EMC Question of the Week: June 28, 2021
At a given frequency between 30 MHz and 1 GHz, what is the lowest common-mode current on a cable attached to a table-top product that is capable of causing an FCC or CISPR Class B radiated emissions failure?
- 5 A
- 5 mA
- 5 μA
- 0.5 μA
Answer
The best answer is “c.” In any given frequency band, a product needs to radiate about 1 nW of power to exceed the FCC or CISPR Class B radiated emission limit. The worst-case radiation resistance for a resonant monopole or dipole antenna over a ground plane is typically between 36 and 40 ohms. This corresponds to an rms current of approximately 5 μA. If the maximum common-mode current on a cable is less than 5 μA, then the cable is very unlikely to be the source of a radiated emissions failure.
On the other hand, if the common-mode current on a cable is orders of magnitude higher than 5 μA, emissions from the cable an easily cause radiated emissions failures.
It's interesting to note that the critical current is not proportional to frequency or cable length between 30 MHz and 1 GHz. It's only necessary that the cable be resonant at the frequency of interest.
A similar question that occasionally appears in EMC texts and application notes is, "How much current on a 1-meter cable is required to exceed the FCC or CISPR Class B radiated emission limit at 30 MHz?" Since the cable is electrically short at 30 MHz, we can use the equation for radiated emissions from an electrically-short wire segment to show that the answer is about 8 μA. Applying the same equation at higher frequencies suggests that even much smaller currents can cause failures. However this is misleading, because the current distribution will not be uniform at higher frequencies. Ultimately, if the highest current on the cable is less than 5 μA, radiated emissions from the cable will not exceed the FCC or CISPR Class B radiated emissions limit.
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