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In some cases, we need to length match several different nets between two ICs, based on the specifications of the driving or receiving IC. Sometimes, some of these nets have a series termination resistor. Is there a rule/guideline that helps to account for the series resistor, when determining the length of the routed net?

To help describe with a "picture":

                       10mm
netA:   SrcVia---------------------------RcvVia

                 4mm                      6mm
netB:   SrcVia---------Via [R33] Via-------------RcvVia

I want to length match netA and netB. NetA is 10mm. Should netB considered 10mm, or should it account for the physical package of the resistor (0201, 0402, 0603, etc) ?

Also, notice that, the signal in netB also has 2 additional vias on the path in order to go to the top or bottom layer, to place the resistor. Those 2 vias should also contribute to the effective "length", I think.

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    \$\begingroup\$ Why aren't all the matched-length nets terminated the same way? \$\endgroup\$
    – Dave Tweed
    Dec 12, 2012 at 16:32
  • \$\begingroup\$ Because some nets are bi-directional and terminated within the IC. As an example, consider a DDR2 chip, which has an ODT (on-die-termination) feature. The data signals can be driven by both a memory controller, and the DDR2 chip, so they are bi-directional. The termination for these signals are done on-chip. However, address lines of a DDR2 chip do not have the ODT feature, so they must be explicitly terminated. \$\endgroup\$
    – SomethingBetter
    Dec 12, 2012 at 16:41
  • \$\begingroup\$ But it would be unusual to require that the clock/command/address signals (unidirectional) to be length-matched to the data/dqs signals (bidirectional) in a DDR application. Usually, only the signals within each of the two groups are length-matched to each other. \$\endgroup\$
    – Dave Tweed
    Dec 12, 2012 at 16:58
  • \$\begingroup\$ The rules state that, DQ, DM and DQS need to be length matched within a byte-group. This is the tightest requirement. Then, each byte should be length matched, with a less-strict requirement. Then, there is a CLK to DQS requirement, followed by a CLK to ADDR requirement. So in effect, they are all length matched, with a varying degree of strictness. \$\endgroup\$
    – SomethingBetter
    Dec 12, 2012 at 17:09

2 Answers 2

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The answer is simple: Use the same termination on both signals.

If you are length matching, then timing is important.

Termination can effect rise/fall times which is essentially a timing variation. Since we have established that timing is important, the easiest way to keep timing on the two signals the same is to make the two signals the same. This means that they have the same termination, the same trace lengths, the same-ish signal routing, same trace impedance, etc.

It is possible to have termination on one signal and no termination on the other and still get the same propagation delay. But to do so is super complex. You would have to model the driver, the receiver, and everything in between. You'd look at edge rates and logic voltage thresholds. You'd look at hysteresis, and pcb trace capacitance, etc. But in the end, a seasoned professional would just terminate the two signals the same way and save a weeks worth of time.

Another thing to consider: If timing is important, then the signal speed is high. If the signal speed is high, then why would you not terminate both signals? The point is, you probably need to terminate the other signal anyway.

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  • \$\begingroup\$ Thanks David, please check my response to @DaveTweed's comment in the original post. That's the reason why the two different lines have different terminations. \$\endgroup\$
    – SomethingBetter
    Dec 12, 2012 at 16:43
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Should netB considered 10mm, or should it account for the physical package of the resistor (0201, 0402, 0603, etc) ?

Yes. However it is more important to put the resistor near the driving end. Since the resistor will be a change in resitance vs. the trace, placing the resistor near the driver lets any reflections die out quickly.

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    \$\begingroup\$ We try to do that, but it is often impossible when driving 16-lines from a 0.8mm pitch BGA. We have to get the lines out from under the BGA and only then place the resistors. \$\endgroup\$
    – SomethingBetter
    Dec 12, 2012 at 17:10

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