I was reviewing the schematic of a motor driver board and noticed that the digital and analog grounds are tied together with a 3.9 ohm 1/4 watt resistor. I am still trying to find the gerber files to see how the ground plains are laid out, but does anyone know why this would be done instead of using a 0-ohm resistor or just connecting them in the copper?

My first thought was that it was an attempt at some sort of noise suppression but this doesn't seem like something that would be very effective at doing that. Feel free to speculate, because for now I'm at a loss.

EDIT: The analog ground feeds back to a 24V supply off the board and the digital ground feeds back to a 5V supply off the board. At each of the power connectors there are several decoupling capacitors. However the resistor is by noneof the connectors and the respective capacitors.

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    \$\begingroup\$ Are there any decoupling capacitors near the resistor? They will form an RC filter with it. Is it a through-hole resistor on an otherwise SMT board? \$\endgroup\$ – pjc50 Mar 14 '14 at 16:20
  • \$\begingroup\$ It is a through hole resistor and most of the other passive components are SMT. There are decoupling capacitors by the power connectors. The digital ground is fed of and returns to a 5V power supply and the analog goes to a 24V supply. However the resistor is relatively far from the decoupling caps. \$\endgroup\$ – Gaddiel Mar 14 '14 at 18:17
  • \$\begingroup\$ Take a peek at fig 7.68 in Horowitz and Hill, that shows high level circuits power separated from a low level circuits power through an RC w/ 10 ohms, and grounds also separated by 10 ohms \$\endgroup\$ – Scott Seidman Mar 14 '14 at 18:57

It could be that a shorted resistor would be equally good as a 3R9 resistor or it may not. There are a few things to consider. Firstly, does the analogue circuit make an external ground connection. If it does and so does the digital side then putting a resistor in to bridge the two planes makes sense because digital currents in the digital plane will be less inclined to take the more arduous route thru the analogue plane. This reduces the likelihood of analogue noise pick up due to digital current: -

enter image description here

If the 3.9 ohm resistor were shorted out there will be a possibility of some digital current from [A] passing onto the analogue ground plane and passing through a sensitive area [B] before returning to the correct grounding point. With a 3R9 present, the digital current will be reduced because it will be more inclined to take the "easier" route to the power grounding point. Because also the resistor is leaded, it will have a significant inductance that might help this. Consider also that the 3R9 resistor may actually be an inductor and that the OP has misread what it is. Easy to do on some inductors.

There could be other reasons but I'm less inclined to speculate because I could be incorrect. Maybe supply pictures and a schematic.

  • \$\begingroup\$ That was my thought too, but to prevent noise pick up a 0-ohm resistor flanked by a capacitor at each end would work just as well or better. The analog ground does have it's own external ground connection, however it does feed back to the same power supply as the connection for the digital ground. Unfortunately the schematic and any images I may have are proprietary so I can't share them. \$\endgroup\$ – Gaddiel Mar 14 '14 at 18:25
  • \$\begingroup\$ @Gaddiel Because of the possibility of digital currents finding their way through a directly connected analogue ground plane back to the power supply, it is better to have a resistor. Capacitors near the resistor would not prevent this and, where would they "ground" to - the ground-plane is the grounding point so they can't really connect to anywhere. \$\endgroup\$ – Andy aka Mar 14 '14 at 18:30
  • \$\begingroup\$ @Gaddiel - are you convinced the 3R9 is a resistor and not an inductor? \$\endgroup\$ – Andy aka Mar 14 '14 at 18:43
  • \$\begingroup\$ I have the bill of materials. If it is not a resistor on the board something is very, very, wrong. But your explanation of preventing the digital current returning through the analog return does make sense. Like I mentioned before, I am still trying to obtain the Gerber data to see what the layout of the grounds actually looks like. They may end up shedding some light. \$\endgroup\$ – Gaddiel Mar 14 '14 at 19:04

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