Help me understand a paragraph from a book (001)

Discussion in 'General Electronics Chat' started by samy555, Sep 10, 2014.

  1. samy555

    Thread Starter Active Member

    May 24, 2010
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    Hi
    From this book:
    [​IMG]
    I read the following in page (2-11):
    [​IMG]
    I did not understand the sentences that are underlined.
    For the first red sentence, I do not understand where the value 7.9 ohm is came from?
    The same for the second blue sentence 43 ohm?
    [​IMG]
    thank you
     
  2. Veracohr

    Well-Known Member

    Jan 3, 2011
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    It's the output resistance of the emitter follower, calculated from the intrinsic emitter resistance (dependent on the emitter current), the emitter resistor, and the 3 resistors on the input of the transistor (including the 200 ohm source resistance) divided by the transistor beta.

    Since it's supposed to drive a 50 ohm circuit, it must be for the purpose of impedance matching (7.9 + 43). I suppose this would be to reduce signal reflection or something, since there's not a lot of power involved.
     
  3. samy555

    Thread Starter Active Member

    May 24, 2010
    116
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    Please show me how the 3 resistors on the input of the transistor (including the 200 ohm source resistance) divided by the transistor beta = 2 ?? (Let beta = 100)
    Thank you very much
     
  4. Veracohr

    Well-Known Member

    Jan 3, 2011
    549
    75
    (200||3.3k||3.3k)/100 = 1.8. Close enough.
     
  5. samy555

    Thread Starter Active Member

    May 24, 2010
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    Thank you
    3.3k||3.3k is ok, but I think that the 200 is in series with them!!! from where I look at the base??

    Why many of basic electronic books dont mention that?
     
  6. Brownout

    Well-Known Member

    Jan 10, 2012
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    Actually, the 200 ohm resistor is in parallel with the 3.3k resistors. A good estimate of resistance is to consider the 200 ohm resistor by itself, since it's resistance is so much lower than the others. The reason books don't mention it is because source resistance is usually so high that it doesn't need to be considered.
     
  7. samy555

    Thread Starter Active Member

    May 24, 2010
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    I can not imagine it in parallel. Can you help me so that I can be convinced it is in parallel?
    It's resistance is not so much lower than the others: 1650/200 = 8.25 . In electronics, to be so much it must be ten times lower!
    200/100 = 2 ohm added to 5.9
    if it is usually so high In this case, it becomes unreasonable ignored. I mean, it's become big to the point that ignoring it makes the calculations inaccurate
    thank you very much
     
  8. Brownout

    Well-Known Member

    Jan 10, 2012
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    In the analysis, set all independent sources to zero. Redraw the circuit, and you'll see why it's in paralle.
    The other two resistors are 3.3k. 3.3k/200 = 16.5

    Source resistance is typically several tens to hundreds of K ohms. But is this case, 200 ohms must be considered.
     
  9. samy555

    Thread Starter Active Member

    May 24, 2010
    116
    3
    Yes, now I understand, thank you

    The other two resistors are (3.3k//3.3K) = 1.65K
    1.65k/ beta = 16.5
    16.5//(200/beta) = 16.5//2 =1.8 ,,,,,,,,,, it is ok
    Let Source resistance (Rs) = 6k
    (3.3k//3.3K//6k) =1.294k
    1.294k/beta = 12.9 ohm, so the output resistance = 12.9 + (re//RE)........ is it true?
    thank you very much
     
  10. Brownout

    Well-Known Member

    Jan 10, 2012
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    Close, it's (12.9 + re) || Re.
     
  11. samy555

    Thread Starter Active Member

    May 24, 2010
    116
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    thank you very much
    [​IMG]
    What about the addition of the 43 ohm resistor to the output, do they actually do that in real circuits?
     
  12. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
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    They do when output impedance needs to match load impedance.

    BTW, in post #9, you wrote:


    But (1.65k/beta)/(200/beta) = 1.65k/200 = 8.25. Further 3.3k || 3.3k || 200 = 178.4. The error in estimating it at 200 is a little over 10%.
     
    Last edited: Sep 11, 2014
    samy555 likes this.
  13. samy555

    Thread Starter Active Member

    May 24, 2010
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    Thank you very much

    Can you please show me how to connect the 43 ohm resistor to the output?
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    Simply put a 43 ohm resistor in series with 100nF output capacitor.
     
  15. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    50 ohm.PNG
     
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  16. samy555

    Thread Starter Active Member

    May 24, 2010
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    Thank you very much
    Still I have a last question:
    Is adding a series 43 ohm resistance is the only solution to accomplish the impedance matching?
    If not, the only solution, does it the best solution ?
     
  17. ISB123

    Well-Known Member

    May 21, 2014
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    No reason to complicate things so just use the resistor.
     
  18. samy555

    Thread Starter Active Member

    May 24, 2010
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    Who said I want to complicate things!!!!
    I just put a question, you have the option to either answer or not answer.
    Thank you
     
  19. ISB123

    Well-Known Member

    May 21, 2014
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    That was proper answer.In practice you would go with something that is simple but at the same time does the job properly.That's why IC exist.
     
  20. samy555

    Thread Starter Active Member

    May 24, 2010
    116
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    Ok thanks
     
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