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Op-amps and Transistors

Discussion in 'Electronic Basics' started by zalzon, Sep 7, 2004.

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  1. zalzon

    zalzon Guest

    Without long explainations, is it fair to say :

    1) A transistor uses change in voltage to control a large flow of
    current.

    2a) An op-amp amplifies small differences in voltages across its 2
    inputs.

    2b) An op-amp using negative feedback tries to make the amplified
    output signal more like the input signal.

    2c) The amplified voltage of an op-amp cannot be greater than the
    supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
    to more than -2.5 to +2.5V.


    Is all of the above right?
     
  2. Yes.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  3. The word, 'large' adds no meaning. The rest is pretty iffy, also.
    This is good enough to be useful.
    No. It uses negative feedback to make its two inputs more the same.
    Yes. The output voltage is the result of some sort of voltage divider
    operation across the supply.
    I wouldn't go that far.
    Besides, simple generalities are almost never 'right'.
    But sometimes they are close enough to right to be useful.
     
  4. A bit of a nit pick me thinks.
    Depends on how one reads "more like the input signal". Your actual
    statement here is obviously correct, but I read this as negative
    feedback makes the "shape" of the output more like the shape of the
    input, i.e. feedback reduces distortion errors, which is indeed what
    feedback does. So, without further information on what the poster really
    means, it seems that his description, although crude is ok.
    Well, I though he had a reasonable handle on it.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  5. What if the feedback includes diode junctions, zeners or other
    nonlinear effects?
    With some unstated assumptions, yes.
    I figured he would learn more if I nibbled at the edges of what he
    knows.
     
  6. Bob Myers

    Bob Myers Guest

    Yes, but that's not a sufficient definition for an "operational
    amplifier" specifically. The above definition would apply
    to any amplifier with differential inputs; to qualify as an
    "operational amplifier," there are generally the additional
    requirements of very high open-loop gain, very high input
    impedance, and very low output impedance.
    No. The operational amplifier itself does not do this.
    Many useful circuits USING op-amps DO employ
    negative feedback, but not all.

    This is true of any amplifier; more generically put,
    the output signal swing cannot exceed the limits
    of the "rails" (the positive and negative supplies, or
    the supply and the reference or "ground" point).
    In practice, it can only approach these limits, and the
    output waveform will begin to distort as the limits
    are closely approached.

    Bob M.
     
  7. Bob Myers

    Bob Myers Guest

    Ahem. Methinks you both missed that fact that the
    op-amp ITSELF does not necessarily employ negative
    feedback in this manner AT ALL.

    Bob M.
     
  8. Maybe, but I major that's not what the poster cared about. One is trying
    to answer the question as one believes the answer should be. One has to
    make assumptions as to what the *main* point of the question is.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  9. Ahmmm.... now your gettig a tad complicated....imo...
    I tried to take a somewhat different approach this time, you know, "Yes"
    instead of my usual...

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  10. Bob Myers

    Bob Myers Guest

    Maybe, but if the original poster leaves thinking that
    this thing in the data book called an "op amp" behaves
    as described, I would submit that they are in for a
    rather rude awakening later on.

    Bob M.
     
  11. Again, I disagree. Sure, an op-amp can be used for many things, but the
    poster gave a pretty reasonable statement on what an op-amp *can* be
    used for. The fact that the op-amp has other uses is not really
    relevant. One has to start somewhere. What he stated was indeed,
    essentially correct.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  12. andy

    andy Guest

    The point of op-amp circuits with negative feedback is that if the amp is
    able to bring the difference between its inputs back to zero by changing
    the output voltage, then it will do. Which means that the output voltage
    depends (almost) only on the inputs and the feedback network you're using,
    rather than the circuit that's been used to build the amp.
     
  13. Bob Myers

    Bob Myers Guest

    I think we're still disagreeing over the original poster's
    confusion between "op-amp" and "op-amp-based circuits".
    The statement that an op-amp, alone, with no further qualfication
    or description, uses "negative feedback" to achieve such-and
    -such a result is incorrect on the face of it. From the original
    statements, I could easily see the original poster believing that
    an op-amp, all by itself, can act as a low-gain linear amplifier,
    and that's simply wrong.

    Bob M.
     
