Help rearranging an equation

Discussion in 'General Electronics Chat' started by Ronscott1, Nov 10, 2009.

  1. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    Here is the original equation:
    Ie = Vcc - Vbe / ((Rc+Re+Rb) / Bdc)
    I need to solve for Rb. Could you guys help me?
    Best,
    Ron
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
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    Rb = [(Vbe*Bdc)/(Vcc - Ie)] - Rc - Re
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think your starting equation is incorrect - probably just misplaced parenthesis.

    Assuming you have made the approximating assumption Beta = (1+Beta)

    I would have written

    Ie = (Vcc-Vbe)/(Rc+Re+(Rb/Beta))

    or more correctly (i.e. not using the approximation)

    Ie=(Vcc-Vbe)/(Rc+Re+(Rb/(1+Beta))

    To re-arrange the approximate version

    Ie(Rc+Re+Rb/Beta)=Vcc-Vbe

    Rc+Re+Rb/Beta=(Vcc-Vbe)/(Ie)

    Rb/Beta=(Vcc-Vbe)/Ie-(Rc+Re)

    Rb= Beta(((Vcc-Vbe)/Ie)-Rc-Re)
     
  4. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    Thank you someonesdad
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    While someonesdad has correctly re-arranged your equation it should be pretty obvious that the term

    [(Vbe*Bdc)/(Vcc - Ie)]

    is not a resistance - it's a nonsense.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    I'll write it nicely...


    I_e = \frac{V_{cc} - V_{be}} {{R_c+R_e+} \frac{R_b}{1+\beta}}
    ...
    Hand Waving here.... (See t_n_k's post)
    ...
    Rb=\beta (\frac{V_{cc}-V_{be}}{I_e})-Rc-Re

    Readable is good. Click on the Ʃ in reply box to get the Tex Editor/helper.
     
  7. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    T_N_K,
    When I set my work up using your equation I am getting very different results from what I calculated before and what multisim is telling me. However when I drop that pair of parenthensis I am getting similar answers. I do not have a scanner so I cannot scan my schematic or calculations
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    How about ....???

    Rb=\beta(\frac{(Vcc-Vbe)}{Ie}-Rc-Re)
     
  9. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    I will post my work tomorrow when I get to school and have access to a scanner but thank you guys so much for you help
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Is this what we are imagining?
     
  11. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    My bad for getting the parenthesis wrong:

    \large{Rb=\beta (\frac{V_{cc}-V_{be}}{I_e}-Rc-Re)}

    It looks correct to me. :confused:
     
  12. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    t_n_k,
    That is exactly what I am working on that same exact circuit. I was given the problem to design the resistor values Rc, Rb, and Re to meet the following specifications:
    Vcc=9V
    Vce at midpoint
    Ic=1 mA
    βdc=300

    Here is my work until the point we are at now:
    Ve=0.1*Vcc
    Ve=(0.1)(9V)
    Ve=0.9V
    Re=Ve/Ie Since Ie≈Ic
    Re=0.9V/1mA
    Re=900Ω
    Rc=4Re
    Rc=4(900Ω)
    Rc=3.6kΩ
    And here is where I got stuck. I am following VDB guideline from Albert Malvinos Electronics Principles Book.
    Is there something that I am doing wrong? If so how should I go about calculating these resistor values?

    Best,
    Ron
     
  13. Ronscott1

    Thread Starter New Member

    Nov 5, 2009
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    In other words, Are there certain steps that someone should take when designing a collector-emitter feedback amplifier? Please help
     
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