This is a PCB with few stepper drivers.

I have a question about the ground plane. I have this idea to split the ground plane for each driver, so, I force the return path to pass through the capacitor.

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What is your opinion about this? the cut may create some noise, and it will affect the thermal dissipation.

thank you!

  • 2
    \$\begingroup\$ sorry, not sure what you have in mind; can you draw in a schematic that shows what these capacitors connect? Why do you want to enforce that? If you want to enforce that, why not put the capacitor at the motor connector instead of somewhere else? Really confused. \$\endgroup\$ Commented Nov 15, 2018 at 11:14
  • \$\begingroup\$ the current goes to the driver... and all the datasheets recommend the capacitor on the driver input. \$\endgroup\$ Commented Nov 16, 2018 at 12:44
  • \$\begingroup\$ Not getting any clearer sorry. Please draw a schematic. \$\endgroup\$ Commented Nov 16, 2018 at 12:45
  • \$\begingroup\$ this is a board with 3 stepper motor drivers , and I would like to optimize the ground plane for all the powe zone of the board. \$\endgroup\$ Commented Nov 16, 2018 at 12:46
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    \$\begingroup\$ done! i will never make al living with my drawing skills \$\endgroup\$ Commented Nov 16, 2018 at 12:56

2 Answers 2


I would prefer the solid/whole board for two reasons:

  1. The motor connections will generate common-mode emissions. Better to treat the whole board as one solid plane for EMI purposes. That is: expect similar noise levels (to what's going to the motors) at the inputs or comm ports or whatever else is connecting to this board.
  2. Better thermal performance.

Consider filtering on the motor outputs, even very basic (e.g. ferrite bead or CMC), to help reduce emissions. Depending on severity and requirements, this can be both a regulatory and functional problem.

For more in-depth filtering, preferably treat each output terminal as its own inverter (half-bridge) filtered with respect to ground, so, an LC filter to GND from each pin. Note the C needs to be dampened, because the motor does not present a terminating resistance for the filter (it's inductive). To solve this, we typically use an Rs+Cs in parallel with the C, with Cs > 2.5 C and Rs = \$\sqrt{L/C}\$.


consider this


simulate this circuit – Schematic created using CircuitLab

The bottom circuit get the motor currents from the global 1,000uf (1mF); the magnetic fields are enormous, and ringing is uncontrolled. The large loop areas will store lots of energy, and switching will be slow because the stored energy has to build up and then later dissipated. Because of the large loop area, the field will decay as 1/Distance until you are far away.

The top circuit, with 100uF right at the pins of the Power Driver, has minimal loop area and thus minimal stored energy, thus fastest switching and lowest power dissipation (coolest operation). The 1uH passes very little high frequency to the Global 1,000uF cap, thus there is little radiation of high frequency energy and easiest task in passing ElectroMagnetic Compatibility tests; the 1 Ohm dampens the L+C (1uH and 100uF), tho 0.1 ohm may be more optimal (Rdamp = sqrt(L / C)). Notice the use of Twisted Pair wiring to the motor. You could also used Twisted Pair from the Global 1,000uF to the local 1uH and 100uF. Because of the small loop areas (100 uF to the Power Driver, the magnetic fields will be smaller and will attenuate faster.


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