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How to bootstrap

Discussion in 'Electronic Basics' started by amdx, Mar 28, 2013.

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  1. amdx

    amdx Guest

    I see the term bootstrapping, used when the designer wants a high
    impedance and usually low capacitance. I know it involves feedback
    but that's all I know.

    I want to learn enough to build a bootstrapped input with 10Meg/2pf
    impedance. Those are ballpark numbers, the end use would be used
    to measure voltage on a high Q coil and NOT load it. Frequency 100khz up
    to 2Mhz, but 10 Mhz adds more uses. (crystal radio stuff)

    I would prefer a transistor circuit, that way I'll learn something,
    but if there is an obvious IC circuit, I would like to know.
    Mikek
     
  2. Phil Allison

    Phil Allison Guest

    "amdx"

    ** I think you have been told this already - but a single JFET ( wired as
    a source follower ) is ideal.

    BTW: a 2pF cap has an impedance of 8000 ohms at 10MHz.


    .... Phil
     
  3. Phil Allison

    Phil Allison Guest

    "Jim Thompson"
    ** Another totally pedantic WANK.

    " Jim Thompson = Autistic Pig "



    .... Phil
     
  4. amdx

    amdx Guest

    Don't recall being told about the JFET source follower.
    With your reminder, I do recall the circuit I built before
    had 0.3pf. So forget the 2pf, need 0.3pf or less.

    Can you give me incite to this;

    I have a High Q LC circuit, I put my bootstrapped measurement device
    in parallel, the C of the circuit adds to the C I used to
    resonate the L. So does the C of my measurement device load the circuit?
    Or just change the resonant frequency?
    Assuming a high Q C in my measurement circuit.
    Mikek
     
  5. amdx

    amdx Guest

    Now Jim, if I could do that I wouldn't be asking such a question.
    I'll Google delta later.
    Mikek
     
  6. Phil Allison

    Phil Allison Guest

    "amdx"
    ** Don't invent silly specs.

    It's the mark of an utter fuckwit to do that.
    ** FET follower OK.

    No bootstrapping needed.

    ** It will probably do neither.

    Cos you circuit is a fantasy wank anyhow.

    Piss off.
     
  7. Phil Allison

    Phil Allison Guest

    "amdx"


    ** What the smug with delusions of grandeur is attempting to point out is
    that if YOU organise things so that the same voltage appears at both ends
    of a resistor - it passes no current.

    Makes an input bias resistor virtually disappear.



    ..... Phil
     
  8. The easiest way (for me) to make a bootstrap is an opamp unity gain
    buffer the output drives a shield 'capacitance'. But as Phil said the
    jfet follower is a classic. A bootstrap in a High Q LC circuit is
    hard to understand. What are you doing?

    George H.
     
  9. amdx

    amdx Guest

    Not really part of the circuit, it would me an instrument.
    Several things. Measure Q (3db points), signal strength meter
    that doesn't load the coil. As a start.
    Here's the circuit I have built, just curious about a bootstrapped
    input. http://www.crystal-radio.eu/enfetamp.htm


    Mikek
     
  10. amdx

    amdx Guest


    Here's your fantasy,
    http://i395.photobucket.com/albums/pp37/Qmavam/inside.jpg
    I didn't design the circuit. Did layout the pcb. I'm looking for a
    different input circuit. I'm not happy with the mechanical part that I
    put together. I used polystyrene with 3/4" hole. Not good mechanical
    strength and it doesn't like the heat of soldering.

    Mikek
     
  11. Jon Kirwan

    Jon Kirwan Guest


    Okay, mikek. Here's your bog standard bootstrapped
    degenerative amplifier using a single BJT. I'll try and look
    over it according to my poor hobbyist mind. I will assume you
    understand much of it, already. But ask about stuff, where
    I'm wrong about that.

    +V
    |
    +V |
    | \
    | / Rc
    | \
    \ /
    / R1 |
    \ |
    / +----Out
    | C1 |
    | || |
    In----------------||--, |
    | || | |
    | R3 | |/c Q1
    +----/\/\-----+--|
    | |>e
    | |
    | C2 |
    | || +------,
    +---------||-------+ |
    | || | |
    | | \
    | | / Ra
    \ | \
    / R2 \ /
    \ / Re |
    / \ |
    | / |
    | | --- Ca
    | | ---
    gnd | |
    gnd |
    gnd

    (The simpler pieces missing from the above diagram are RS,
    the source resistance, and RL, the load resistance. Ignore
    them for now.)

    As you probably know already, the biasing pair of resistors
    R1 and R2 (in a non-bootstrapped case) are supposed to be
    stiff enough for the task of keeping Q1's bias point from
    moving much. It's common to read suggestions that the current
    through R1+R2 should be about 1/10th of Iq, which is the
    current through Rc in the quiescent state. Making them that
    stiff also means that they load the source (by their shunting
    effect.)

    I liked the description I saw from Phil Hobbs. It's exactly
    how I learned to see this, as well. He said that if you can
    make the swing across a resistor (he said admittance) to be
    identical (or as close as possible to that), that this means
    no current is drawn.

