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question about power dissipation

Discussion in 'Electronic Design' started by Geronimo Stempovski, Feb 7, 2007.

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  1. I just read an interesting paper about high-speed I/O's power dissipation.
    Unfortunately there is an equation I don't quite understand. Maybe someone
    is in the mood for discussing and explaining the correctness of the equation
    to me.
    The formula I am talking about is (1) in the paper [
    http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

    For high-common mode signaling (which standard would that be, anyway? TTL?
    CMOS? SSTL?) it is assumed
    P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

    For low-common mode signaling (LVDS? CML? LVPECL?) it states
    P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

    What I don't understand is the factor 2 (2*Z0) in the calculation of the
    low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
    correct.

    Any help is highly appreciated! Thanks a lot in advance!

    Regards, Gero
     
  2. operator jay

    operator jay Guest

    I don't know the topic or the equation in question, however,

    P = Vpeak/sqrt(2) * Vpeak/sqrt(2) / R = Vpeak^2 / 2 * R
    would give power for a simple AC voltage applied to an R. Here the
    factor of 2 is 'changing' the voltages to their RMS values. Could it
    be similar in the equation you are looking at?
     
  3. Thank you, Jay. Unforutnately not. Could I mail you the paper and you have a
    look at it? What's your email adress? My email adress is valid, by the way.
     
  4. Noway2

    Noway2 Guest

    NOTE: Cross posting linker were removed and this response was posted
    only to sci.electronics.design

    I am not familiar with the terms high-common mode and low-common mode,
    but the methods you refer to in parentheses seem to suggest single sided
    transmission with a common ground return versus differential.

    If this is the case, then the 2X factor could be due to that you have
    differential signaling and hence two conductors with currents of equal
    magnitude and opposite direction flowing in both conductors?
     
  5. That does seem mangled.
    When in doubt, check the dimensions of the answer ?

    Normally where frequency is used in power calcs, it is of the form
    of power dissipation capacitance : W = Fo * Cp * Vcc^2

    -jg
     
  6. Terry Given

    Terry Given Guest

    it looks like its analagous to an RMS calculation, but made on an
    unknown dataset of a particular (hardware) flavour. its a mere proof by
    blatant assertion, so it'd take a fair bit of digging or maths to figure
    out where it came from.

    ya gotta love professional comics.

    Cheers
    Terry
     
  7. Fred Bloggs

    Fred Bloggs Guest

    The background presentation in the paper is somewhat abbreviated because
    the authors are implicitly assuming the reader is participating in an
    ongoing discussion of fundamental limitations of I/O throughput due to
    channel, architectural, and power efficiency considerations. Both power
    equations are of the form V*I where V is DC power supply for the
    individual channel circuit driver and I is the current switched into the
    line termination *at the destination* , making this is a calculation of
    power supplied by the DC supply that is dissipated in the termination at
    the destination. In the case of HCM the channel driver is powered by
    Vdd, making V=Vdd, and the current switched from it into the line is
    I=Vswing/Zo, where it is assumed the line is terminated in a matched
    resistive impedance Zo, so that P=Vdd*Vswing/Zo. In the case of LCM, the
    channel driver is powered by a DC supply of magnitude Vswing and the
    driver is a series terminated circuit, by the MOSFET triode region
    channel resistance, putting Vswing/2 on the Zo termination at the
    destination. Therefore, for LCM, V=Vswing and I=(Vswing/2)/Zo. This may
    seem inconsistent with the HCM equation but it is not when you
    understand that the assumption in both cases is that the receiver
    equalization makes I*Zo=Vrx/H(f). This confusion is just a scale factor
    anyway and does not affect the calculation in section II for minimum
    energy per bit versus data rate gamma*f over channel H(f).
     
  8. Guest

    This is a nice explanation but still a bit puzzling. Especially
    confusing is the fact that
    Vswing has a different definition in the cases of LCM and HCM. Also,
    according to the paper, the signaling power is that dissipated by the
    termination and the lossy transmission line. If Vswing stands for the
    voltage measured at the output pin of the transmitter and the line is
    lossless, wouldn't the power always be Vswing^2/Z0 ?

    Also, in the case of HCM, Vdd is not really the voltage `across the
    channel' so the product V*I is not really the power dissipated by the
    channel+termination. If I have two equal resistances in a series and
    put V volts across both resistors and try to measure the power
    dissipated by the bottom one, it will not be V*V(across R)/R=V^2/2R
    but rather V^2/4R. Why is this not an accurate model of the HCM
    transmitter? Or is the load resistor not considered part of the
    driver? Or (more likely) they just use their expression because it has
    the dimension of power and is related to the actual power dissipated
    at the receiving end?

