Electronic – Why does radiated emission of a PCB decrease as the frequency of the signal increases

emcgroundloopspcbpcb-designRF

I was just reading Henry Ott's book "Electromagnetic Compatibility Engineering" and chapter 16.3.3 where he discusses how changing reference planes is detrimental to EMI.

Quoting from the book:

When a signal trace changes from one layer to another, the return current path is interrupted because the return current must also change reference planes

which he explains increases loop area and radiated emissions.

Then I then came across this statement:

At 247 MHz (diamond marker in Fig. 16-9B), the emission is almost 30 dB greater for the case where the signal transitions from the top to bottom layer, versus the case where the signal is routed on a single layer.

That is in perfect accordance with the previous statement, but:

"Above about 2 GHz, the interplane capacitance is sufficient to reduce the impedance of the return path, and hence, the radiation in both cases are about equal"

Figure 16-9 from Electromagnetic Compatibility Engineering

And indeed by observing the graphs, above 2 GHz emissions in Figures 16-9 A,B are approximately the same.

So my question is:

Does designing for frequencies above 2 GHz have less restrictions, in that you don't have to take such issues as changing reference planes into consideration? That would be counterintuitive to what I've considered so far about higher frequency signals and how they are more susceptible to EMI.

Thanks

Best Answer

It is not easier to design for higher frequencies and the confusion you are experiencing is caused by the misconception of thinking that the two solutions compared in the book would be as good at 2 GHz when they are actually as bad.

In other words, traditional signal and power traces are great antennas anyway at those frequencies, but usually the signals spectrum has already faded so low that it is not a problem anymore. This can be easily understood if you think for example the envelope of the spectrum of a trapezoid which is quite common signal waveform. The amplitude decreases 20 db/dec after the frequency of \$\frac{1}{\pi \cdot T}\$, where T is the signal on-time and 40 db/dec after the frequency of \$\frac{1} {\pi \cdot t_r}\$, where \$t_r\$ is the rise time of the signal. Therefore, for example the spectrum of a signal with risetime of 10 ns has a second knee at 32 MHz whichs is almost two decades under the frequency range of interest. Therefore, the signal amplitude has faded to about 1/10000 because of this knee alone at the frequency of 2 GHz.

If we were studying discontinuties in structures designed for signals having considerable spectral content at 2 GHz, it would be clear that the circuit with discontinuties would behave worse.