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I want to use LoRa RA-02 based on SX1278 with 433MHz.

In some country I can send data in 10db max and I will connect 1/4wavelength (17cm) of antena.

In RA-02 write high sensitivity of over -148dBm

How can I calculate the maximum distance between transmitter and receiver?

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  • Any theoretical calculated range would still need to be tested in real world conditions.
    – hardillb
    Aug 18, 2022 at 7:55
  • Ofcoues, but how can I calculate it theoretically?
    – Kokomelom
    Aug 18, 2022 at 9:54
  • I'm not sure low level RF propagation maths is on topic for here, what research did you do before asking?
    – hardillb
    Aug 18, 2022 at 10:44

1 Answer 1

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First and foremost, any theoretical calculations are only valid if you have line-of-sight (LoS) between transmitter and receiver. As soon as you have an obstacle, it is nearly impossible to compute it, you'll only be able to measure it to take into account the related losses.

Second, you need not only direct line of sight to be free of obstacles, but you need the Fresnel zone to be free as well. At 433 MHz, the radius of the Fresnel zone at 1 km is 13 meters. At 100 km it is over 130 meters.

Taking into account this and the curvature of the Earth, that means that for any significant distance, even on flat terrain without any obstacles, you need one or both of the devices to be quite high above ground. All LoRa distance records have involved either balloons or devices very high above ground (on top of mountains or towers).

Now, for the calculation...

What you need is a link budget calculator, which uses the link budget equation to compute the result. There are plenty on the web, though they may not work in the direction you need (they are often distance -> budget rather than budget -> distance), but you can always do a binary search.

The inputs you will need are:

  • The TX power (10 dBm apparently)
  • The losses in cabling etc. on the TX side (unknown)
  • The TX antenna gain (unknown, but it's probably an isotropic antenna, so about 2.1 dB)
  • The RX antenna gain (the same as the TX antenna?)
  • The losses in cabling etc. on the RX side (unknown)
  • The RX sensitivity (-148 dBm)

Counting 0 for the losses in cabling etc, the link budget without free space path loss is 10 + 0 + 2.1 + 2.1 + 0 - -148 or 162.2 dB

So your max distance is where the free space path loss (FSPL) is 162.2 dB. FSPL in dB is 20log10(d) + 20log10(f) - 147.55 (with d in meters and f in Hz)

With f = 433 MHz, FSPL = 20log10(d) + 25.18, which means d = 10^((FSPL - 25.18)/20) or a bit over 7000 km.

Note that there are quite a few assumptions here:

  • That misc losses are 0. Most probably not true
  • That there are no obstacles in the line of sight but also in the Fresnel zone.
  • That sensitivity is actually -148 dBm (this is usually only achieved for the very very very slow data rates)
  • That only RSSI is relevant, and not SNR (in which case noise becomes an issue).

In practice, what will determine the max distance is most likely to be the terrain. There are tools which allow you to enter points on a map and see if a given link is feasible. Other tools allow you to see the theoretical coverage from a given point.

Note, again, that all this is only valid if both devices are outdoors with strictly no obstacles between them. Anything where one or both devices are indoors, or there are any obstacles (terrain, trees, buildings) will very, very, very significantly reduce the max distance (it can go down to a few dozen meters!).

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