Bipolar junction transistor analysis at emitter.

Discussion in 'Homework Help' started by lam58, Jan 12, 2014.

  1. lam58

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    Jan 3, 2014
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    Hi for the question in the attached image below, would

    Ve/Vs = [hfe*Re]/[Rb + hie + (hf'e*Re)], with hf'e being the value I found for Vc/Vs?
     
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  2. t_n_k

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    I would rather have:

    \text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}

    where hfe is the value per the original problem statement.
     
  3. The Electrician

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    What's the difference between hfe and hf'e? I never did see an explanation for that in the other thread.
     
  4. Efron

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    Oct 10, 2010
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    t_n_k is right.

    \text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}
     
  5. lam58

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    Jan 3, 2014
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    Ok but would Vc/Vb stay as (-hfe*Rc)/{Rb+hie + (Re*hfe)} or would this also be Re*(hfe + 1)?
     
  6. Jony130

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    Do you know why we use (hfe+1) ??
     
  7. lam58

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    Jan 3, 2014
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    Not entirely, I think it's because it's the resistance Re + the resistance of the emitter Re affected by the hfe of the junction?
     
  8. Jony130

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    We use hfe+1 because the emitter current is equal to
    Ie = Ib + Ic = Ib + Ib*hfe = (hfe + 1)*Ib
    And this is why Re resistor is seen at the base as (Hfe +1)*Re
     
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  9. lam58

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    Jan 3, 2014
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    So to clarify my initial question is yes it would be (Re*hfe+1)/{Rb+hie+(Re*hfe+1)}? :)
     
  10. Jony130

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    Check your parentheses
     
  11. lam58

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    Jan 3, 2014
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    Sorry just spotted it (Rc*hfe)
     
  12. Jony130

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    But you in post 9 show equation for the voltage gain

    The correct one is Ve/Vb = (Re*(hfe+1))/{Rb+hie+(Re*(hfe+1))

    But now Vc/Vb = ?
     
  13. The Electrician

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    Sometimes you are asking about Ve/Vs and Vc/Vs, but other times it seems to morph into Ve/Vb and Vc/Vb.

    What do you understand to be meant by Ve/Vb as opposed to Ve/Vs? Wouldn't Vb mean the signal voltage at the base directly, whereas Vs is the signal at the left end of Rb? So, Ve/Vb is a different gain ratio than Ve/Vs?

    Please be absolutely clear which you are asking about.
     
  14. lam58

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    Jan 3, 2014
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    It's supposed to be Vc/Vs, it's just a typo.

    Vc/Vs= (hfe*Rc)/{Rb + hie +(Re*1+hfe)}

    Although technically wouldn't Vc/Vs = Vc/Vb since R1, and R2 are ignored?
     
    Last edited: Jan 13, 2014
  15. lam58

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    Jan 3, 2014
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    If there were a capacitor across Re that grounded/shorted Re, would Ve/Vs still = \text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}} ?
     
  16. Jony130

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    OK, now every thinks looks good except one small detail.
    The voltage gain Vc/Vs is negative because transistor in common emitter configuration have 180 degree phase shift.
    When the base current increases the collector current also increases.
    [​IMG]
    As you can see when input voltage is increasing ( base current also) the Vce voltage decrease. So we have 180 degrees out of phase (negative gain).
     
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  17. lam58

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    Jan 3, 2014
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    Ahh ok, I see. Thanks:)
     
  18. Jony130

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    No ,try to use small signal analysis and you will know.
     
  19. lam58

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    Would there be no gain since Re is grounded meaning Ve/Vs would = Vs?
     
    Last edited: Jan 13, 2014
  20. Jony130

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    Yes, how can we talk about the voltage gain in circuit in which we short our output terminal to ground.
     
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