Below single layer PCB is a constant current sink circuit (electronic load), it's pure analog supplied with ±12 V and consumes very low current (~50 mA) except the transistor control part (which indicated with red lines) that's going to carry 10A max.
There's two op amps in the circuit, one configured as differential (measuring the current) and other as integrator (driving the darlington pair), both are dual supply with decoupling capacitors close to the IC.
On the left is a fan drive circuit that is being fed from the same transformer but with another bridge rectifier (24 V), there's an small RC filter on the output of both bridge rectifiers to avoid the fan noise coming to the right circuit.
Each bridge have their own linear voltage regulator, the space between two grounds is 1.5mm.
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Do I have any ground loops on the right circuit and how do I avoid/eliminate them?
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Is the separate bridge rectifier + RC filters + linear regulator ripple rejection enough to eliminate the fan ~70 Hz noise showing up in the right circuit?
Best Answer
This probably isn't the right question to ask, since you would need a very high magnetic field (one that isn't achievable on a PCB with a loop antenna diameter in the cm range and a very low resistance of the ground plane).
What you do have to worry about is common mode noise from resistance of the ground plane as current returns back to the source through the ground plane.
The a good designer knows that ground isn't really ground, on a schematic we think of it as zero, but ground is never truly zero because every material has resistance.
I've drawn return currents on the picture below. Any pin with a ground that has current returning to ground (from a high voltage) will flow back on the ground plane to the GND IN point. The current doesn't follow one path because each path has resistance, but it spreads out and starts flowing to ground.
The resistance can be estimated, there are calculators out there but to give an idea for what the resistance is I'll estimate it. For the two orange boxes I've drawn on the left in the image below if the copper plane is 0.5oz then the resistance for a 3"x 0.15" square of copper would be about 5-7mΩ. For the single box on the right would also be about 5-7mΩ. The rest of the copper will also add some resistance maybe on the order of 10mΩ.
So lets see what 10mΩ does to a signal
At 10mA it would be 10mA*10mΩ= 0.1mV
At 100mA it would be 1mV
and at 1A it would be 10mV
If the current were a switching load, then many of the pins (especially the purple ones in the back) would see the switching from other currents would be seen by many of the loads on the ground plane. The shift in the ground also could also mean a shift in the analog signal. (It gets really complicated from here because if all of the analog signal chain sees the same shift and it is the same for all then it may not be a problem, however it would take more time and effort to simulate and calculate the system with all parasitics then to actually build one so in many cases its better to just build the PCB and test it )
0.1mV or 1mV of shift in the ground may not be acceptable for many analog applications if it's not acceptable for your application, then you'll need a ground plane redesign. Most digital applications will be tolerant of ground shift into 10mV range (but starts to become a problem at 50mV)
A good ground plane design makes the plane as continuous as possible. Also multiple vias are also helpful to keep common mode noise low. Avoid the orange box areas, maybe consider moving to a 4 layer design (the cost has come way down). A good reason to use a 4 layer design is then you don't have to worry about ground plane noise, because a continous ground plane is used, and that is as good as it gets.
The fan current should not cross the analog electronics, if the fan ground were one of the purple circled vias, then the current would create common mode noise for all of the circuits. So use proper placement of the fan to avoid problems. Make sure the linear regulators are placed near the analog electronics so they see the same ground voltage as the analog electronics.