I'm building a dipole antenna for 433 MHz, first one ever. I will have to use a ferrite bead to balance the feed. How many decibels of attenuation should I expect from the ferrite bead?
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2$\begingroup$ Before I edit it, I feel like your question is "how much attenuation is expected" rather than "how much attenuation is needed". Do I read this correctly? $\endgroup$– David Hoelzer ♦Oct 17, 2021 at 14:51
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2 Answers
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If you do it right, the bead(s) shouldn't get hot at all, and shouldn't add much loss.
The impedance of the bead(s) should be about 10 x the impedance of the dipole. The current on the feedline with no balun might be about half the antenna current. With the beads on it it'll be less than 1/10 of the curent, or 1/100 of the power. Small beads (3.5 mm inside, 6 mm outside, 12 mm long) can dissipate about 5 W, but in this case they'd dissipate 50 mW if you transmit 50 W, so they'll stay cool.
Remember to run the feed line perpendicular to the antenna for some distance.
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Different beats have different attenuation at different frequencies. My gut tells me you want at least 6 dB, preferably more. Guys seem to wind their own inductors as RF chokes but not sure how much attenuation at the desired frequencies.
