Help with Transistor analysis

Discussion in 'Homework Help' started by DYLH, Sep 14, 2013.

  1. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    For the problem in the attached image, I'm able to find the Vo:

    -5 + 10000 Ib + Vbe + 200 Ie = 0
    Ic = 100 Ib
    Ie = Ib + Ic

    -5 + 10000 Ib + 0.7 + 200 (Ib + Ic) = 0
    10000 Ib + 200 (Ib + Ic) = 4.3
    10000 Ib + 200 (Ib + 100 Ib) = 4.3
    Ib = 142.38 uA
    Ic = 14.238 mA
    Ie = 14.38 Ma

    Vo = Ie * R
    Vo = 14.38 mA * 200
    Vo = 2.876 V (matches given solution)

    I'm stuck on solving for Vce, I'm comming up with:
    Vo - 12 + Ic (500) + Vce = 0
    2.876 - 12 + (14.238)(500)(10^-3) = -Vce
    Vce = 2.005 (doesn't match solution of 1.984V)

    where am I going wrong?


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  2. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    12 - 500 Ic = Vc (4.881)
    200 Ie = Ve (2.876)
    Vc-Ve = Vce (2.005)

    Ima 'fraid it doesn't match my answer, either.
    But then, I'm surprised I could read your math, this being 37 years since I was in a school room and they keep changing how they teach things.:rolleyes:

    Another helper will be along soon enough and double check our math.
     
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Where you calculated:

    Vo - 12 + Ic (500) + Vce = 0

    I think they calculated:

    Vo - 12 + Ie (500) + Vce = 0

    The book made a mistake.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    One of the nice things about most engineering problems is that you can determine the validity of an answer from the answer itself -- whether it's YOUR answer or the AUTHOR's answer.

    So take their answers:

    Vo = 2.876V
    Ie = Vo/200Ω = 14.380mA

    Vce = 1.984V

    Vc = Ve + Vce = Vo + Vce = 2.876V + 1.984V = 4.860V
    Ic = (12V-Vc)/500Ω = (12V-4.860V)/500Ω = 7.14V/500Ω = 14.280mA

    Ib = Ie - Ic = 14.380mA - 14.280mA = 100μA

    β = Ic/Ib = 14.280mA/100μA = 142.8

    It would not appear that the author's solution is not consistent with the problem given.

    Now, keep in mind that the difference between their answer and your answer is only about 1%, so it is very possibly due to roundoff error on their part. But since THEY choose to use answers with four sig figs, it was THEIR responsibility to provide an answer that was correct to four sig figs.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    That sounds very likely -- and since β=100 would throw the answer off by 1%, that matches well. So they took the often used approximation that Ic≈Ie but then felt justified in reporting answers to four sig figs when a β of 100 only means that Ic matches Ie to two sig figs. Sloppy.

    But we all make sloppy mistakes, so don't judge the book or the author based on an occasional sloppy mistake.
     
  6. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    Thanks, all!

    FWIW, this is for Fundamentals of Electric Circuits by Alexander and Sadiku 5th edition, Practice Problem 3.12... in the event someone else searches.
     
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