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Finding roe

  1. Oct 28, 2008 #1
    Hey everyone,

    In my EET program we're going over AC equivalent transistor amplifiers. We've gone over how to do some of them in class but I can't seem to figure out how to find roe. I have the output characteristic of our 2N2222 transistor and made the DC load line but I'm not sure were to go from there. Am I supposed to take the change in voltage and divide it by the change in Ib (from the Q point)?

    Thanks for the help!
     
  2. jcsd
  3. Oct 29, 2008 #2
    I would think that roe would be the ratio of a small change in collector voltage to the small change in collector accompanying it; Ib is not involved. It should be fairly large, as you can see by looking at the characteristic curves; notice how they are almost horizontal.
     
  4. Oct 29, 2008 #3

    Integral

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    roe is fish eggs.

    Perhaps you mean [itex] \rho [/itex] or rho.
     
  5. Oct 29, 2008 #4

    rbj

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    no, i think he means one of the "h-parameters" called [itex]h_{oe}[/itex] (or the reciprocal of it).

    where the quiessent point is (the "Q-point") in the IC vs. VCE, the slope of the curve at that point is your [itex]h_{oe}[/itex] parameter. that slope should be pretty small for a decent BJT and is physically a conductance quantity. the reciprocal of that slope is the output resistance of the Norton or Thevenin equivalent of the dependent source that models the transistor.

    [tex] r_{oe} = \frac{1}{h_{oe}} [/tex]

    that's what i think the OP means.
     
  6. Oct 29, 2008 #5
    That's it, thanks rbj.

    Sorry about the confusion; I didn't realized the formula function on here.
     
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