• Z-Wave operates at 868MHz (in EU) and has up to ~100-meter range
  • LoRa operates at 868MHz (in EU) and has up to ~15 km range

My knowledge says that the lower the frequency = the higher the range/penetrability.

However, both these technologies operate at the same frequency, so how come LoRa is able to achieve such a long-distance?

Furthermore, I'm a bit confused, as to how power consumption plays a role in the above situation. I'm assuming a high-powered signal means that the waves themselves have a higher amplitude, thus making it harder for other signals to interfere.

With that assumption, does that mean that LoRa uses a lot of power to reach such a long-distance?

  • $\begingroup$ I don't know, but did you look at the power output limits for each technology? $\endgroup$
    – Transistor
    Jun 27, 2022 at 22:29
  • $\begingroup$ what is the data rate for each of those? $\endgroup$
    – jsotola
    Jun 28, 2022 at 0:10

1 Answer 1


Three factors:

  1. Transmitter power: LoRa, for example, typically uses about 160x the power of Z-Wave. However, this alone cannot explain a 150x increase in operating range given that signal strength falls off with the square of distance (so you would expect that extra power to get you to perhaps 1250 meters of range).

  2. Beam-forming: Some protocols are able to directionally transmit signals instead of sending them in all directions. This increases the range by concentrating the power.

  3. Error correction: (Over simplifying here because this is a complex topic) Some protocols send the same data many times to ensure it reaches it's destination even if there is a lot of interference (which increases with distance). Others send the data only a few times, which is faster but limits the range/reliability. Many protocols dynamically adapt how many copies of the data they send based on operating conditions.

  • 1
    $\begingroup$ Hi Emily, thank you the elaborative answer :) $\endgroup$
    – Jazerix
    Jun 30, 2022 at 12:30
  • $\begingroup$ Always glad to help! It's a complex field and there are probably other factors at play too - I'm far from an expert! $\endgroup$
    – Emily Conn
    Jun 30, 2022 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.