CC amplifier configuration.

Discussion in 'General Electronics Chat' started by alphacat, Aug 3, 2009.

  1. alphacat

    Thread Starter Active Member

    Jun 6, 2009
    186
    0
    Hey,
    I ran into this configuratuin of CC amplifier (output is taken from Emitter) and wanted to know what is the advantage of using such configuration?

    The input resistance of this amplifier is approximately RE||RL, which in normal CC configuration, its β times larger: β*(RE||RL).

    Since CC amplifier's gain should be as close to unity as possible, then i dont understand why would one want to use such configuration.

    Thanks.

    [​IMG]
     
    • CC.JPG
      CC.JPG
      File size:
      6.2 KB
      Views:
      128
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Common Collector amplifiers are useful in buffering the signal from a previous amplifier stage (Typical one with fairly high output impedance) so that the load resistance does not appreciably disturb/load the output of the previous amplifier stage.

    hgmjr
     
  3. alphacat

    Thread Starter Active Member

    Jun 6, 2009
    186
    0
    Hello,
    Thank you for your answer.

    In your example, lets call Rout1 the output resistance of the first stage, and Rin2 the input resistance of the second stage - the CC amplifier.

    As you said, Rout1 is usually too large, and we want to decrease it, therefore we use the CC amplifier, which has a pretty low Rout.

    But if Rin2 is low, and Rout1 is large, then the output voltage of the first stage will be degraded pretty badly - the attenuation factor will be Rin2 / (Rin2 + Rout1) .
    So wouldnt you rather Rin2 to be as large as it can be?

    (As i said, this CC configuration causes Rin2 to be β times lower than its value in the natural CC configuration, thats why i was asking what is so special about this configuration that would make us wanna use it).

    (When saying natural CC configuration, i meant this:)
    [​IMG]
     
  4. davebee

    Well-Known Member

    Oct 22, 2008
    539
    46
    Sometimes you have a signal that has plenty of voltage already, but is at too high of an impedance to use directly.

    An example might be pulling a radio signal from a tuned loop antenna - this signal can easily be several volts with no load on the loop, but at a very high impedance.

    If you connect a relatively low-impedance input amplifier to the loop, the load will detune the loop and the signal voltage will drop (plus you'll change the frequency response of the loop), but a high-impedance input amplifier can be used.

    You still might need to amplify the signal past this buffer stage, but the input impedance requirements for the second stage of amplification are much less restrictive.
     
  5. alphacat

    Thread Starter Active Member

    Jun 6, 2009
    186
    0
    Hey Dave,

    So do you agree that the #2 CC circuit which i've uploaded - which has Rin ≈ β * (R_EMITTER || R_LOAD) - is better than the #1 CC circuit - which has Rin ≈ (R_EMITTER || R_LOAD) ?
     
  6. davebee

    Well-Known Member

    Oct 22, 2008
    539
    46
    Sorry, I have a general knowledge of transistor design, but I don't know enough of the engineering details to help with that question. Maybe others can help with this.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    If you want help in understanding this, you need to be consistent in the names of your resistors. The names in your schematics do not match those in your equations.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    The CC amplifier (emitter follower) is a current amplifier not a voltage one.
    If we assume we have the input signal that has 1V and 1K internal resistance. And our load will have RL=600Ω.
    So if we connect the input signal direct to the load, the voltage on the RL will be equal
    UL=1V*(600Ω)/(1kΩ+600Ω)=0.375V
    If we add emitter follower between the input source an the load
    Then resistance that is seen by input source will by (β+1) larger then RL.
    If β=99 then
    RL seen be a input source:
    Rin=100*600Ω=60KΩ so the voltage on a RL will by
    0.983V


    In a real circuit
    [​IMG]
    Rin=Rb || [ (β+1)*( (Re+re) || RL ) ]


    http://forum.allaboutcircuits.com/showpost.php?p=149702&postcount=13
    http://forum.allaboutcircuits.com/showpost.php?p=153702&postcount=8
     
    Last edited: Aug 4, 2009
  9. alphacat

    Thread Starter Active Member

    Jun 6, 2009
    186
    0
    Hey Jony,

    The CC amplifier has a large Rin and low Rout, therefore it would make a very bad current amplifier, but excellent voltage buffer.
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    You could put the transistor in a linear region by making RB very large, and using the input resistance of the base to ground with RB as a voltage divider, but you would be better off having a voltage divider instead of RB.
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Well maybe I should say "current booster" or " impedance transformer".
    But it's true, an emitter follower has a current gain.
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    A minor correction:

    Rin=Rb || [ (β+1)*(re + (Re || RL )) ]
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Iload>>Isource. That's a current amplifier.
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    In this circuit
    [​IMG]

    Ig=181uA and IL=3mA
    Ki=16[A/A]
    So we have the current gain, So this isn't a current amplifier?
     
  15. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    It is. You just offered a concrete example of what I said.
     
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Yes you're right, don't understand what you mean in your answer.
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    I said:
    The symbol ">>" means "much greater than". So, Iload is much greater than Isource.
    You said:
    In other words, IL>>Ig. Which is equivalent to what I said.
     
Loading...