    Look at R3 for a moment. Assume that there is a DC bias
    across R3 providing the necessary Q1 bias current for its
    base. Now imagine keeping that DC bias, but asking yourself
    what would happen if the biasing pair node can be made to
    move up and down (AC wise) in exact lock-step with the base
    of Q1. If that could happen then the bias current would
    remain intact, but there would be no "change" in the current
    with AC changes. So no AC loading. And the DC loading can be
    increased because, although R1 and R2 still have their
    Thevenin equivalent shunting effect, now you can add R3
    straight away to that, so that the DC loading is much lighter
    than before despite a stiff bias pair.

    So more degrees of design freedom.

    What C2 does is to implement that requirement that the
    biasing pair node moves up and down in concert with Q1's
    base. At AC, the emitter is "following" the base (with very
    slightly less than 1 gain, Q1's alpha.) Assume the gain is 1
    for now. If C2 is designed to be a 'short' at AC frequencies
    of interest and if the emitter of Q1 is a good, low impedance
    "source" of these AC changes (it is such a good source), then
    C2 will bypass emitter fluctuations directly to the biasing
    pair and it will do so at low impedance, easily driving the
    biasing node up and down in concert with the base signal.
    What's neat about this is that it uses an existing low
    impedance replica of the input signal that is isolated
    (mostly) from the input by the beta of Q1. And it uses it to
    force the biasing pair node up and down in phase with the AC
    signal. Since it is obvious from the circuit that Q1's base
    itself is moving also up and down by the same amount (almost)
    then it follows that the AC signal on both sides of R3 is the
    same (almost.) So at AC signals, R3 "looks" like an infinite
    impedance. And at DC, R3 greatly reduces the bypass loading.

    Okay. So some analysis. Miller's theorem says that the
    impedance between two nodes can be resolved out into two
    different components: z/(1-k) and z*k/(k-1). In this case, k
    is the voltage gain, k = Av = Vo/Vi, and R3 is z. Since we
    are tapping off of the emitter, not the collector, Vo/Vi is
    nearly 1. For example, if Av=.99 and R3=200k then the
    effective resistance is 20M Ohm.

    Full analysis is, of course, more algebra and work. But I
    thought I'd just get this out there for you to get the main
    point across. I'd use the term bootstrapping anytime this
    particular kind of approach is being taken.

    Jon
     
  12. amdx

    amdx Guest

    Thanks for that Jon,
    I'm at work now, but I'll print it and look it over when I'm home.
    Mikek
     
  13. Jon Kirwan

    Jon Kirwan Guest

    No problem. The basic idea boils down to this:

    If you can keep both sides of R3 moving around in lock step
    then in effect R3 has infinite impedance (from the AC point
    of view) and completely isolates the Q1 base from the divider
    network. So the biasing pair (via R3) gets to perform its DC
    function of biasing Q1 but, at AC, R3 works to decouple the
    Q1 base from the biasing pair.

    Of course, in reality it's not perfect and so the isolation
    is imperfect as well.

    Jon
     
  14. What delta - just a change on the input? If gain is a function of that,
    even with the minus sign, I'd think it was an exponential amplifier rather
    than 'ideal'.
     
  15. Phil Allison

    Phil Allison Guest

    "Phil Hobbs"
    ** One issue with bootstrapping input resistors is that it can dramatically
    increase the noise compared to simply using a large value resistor.

    30 odd years ago I attempted to build a JFET pre-amp for a condenser mic
    capsule and not having any 1Gohm resistors handy tried bootstrapping a
    10Mohm one. The pre-amp tested fine, with an effective input resistance
    close to 1Gohm - ie response was flat across the audio band when driven
    via a 22pF cap simulating the capsule.

    When the capsule was tried, the background noise ( hiss) was about 20dB more
    than with a commercial pre-amp and quite unacceptable for studio work.

    Thing is, with a 1Gohm ( gate bias) resistor, 22pF is enough to shunt nearly
    all the audio frequency ( Johnson ) noise away - not so with 10 Mohms and a
    bunch of positive feedback in place.


    .... Phil
     
  16. Ok I can solve for that, if I treat delta like a constant of, say, 0.01.

    But why do you call it delta? What change does it refer to?
     
  17. Phil Allison

    Phil Allison Guest

    "Tom Del Rosso"
    ** One is reminded of Humpty Dumpty's declaration to Alice:

    " When I use a word ... it means just what I want it to mean, neither more
    nor less"




    .... Phil
     
  18. Phil Allison

    Phil Allison Guest

    "Phil Allison"

    ** Another sort of bootstrapping involves the output stage of an audio power
    amplifier.

    See low budget Germanium output stage typical of the late 1960s:

    http://1.bp.blogspot.com/_7q93wN0Pq1k/TRcwk1PjNZI/AAAAAAAAAA0/BvJcYYW0-FA/s1600/Picture+012.jpg

    The 180ohm collector load of the BFX88 is bootstrapped direct to the
    loudspeaker. Doing this improves both drive and drive linearity - as the
    speaker signal swings to 5 volts below the ground rail.

    Note how if the ( 3 ohm) speaker is disconnected, the OP stage is disabled.

    There is a double dose of bootstrapping going on at the input too.


    ..... Phil
     
  19. Jon Kirwan

    Jon Kirwan Guest

    Very bad wording. What I should have written (and intended to
    write) was:

    "And the DC load impedance as seen by the input increases..."

    I think it's clear if you take the whole context in... but I
    wrote too abruptly and it could have changed the meaning.
    Sorry about that.

    Jon
     
  20. Phil Allison

    Phil Allison Guest

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