I haven't come across any mentions of signals running in parallel with power traces. Everyt mention thus far has been two signals to each other.

If I have a low amplitude signal (sinusoid < 10khz) (10 mil trace), lets just say its 1V, and its running in parallel (10 mils) with a 5V power trace (30 mils) and there is an ground plane directly below. What happens to my signal ? What happens to my power ?

A few scenarios

  1. Power trace feeds IC's directly
  2. Power trace feeds decoupling caps directly which were chosen as the typical 100nF value.
  3. Power traces feed decoupling caps directly which were chosen to have the largest capacitance possible for its footprint ? (This can be anything you feel is realistic)
  4. Power trace feeds local reservoir caps and each reservoir cap distributes its power for a particular region of the board. All the components within that area have decoupling caps being fed from their local reservoir.

The order of the scenarios should be that would cause the power trace to go from most demanding to least demanding.


Added signal type - its a sinsusoid

  • \$\begingroup\$ You do have decoupling on the board, right? \$\endgroup\$ Oct 31, 2014 at 21:43
  • \$\begingroup\$ only changes in voltage induce changes in nearby traces, I guess very small (sub-millivolt, maybe a few millivolts max) fluctuations will be picked up on the power trace due to the signal running nearby. Depends on the behaviour of your 1V signal though, if it's nice and slow, basically DC, then it's not an issue. Even then, power traces are not really an issue for induced noise, especially if the regulators attached to them have good Power Supply Rejection Ratio (PSRR) \$\endgroup\$
    – KyranF
    Oct 31, 2014 at 21:43
  • \$\begingroup\$ I was actually editing the question before you even asked. Give me a few mins. \$\endgroup\$
    – efox29
    Oct 31, 2014 at 21:44

1 Answer 1


Yes there will always be crosstalk. This is caused by capacitive and inductive coupling of the traces. In most cases this will however not be notable because your signal is large compared to the signal that is coupled in.

Furthermore ripple on your power supply and large dI/dt in your power trace would couse much more noise to be coupled in your signal trace than a clean power trace without ripple.

The reason crosstalk is mostly known of signal traces is because signal traces have fast changes in voltage (high dV/dt) and/or fast changes in current (high dI/dt), but depending on your application this can be the same case for your power trace.

Edit: To answer the added question: Decoupling is one way to solve the problem of high currents trough your power traces, thereby minimising crosstalk to nearby signal traces.

However it is not (always) good to have a higher value as decoupling capacitor. The smaller values (say 100nF) decouple higher frequencies (around 10MHz), whereas the large capacitors decouple the lower frequencies. For your scenarios this would mean that case 3 is in most cases worse than case 2.

  • \$\begingroup\$ i updated the question. Signal is not a digital signal. With a digital signal - who would be the aggressor/victim or its hard to say ? \$\endgroup\$
    – efox29
    Oct 31, 2014 at 21:59
  • \$\begingroup\$ But if you can control your power distribution, then would it matter to decouple any frequency since the caps have the energy to meet the demand and the distance to refill is shortened ? I guess this applies to scenario #5 because of the local bulk reservoir. \$\endgroup\$
    – efox29
    Oct 31, 2014 at 22:01
  • \$\begingroup\$ You can probably see the signal trace as the victim. Say it conains data, or when analog a value, any noise coupled in can/will cause the data to change. The same thing is true for your power supply, but normally you would filter, decouple etc. your power supply to solve this problem. Furthermore the power supply is most likely a large voltage than the signal trace and doesnt contain data... \$\endgroup\$
    – Douwe66
    Oct 31, 2014 at 22:02
  • \$\begingroup\$ Makes sense. So if crosstalk is only present with a di/dt or dv/dt, then if power is smooth, with all the necessary caps to keep it from working too hard, then crosstalk is negligle. Would that be a correct statement ? \$\endgroup\$
    – efox29
    Oct 31, 2014 at 22:06
  • \$\begingroup\$ Yes. (except for the crosstalk from your signal trace to the power trace of course). \$\endgroup\$
    – Douwe66
    Oct 31, 2014 at 22:09

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