Connect with us

The secrets of shot noise

Discussion in 'Electronic Basics' started by Christian Rausch, Aug 30, 2004.

Scroll to continue with content
  1. Help!

    Shot noise, as stated in Horowitz&Hill, ch7.11, p.432, shows a noise current

    Inoise(rms) = sqrt(2*q*Idc*B)
    with q:electron charge, B:bandwidth and Idc:the DC current,

    but this formula "assumes that the charge carriers making up the current act
    independently. This is indeed the case for charges crossing a barrier, as
    for example the current in a junction diode...but is not true for the
    important case of metallic conductors, where there are long-range
    correlations between charge carriers..."

    My questions are the following, and I hope that there is somebody out there
    who can answer them:

    Assume we have a battery with a resistor connected between the poles. Is
    there a shot noise current flowing in the resistor? Is the battery a
    potential barrier like the one mentioned by H&H?

    What about a (charged) capacitor discharging into a resistor? Shot noise
    current flowing or not? The dielectric between the plates seems to be quite
    a huge barrier, but there are no charge carriers crossing it; only a
    displacement current is flowing?

    Assume a battery which is shorted for AC via a large capacitor, that means
    that any shot noise current that the battery may produce runs through the
    capacitor and not elsewhere. Across it is a resistor. Does it really depend
    on the composition of this resistor (metallic conductor, i.e. metal film vs.
    metal oxide, thick film etc.) if there is a shot noise current flowing or

    At the bottom of the first column of p.432, H&H mention that the standard
    transistor current source runs quieter than shot-noise-limited. Anybody out
    there who knows a little more (literature, math) about this?

    Thanks for any advice!

  2. Fred Chen

    Fred Chen Guest

    I have been studying shot noise characteristics in transistors myself.
    Also came across the cited passage above.
    It depends on the resistance, but a metallic wire resistor falls in
    line with the example of Horowitz and Hill. The way to think of it
    here is the electric field inside the metal wire is not large compared
    to the interactions between the electrons. These interactions
    therefore correlate the electrons fairly significantly, so there is
    negligible shot noise.

    If the wire were a semi-insulator, the correlations would not be so
    strong compared to the electric field responsible for drift current.

    Here is a reference to study further:
    It's the same as above. We're considering the shot noise of the
    current in the current-flowing parts.
    So your current consists of electrons going thru the resistor. As in
    #1, if the resistance is high enough, the field inside will make the
    correlations between the electrons relatively negligible. This would
    allow a shot noise description.
    According to the passage, the feedback in the current source circuit
    helps to quiet the shot noise. The base junction is usually so thin it
    is effectively ignored, because the base current is negligibly small,
    even though a shot description is appropriate there. The
    collector-emitter current path is considered almost like a single wire
    (current~constant), so probably in that sense, the shot noise is
  3. No, and I believe that AofE (H&H) states this somewhere. Resistors do have
    several other sources of noise, one of which operates without regard to the
    fabrication itself and is tied directly to the resistance itself, Johnson noise.
    AofE talks about this, too.
    I can't say there is no noise process in battery chemistry. But I don't believe
    that any noise process in battery chemistry is similar to that across a PN
    junction. If there is, I don't think it wouldn't be described by this same
    equation. In the case of the PN junction, it's a statistical process where
    electrons move from valence into the conduction band (and back) and there is a
    small 'step' to it, in the neighborhood of 1+ eV, or so. Perhaps someone can be
    more 'sure' about this or correct me.

    My 'understanding' of shot noise is that it is always associated with cases
    where discrete quanta of charge (e) or quanta of energy (hv) are emitted
    randomly in time by some source. The 'independence' part that AofE talks about
    here is that each of these individual, elemental 'emitters' do so (roughly)
    independently of each other. If there's a large number of these emitters and
    there is a very small probability of emission then the overall emission rate is
    (or, at least, theoretically should be) Poisson distributed. It's this that is
    talked about when saying 'shot noise,' I think.
    The charge on the capacitor has an energy distribution that has to obey
    Boltzmann statistics for the probability density.

    There is some noise associated with capacitance, kT/C, based on Boltzman
    statistics. In fact, from this you can derive the Johnson noise equation for
    resistance. In Boltzmann's equipartition theorem, the mean internal energy
    associated with each degree of freedom in a system is (1/2)kT (Joules.) In the
    case of electrostatic energy in a capacitor, this becomes:

    (1/2) C V^2 = (1/2) k T

    or, the geometric mean of the V^2's is:

    V^2 = kT/C

    Or, another way... The energy distribution [E=(1/2)*C*V^2] must obey Boltmann
    statistics for the probability density, such that:

    p(E) = (1/(k*T))*e^(-E/(k*T))

    Then, the probability density of V^2 is:

    p(V^2) = p(E) * dE/d(V^2)

    Since dE/d(V^2) of [E=(1/2)*C*V^2] is just (1/2)C, we can substitute for E and
    for dE/d(V^2) to get:

    p(V^2) = (C/(2kT))*e^(-CV^2/(2kT))

    By this, you can arrive at the geometrical mean of <V^2> must be 0 or kT/C.

    Or it can be taken from the point of view of a statistical thermodynamics
    theorem from Einstein, which is that "every state function A that can be
    expressed in terms of the system entropy S has a mean value <A> that is obtained
    by finding the maximum of S, solving for dS/dA=0 (well, often the partials
    instead of the derivatives.) By solving this, we also find that the variance is
    again, kT/C.

