bjt - two-port network

Discussion in 'Homework Help' started by incognita, Jun 15, 2010.

  1. incognita

    Thread Starter New Member

    Mar 26, 2009
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    Hi.


    I hope you can clarify 2 doubts I came across when I read this sentence in 6th edition SEDRA and SMITH's book "Microelectronic circuits", in page 359: "From this model we observe that if transistor is used as a two-port network with the input between E and B and the output port between C and B, then the current gain observed is equal to alfa." (see the model image attached, please)

    1) Why don't they consider the INPUT being between C and B and the OUTPUT being E and B? This is confusing me..

    2) Is correct to tell that: the current I(c) is drawn as if it was there a current-source controlled by the emissor current?? i.e. bigger I(e) leaves to bigger I(c)?

    3) Just more one addendum for now: I have read in some electronic websites and some books and it is constantly referring the "large-signal models" and "small-signal models" on transistors. Is it correct to tell that the former is referred to the DC signals? And for small-signal models it is usually (always?) referred to AC signals that we want amplify in the transistor? Mine idea is: if we want to amplify a small AC signal, we first must biasing the signal with a DC component so we can use the linear region and in this way the small AC signal will not saturate.. (i.e. having clips in their peaks..). Is it correct? Please feel free to fix any errors. Thank you.


    Thanks in advance for your attention and patience!



    ps This is not my native language. Sorry for any mistakes...
     
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  2. Georacer

    Moderator

    Nov 25, 2009
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    This is funny: I was examined on this book this very day!
    Now, on the topic at hand:
    1) The transistor is a 3 pin component. You can interact with any one of its pins and measure the changes in any of the remaining two. In this case the transfer function is Ie/Ic=a. Don't be restrained by the direction of the current.
    2)In this very book there are examples where in order to bring the transistor to the active region you use a current source in the emitter to bias it, so, yes, it is totaly feasible but not practical.
    3)Large Signal models are usually used in DC analysis. I say usually because even a large signal model cannot explain accurately what is going on inside a transistor when you factor for many things. But for now, yes, large signal models are for DC. Personally though, as far as DC analysis goes, I leave the transistor component as is and use my equations to figure things out.
    As AC is concerned, small signal approach does use a small signal model, but a small signal model isn't suitable for just any AC signal (it needs to be a small one, as the name suggests).
    When you want to amplify a small signal you always want to bias the transistor, but that bias signal doesn't nessessarily have to come along with the small signal. It can be induced by a positive/negative power supply or a current source as we saw earlier.

    I hope that covers it all.

    P.S. Where are you from?
     
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  3. incognita

    Thread Starter New Member

    Mar 26, 2009
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    Transfer function = Output function / Input function...

    How can a current source biasing the transistor?

    Thanks.





    "P.S. Where are you from?"
    Spain
     
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Let's revise some of the things I said:

    I guess you have certain doubts about the definition of the transfer function. Actually a transfer function if Xo(s)/Xin(s), that is how the output measure (wich can be pretty much anything) is related to the input measure. By working the relations between those two measures we can extract the transfer function.

    As for the current source biasing the transistor, why couldn't it? A bias signal is a voltage or current that will lead in a collector current that causes the transistor to operate in the active region. If it's a voltage source that causes that bias current, or an actual current source, what does it matter?

    It seems Sedra/Smith have a knack in the mediterranean. I' m from Greece. (As you can read above all my posts, duh!)
     
  5. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You do have practical biasing with current sources, usually set up as some variation of a current mirror (they're in the mid chapters of S&S and are introduced briefly in the BJT chapter, I think the MOSFET one as well). It's especially popular in IC design.

    A current source tries to make sure its current is at some given point. If it's in the emitter of a BJT then it will pull down the emitter voltage to get that current.
     
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  6. incognita

    Thread Starter New Member

    Mar 26, 2009
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    @Ghar: thank you a lot!! You was very clear. All my doubts on this issue were vanished! Thanks!

    @Georacer: i know the meaning of transfer function. We use Laplace transform to calculate it for continuous-time systems, and z transform for discrete-time systems.
    I would be more right if I told that transfer function is the ratio of the output Laplace over the input Laplace. (for continuous-time systems)
     
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