# About the Series Termination for matching the characteristic impedance

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

1. ### 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
SDRAM?

Thank a lot!

Jack. X

2. ### Andrew HolmeGuest

See Fig. 11 in http://www.analogzone.com/iot_0919.pdf

3. ### BobGuest

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
happen:
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
driver

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.

Bob

4. ### BobGuest

Oops...one 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.

Bob

5. ### X.Y.Guest

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

Jack. X

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