I have 3 main questions that I want to know. I mostly worked with computer science but recently I got an arduino to play around with. I also extracted some motors from an old rc quadcopter that i had. I figured that with these parts, I could build my own arduino quadcopter as a learning project. Anyway, I did some research on RF transmitters and receivers, but there were many types, and I wasnt sure which one to get. Which leads to my questions

  1. Does power determine range? There are many transmitters with varying degrees of power, and it seems pretty consistent that more wattage = longer range. I was wondering if this is in fact the case, and I should be looking for a transmitter with higher power intake.
  2. Does 2.4 ghz go through objects? I was reading about most RC transmitters switching to 2.4 ghz because there is less interference, but i also read that you need line of sight. But then looking up line of sight, it turns out that you only need radio line of sight, which looks more like an elipse than a straight line. So I was wondering if 2.4 ghz will work for me, since I eventually want to fly it where I cannot see it, and control it with an FP camera.
  3. What gives me channels? https://m.ebay.com/itm/1-pair-of-RF-433Mhz-Transmitter-and-Receiver-Module-Kit-for-Arduino-Raspberry-Pi-/311858165335 This module for example has on the receiver 3 terminals, VCC, GND, and data. Does this mean it only has 1 channel? Does this mean that I can only control 1 function (ie. Rotation, forward/backward) with this module?

So in summary, i was hoping someone could point me to a decent transmitter and receiver combo with 1000m range and multiple channels. Not looking for prebuilt, just the circuits. Hope everything made sense and thanks

  • \$\begingroup\$ Asking for product advice is off-topic but as you ask some other questions as well I don't vote to close but recommend that you remove the part about specific product recommendation and change the title according to the rest of your question. \$\endgroup\$
    – Arsenal
    Commented Jan 2, 2018 at 14:16
  • \$\begingroup\$ What's important for RC toys would be which radio bands there are freely available in your country. Looking at your user picture, I'm guessing you live in USA and then 433MHz isn't really an option. 2,4GHz is however a free band, but 2,4GHz radio will have inferior range compared to 433MHz. \$\endgroup\$
    – Lundin
    Commented Jan 11, 2018 at 9:35

2 Answers 2


The range in RF communication is primarily determined by the Signal-to-Noise (SNR) ratio of the received signal and the sensitivity of the receiver chipset. SNR is directly proportional to the received power \$P_r\$.

  1. Does power determine range?

Yes, more power makes the range longer. But it is not a good idea to increase range by merely increasing the power as there are FCC regulations over max power that can be transmitted as well as more power may increase the interference in the nearby devices. Also, in order to increase the range two times you need to increase the power more than 4 times because of inverse square law. You need to study Friss transmission equation in more detail.

\$P_r=\frac{P_tG_tG_r\lambda^2}{4\pi R^2}\$

From Friss equation, you can see that there are other factors which we can tweak to increase the range (eg. antenna Gain and frequency).

  1. Does 2.4 ghz go through objects?

Frequencies such as 2.4 GHz or 5 GHz get highly absorbed by bodies with high water content. So trees will tend to attenuate the signal significantly (if they are present in the line-of-sight(LOS)).There are drones which use these frequencies for remote control or video FPV transmission, but they employ directional antenna(antenna with higher gain) at the stationary side. You can do the same or use lower sub-GHz frequencies (such as 868 MHz/915 MHz ISM band or 433 MHz ISM band depending on your country). These lower frequencies have lower path loss and thus tend to penetrate longer distances with the same amount of transmitted power and antenna gains.

  1. What gives me channels?

Channels are generally regions in the frequency domain which Tx-Rx uses for communication. Your Tx-Rx pair seems to use only one frequency of 433 MHz and thus has only one channel. But you can put any digital data (corresponding to the rotation, forward-backward movement) over this channel depending on how you design your data protocol. You don't need separate channels for transmitting different types of data. [Note: There will be limitation over the max data-rate that can be achieved over this 433 MHz frequency and that rate won't be sufficient enough for your video transmission.]

Also in multichannel devices, you can use only one channel at one time. The advantage of multichannel devices is that if some channel is congested then you can switch to other channels.

There are many ready-made FPV Tx-Rx devices based on different frequencies. If you want to build it on your own then you need to first decide which frequency band you want to operate in and then look for the corresponding chipsets/modules which can give you desired data-rate. Only after you do the RF link budget analysis you go for actual implementation.

I hope it clarifies your doubts.

  • \$\begingroup\$ Thanks it helped a lot. But when you say that using 2.4 ghz, most drones use a higher gain antenna, does that mean the antenna is a seperate component to the Tx-Rx module? I thought that the antenna was built in, but can you modify it? \$\endgroup\$ Commented Jan 2, 2018 at 16:01
  • \$\begingroup\$ You can not use a built-in antenna. It won't be directional (i.e. high gain). So you should select a module to which you can attach an external antenna through connectors like U. FL or SMA. \$\endgroup\$
    – Ashutosh
    Commented Jan 2, 2018 at 16:43
  • \$\begingroup\$ sparkfun.com/products/690 would this transmitter count as having multiple channels? It says "Software selectable channel from 2400MHz to 2525MHz (125 Selectable channels)". And looking at the data sheet, it has a maximum power of 0 dbm, so in this case, how much range could I get with a 15 dbi antenna? \$\endgroup\$ Commented Jan 3, 2018 at 3:52
  • \$\begingroup\$ In addition, higher frequencies makes it harder for the radio waves to bounce on surrounding objects. 2,4GHz and above have horrible characteristics if you put a large metal object in between tx and rx. Such as buildings made of reinforced concrete. \$\endgroup\$
    – Lundin
    Commented Jan 11, 2018 at 9:40

Here is link analysis for 3,000MHz link, using Omni-directional antennas for TX and RX, with 10dB SignalNoiseRatio (this will require error-detect-correct for a robust line), 1MegaBit datarate, first Receiver amplifier Noise Figure of 3dB (a very easy param to satisfy)

-174 dBm/rootHertz the Boltzmann/Johnson/Nyquist/random noise floor at 290K

  • 60 dB for the 1MegaBit (close to 1MegaHertz bandwidth)

  • 3 dB for Low Noise Amplifier noise figure

  • 10 dB for usable system SNR (before EDC)

  • 1 dB matching losses from Antenna to LNA

-174 + 60 + 3 + 10 + 1 = -174 + 74 = -100dBm (which is 6.3uVpp across 50 ohms)

Assume TX power of 1 watt; the energy spreads out in the far field (anything further than 1 wavelength is very close to pure far field) in a Range^2 behavior; a bit of math involving hemispheres produces a 22dB loss. The pathloss formula becomes

PathLoss = +22dB + 10 * log10[ (distance/wavelength)^2 ]

wavelength is 0.1 meters at 3,000MHz

Assume distance is 1,000 meters, thus distance/wavelength = 10,000

Pathloss = +22 + 10 * log10(10,000^2) = +22 + 10 * log10(100,000,000)

Pathloss = +22 + 10 * 8 = 22 + 80 = +102dB.

Our TX power is 1 watt, or 0dBw or +30 dBmilliwatt or +30 dBm

+30dBm - (PathLoss of 102 dB) = -72dB.

And our first part of the analysis showed we need at least, or stronger than, -100dBm

Thus we have, we think, -72 - (100) = +28 dB margin.

As some others' answers indicate, losses kick in. Have fun.

  • \$\begingroup\$ I have almost no background in radio communications so most of it went over my head. Could you explain it a bit more? \$\endgroup\$ Commented Jan 3, 2018 at 3:45

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.