    In the case of an RC discharge, the noise is just Johnson and not shot. In
    fact, it's the kT/C realization that generates the formula for Johnson noise, as
    any stray capacitance C that is inevitably present across an R, gives a cutoff
    frequency that is B=1/(2*PI*RC) and the voltage's spectral density must be:

    Integral(Sv(f) df) = kT/C

    substituting Sv(f) with Sv0/(1+(f/B)^2), where Sv0 is the DC value of spectral
    density, we can then combine to yield:

    kT/C = 2*Integ(Sv0/[1+(f/B)^2] df) = 2*Sv0*|atn(f/B)| at 0,B
    or, kT/C = 2*(Sv0/(2*PI*RC))*(PI/4)

    Solving for Sv0, we find that Sv0 = 4kTR!
    assumes a fact not in evidence -- is there shot noise in batteries?
    The composition of resistors does have an impact on 1/f noise (which is
    sometimes called 'shot noise' in SPICE simulators, I gather), but the shot noise
    I've been talking about is white, not 1/f. White shot noise doesn't depend on
    the composition. But I'm not sure of your question.
    Well, perhaps Win can answer what he meant here (more likely if you ask in, I think.) But I'd imagine as a hobbyist-guess that the
    "independence" mentioned earlier isn't entirely true for this case -- some of
    the current is dependent in some way or the probability of emission is high, so
    the integral over all the behavior is no longer quite Poisson.
    Wish I could have done better. But that's all I can do without putting effort

    Might want to ask these questions in I'm no expert, but
    there are some over there.

  4. Thanks, Fred and Jon,

    your answers were very enlightening.
    Jon, I will follow your suggestion and ask my current-source-question in

  5. Fred Chen

    Fred Chen Guest

    Another shot noise came to mind in the last couple of days. This is
    associated with charging of capacitors. Specifically, this is Poisson
    noise from electrons randomly arriving at the capacitor as it charges
    up. This is different from the formal shot noise in the other
    discussion but has the same origin in the charge discreteness. It is
    essentially counting (sqrt(N)) noise.
  6. Roy McCammon

    Roy McCammon Guest

    It might seem that way, but there is no abrupt change
    from the wire to the capacitor plate. The influence
    of the next electron on the charges already on the
    capacitor and the influence of those charges on the
    next electron are felt all along the way. However,
    if the electrons are being shot through a nonconductor
    like air or vacuum, then you might well hear some shot
  7. Fred Chen

    Fred Chen Guest

    I am just wondering aloud. My thinking was that the electrons will
    have interactions but they also scatter along the way to the
    capacitor. This is what makes the exact arrival times random. Even if
    the flux or field varies as a well-known function of time, that
    describes the average electron behavior rather than the individual
    motion. The well-known behavior j(t) or E(t) or V(t) is independent of
    the scattering events. Even with feedback (thru heating or electric
    field) to V or E or j, you have to consider that feedback a
    macroscopic, average phenomenon, rather than describing the actual
    counting of electrons passing by. Q(t), charge on capacitor, is also
    macroscopic and well behaved, e.g., in an RC circuit, but there is
    also a microscopic level. That is where I am pondering.
  8. Jonathan Kirwan wrote...
    It's simple enough, but a rather powerful result nonetheless. Ohms law says
    that current is voltage divided by the resistance, so current-source noise
    density is given by ac voltage-noise density divided by the ac resistance,
    namely i_n = (e_n + sqrt(4kTR)) / (R + r_e). Manipulating equations on AoE
    page 436, we see that e_n = (4kT r_e/2)^1/2 (ignoring r_bb). So if the dc
    voltage across the emitter resistor, Ie*R, is greater than 50 to 100mV, so
    I*R >> kT/qI, then the current-source noise density is largely determined
    by the bias resistor's Johnson noise density, i_n = sgrt(4kT/R), and not
    the transistor's shot noise. This can be used to create a nearly-perfect
    quiet current source, using a moderate to high bias voltage (even 100 to
    500V), regulated from a modestly-quiet voltage source.
  9. Roy McCammon

    Roy McCammon Guest

    Their "arrival" times may be random, but their effects
    do not arrive all at once. The "arrival" may be discrete,
    but the effect is not discrete.
  10. Phil Hobbs

    Phil Hobbs Guest

    The shot noise gets suppressed by 6 dB when the emitter resistor drops
    25 mV, so using a 100V supply ought to suppress it by 4000 times (72
    dB). Base current shot noise will be a problem, though, so there isn't
    much point in making |Vbias| more than a few times beta*kT/e.
    Darlingtons and FETs don't help much with this.

    Another point is that for situations like current mirrors or current
    dividers, where you want to accommodate a wide current range without
    adding full shot noise, you can use diode-connected transistors as
    emitter degeneration. The shot noise contributions add in RMS, whereas
    the transconductance goes down linearly, so N diode-connected
    transistors suppresses the added shot noise by 10*log(N) dB. I needed
    to do this once for a variation of the laser noise canceller that
    suppresses the photocurrent excess noise by an additional 2 dB over the
    regular noise canceller.


    Phil Hobbs
  11. Jim Thompson

    Jim Thompson Guest

    It can also been shown that current mirror accuracy dramatically
    improves as the voltage drop greatly exceeds kT/q (compensates offset
    mismatch between transistors).

    ...Jim Thompson
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day