BIASING

Discussion in 'General Electronics Chat' started by koley, Mar 3, 2006.

  1. koley

    Thread Starter New Member

    Mar 3, 2006
    1
    0
    Could someone kindly help me with this assignment.


    The diode current, i terms of voltaveV, is approximately given by
    I=I.{exp(eV/kT)-1} where e=1.6^-19 coulomb is the electronic charge, k=1.38^-23joule k=boltzmann constant and T-temp. in K
    if the saturation current I. is 1ηA, cal the forward and reverse resistances at V=0.2 and T=300K[/FONT]



    Thanks a lot.
     
  2. Papabravo

    Expert

    Feb 24, 2006
    9,903
    1,723
    I'm going to assume that V(reverse) = -V(forward) = -0.2 Volts

    So
    R(forward) = V(forward)/I(forward) = +0.2/I(forward)
    R(reverse) = V(reverse)/I(reverse) = -0.2/I(reverse)

    does that help?

    By the way is this a germanium diode?
     
  3. windoze killa

    AAC Fanatic!

    Feb 23, 2006
    605
    24
    Aren't germanium diodes dead and buried and pushing up daisies?????

    Or should that be pushing up geraniums?????

    Just some sick humour. Please forgive me and even ignore me.

    Sorry. Couldn't resist.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    9,903
    1,723
    Apparently not
    Germanium Diodes
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Just bought some 1N34's - you just never know when they'll come in handy. Might even save an oatmeal box and wind a coil some day.
     
  6. aac

    Active Member

    Jun 13, 2005
    35
    0
    The small signal equivalent conductance can be found by taking the derivative of the current equation with respect to V.

    dI.{exp(eV/kT)-1}/dV = I.{exp(eV/kT)}*(e/kT)

    Just put .2 and -.2 in calculate the conductance and divide into 1 to get resistance.

    Alternately you could start with

    V = (kT/e)ln(I/I.+1)

    and take the derivative with respect to current to get resistance directly.
     
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