It is quite common to read posts on various "Part 15" boards about the unlicensed use of FM transmitters having 25 mW output power (and more) with 1/2-wave dipole antennas. But if compliance with FCC Part 15.239 is needed/desired, then such configurations are unlikely to provide it. Here's why.
The FCC limit in Part 15.239 is a 250 µV/m field measured 3 meters in any direction from the transmit antenna. Physics shows that a 1/2-wave dipole will radiate that peak legal field when a matched power of only 11.43 nanowatts is applied to its feedpoint.
A matched system using a transmitter delivering 1 milliwatt or more at the far end of a coax cable connected to a 1/2-wave dipole will radiate a peak field that is at least 295 X greater than permitted by FCC 15.239. Here is the math:
SQRT(1 mW / 0.00001143 mW) = 295.78...
The reason that the square root applies is because field intensity varies as the square root of the power.
The loss in 100 feet of RG-6 coax cable is about 2 dB at 108 MHz, when the cable is properly terminated. For a matched system this means that if 25 mW is applied by the transmitter to one end of 100 feet of that coax, about 15.8 milliwatts is delivered to the other end.
That 15.8 milliwatts produces a peak field 3 meters from a 1/2-wave dipole that is 1,175 times greater than permitted by FCC 15.239.
With such performance from unlicensed systems in the 88-108 MHz band in the US, FCC action is not an impossibility -- and it would be easy for them to detect non-compliance.
Of course those who operate where they aren't subject to Part 15.239 have nothing to worry about from the FCC.
The FCC limit in Part 15.239 is a 250 µV/m field measured 3 meters in any direction from the transmit antenna. Physics shows that a 1/2-wave dipole will radiate that peak legal field when a matched power of only 11.43 nanowatts is applied to its feedpoint.
A matched system using a transmitter delivering 1 milliwatt or more at the far end of a coax cable connected to a 1/2-wave dipole will radiate a peak field that is at least 295 X greater than permitted by FCC 15.239. Here is the math:
SQRT(1 mW / 0.00001143 mW) = 295.78...
The reason that the square root applies is because field intensity varies as the square root of the power.
The loss in 100 feet of RG-6 coax cable is about 2 dB at 108 MHz, when the cable is properly terminated. For a matched system this means that if 25 mW is applied by the transmitter to one end of 100 feet of that coax, about 15.8 milliwatts is delivered to the other end.
That 15.8 milliwatts produces a peak field 3 meters from a 1/2-wave dipole that is 1,175 times greater than permitted by FCC 15.239.
With such performance from unlicensed systems in the 88-108 MHz band in the US, FCC action is not an impossibility -- and it would be easy for them to detect non-compliance.
Of course those who operate where they aren't subject to Part 15.239 have nothing to worry about from the FCC.