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What is the maximum range of a 27 MHz ISM band when transmitting at full power? I am interested in general situation for US and EU but that will go back to the power limits allowed. My country (CZ in EU) allows 100 mW on "26.995; 27.045; 27.095; 27.145; 27.195 MHz" with a "channel width" (or spacing?) of 10 kHz.

The system should transmit only in one direction - it should broadcast GPS RTK corrections from base station to rover. Throughput needed is only 1 kB/s (= say 9600 bps). Desired range is 10-20 km but I am afraid I won't get that much - few kms is still very good though.

Will antenna size (not power) influence the range? I consider a case with the transmitter being a permanent base station (can have bigger antenna, but not huge) and the receiver being a mobile rover (small 10" rubber duck antenna). Both antennas will be omnidirectional.

I was considering using ham radio but that is most likely not legal ("broadcasting" for "general public").

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    \$\begingroup\$ There is a presentation from the ITU that lists the 26.96-27.28 MHz band with a permissible level of "0.5W to 3W e.r.p. /42 dBμA/m @ 10m" but makes a few remarks about the math not adding up. \$\endgroup\$
    – Simon Richter
    Commented Apr 12, 2016 at 1:28
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    \$\begingroup\$ Not sure if it is still the case, but 27 MHz used to be full of potentially interfering signals such as RC aircraft and garage door openers and whatnot. They may have all migrated to other bands, now, though. \$\endgroup\$
    – user57037
    Commented Apr 12, 2016 at 5:49

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What is the maximum range of a 27 MHz ISM band when transmitting at full power?

It depends on power emitted, antenna types at both ends, temperature, frequency, terrain, interference, bandwidth of data and (sometimes) a bit of good luck.

What you should consider doing is following the general rules for estimating link-loss i.e. how many dB is lost between transmitter and receiver in a worst case situation. You can start this process by using the perfect scenario of free-space (i.e. earth to moon transmission) then apply antenna gains and more earthly constraints like fade margin.

Link Loss (dB) = 32.4 + 20\$log_{10}\$(F) + 20\$log_{10}\$(d)

Where F is MHz and d is distance between the two antennas (kilometres).

So, for an example of 27 MHz and 10 km, the link loss is 32.4 dB + 28.6 dB + 20 dB = 81 dB i.e. not really a problem in free space.

Your antenna types probably won't bring anything more than about 4 dB to the party so link loss becomes maybe 77 dB. Fade margin might be 30 dB worst case for over 99.9% of the time so this takes the link loss to 107 dB.

How much power do you need to receive to form an adequate signal that can be reasonably decoded with an acceptably low bit error rate? A useful formula is: -

Power required in dBm is -154dBm + 10\$log_{10}\$(data rate) dBm

So if your data rate is 1 kbps and you have your receiver designed accordingly to have a limited bandwidth that suits this data rate, your receiver, at ambient temperature requires -154 dBm + 30 dBm = -124 dBm.

Looks like 10 km is doable to me (with 100 mW aka 20 dBm) providing you don't get significant corruptions from anyone else using the band. Unfortunately I cannot help you on this matter other than to suggest you transmit everything two or three times (or more). You could also take steps to add check-sums and maybe error correction codes. Some techniques redistribute the normal sequence of data bits in a packet to avoid pulsed interference killing off what would have been several sequential bits - this allows some error correction methods to work better with interference that is more "man-made".

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  • \$\begingroup\$ Thanks for valuable answer! The required speed is actually 1 kB/s (say 9600 bps), dont know if it changes the game. I added it to the question. \$\endgroup\$
    – Kozuch
    Commented Apr 12, 2016 at 8:13
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    \$\begingroup\$ @Kozuch - do the math - power required is -154 dBm + 40 dBm = -114 dBm. Link loss plus fade margin makes minimum transmit power -114 dBm + 107 dB = -7 dBm. If you have 100 mW (20 dBm) it should be still doable. \$\endgroup\$
    – Andy aka
    Commented Apr 12, 2016 at 8:23

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