differential pair radiation

Question

For experienced designers probably it is easy question but I am reading "Signal and Power integrity, E. Bogatin" and don't understand one thing.

"The two most common sources of EMI are the conversion of some differential signal into a common signal which eventually gets out on an external-twisted pair cable and ground circuit board generating common currents on external single-ended shielded cables."

My questions are
- why differential lines can be such big problems ? It is good in communication lines because it provides immunity to environment noise (like rs485) but why in opposite side it is bad for EMI ?
- ground common currents on shielded cable - this is about current in the shielding which emmits radiation ? If shielding is connected on both sides the problem doesn't exist ?

Sorry if that's stupid questions but it is my begining in signal integrity and EMI field.

Answer 1

why differential lines can be such big problems ?

That's not what it says. It says "conversion of some differential signal into a common signal" can lead to a problem.

I might word it as "unintentional common-mode signals on conductors (or in circuits) designed for differential signals" that cause the problem. These can in fact be generated from differential signals, for example when the trace lengths of the differential tracks are mismatched.

ground common currents on shielded cable - this is about current in the shielding which emmits radiation ?

The current is common when it flows on both the center conductor and the shield in the same direction.

Return currents in the shield (for singly-shielded cable, like coax) opposite the signal current are expected. Currents in the same direction as the signal current are undesirable and lead to radiation.

If shielding is connected on both sides the problem doesn't exist ?

No, even if shielding is grounded on both sides the shield can form a loop antenna and cause radiation.

It's often preferable not to connect the shield on both sides, because doing so will produce shield currents due to the difference in ground potential at the two ends.