Bit confused about return currents on PCB

[FETlife]

I've realised I don't fully understand PCB return current. Lets assume for now that the return current will take the path of least resistance, as it does for DC. Please check the example below using a ground plane, a source, a potentiometer (as potential divider to ground) and a load.

Will the return current follow the yellow path and basically all of it go from one ground connection to the next in sequence, following the path of least resistance between each ground connection? Or is it more like the blue path where some (the majority?) of the return current finds it's way directly to the source with the least resistance but some of it will return to the intermediate ground connections on the way? Or none of the above?

Also, kind of unrelated but with an inverting op amp summing bus with long pcb traces from each source to the virtual earth point, from a noise perspective, is it better to have the sum resistors at the individual sources or all together at the inverting input? (Do we want the long traces to be at the impedance of the summing resistor or do we want the virtual earth to have the long traces?)

Thanks.

 

[JohnRoberts]

I've realised I don't fully understand PCB return current.

Welcome to the club

Lets assume for now that the return current will take the path of least resistance, as it does for DC. Please check the example below using a ground plane, a source, a potentiometer (as potential divider to ground) and a load.

Will the return current follow the yellow path and basically all of it go from one ground connection to the next in sequence, following the path of least resistance between each ground connection? Or is it more like the blue path where some (the majority?) of the return current finds it's way directly to the source with the least resistance but some of it will return to the intermediate ground connections on the way? Or none of the above?

I'm not sure I understand your drawing/question. One mental exercise I use when trying to suss out ground current flows is imagine that every path has a series resistance associated with it. These resistances will define how current flows and even predict voltage drops across the (should be very small) resistances.

Also, kind of unrelated but with an inverting op amp summing bus with long pcb traces from each source to the virtual earth point, from a noise perspective, is it better to have the sum resistors at the individual sources or all together at the inverting input? (Do we want the long traces to be at the impedance of the summing resistor or do we want the virtual earth to have the long traces?)

Thanks.

In an ideal world you want the long traces to be the low impedance signal sources feeding the summing resistors so resistors close to the summing amp with short virtual earth bus.

In the real world that is not always an option.

JR

 

[ruffrecords]

In very basic terms, at dc the current will flow in the path of least resistance. In ac terms it will tend to flow in the path of least inductance, which at higher frequencies tends to be underneath the sending track (if possible). We have discussed ground planes and their properties here several times. Do a search for ground planes and you will find lots of useful info about your question.

Cheers

ian

 

[ccaudle]

Or is it more like the blue path where some (the majority?) of the return current finds it's way directly to the source with the least resistance

That is the case at DC, the return current would flow directly back to the source because that is the shortest path, hence lowest resistance.

For AC signals the return current follows the path of least inductance, so it depends on the frequency content of the signal. There would be a current density distribution across the ground plane, you would have to view it as a 3 dimensional graph, not just a simple line. Each 3d graph would be valid for a particular frequency, and as the frequency increased the current density would begin to concentrate directly under the signal line.

 

[trobbins]

Also don't forget to appreciate that the pot is a load to the source, and the loop current resulting from source and pot and driving signal may be much larger than the current loop that includes pot wiper to its load.

 

[Newmarket]

I've realised I don't fully understand PCB return current. Lets assume for now that the return current will take the path of least resistance, as it does for DC. Please check the example below using a ground plane, a source, a potentiometer (as potential divider to ground) and a load.

Will the return current follow the yellow path and basically all of it go from one ground connection to the next in sequence, following the path of least resistance between each ground connection? Or is it more like the blue path where some (the majority?) of the return current finds it's way directly to the source with the least resistance but some of it will return to the intermediate ground connections on the way? Or none of the above?

In the terms you describe (treat as DC) then the current will split between the paths in (inverse) proportion to the ratio of the path impedances.
Just consider the paths as resistances (see JR above) and Mr Ohm gives you the result.

 

[FETlife]

That is the case at DC, the return current would flow directly back to the source because that is the shortest path, hence lowest resistance.

For AC signals the return current follows the path of least inductance, so it depends on the frequency content of the signal. There would be a current density distribution across the ground plane, you would have to view it as a 3 dimensional graph, not just a simple line. Each 3d graph would be valid for a particular frequency, and as the frequency increased the current density would begin to concentrate directly under the signal line.

So I guess what I don't understand is why the pot isn't a source to the load in this case?

And suppose there were an active gain directly next to the pot, would that change the current path or would current at the load still flow directly back to the original source (far left on the picture).

 

[ruffrecords]

So I guess what I don't understand is why the pot isn't a source to the load in this case?

It is. It does not affect what the current will do.

And suppose there were an active gain directly next to the pot, would that change the current path or would current at the load still flow directly back to the original source (far left on the picture).

The thing to remember is that current originates from and returns to the power supply so you must include those two in your overall current loop.

Cheers

Ian