1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding current and voltage amplification via load line graph plotting

  1. Aug 2, 2013 #1
    1. The problem statement, all variables and given/known data
    This is a question refer to a transistor in a common emitter configuration.

    And this is the table with data of Vce and Ic at certain range of Ib stored in microsoft excel, and the graph is plotted within the file attached(in another tab)
    https://www.filesanywhere.com/fs/v.aspx?v=8b6a6b8f595e70a970a0

    The question also stated load resistance is 2k Ω

    The base current is suppose to be in micro Amp, forgot to include this in the data table..

    3. The attempt at a solution

    no other pictorial information about how circuit, so i suppose this common emitter, have a load resistance of 2k Ω before collector and a Base resistor

    so i come out with my load line equation,

    Ic = -Vce / 2k + 4m
    I induced this load line in my attachment,

    So lets say i choose my quiscent operating point at
    Ib = 40 uA, and my output current and voltage peak to peak is :

    Vpp = 5.8 - 3.1 = 2.7V
    Ipp = 2.45 - 1.2 = 1.25 m A

    so my current amplification is
    Ai = 1.25m / 40u = 31.25



    and if the input resistance is 1.5k Ω,



    my voltage amplification is,

    Av = (Iout)(Rload) / (Iin)(Rinput)
    =Ai x (Rload)/(Rinput)
    = 31.25(2/1.5)
    = 41.667
     
  2. jcsd
  3. Aug 3, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Isn't your Ib varying from the quiescent point to produce the change in Ic (A signal superimposed upon the quiescent operating point)?
    Presumably this is a given value?
     
  4. Aug 3, 2013 #3
    yes

    And the given value is the output resistance(2k ohms), input resistance(1.5k ohms)
    And a table of Ic and Vce value at different Ib(20u, 40u, 60u, 80).

    In the attachment file i posted in the my first post, the right most column is the value i compute with
    Ic = -Vce / 2k + 4m, is not actually the value given in the question.
     
  5. Aug 3, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    So shouldn't the actual ΔIb be used to calculate the current gain rather than the fixed quiescent current?
    Okay. But your table claims 68 μA for one of the Ib values... Is that a typo? should it in fact be 60 μA? If it is, it would explain your use of 40 μA in the current gain calculation.
     
  6. Aug 3, 2013 #5
    yeah my bad, it is 60u...my typo, and the question also stated the Ib swing at peak to peak value of 40u, forgot to include this.

    Vcc = 8v
     
  7. Aug 3, 2013 #6

    gneill

    User Avatar

    Staff: Mentor

    Okay. So then it looks like your calculations are fine.
     
  8. Aug 3, 2013 #7
    Thanks you very much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding current and voltage amplification via load line graph plotting
  1. Current amplification (Replies: 2)

Loading...