  14. Maybe, but I think its more of understanding what the poster thought he
    wrote, rather then what he actually wrote. Most beginners can't
    articulate what they mean correctly.
    I still disagree. By itself, with a reasonable assumption, the statement
    is completly reasonable. There seems to be some implications here on the
    difference between necessary and sufficient, and other like terms. An
    op-amp circuit usually uses negative feedback to achieve such and such a
    result. This is an indisputable fact, so I simply don't have a major
    problem with that statement. The notion that there may be *additional*
    factors as well to *complete* the description is not necessarily
    relevant. Of course, the poster could believe how you interpret it.
    Maybe he/she will renter the discussion.
    Maybe, if, so, the posters views would be incorrect, but I simply don't
    read that as what the poster is meaning, although that may be the case.
    I am filling in the blanks with the assumption that the poster has some
    idea that one connects up other bits and bobs to get actual op-amp
    circuit to work.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  15. Ingvar Esk

    Ingvar Esk Guest

    Depends how to interpret your questions. Litterally or trying to understand
    what you "really mean".

    No. A transistor amplifies current (hFE). But by "overstearing" the
    transistor you can have it act as a current switch.
    In "normal" circuits you provide negative feedback to limit the
    amplification. Often a 100,000+ times amplification is to much.
    Yes, but your conclusion is wrong. If you have 5V supply, the output can
    swing between 0V and 5V (allmost). If you have +/-5V supply the output can
    swing between -5V and 5V.
    Sorry, no.

    /Ingvar
     
  16. No. You are wrong. The orginal claim was *100%* correct. Its not
    debatable.
    Indeed it does as a side effect, however the transistor is absolutely
    and fundamentally a *voltage* controlled and operated device
    (http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html).

    Applying a *voltage* to the base emitter injects carriers into the base
    region. These carriers are then swept up by the collectors accelerating
    voltage. The fact that a few leak away through the base is just a
    nuisance. Ideally, there would be no base current at all.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  17. Ingvar Esk

    Ingvar Esk Guest

    Well, as you state in your paper "Despite much literature that implies other
    wise..." I think it must be debatable, otherwise I can't see that "much
    literature" would imply so :)
    From my point of view. If I feed 1mA into the base the Ic would be some
    hundreds (hFE) mA (if available), even though I know that hFE is not very
    accurate. On the other hand if I supplied 0.1V to the base, nothing much
    will happen at the collector. I can accept that it can be seen as a Voltage
    amplifier, but only around its working point (Vb ~0.6-0.7V).

    Ingvar Esk
     
  18. Its not debatable in the sense that it is accepted by any physicists
    that understands how transistors actually work. Its simply impossible to
    derive and sensible device equations on the assumption that base current
    controls emitter current. The most simple derived equation is:

    Ie = Io.(exp(Vbe/Vt) - 1)

    Its called the diode equation. It is derived/shown in *any* standard
    text book devoted to the device physics of the transistor. Its based on
    applying a potential to the junction. There is no iffs or butts about
    it.

    Unfortunately there are rather a lot of non or semi technical books that
    use explanations equivalent to water down a pipe. The sooner novices get
    to grips with the fact that the ic=ib.hfe model, is a *grossly*
    simplified model, with limited use, the better. It causes never ending
    confusion.
    You are looking at this from a way too naive point of view.

    The transistor is a transconductance device. It outputs a current based
    on its input voltage. It can be accurately exponential over 6 decades.
    Sure, at low voltages, there is only a small current, but this doesn't
    change how the transistor operates.

    Its not a mater of what you want to accept. Its how it is. You can't
    disagree with all the semiconductor physicista in the world, well not
    unless your name is Einstein anyway-)

    What I am explaining is *THE* *standard* *accepted* physics of the
    situation. *Only* those *without* the academic background have
    transistor operation mistaken. Go and have a read of *any* standard text
    on semiconductor physics.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  19. Bob Myers

    Bob Myers Guest

    Sorry - that's one model you can use, and it's often a
    useful one. But with respect to the actual physics of the
    device, it is more accurate to say that slight VOLTAGE
    variations across the base-emitter junction control the
    current "into" the collector. The standard "hybrid pi"
    model of the transistor is best for showing this in simple
    form.


    Bob M.
     
  20. Bob Myers

    Bob Myers Guest

    But you can't "supply 0.1V to the base" - voltages do not
    exist at isolated points, but only with respect to a
    reference. In this case, it is the voltage across the base-
    emitter junction (Vbe - the "v" should normally be written
    lower-case in this context, but I'm afraid that would look
    confusing here) which controls, through the transistor's
    "transconductance" (Gm, again normally a lower-case G
    with a subscript m), the collector current. I am afraid
    you are confusing the DC bias conditions (and the
    resulting DC forward drop across the B-E junction,
    VBE) with the small-signal voltages. With the transistor
    biased in the active mode, and that 0.6-0.7V drop you
    mentioned established, yes, a relatively small *AC*
    signal across this junction DOES control the collector
    current.

    Bob M.
     
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