BJt, Forward Active

Discussion in 'General Electronics Chat' started by JoyAm, Sep 8, 2014.

  1. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Hello everyone, there are somethings that i dont understand in this BJT circuit http://prntscr.com/4kx1ub
    first thing to notice is that the way i see it the current that goes down the Rc resistance is IC+IB this sum gives us the IE current so i suppose they are the same, arent they ?
    After that i made some sort of analysis here http://prntscr.com/4kx3vm (when i write f.o. i actually mean f.a )
    The reason i bother myself with that circuit is that it is not dependent by temperature --> B (beta)
    we know that α( alpha) =(B+1)/B which we can say that is a bit less than 1. but we still have that RB/B factor, how do we solve that confusion ? Thanks in advance :)
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    The first thing you need to learn is that when we analyse such circuits we do two analyses.

    A DC or biasing analysis

    An AC or signal analysis

    The circuit parameters will not in general be the same for AC and DC.

    It is usual to start with the DC biasing analysis and set the transistor up at its working or Q point first.

    Is this what you would like to do?
     
  3. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Yes i planned to make the small signal analysis after ( i am actually doing it atm ) can it lead me to wrong results if i do the dc analysis first ?
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    No you should do the DC analysis first.

    Have you looked up Q point?
     
  5. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    No i havent but i believe that this circuit is self-biased so there is no need for such thing, right ?
     
  6. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    So i also made the ac analysis but i am not sure if my results are right, and even if they are i dont know how to interpret the result
    http://prntscr.com/4l4epk
    http://prntscr.com/4l4eu1
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I don't understand what you are trying to say here? What confusion? The β ---> is a current gain = Ic/Ib. β varies with Ic current and also with temperature.
    Vbe is also current and temperature dependent. As for the solution, looks good.

    No. You always need to do a DC analysis. Because you don't know in witch operation region transistor is. The BJT can be in saturation, linear region or near cut-off. So you need to check this first.
    Can you next time use white background?? And try to use a forum upload manager.
    And you made a error.
    Ix = Ic - IRb = h21*Ib - IRb and
    IRb = ?
     
  8. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Thanks for your reply sir, what i would like to do is have an expression without the β factor at all ( using approximations of course, is that possible ?
    As about the IRB maybe its equal with (β-1)iB if we look at this as a mesh ? i assume rπ and RB are not parallel.
    I think that what i write is best seen when i turn it this way, i wont do it again if its easier on white background.
    I haven't tried the sites upload manager, is there any how to tutorial ?
     
    Last edited: Sep 9, 2014
  9. crutschow

    Expert

    Mar 14, 2008
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    No. The DC bias point for your original circuit is affected by the β, so you can't ignore it.
     
  10. Jony130

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    Feb 17, 2009
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    But why you want an expression without the β factor?
    And you always can assume that β = infinity. Which is not true, because in real life β is in range 100..800 for a small-signal BJT.

    IRb = (Vin - Vout)/RB

    Simply click on "upload a file"
    n.PNG
     
  11. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Can we assume that RC and RB have the same "magnitude" ( i am not sure if this is the right term) and so RB/B<<RC so we practically ignore the RB/B ?

    @jony 130 IRb = (Vin - Vout)/RB , yes thats right, thank you
    and thanks for the tip about how to upload a file :)
     
  12. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Actually the way i see it its Ix=IRb-Ic is that right ?
    Screenshot_4.png is this result right ?
     
  13. crutschow

    Expert

    Mar 14, 2008
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    The circuit has negative feedback from the collector to base, which determines the bias point, and the amount of base current at the equilibrium point is determined by the transistor β, so any expression for the bias point that doesn't include β is not correct.
     
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  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you agree that Ib*h21 = Ix + IRb ?? So Ix = Ib*h21 - IRb
    And don't forget that

    Vout = -Ix*Rc (1)

    thus

    Ix = Ib*h21 - IRb
    and IRb = (Vin - Vout)/Rb

    so

    Ix = h21*Ib - (Vin/Rb - Vout/Rb) = h21*Ib - Vin/Rb + Vout/Rb


    And Ib = Vin/h11

    thus

    Vout = -((h21*Vin)/h11 - Vin/Rb + Vout/Rb)*Rc

    now simply solve for Vout/Vin
     
    Last edited: Sep 10, 2014
  15. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    what is h21 and h11 ?
     
  16. Jony130

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    Feb 17, 2009
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    h21 = hfe = beta
    h11 = hie = r_pi = rin
     
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