Using Ferrites

"I recently attended your class on high speed digital design. I found it very useful and have an opportunity to immediately put it to use with a design using an IC that has caused us some grief in the past. My specific question is when using ferrites to filter power to noise sensitive pins, if two or more ferrites are placed parallel and close to each other will this result in crosstalk between them?"

Thanks for your interest in High-Speed Digital Design. In answer to your question, yes, the inductors will interact. Here's how the crosstalk is calculated: (see p. 421 in High-Speed Digital Design) "Mutual Inductance of Two Loops"

Assume we have two beads, labeled A and B. Assume the center conductor in each bead is 0.05" above ground. Assume each the bead is 0.1" long. The exposed loop area of each bead is:

AREA = 0.05 x 0.10 = 0.005 square inches

Assume the centerline spacing between beads is:

r = 0.2 in.

The mutual inductance between the loops is:

LM = 5.08 x (AREA^^2 ) / (r^^3) = 1.5 nH

The mutual inductance at nominal separations, away from the ferrite material, is just a function of the wire geometry and really has little to do with the ferrite material. The ferrite permeability doesn't enter into the approximate equation used above.

The voltage induced in loop B, due to aggressor current flowing in loop A, is:

VB = LM * dIA/dt[1]

Now we need to find the dIA/dt in order to finish off the calculation.

Assume this is a power supply filtering application, and that the filter A is doing a good job.
Let's say circuit A has 300 mV of Vcc noise (VNOISE) on its noisy side, and no Vcc noise on the other. (There is a capacitor to ground on either side of the inductor).

The voltage acrss the inductor is, therefore, just VNOISE. The voltage across the inductor and the current through it are related by this fundamental equation:

VNOISE = LA * dIA/dt[2]

(where LA is the inductance of the bead)

Combining equations [1] and [2] you can see that the ratio of the voltage induced in circuit B to the power supply noise in circuit A is the same as the ratio of the mutual coupling of the beads to the primary inductance of bead A.

This ratio is the coupling coefficient (K) for a weakly coupled transformer system.


Assuming the bead has about 80 nH of inductance, the ratio K, and therefore the percentage crosstalk between circuits, will be:

K = LM/LA = 1.5/80 = 0.019

Therefore, if there is 300 mV of noise being removed by filter A, the loop in filter B will pick up about 0.019*300mV = 5.7 mV of crosstalk.

Note that for short, squat loops, the crosstalk falls off as the **CUBE** of separation. (This is as opposed to the crosstalk between two long, skinny traces for which crosstalk falls off as the **SQUARE** of separation.)

Separating the loops will dramatically improve the crosstalk.

Best regards,
Dr. Howard Johnson