DC analysis on this NPN BJT

Discussion in 'Homework Help' started by u-will-neva-no, May 14, 2011.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Hello everyone. Would it be possible to check my analysis for my attached circuit?

    Here is my analysis:

    The question states that B is 200 and that Vbe = 0.7v.

    From the graph, vb = 0v, because it is grounded. Therefore:
    Vbe = Vb - Ve
    0.7 = 0-Ve
    Ve = -0.7V

    Now to find IE. using ohms law I have:
    Ve - (-10v) = IE*RE
    (-0.7)+10 = IE *10K
    IE = 0.93mA

    finding IC by using the formula IC = αIE
    α=β/(β+1)IE
    α = 200/201* 0.93mA
    IC = 0.925mA

    Finding Vc using omhs law again
    Vcc -Vc = ICRC
    10-Vc = 0.925mA * 10k
    Vc = 0.75V

    I would just like to know if my method is correct, because my teacher has completely different values! Thanks.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You forgot about RB resistor in you analysis and the base current.
    VEE - Vbe = Ie*Re + Ib *Rb
    And for me the BJT will be in saturation

    EDIT
    I do simply calculation
    And I got
    Ie = 886uA; Ic = 881.6uA;
    And the BJT for sure is in active mode
     
    Last edited: May 14, 2011
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  3. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    I am having difficult trying to visualise how you obtained that equation. I accept that your correct though!
     
  4. jegues

    Well-Known Member

    Sep 13, 2010
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    You are confusing the where the voltage at the base is.

    I've attached a figure to help clear up your confusion.

    You can write a KVL in the BEJ loop and solve for Ic, Ie or Ib, whichever you prefer. (Exactly like what Jony130 has demonstrated for you)

    If you find that they are nonzero then certainly VB is nonzero as well.
     
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  5. Jony130

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    Feb 17, 2009
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    I wrote KVL for this loop

    [​IMG]
     
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  6. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Ok I think I understand it now. I am going to try to explain what I think is going on, so that I can apply it to different examples in the future.

    Just to explain some of my thoughts to different problems, say I wanted to calculate the current flowing through the collector, i would do something like Vcc - Vc = ic* Rc and rearrange. If I wanted to find the current ie, i would do something like Ve - VEE = ie*Re.

    from these two approaches, i would do VEE - Vbe = Ib + Ie (1). Am i correct to say that there is a pattern, such that you must go in a clockwise direction for the Vcc - Vc,
    Ve - VEE and VEE - Vbe part?

    Now, im picturing that we are interested in the current Ib and current Ie, hence why we get the Ib + Ie part in equation (1)? Let me know if this is the correct way to visualise it :)
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For the BJT we always need to find Ib or Ie first.
    Then if we know Ib we can find Ic form
    Ic = β*Ib and Ie = Ib + Ic

    You equation are wrong
    How can current be equal to voltage (mismatch in units) ??
     
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  8. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Woops, I messed up...I meant to have IeRe +IbRb!
     
  9. jegues

    Well-Known Member

    Sep 13, 2010
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    I don't think this is true.

    You can solve it in whatever order/manner you choose (I could solve for Ic first if I choose to do so) as long as your results are correct.

    After doing so, you must check to see that any assumptions you've made during the initial analysis are valid. (i.e. assuming active mode)
     
  10. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    @Jony130, using the formula VEE - Vbe = IeRe IbRb, and also, Ib = Ie/(B+1), i get:

    VEE-Vbe = IeRe + (Ie/(B+1))*Rb

    -10 -0.7 = Ie (Re + Rb/(B+1))

    Ie = -1.02mA....Im clearly doing something wrong...
     
  11. jegues

    Well-Known Member

    Sep 13, 2010
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    When I write a KVL I find that,

    I_{b}R_{b} + V_{be} + I_{e}R_{e} + V_{ee} = 0

    \Rightarrow I_{e} = \frac{-V_{ee} - V_{be}}{\frac{R_{b}}{\beta + 1} + R_{e}}

    But it is not clear how the Vee source is hooked up, so it could change by a negative sign.
     
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  12. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    jegues, your answer is correct. My problem is that the formula is different to Jony130's which is what I understood. Could you explain me how you obtained your KVL?

    Jony130's formula would work if i took my Vee value (-10v) as just 10v, but i dont think i should be doing that...
     
  13. jegues

    Well-Known Member

    Sep 13, 2010
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    It was a simple application of KVL.

    Why don't you make an attempt of drawing the voltages and their respective polarities around the BEJ loop.

    If there is any problems, someone will correct you.
     
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  14. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Oki doki, my picture is attached. Its practically exactly the same as Jony130's representation (Just a lot messier). Anyway, from my diagram, I have VEE in a clockwise direction whereas all the other voltages are in an anticlockwise direction. Therefore, i am not getting the same formula as Jegues..
     
  15. jegues

    Well-Known Member

    Sep 13, 2010
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    Here is my drawing of my KVL.

    I cannot make it any clearer than this. This is nothing but a basic application of KVL.
     
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  16. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Thank you to everyone for your help, much appreciated!
     
  17. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For your circuit
    http://forum.allaboutcircuits.com/attachment.php?attachmentid=30547&d=1305406350

    Vee = -VRB + Vbe - Vbe

    But you have VEE in a clockwise directions so Vee = 10V
    You are measure the voltage across Vee withe respect to minus terminal of a Vee and Vbe = - 0.7V

    Ie = Vee + Vbe/ ( Re + Rb/(β+1) = (10V + (-0.7V))/ (10K + 497.5Ω) = 9.3V/10.497K = 886uA

    BJT is a current control current source, so to find Ic you need to know Ib or Ie first.
     
  18. Jony130

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    Feb 17, 2009
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    And let me remind you what negative voltage is in real life
    [​IMG]

    [​IMG]


    [​IMG]
     
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  19. jegues

    Well-Known Member

    Sep 13, 2010
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    Yes I agree it acts as a current controlled current source, but as far as the analysis is concerned I could solve for Ic first if I choose to do so.
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    In principle the same equation can be translated to immediately solve for any of the three interdependent values Ie, Ic or Ib. The original equation itself explicitly accounts for the functional controlled source.

    So, per Jony130's notation

    I_e=\frac{V_{ee}+V_{be}}{R_e+\frac{R_b}{1+\beta}}

    I_b=\frac{V_{ee}+V_{be}}{(1+\beta)R_e+R_b}

    I_c=\beta \frac{V_{ee}+V_{be}}{(1+\beta)R_e+R_b}

    Certainly in that sense one may immediately solve for any of those values - as desired. Obviously, one must firstly form the correct algebraic equation for the solution to be valid.
     
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