RF wireless transmission distance

Jacktang

RF wireless transmission distance [Copy link]

The transmission distance is related to the transmit power of the sender. The higher the transmit power, the farther the data can be transmitted.

The transmission distance is related to the receiver's receiving sensitivity. The lower the sensitivity value, the more conducive it is to receiving weaker signals.

The transmission distance is also related to the radio wave frequency (433MHz, 315MHz, etc.).

The transmission distance is also related to the attenuation and absorption of radio waves by the transmission medium (air quality, material of obstructions, etc.).

The transmission distance calculation formula is as follows:

Example:

Transmitting power = 12dBm, receiving sensitivity = -127dBm, dielectric loss = 35dBm, wireless frequency = 433MHz, then the theoretical transmission distance is:

That is: the approximate communication distance is 8.7 kilometers.

Transmitting power = 5dBm, receiving sensitivity = -118.5dBm, dielectric loss = 45dBm, wireless frequency = 433MHz, then the theoretical communication distance is: 0.46 kilometers.

Transmitting power = 20dBm, receiving sensitivity = -148dBm, dielectric loss = 45dBm, wireless frequency = 433MHz, then the theoretical communication distance is: 13.82 kilometers.

About receiving sensitivity (taking SX1278 as an example)

The receiving sensitivity is related to the spreading factor SF and the channel bandwidth BW. The calculation formula is as follows:

Example:

Spreading factor SF = 8, channel bandwidth BW = 125kHz, then the receiving sensitivity is:

Spreading factor SF = 7, channel bandwidth BW = 500kHz, then the receiving sensitivity is:

Spreading factor SF = 12, channel bandwidth BW = 7.8kHz, then the receiving sensitivity is:

About dielectric loss

Based on the estimated values of the usage environment, the following can be used as reference

Visible: open space 25dBm, rural 30dBm, small town 35dBm, large city 40dBm

Wall: Steel structure 15~25, Concrete 13~18, Hollow brick 4~6, Gypsum board 3~5, Glass 3~5, Coated glass 12~15

Others: Cars 8~12, Train carriages 15~30, Elevators about 30, Forests 10~20

In practical applications, the radio signal strength loss between the sender and the receiver is often the cumulative sum of multiple dielectric losses, that is, on the basis of 35dBm, the absorption loss of other materials between the sender and the receiver is added. The total dielectric loss is more reasonable at 45dBm.

From the transmission distance calculation formula, we can know that when the transmission power, receiving sensitivity or dielectric loss changes by 6dBm, the transmission distance will double. Therefore, we need to pay special attention to the influence of objects such as windows, doors, glass, and walls, and install the sender and receiver antennas in the most advantageous locations.

okhxyyo

Thank you for sharing~~

alan000345

Learned, very useful

btty038

The formula notes are simple and clear. Support