I'm buying FM transmitter 0.2 watt at Lazada to broadcast for my village community at my area. Not a normal small FM transmitter for car or portable low watt (about 10mW-100mW) which broadcast up to 10 metres. Is it possible to broadcast up to 500 metres? Because I don't want to get trouble to broadcast and get caught. I want to broadcast for my community only and play my own music without any ads (like what commercial radio does). P.S. I'm from Malaysia and planned to broadcast up to 500 metres. Not long range like what commercial radio and government radio does.
OK lets provide a line-of-sight non-multi-path (no reflected energy that may cause peaking and nulling, as airplanes may do, or large drones may cause) link analysis.
you want good FM audio, so at least 20dB SNR at the receivers
gain of 1 (0dB) receiver antennas ( easy to get 5 or 10 dB more gain, if needed by the more distant homes, with fancy antennas)
bandwidth of 200,000 Hertz in the receivers
losses between antenna and the first receiver amplifier: 0dB; there will be some, but a bit of PI matching will compensate
noise figure (thermal noise) of the first receiver antenna: 3dB (should be real easy to achieve at 100MHz)
Now compute the required energy out of the receiver antenna
-174dBm / Hertz bandwidth
+53 dB for the 200,000Hz bandwidth
+20 dB for the signal-noise-ratio for clean music
+0 dB for losses/matching-up-transform(PI)
+3 dB Noise Figure of the first amplifier
-174 +53 +20 +0 +3 = -174 + 76 = -98 dBm <<< Minimum Received Power
Now for the PathLoss for a line-of-sight path, no rain attenuation, no multipathing, good antenna alignment
At 100MHz (wavelength = 3 meters), and a 300 meter path, with the energy spreading out into a hemisphere with 4*pi*(range)^2 surface area, we use this formula
PathLoss = +22dB + 10 * log10[(Distance/wavelength)^2]
PathLoss = +22dB + 20 * log10(Distance/wavelength)
and for Distance = 300 meters , and wavelength = 3 meters, we have
PathLoss = +22dB + 20*log10(300/3) = +22 + 20*2 = 62dB PathLoss
What must the Transmitter power be?
Simply add the PathLoss to the Minimum Received Power
+62 dB + (-98 dBm) = -36 dBm <<< Minumum Transmitted power if no rain nor multipath nor antenna bad alignment
Helpful numbers: 0dBm is 1 milliWatt. If -36dBm, you'd have 3 factors of 0.1x (each of 10dB), and 2 factors 0.5x (each of 3dB), computed like this
0.1 * 0.1 * 0.1 * 0.5 * 0.5 = 0.001 * 0.25 = 0.00025x
which is 1milliWatt * 0.00025 = 0.25 microwatt.
If you run a suitable vertical antenna up a tree, not against the tree trunk, but hanging from a limb with coax running from the TX circuit to the antenna, and feed that antenna with 1 milliWatt (about 0.223 volt RMS or 0.632 volt PeakPeak) you will have 30 dB or more margin.
Now----are you understanding the regulations in your state?
What about lightning strikes, onto the antennas?
Notice you may detect large drones with the multipathing caused by reflections off the wiring from central body to the rotor motors.
You might even place FM-resonant dipoles around in your village, to collect and re-radiate FM energy. This re-radiation of RF energy will contribute to the richness of the multipath environment. By quenching and then un-quenching (un-dampening?) each of your dipoles on a fixed schedule, you may better detect the drone presence. In essence, you are using these dipoles (with no receiver attaced, just a CMOS switch and dampening resistor) as "radar" Transmitters that cause changes in the RF FM amplitude and phasing. Since YOU are controlling the "radar" timing, you get to use the amplitude/phase changes as information to detect the drone.
By the way, there is/was a 67KHz subcarrier that is often used for TEXT/stock_quotes/etc.
You can use that for sending messages out to all the homes in the village, if there is no cellphone service.
Just on general principles, if a 10 mW to 100 mW transmitter will only transmit to 10 yards, then a .2 W (200 mW) unit won't go much farther.
However, I suspect your figures are wrong. Amazon carries a long-range FM transmitter which is claimed to have 0.5 watts output and a range of up to 3 miles. Even if the 3 mile number is under unusual circumstances, 500 feet from 0.2 watts seems entirely reasonable.
EDIT - As has been pointed out, the desired distance is 500 meters, not 500 feet.
However, that doesn't change things. Signal power goes as the inverse square of distance. 3 miles is about 5 km, so dropping the transmitter power to .2 watts should drop the maximum range to 3 km and this is a factor of six more than required.