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About the Series Termination for matching the characteristic impedance

Discussion in 'Electronic Design' started by X.Y., Dec 11, 2006.

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  1. X.Y.

    X.Y. Guest

    Hi, everyone:

    I need to layout a PCB of DSP C6000, it's EMIF which connects it and
    SDRAM is 133MHz, it is suggested that I should set a series resistor of
    33 ohm for matching impedance. However, should I put the match resistor
    more close to the DSP chip or close to the SDRAM? Why?

    In this case, which one is driver/source and which is receiver? DSP or

    Thank a lot!

    Jack. X
  2. Andrew Holme

    Andrew Holme Guest

    See Fig. 11 in
  3. Bob

    Bob Guest

    Another poster gave a link to a pdf describing the theory and application of
    termination techniques. It would be in your best interest to read and
    understand this.

    Here's the short version for so-called 'source termination'.

    First, assume the fastest case where the entire edge of your signal fits
    (timewise) into the transmission line (pcb trace).

    The basic idea is that you want to launch a half-voltage signal into the
    trace. This will happen if your net source impedance (of the driver and
    corresponding series R) is equal to the characteristic impedance of the
    trace. That is, if your driver would generate a signal from zero volts to V
    volts when it's not connected to anything then you will get a zero to V/2
    signal, into the (long) trace, if you drive the trace as stated.

    When this V/2 edge hits the open trace (at the end) the following will
    1) A same phase signal, but double in amplitude (2x as compared with what
    was traveling down the trace), will appear at the end of the trace
    2) An opposite phase (-1x) signal will be reflected back toward the source

    Since a V/2 edge is traveling in the trace (in the forward direction) and it
    will be doubled (2x) at the end, you'll end up with an amplitude of V at the
    end -- which is exactly what you want (for source termination).

    The -1x reflection (in this case -V/2) heading back toward the source will
    be completely absorbed by the source termination (assuming it's close to the
    source driver) and no further reflections (of this particular edge) occur.
    Life is good.

    Please read that pdf and/or other related material so you'll really
    understand this stuff.

  4. Bob

    Bob Guest correction.

    In #2 above, it's a same phase 1x signal that is reflected back toward the
    source driver, not a -1x, so V/2 is reflected back toward the source. A -1x
    is reflected only if it runs into a short circuit.

  5. X.Y.

    X.Y. Guest

    thanks for your help, I will read the pdf file carefully.

    Jack. X
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