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Current and density

  1. Oct 16, 2005 #1
    A pn junction is formed from two different semiconducting materials in the form of two identical cylinders placed circular face to circular face. The radius of the cylinders is 0.065mm. In one application, 3.50 x 10^15 electrons per second flow across the junction from the n to the p side while 2.25 x 10^15 holes per second flow from the p side to the n side. (A hole acts like a particle with charge +e)
    Find a) Current b) Current Density

    For the current
    i = dq/dt = [(3.50-2.25) x 10^15] (1.6 x 10^ -19) / 1 sec = 2 x 10^-4 C.
    For the current density
    J = i/A = (2 x 10^-4) / [pi (0.065 x 10^-3)^2 ] = 1.50 x 10^4 C/ m^2
    is this correct??

    A copper wire and an iron wire of hte same length have the same potential difference applied to them.

    a) what must be the ratio of their radii if hte current is to be the same?
    [tex] i = \frac{A \Delta V}{\rho L} = \frac{\pi r^2 \Delta V}{\rho L} [/tex]
    equating for both the copper(C) and the iron(I)
    [tex] \frac{r_{I}}{r_{C}} = \sqrt{\frac{\rho_{I}}{\rho_{C}} [/tex]
    This part should be fine

    b) Can the current density be made the same by suitable choices of radii?THis is the part i'm having doubts with. SInce J = i/A, would this not be possible? Please do advise!

    Thank you for your help!
  2. jcsd
  3. Oct 17, 2005 #2


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    You wrote correctly that i=dq/dt. The positively charged holes flow from the p side towards the n side, tending to increase the charge of the n side and their flow determines the direction of current.
    What happens with the charge of the n side when negative charges (electrons) flow away from it? As the result of leaving negative charges, the n side tends to become more positive, so the contribution of the electrons to the current is positive.
    In principle you get the net current arising from the drift motion of different charged particles as the summ of terms qvn where q is the charge of the particle, v is its drift velocity and n is the concentration of that kind of particle. If both q and v are negative you get a positive contribution.
  4. Oct 17, 2005 #3
    so in other words i should be adding the numberof holes with the electrons since the holes along iwth the electrons increase the amount of current flowing?
  5. Oct 17, 2005 #4


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    Refer to this about holes and conduction -




    Holes are formed when an electron leaves an atoms so there is net positive charge. A hole moves when an electron from one atom (lattice site) moves into the hole 'filling it' so that the atom (lattice site) from which the electron originate now has the hole. So holes move as fast as the electrons can fill them.

    More generally - http://hyperphysics.phy-astr.gsu.edu/hbase/solids/band.html
  6. Oct 17, 2005 #5


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    You've written down the expression for the current, i. Now write the expression for the current density, j = i/A. Look at this expression and decide if j can be varied by changing the radius.
  7. Oct 18, 2005 #6
    sooo is the answer to the first one correct or not? Since the electrons flow they create a current and since the holes do they ADD the current??

    Please advise

    Thnak you so far for the advice!
  8. Oct 20, 2005 #7


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    Yes, they add to the current, as you correctly said in post#3.
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