Bipolar common emitter voltage

Discussion in 'Homework Help' started by ham3388, Oct 15, 2016.

  1. ham3388

    Thread Starter Member

    Jul 3, 2012
    97
    3
    Hi again dear friends. , hope all are doing well.
    Please could some one evaluate my work and advise me in case any correction isrequired.
    Have a good day Screenshot_2016-10-15-11-54-09.jpg 20161015_113525.jpg 20161015_113535.jpg 20161015_113642.jpg
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,792
    What steps have you taken to check your results?

    Which do you expect to have the higher Vbe, the answer to (i) or the answer to (ii)?
     
  3. ham3388

    Thread Starter Member

    Jul 3, 2012
    97
    3
    I think the answer to (ii) sould have higher Vbe since Ic is more
     
  4. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    See
    Draft342.png
     
    Last edited: Oct 18, 2016
  5. ham3388

    Thread Starter Member

    Jul 3, 2012
    97
    3
    Thanks Bordodynov...
    The Vbe voltage is not matching with the values that obtained by me...

    So what has gone wrong, I'm wondering.

    Can you suggest where is my mistake?
     
  6. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    i. 25e-3*ln((5e-3/0,98e-13)+1) = 0,6163872887428
    exp(Vbe/25mV)-1=5e-3/0,98e-13
    exp(Vbe/25mV)=5e-3/0,98e-13+1
    Vbe/25mV=ln((5e-3/0,98e-13)+1)= 24,65549154971
    24,65549154971*25e-3 = 0,6163872887428
    I shall understand with program LTspice, why program gave the value-added result.
    ii. if Vce=const=5V ==>
    Vbe(9.95mA)=25e-3*ln((9,95e-3/0,98e-13)+1) =0,633590654711
    but Vce != 5V Vce=(10V-1000*9.95)=50mV Vbe-Vbc=50mV !
    That must be taken into account for the calculation of both terms of the equation
    Make the equation. Unfortunately for accurate calculation of the gain "ar" needed.
     
  7. ham3388

    Thread Starter Member

    Jul 3, 2012
    97
    3
    Thank pal for your time
    I appreciate you
    Please see below
    I don't understand the hi lighted once.
    Which formula you have used.
    Can you explain this for me because I'm confused little bit since they are not according to my formula


    Screenshot_2016-10-18-19-32-28.jpg
     
  8. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    See
    Draft341_1.png Draft341_2.png
     
  9. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    I figured out with LTspice.
    With regard to my calculations. In the beginning I wrote the final result, and then brought the detailed steps.
    Do not pay attention to the second calculation result.
    In the case of 9.95 mA compose equation Ebers-Mole. Take as I ar = 0.5 (af = 0.98). Drop unit in parentheses (because of their smallness in comparison with the exhibitors). Add the equation Vbe-Vbc = 50mV. Use the properties of the exponential exp (A-B) = exp (a) * exp (-b)
     
    ham3388 likes this.
  10. ham3388

    Thread Starter Member

    Jul 3, 2012
    97
    3
    Thanks very much Bordodynov
    You are great
    Finally I got the correct values
    Again I say thank you very much since you paid attention to my thread.
    Godby for timbieng.
     
Loading...