Bipolar Junction Transistor analysis.

Discussion in 'Homework Help' started by lam58, Jan 5, 2014.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
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    Hello, could some check if I've got what I've already done is right and help me find Rin.

    The question asks to find the voltage gains (Vc/Vs) and (Ve/Vs) and the equivalent input resistance as seen by the signal source for the BJT circuit which is biased to operate as a linear small signal ac amplifier, pictured below.

    It is assumed that Cb, R1 and R2 are sufficiently large that they can be ignored.

    The hybrid parameters for the BJT are;

    hfe=50
    hie=1.5kΩ
    hre=hoe=0

    And external resistors are:

    Rb = 500Ω
    Re = 50Ω
    Rc = 6kΩ

    (
    My answer:

    Vc/Vs = (-hfe*Rc)/(Rs + hfe*Re) = (-50*6k)/(500 + 50*50) = 100

    Ve/Vs = (hfe*Re)/(Rs + hie + hfe' *Re) = (-50*50)/(500 + 1.5k + 100*50) = 0.8333

    Rin = I'm not quite sure how to work this out.

    EDIT: also this won't be marked it is merely a tutorial.
     
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  2. Efron

    Member

    Oct 10, 2010
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    If this is a real Common emitter small signal ac amplifier, and assuming correct biasing, CE will/should be seen as a short-cut by the ac signal.

    In that case, you should consider RE as being short-cut to ground G in your ac signal analysis.

    As a consequence, your Vc/Vs becomes (-hfe*Rc)/(Rs + hie) = -150.

    In your formula you're using hfe*Re. This would be correct if there would be no CE. But you're still forgetting hie.

    hie represents the equivalent resistor of the base-emitter PN junction seen by a small ac signal.

    So, draw the equivalent circuit for ac signal and try to get Rin, which is the input resistor seen by Vs.

    In ac signal analysis, you should consider all dc voltage source as short-cut. You can do this because of the superposition theory of linear circuits.

    What is DISTURBING me is the output capacitor Cp. If this is high value it would be considered as a short-cut by the ac signal, and so short-cutting the output of the amplifier. I would have expect to find Cp in serial with the output, as it is the case of CB.

    Still Cp could be the effective load of the amplifier, but you're not giving any data on it.
     
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  3. WBahn

    Moderator

    Mar 31, 2012
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    My guess is that Cp is there to establish an upper cutoff frequency to the response. Thus, well within the passband, Cb and Ce are shorts while Cp is an open.
     
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  4. lam58

    Thread Starter Member

    Jan 3, 2014
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    Would Rin then = (hfe+1) * Re = (150+1) * 50 = 7.55kΩ
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You originally stated hfe as 50....????

    What about Rb (500 ohms)?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    BTW - That Rb is a bit of an anomaly. Can you see why?
     
  7. lam58

    Thread Starter Member

    Jan 3, 2014
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    Sorry calculation error.

    Not sure. :confused:
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps ask the questions:

    1. What purpose, if any, is there for having Rb in the circuit?
    2. What loss of performance is incurred, if Rb serves no useful purpose?
     
  9. Efron

    Member

    Oct 10, 2010
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    DRAW the schematic for ac signal.

    Without doing that, you'll not get to visualize Rin?

    Tip: Rin is the equivalent resistor seen by Vs. Draw the equivalent circuit in ac starting from Vs (from left to right).
     
  10. lam58

    Thread Starter Member

    Jan 3, 2014
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    Ok, I think it Rin then might be hie + Rs + (hfe+1 * Re) = 4.55kΩ
     
  11. LvW

    Active Member

    Jun 13, 2013
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    (hfe+1 * Re)

    Looks a bit uncommon. Do you mean (hfe+Re) ?
    Watch the units!
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Yep, watch the units! :eek:

    How can we add hfe to Re?

    My guess would be that he meant (hfe+1)*Re, though I haven't looked at it to see if this is correct.

    Did you mean to type (hfe*Re)?
     
  13. Efron

    Member

    Oct 10, 2010
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    Lam58,

    Again, DRAW the equivalent circuit in ac signal and show it to us.

    By doing that, you will see that Re is short-cut by Ce (at medium frequencies) so Re has no influence at all in Rin.

    The only resistors that have influence in Rin are:
    1/ R1 // R2, but assumed to be high enough to not be considered.
    2/ Rb

    Guess what Rin will be?
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I agree with Efron. Clarify the problem with a small signal ac equivalent.

    The sloppy indiscriminate use of formulas is questionable.
     
  15. LvW

    Active Member

    Jun 13, 2013
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    ...The only resistors? And what about hie?
     
  16. Efron

    Member

    Oct 10, 2010
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    Ups, hie also :), thanks LvW

    1/ R1 // R2, but assumed to be high enough to not be considered.
    2/ Rb
    3/ hie
     
  17. lam58

    Thread Starter Member

    Jan 3, 2014
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    Ok, my last attempt before I jump off a high height :p

    So my equivalent circuit is below, if Re is short because of Ce then Rin = Rb + hie = 550Ω.................? :confused:
     
  18. Efron

    Member

    Oct 10, 2010
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    GOOOOOOD ! :) for the schematic and logic.

    :mad: But you went wrong with the maths! Re-read your assumptions in your post #1.

    hie is 1.5KΩ, so Rin = 0.5 + 1.5 = 2KΩ.

    Additional note: note that the current source given by the BJT (assumed linear behaviour) has no influence on Rin because of Ce. This is important to retain.

    If there was no Ce, Re and the current source would have impact --> Additional equivalent resistor until ac ground = Re*(1+hfe). So, in this case, total Rin would be Rb + hie + Re*(1+hfe).

    Hope you understood the importance of making simple drawings in electronics.
     
    Last edited: Jan 8, 2014
  19. LvW

    Active Member

    Jun 13, 2013
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    Lam58, I am sorry to disappoint you, but in your eqivalent circuit diagram the orientation of the current source is not correct.
    The source Ic=hfe*Ib must not drive a current through Rb and the input source.
     
  20. Efron

    Member

    Oct 10, 2010
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    LvW, where do you see the orientation of the current source in the schematic? Is it from the Vo arrow?

    Should be noted that in BJT transistors in the linear area, the current goes from collector to emitter. In this case, the emitter is directly grounded because of Ce.

    Simplified circuit is shown below.

    But yes, in a common emitter transistor amplifier, the ac output signal is inverted.
     
    Last edited: Jan 8, 2014
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