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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.

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closed as off-topic by Chris Stratton, Dmitry Grigoryev, RoyC, Elliot Alderson, Finbarr Oct 18 '18 at 9:38

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  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Chris Stratton, RoyC, Finbarr
If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ That's a legal question, not an engineering one. \$\endgroup\$ – Chris Stratton Oct 17 '18 at 0:59
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    \$\begingroup\$ It's a technical question, can a 0.2W radio on the FM bands transmit out to 500 meters, or be limited to 500 meters. It's not a legal question. \$\endgroup\$ – Passerby Oct 17 '18 at 1:20
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    \$\begingroup\$ I thought Choo Khay Senn was asking if he/she was allowed to transmit FM radio up to 500 metres, not whether 0.2W was enough. \$\endgroup\$ – immibis Oct 17 '18 at 1:24
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    \$\begingroup\$ it is unclear what you are asking .... are you asking about the transmitter? .... are you asking about the legality of transmitting? .... what does this mean? I don't want to get trouble .... we have no information about the laws in your country \$\endgroup\$ – jsotola Oct 17 '18 at 1:45
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    \$\begingroup\$ Notice that, in the answers, 0.2W is much more than you need. It is good practice to only use as much power as required. If you can adjust the output power, you should start low and increase it until you get the range you want. Not only does this minimize the power absorbed by nearby people, but also helps keep your transmissions limited to your smaller local area. By the way, it is likely illegal to do what you are suggesting. I recommend that you look up the laws before proceeding! \$\endgroup\$ – bitsmack Oct 17 '18 at 15:15
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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.

Assume:

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]

or

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.

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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.

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  • \$\begingroup\$ Nice answer. Just For curiosity, where does the +53dB for 200Hz BW come from? \$\endgroup\$ – Linkyyy Oct 17 '18 at 5:30
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    \$\begingroup\$ @ Linkyyy Noise power (and we are computing power here), if the noise is white (flat energy over frequency spectrum), will increase linearly with the bandwidth. 1 cycle-per-second bandwidth is that -174dBm. 2 cycle-per-second (double the bandwidth) is -174 + 3 = -171 dBm. 4 cycle-per-second(4X the bandwidth is -174 +3 +3 = -174 + 6 = -168 dBm. For 10 Hertz BW, add 10 dB (because 10*log10(10) = 10*1 = 10dB), to -164 dBm. For 100Hz BW, add 20dB. For 100,000Hz BW , add 50 dB. And given 2X BW adds 3dB more noise power, we end up with -174 + 50 + 3 = -174 + 53 = -121 dB. etc etc \$\endgroup\$ – analogsystemsrf Oct 17 '18 at 5:41
  • \$\begingroup\$ Ah alright, that is some quick and handy calculations. So basically the noise floor for a single frequency bandwidth brick wall filter is -174dBm? Assuming only white noise. \$\endgroup\$ – Linkyyy Oct 17 '18 at 8:28
  • \$\begingroup\$ Centuries ago, Maxwell and others investigated fluctuations in molecules, viscosities, etc. Boltzmann et al defined the energy stored in one-degree-of-freedom of a vibrating particle; this is temperature dependent. In the 1920s, Johnson investigated the random noise in vacuum tubes, described his findings to Nyquist, and Nyquist realized the noise power was just K * T, or just Boltzmann's constant * Kelvin temperature. At 290 degrees K (17 C), this product is close to 4.00 * 10^-21 watts/cycle-per-second, later changed to 4.00 * 10^-21 watts/Hertz. Yes---you are correct in your assumption. \$\endgroup\$ – analogsystemsrf Oct 17 '18 at 15:34
  • \$\begingroup\$ Notice the math: 4 * 10^-21 watts is also 4 * 10^-18 milliWatts; using log10, this becomes -18 + 0.6 = -17.4 and we multiply by 10 to covert to deciBels, with the answer of the very useful system design coefficient being -174 dBm per Hertz bandwidth. You will find the physically-realizable filters, if a single-pole rolloff, have bandwidth of PI/2 larger than the F3dB; this PI/2 is not a big factor/error, but be aware of it; sometimes a couple dB is a big deal. \$\endgroup\$ – analogsystemsrf Oct 17 '18 at 15:42
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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.

Oops.

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.

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  • \$\begingroup\$ 500 meters is roughly 1500 feet. \$\endgroup\$ – Passerby Oct 17 '18 at 3:33

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