BJT Beta

Discussion in 'Homework Help' started by Agonche, Aug 31, 2012.

  1. Agonche

    Thread Starter Member

    Aug 26, 2011
    30
    0
    If \beta =30

    * Find V_B, V_E, V_C.

    * If R_B changes to 270k\Omega, find V_B, V_E, V_C.

    * For what value of \beta will voltages V(B),V(E),V(C) get back to previous values.

    [​IMG]

    ==================================================

    Here's what I got.
    I_B=0.074977mA
    I_C=2.24932mA
    I_E=2.3243mA

    V_B=2.0244V
    V_C=-2.9268
    V_E=6.65757

    If R(B) changes to 270k:

    I_B=0.02347mA
    I_C=0.70398mA
    I_E=0.72745mA

    V_B=6.3358V
    V_C=-7.09924
    V_E=7.03588

    Here comes the 'problem', finding Beta...
    I calculated I_B using the previous (first) V(B) value.
    I_B=0.0074977mA
    So If V(C) gets back to the previous value, I(C) should be the same.
    And \beta = \frac{I_B}{I_C}=300.
    But I realized I can find Beta with another method, using K-laws.
    9-2.7kI_E-0.7-270kI_B=0
    using:
    I_B=\frac{I_E}{\beta +1}
    and the value of I(C) found first, I get:
    \beta =309

    Can anyone tell me if any of the solutions is right or wrong and why.
    Thanks in advance.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you use a simulation program to find Ib,Ic and Ie ??
     
  3. Agonche

    Thread Starter Member

    Aug 26, 2011
    30
    0
    No.
    I_B=\frac{9-0.7}{27k+2.7k\cdot 31}
    Ic=Beta*Ib
    Ie=(Beta+1)*Ib

    =)
     
  4. mlog

    Member

    Feb 11, 2012
    276
    36
    Check your original VE. It's inconsistent with your emitter current. The emitter voltage should be ~0.7 V above the base voltage.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    For

    Hfe = 30 and Rb = 27K

    Ib = (9V - 0.7V)/ ( 27K + 31*2.7K) = 74.977μA

    Ic = 30 *Ib = 2.249mA

    Ie = 31 *Ib = 2.324mA

    Ve = 9V - Ie*Re = 2.725V

    Vc = Ic*Rc - 9V = -2.927V


    Now for Rb = 270K and β = 30

    Ib = (9V - 0.7V)/ ( 270K + 31*2.7K) = 23.466μA

    Ic = 30 * Ib = 703.986μA

    Ie = 31 * Ib = 727.452μA


    Ve = 9 - Ie*Re = 7.035V

    Vc = Ic*Rc - 9V = 7.099V


    And I don't see any problem ?
     
  6. Agonche

    Thread Starter Member

    Aug 26, 2011
    30
    0
    @mlog - Yeah you're right. Ve should be 2.724V.

    @Jony130 - Thanks for the confirmations, but I'm having problems with the third part of the problem.

    so with Rb=270k, I have to find the value of Beta so the terminal voltages get their previous values.

    I think Beta should be 300, because Rb changed from 27k to 270k.
    So if Beta was 30 first, it should be 300 when Rb=270k and the terminal voltages get back to their previous values.

    I tried another method (you can see in the first post) and I get Beta=309.
    What am I missing ?
    Is Beta 300 or 309 or none?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    I think that both of your answer are correct.
    Simply it is impassible to get exact voltage for R = 270K.
    The the reason for this is that emitter current is equal
    Ie = Ic + Ib but for R=270K the Ib is ten times smeller than previous values. And this cause the "error".
     
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