    Finally, their analysis does not really care about the 2 in the
    denominator (or any other constants for that matter) but does care
    about the power of H(f) there. I can belive the fact that LCM is more
    sensitive to the frequency response since the HCM transmitter is a
    current source and therefore in V*I, the I part does not depend on
    H(f) (as noone looses current along the way in a transmission line).
    Is this the idea of their analysis (greatly simplified, of course)?

    In any case, thanks for the explanation. I would appreciate any
    clarification on the questions above.
    This part is somewhat unclear to me. Regardless of what happens at the
    receiver end, wouldn't it always be that
    I*Z0=Vout and Vrx=H(f)*Vout so I*Z0=Vrx/H(f) (provided the load looks
    resistive and that's essentially what equalization will achieve)?
    Essentially, equalization will just make the reactive component of the
    load `invisible' to the reciever but this is already assumed in all
    the models above, so how does this relate to the factor of 2 in
    equation (2) in the paper?
    Agreed.

    Thanks again

    Alex
     
  9. Fred Bloggs

    Fred Bloggs Guest

    The definition is the same and the difference between HCM and LCM is
    that HCM powers the channel drivers from the logic level power supply,
    Vdd, and LCM powers its drivers from a low voltage power supply on the
    order of Vswing.
    Lossless or not, the transmitter puts Vswing across the line at its
    output, so the power into the line will be Vswing^2/Zo is correct.
    V*I is the generic calculation of the power *supplied* by the circuit DC
    power source. If the line draws current Vswing/Zo from the driver, then
    the DC supply, Vdd, must provide Vswing/Zo amperes to the driver, making
    the power provided by the DC supply Vdd*Vswing/Zo. The paper is
    interested in computing the power supply requirements as well as the
    driver circuit internal dissipation requirements as a function of data rate.

    They break the total power drawn from the DC power supply into two
    components: the power dissipated by the line and its termination; and
    the power dissipated by the driver electronics and whatever internal
    source termination it uses. The receiver Vrx, the line loss H(f), and
    the characteristic impedance Zo, are all that are necessary to compute
    the power required by the line and its termination at a particular
    Nyquist frequency f. The driver internal dissipation is not so
    straightforward and must be computed on a case by case basis as a
    function of CMOS process and circuit topology.

    Yes, it is all very straightforward. Once again:

    for the HCM case:

    the receiver requires Vrx at its input,

    this means the transmitter must place Vrx/H(f) at its side of the line,

    Vrx/H(f) across the line input requires (Vrx/H(f))/Zo amperes,

    the (Vrx/H(f))/Zo amperes supplied by the driver, into the line,
    ultimately is supplied by the DC power supply Vdd to the driver,

    so the DC power supply Vdd is providing Vdd*(Vrx/H(f))/Zo watts to make
    just the line portion of the bit transmission happen.

    The LCM case is similar.

    First of all, the I at the receiver end is not the same as the I at the
    transmitter end. A transmission line does not obey the laws of lumped
    element circuit analysis because it is entails energy transmission by
    wave propagation. In both LCM and HCM, you have (Vrx/H(f))/Zo amperes
    must be injected into the line at the transmitter side. One difference
    is that for HCM, Vswing=Vrx/H(f), and for LCM, Vswing/2=Vrx/H(f). The
    second difference is that for HCM, the DC voltage supply for the circuit
    is Vdd, and for LCM, DC voltage supply for the circuit is Vswing. The
    resulting power required of the DC supply for HCM and LCM are then
    Vdd*Vrx/(H(f)*Zo), and Vswing*Vrx/(H(f)*Zo), respectively. But because
    Vswing/2=Vrx/H(f) in the LCM case, Vswing=2*Vrx/H(f), making the power
    in terms of Vrx, 2*(Vrx/H(f))^2/Zo watts, and not (Vrx/H(f))^2/(2*Zo) as
    they state in the paper.
     
  10. Guest

    Fred,

    Thanks for your explanation. The last paragraph answered almost all
    the questions I had.

    The only remark I have is that this is not the power that goes into
    signaling: some of it is spent on driving circuitry in the driver, say
    the load resistor in the HCM driver but I guess it is a power metric
    as this is the minimal power the power supply must provide for driving
    the line.
    This is certainly true, however, conservation of charge will guarantee
    that the integral of I is the same at both ends. This, with
    transmitter pre-emphasis (that will take care of phase disparity) is
    enough to make the power computation valid. Now, that I have reread
    it, what you said makes perfect sense, except that you used
    equalization instead of pre-emphasis.

    In any case, thanks.

    Alex
     
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