about Iceo

Discussion in 'General Electronics Chat' started by stupid, May 4, 2010.

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    hi
    i m having trouble to make sense of common emitter config.
    first,
    the book says for a common-base config the base-emitter is forward biased while collector-base is reversed & so with common emitter config.
    & go on to derive Iceo based on above statement.
    Iceo= Icbo/(1-α) where Iceo is collector-emitter reversed current when Ib=0, Icbo is collector-base reversed current when Ie=0.

    shouldnt common emitter config has collector-emitter reversed?

    i think Icbo doesnt belong to common emitter config as that implies Ie=0 which is not the case.

    i dun understand why he used the common base config parameters to derive Iceo that is for common emitter which clearly has collector-emitter reverse biased.

    my assumption for Iceo.
    Ic = αIe + Iceo
    Iceo = Ic - αIe
    =Ic - α(Ic + Ib)
    rearrange, Ic = (Iceo+αIb)/(1-α)

    Iceo= Ic(1-α)-αIb

    when Ib=0
    Iceo= Ic(1-α)

    is that a right assumption?

    thanks
    stupid
     
  2. kkazem

    Active Member

    Jul 23, 2009
    160
    26
    Hi,
    You are totally misinterpreting the meaning of ICEo, which is defined as the current from Collector to Emitter with the Base Open. This is a generally undesirable leakage current and not a parameter that you can even calculate based on the normal transistor biasing. In a common-emitter configuration, the Collector to emitter junction is reversed-biased. And you cannot derrive the ICEo from the ICBo as far as I recall. It is a measured parameter and it's value is measured at a fixed VCE and put on the datasheet based on a statistical mean of many devices measured to get a "typical" value.
    Its main use is in computing how your base drive circuit will work. If it is a very low value and the transistor is a relatively low-voltage device, then when the transistor is off, you can compute the power dissipation in the transistor from VCE x ICEo for the off-period and see it is an acceptable value. If it is too high, you have to not have the base to emitter open, but terminated with a resistor, a transformer winding (for transformer base drive) or even have a reverse base-to-emitter voltage when the transistor is off to such out the leakage current from the base to greatly lower the collector-to-emitter current in the off condition.

    That's about it.
    Regards,
    Kamran Kazem
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Icbo is leakage of the reverse biased C-B junction. It flows whether or not the emitter is floating, as long as there is a complete C-B circuit. The emitter just has to be floating in order to measure it.
    When the base is floating and the emitter is grounded, this leakage current flows into the B-E junction and gets multiplied by β. The resulting emitter current (identical to collector current) is Iceo.
     
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