feb 14, 2015 #17
    Looks like you didn't read pdf in post #6... Yes, overall (total) current consists of drift and diffusion component. When forward biased diode conducts, diffusion component is dominant and only small increase of applied voltage is needed for diffusion and overall current to increase dramatically (Like Shockley's equation shows).
     
  19. Feb 14, 2015 #18
    haha yes. Really that is my last question, the electric field due to resistance is acting on all the current thus on the diffusion right?
     
  20. Feb 14, 2015 #19
    No. Voltage drives the current, and resistance regulates current magnitude. That's Ohm's Law.
    Resistance of diode is highly nonlinear due to properties of materials involved.
     
  21. Feb 15, 2015 #20
    Yes voltage drives the current due to electric field.So electric electric field is driving also the diffusion current to overcome the resistance right?
     
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