EMC Question of the Week: March 6, 2023
A voltage source drives a quarter-wave monopole that resonates at 300 MHz. The radiation resistance is 36 Ω. If a ferrite core (R = 80 Ω at 300 MHz) is placed on the monopole, the radiated emissions from the monopole would be reduced by as much as
- 3 dB
- 6 dB
- 10 dB
- 20 dB
Answer
The best answer is “c.” Without the ferrite, the maximum current is equal to the voltage divided by 36 Ω. With the ferrite, the maximum current is equal to the voltage divided by 80 + 36 = 116 Ω. The radiated emissions are proportional to the current, so the ferrite reduces the radiated emissions by (80 + 36)/36 or approximately 10 dB.
At least that is the simple answer (and the best answer of the four options provided). However, there are some important qualifications. Note that any appreciable source impedance or antenna loss would cause the attenuation provided by the ferrite to be less. And the ferrite would have to be located at the point where the current is a maximum (in this case, at the source). Also, if the ferrite had any appreciable inductance at 300 MHz, it could shift the resonant frequency of the monopole reducing emissions at 300 MHz but increasing emissions at the new resonant frequency.
Two of these ferrites would theoretically provide ((2 x 80) + 36)/36 or 15 dB of attenuation. Three ferrites could provide ((3 x 80) + 36)/36 or 18 dB of attenuation. While ferrites cores can reduce radiated emissions at cable resonances, their benefit/cost ratio is generally low. Clamp-on ferrite cores are typically added as a last-minute fix, when it's too late to do anything else. In general, it's much more effective to address the problem at the source and simply avoid driving the cable with a common-mode voltage in the first place.
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