I own several HM-TRP TTL FSK radio modules, and every time I turn one on, it reports it as being version 2.

Anyways I did read the manual and it has a setting for transmitter power as follows:

0 = +1 dbm
1 = +2 dbm
2 = +5 dbm
3 = +8 dbm
4 = +11 dbm
5 = +14 dbm
6 = +17 dbm
7 = +20 dbm

How do I convert these values to transmission distance in meters if all of the following conditions are met:

A. The transmitter module and receiver module are powered with a steady 3VDC

B. There are no other objects between this transmitter and an HM-TRP receiver

C. The receiver's noise floor is at maximum

D. Both modules have copper coil antennas measuring about 1cm in length
   and the end is connected to the ANT pin on the module

E. Both modules operate at the frequency 915 Mhz.

F. Both modules use a UART data rate of 38400bps 
   and a wireless data rate of 38400bps as well.

So what's the formula?


So I'm trying to understand here... Do I use the formula for path loss and subtract that from the transmission DBM setting I select in my receiver and if the result is not negative then I get a signal?

And I got two different equations for calculating path loss. Which is officially correct?

  • \$\begingroup\$ Yes, regards your edit, transmit power minus path loss is receiver receive power. If you have negative decibels with respect to 0 dBm then the receive power is a fraction of 1 milli watt. That is what the m in dBm stands for. \$\endgroup\$ – Andy aka Jun 9 '18 at 17:32

In free space and assuming an isotropic antenna at each end: -

  • Path loss (dB) = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.

If you know the gains of the antennas you can reduce the path loss by a few dB.

  • Power (dBm) needed by a receiver is -154 dBm + 10\$log_{10}\$(data rate)

This is based on an empirical derivation and assumes an ambient temperature of 300 kelvins and a reasonably low data error rate. This is a baseline value.

But on earth, no matter what the terrain appears to be, there will be added attenuations that are really difficult to account for and describe here. There's a thing called fade margin and this, as a rule of thumb basically says - try to ensure your received power is at least 20 dB greater than its baseline sensitivity - this means that if you designed a receiver requiring -120 dBm you should expect to receive -100 dBm on a good day.

Information taken from my answer here.

  • \$\begingroup\$ I'm trying to turn the logarithms into simpler math but I see tutorials showing where the number is after the subscripts (calculus.nipissingu.ca/tutorials/logarithms.html). I'm lost. \$\endgroup\$ – Mike Jun 9 '18 at 16:11
  • \$\begingroup\$ I'm not sure what you mean. Maybe this: 32.45 dB is a ratio power loss of 1/1758 = 0.00056885 so if F = 1 MHz and distance = 1 km you'll receive 0.56885 mW from a 1 watt transmitter. \$\endgroup\$ – Andy aka Jun 9 '18 at 16:19
  • \$\begingroup\$ If you prefer an equation that does use logs then look at equation 1 here: antenna-theory.com/basics/friis.php. It comes up with 0.56993 mW because the 32.45 dB in my equation is rounded - it should be 32.44 dB. But what's 10 milli dB between dudes. \$\endgroup\$ – Andy aka Jun 9 '18 at 16:26

We'll assume the datarate is exactly the bandwidth (close enough) at 40,000Hertz. The increase in noise power is 3dB per octave of bandwidth, or 10dB per decade of bandwidth, thus 44dB increase in noise floor.

We'll assume 3.5dB losses compared to a quarter-wave antenna, in capturing and moving energy from the antenna into the LowNoise Preamp. We'll assume 0.5dB Noise Figure in the PreAmp; at 915Mhz, this should be possible. We'll assume NO CODING, that is, no Viterbi nor Turbo Coding; we'll assume your BitError rate is 0.001, or 0.1%, and that this is caused by Signal Noise Ratio(SNR) of 10dB (could be 7dB, if single-sided noise power measurement).

Lets compute how strong a signal you need at the (quarterwave antenna).

-174 dBm/rootHertz thermal/Johnson/Boltzmann quantum floor

  • 46 dB the 40,000 Hz bandwidth

  • 0.5 dB Low Noise Amplifier noise-figure

  • 3.5 dB antenna efficiency, and PI-matching into LNA

  • 10 dB SNR needed for 1e-3 (0.1%) bit errors

-174 + 44 + 0.5 + 3.5 + 10

-174 + 60dB = -114 dBm RMS signal

Knowing 0 dBm is 0.632 volt RMS across 50 ohms, or 0.223 volts RMS, and that -120dBm (1/1,000,000 smaller) is 0.223 microVolts RMS, we know our minimum signal level under our stated assumptions is 6dB (2 x 2) larger or about 0.9 microvolts RMS.

If our Transmitter provides 0dBm output power into quarter-wave antenna (our reference gain ~~ 0dB), then we have 114dB for path loss.

The path loss formula I'll let you look up. The math includes 22dB for some area-of-sphere computation.

The Path loss is 22dB + 10 * log10[ (distance/wavelength)^2 ]

We have 114 - 22 = 92dB for our (distance/wavelength)^2 factor.

If the factor is 100dB, we are allowed 10^5 wavelengths between Transmitter antenna and Receiver antenna, assuming no orientation-losses (there will be orientation losses).

At 915Mhz, or Wavelength of 1/3 meter, 100,000 * 1/3 == 30,000 meters or 30 kilometers or about 20 miles.

Your factor of 92dB provides slightly longer range.

  • \$\begingroup\$ "-114 dBm RMS" is a misnomer. Power is not an RMS quantity. \$\endgroup\$ – Andy aka Jun 9 '18 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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