Of course, the ground return current in the reference plane flows along the path of least impedance right under the signal current, which is the path of least inductance for high frequency currents. This is called proximity effect. Nevertheless, this current isn't perfectly limited, but spreads relevantly in a way that makes it very difficult to be controlled in a 24bit application when using an unsplitted ground plane.

The link http://www.national.com/nationale...ticle.html gives a formula for the proximity effect. I would like to make a rough calculation to show you, why I believe, that a 24bit ADC application shouldn't use an unsplitted ground plane. It's no proof for my point of view, of course, more a broad hint.

The proximity formula is

Irp = I / (pi x h x (1 + d^2/h^2))

where "h" is the distance between a signal copper trace and the reference plane (ground plane), "d" is the horizontal distance from the edge of copper trace (representing the spreading), "I" is the current flowing along the signal copper trace and "Irp", finally, is the current density of ground return current in the reference plane. "h" is about 0.5mm for a four layer board.

When d>>h the above formula reduces to

Irp ~ I / (pi x d^2/h)

Now assume that you have found a proper layout in a 12bit ADC application, where the distance "d1" between an interfering digital signal copper trace and the sensitive analog circuitry has found to show just acceptable results. Means, where reducing "d1" would result in an unacceptable increase of noise.

What distance "d2" had to be chosen then if the ADC was an 24bit ADC?

"Irp" must be 2^12 = 4096 times smaller at that distance:

d1^2 x 4096 = d2^2

resulting in

d2 = 64 x d1

So, if d1 = 10mm was an acceptable distance in a 12bit ADC application, an 24bit ADC would require a distance of d2 = 64cm!

Kai