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What's the reason for modifying these amplifier circuits?

  1. Feb 18, 2016 #1
    If I have a bjt amplifier with pure differential input signals like this:
    bjt diff.PNG

    Now let's say I apply pure common mode signals to the same circuit, then why is the circuit above modified to this ?

    bjt common.PNG

    Also when finding the Common Mode Input Resistance the circuit above excludes RE and includes ro, why ?
    small signal.PNG (finding input common mode resistance)
     
  2. jcsd
  3. Feb 18, 2016 #2

    LvW

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    For a pure common mode signal at both inputs both transistors are doing the same: Both will show slight increase of the collector currents.
    OK - if they are doing the same, we can treat both transistor stages individually - however, in the first circuit the emittercurrent was split into two equal halfs. Therefore, each stage must have the same current IE as before: IT/2. And the same applies to REE. Putting both emitter nodes again together we gave 2REE||2REE=REE.
     
  4. Feb 18, 2016 #3
    Ok yes I get that the two circuits are equivalent but then what if I analysed the circuit for the common mode signals without splitting the current source and REE (like first diagram), then for my small signal analysis I would neglect REE because the voltage at x is zero and I would get the wrong equation
     
  5. Feb 18, 2016 #4
    Apparently they are making some simplifying assumptions. What do you get for the common mode input resistance?

    By the way, is this homework?
     
  6. Feb 19, 2016 #5

    LvW

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    Why do you think the voltage at x would be zero? Can you verify this assumption?
     
  7. Feb 19, 2016 #6
    Ricm = Vic/ib
    = (B+1)*ib*(2*REE//ro) / ib.................................// means parallel combination
    = (B+1)*(2*REE//ro)

    This is not homework. This stuff is from the textbook and I just dont understand certain things
     
  8. Feb 19, 2016 #7
    Because the symmetry ?
     
  9. Feb 19, 2016 #8
    Are you just using the simplified circuit shown in the third image in post #1?

    What do you get if you analyze half the circuit shown in the second image of post #1? If I perform a nodal analysis on that half circuit, I get a much different result.
     
  10. Feb 19, 2016 #9

    LvW

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    Is there an ideal current source in the emitter leg? I don`t think so.
    Hence, you have the classical common emiiter configuration with emitter feedback.
     
  11. Feb 20, 2016 #10
    Yes, it's the last circuit in the post. I'm not too sure how to analyse it without the small signal model
     
  12. Feb 20, 2016 #11
    In the book it says that the increase in signal current in one bjt is exactly matched by a decrease in the other hence the signal voltage at x is zero ?
     
  13. Feb 20, 2016 #12

    LvW

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    But his can`t be true for a common mode signal.
    If the increase in T1 is identical to the decrease in T2 we have a differential signal without any common mode portion. Remember: In post#3 you speak abouta common-mode signal.
     
    Last edited: Feb 20, 2016
  14. Feb 20, 2016 #13
    So analyze the half circuit shown as the second image in your first post using the small signal model. What do you get if you do that?
     
  15. Feb 20, 2016 #14
    It's the same answer that I posted above Ricm = (B+1)*(2*REE//ro)
     
  16. Feb 20, 2016 #15
    Ok, why is it not true for common mode signals ?
     
  17. Feb 20, 2016 #16
    I can't tell you why they excluded RE, but when I solve for the common input Z, I get this (I included re in the model):

    ComZ_1.png

    If I delete the effect of various components, I get:

    ComZ_2.png
     
  18. Feb 21, 2016 #17

    LvW

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    Because common-mode signals will cause the same current changes in both transistors (same direction).
     
  19. Feb 22, 2016 #18
    They also neglected re, since it is much less than the parallel combination of 2REE and ro
     
  20. Feb 22, 2016 #19
    Is the voltage at x measured across REE ?
     
  21. Feb 22, 2016 #20

    LvW

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    Have a look into your diagram how and where Vx is defined.
     
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