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output impedance of op-amps

Discussion in 'Electronic Basics' started by cruiser, Mar 20, 2007.

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  1. cruiser

    cruiser Guest

    Hi there,

    Which part of the internal circuitry of an op-amp helps achieve the
    requirement for an op-amp to have low output impedance, and how does
    it do that?

  2. Agreed, and do add "negative feedback" to your searches.
  3. Bob Myers

    Bob Myers Guest

    Do thy own homework.

    Bob M.
  4. Phil Allison

    Phil Allison Guest

    "Charles Schuler"

    ** Not part of the "internal circuit " of an op-amp.

    ........ Phil
  5. Phil Allison

    Phil Allison Guest


    ** The use of a "voltage follower" output stage.

    Low output impedance is due to inherent local negative feedback in the such
    voltage followers.

    Most op-amps use complementary ( NPN + PNP ) transistors with output
    supplied from the two emitters.

    ........ Phil
  6. cruiser

    cruiser Guest

    I have done my research - I just don't understand it... What i've
    found is that the low output impedance is designed in the last stage
    (the output voltage amplifier stage) and it has something to do with a
    BJT in emitter follower configuration, which is loading a class AB
    amplifier. While I understand these two things separately, when they
    are put together, along with the other components in the op-amp, I
    can't follow the direction of the current flow and I don't know what
    the transistors are doing.
  7. Phil Allison

    Phil Allison Guest

    ** Hey, Groper.

    Alter your settings so the person and post you are replying to is quoted.

    Posting " in mid air" is very bad etiquette.

    ** See "Emitter Follower Discussion".

    Then see "Feedback in the Emitter Follower".

    Also, post a link ot the op-amp schematic you are looking at.

    ....... Phil
  8. Emitter followers have the lowest possible output impedance of the three
    configurations. Check into that configuration ... emitter follower (or
    common collector). Google!

    Class AB is simply a means to reduce crossover distortion. Don't let that
    confuse you as to output impedance ... they are separate issues.
  9. Eeyore

    Eeyore Guest

    Look at the internals of a TL071. It should be very clear from that. And indeed
    how to set a specific figure.

  10. Eeyore

    Eeyore Guest

    You need to learn something about circuit theory (and practice) then !


  11. Err.. actually...yes..very low, open loop, output impedance is often
    fundamentally due to internal negative feedback, to wit via the usual Miller
    compensation capacitor.

    Typically, in the raw state, the main gain stage would be a very high
    impedance node indeed, often a cascade at top and bottom (ro=100sMeg?). Even
    driving double source followers, the transformation of this impedance to the
    emitters of the buffers, might still be relatively high. However, the Cbc
    capacitor at the high gain stage instead ensures that the driving impedance
    driving the buffer followers is of the order of 1/gm (say 100 ohms) of the
    gain stage transistor at mid band and above.

    relevant formula:

    zo = rs/hfe + re.

    To wit, you don't get the re of the output buffer if the driving source is
    very high impedance, which it would be if no Miller cap. This one of the
    reasons its used it does several things at once, e.g. pole splitting.
  12. Phil Allison

    Phil Allison Guest

    "Kevin Aylward"

    ** But that is not what the poster who used the phrase " negative feedback "
    was alluding to.

    Or he would have said so.

    ........ Phil
  13. I was referring to the original poster text.

    "Which part of the internal circuitry of an op-amp helps achieve the
    requirement for an op-amp to have low output impedance, and how does
    it do that?"

    It is the Millor capacitor that is part of the internal circuitry of an
    op-amp that helpes achieves an op amps low output impedance, in addition to
    emitter followers of course. In rail-to-rail output amps, the Millar cap at
    the output might be the only key feature that results in a low output
  14. jasen

    jasen Guest

    output transistors, but low output impedance is usuall achived using
    external components.

  15. Phil Allison

    Phil Allison Guest

    "Kevin Aylward"

    ** But I clearly was not.

    And you posted your remarks to me.

    Your error.

    ........ Phil
  16. Dr. Polemic

    Dr. Polemic Guest

    Jane, you ignorant slut!
  17. Eeyore

    Eeyore Guest

    How exactly does a Miller cap help output resistance ?


  18. The output resistance of an emitter follower is:

    ro = Zs/hfe + re., Zs is the driving impedance. If Zs is not low, ro is not
    = re

    That is, the driving impedance gets reflected by hfe to the output.

    Consider a high gain cascode stage. It might be say, Va/Ic.(gm.zx) =
    100/1ma*100 = 10meg
    gm.zx being a gain increases of the raw 1/ho of the transistor, and left as
    an exercise for the reader to work out.

    If there is a feedback impedance from collector to base, we have,

    io = vo . hie/(Zf +hie) . gm

    That is, feeding a signal accross CE results in a potential divider driving
    the base, which turns on the collecter current.


    Zs = vo/io = (1 + Zf/(re.hfe)).re

    Once Zf < hie, vo/io tends to 1/gm, i.e it looks like a diode impedance.
    Even before it gets that low, zo is way lower than, the 10meg raw output
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