I want to create a small battery-powered device that can emit a few bytes per day to as far as possible (100 km?).

While discussing with someone working on sensors placed outside (even in the countryside), he said that they use ~170 Mhz (ultra short waves), with a device such that the battery can last 10 years without any human intervention.

This looks too good to be true, is this really possible? I mean:

  • battery-powered (Lithium Cell Battery CR2032 3V)?
  • long range (100+ km)
  • no GSM signal required, using ultra short waves instead

and let's add (to make it completely incredible):

  • small PCB possible (< 5 x 5 cm), and cheap
  • \$\begingroup\$ You said nothing about reception. Theoretically an electromagnetic wave is extending to infinity. the question how you detect it... \$\endgroup\$ – Eugene Sh. Sep 15 '17 at 14:49
  • \$\begingroup\$ Do understand the Friis Loss calculator , antenna gain , waves bounce and earth is not flat and high performance receiver costs \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 15 '17 at 14:51
  • \$\begingroup\$ 170 MHz is VHF, not "ultra short wave". Propogation is essentially line-of-sight, 100 Km range is highly likely, particularly with low power. Typical range for Marine VHF radio (samefrequency range) with 25 watt transmittters is about 30 Km. \$\endgroup\$ – Peter Bennett Sep 15 '17 at 14:58
  • \$\begingroup\$ try 5 Watts with special antenna and hope troposphere bounce and weather is good that day with signal bounce 90% of the time during summer days only. CR2032, haha no way. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 15 '17 at 15:03
  • \$\begingroup\$ A few bytes a day and 100km range sounds more like a text message or a pigeon. \$\endgroup\$ – Chu Sep 15 '17 at 15:08

I can give you a few pointers as to whether this might be practical.

It is generally accepted that the power needed at ambient temperature to adequately receive data is -154 dBm + 10 log\$_{10}\$(data rate). So, if your data rate is 1 bit(s) per second, a receiver should just about be able to work with an input power from the antenna of -154 dBm. If the data rate was 10 bits per second then the power needed would be -144 dBm.

However, this isn't an unhindered transmission through free space so a general rule of thumb is to add back another 30 dB of power so, you might get away (most of the time) with about -124 dBm.

Of course you need to design the receiver so that it had fairly substantial band-pass filtering and a 1 bit per second data rate would require a 1 Hz filter as a general rule. This does of course mean that your transmitter has to transmit a carrier that is substantially better than 1 Hz accurate or the system just won't work.

So, you design a receive band pass filter that is wide enough to cover the drift of the transmitter and this might mean the wider bandwidth accepts much more noise interference. This is where the formula I wrote above comes into play. It's an empirical formula that assumes you do some good filtering in your reciever and that the transmit carrier remains substantially within the range of that receive filter.

So, realistically you are looking to work with a power that might be about -110 dBm to get good data.

There is another formula invented by Harald Friis that describes how power is thinned out with distance and it also brings in the transmission frequency: -

Loss (dB) = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is MHz and d is kilometres.

So at 170 MHz the attenuation over a 100 km distance is 32.45 dB + 44.6 dB + 40 dB = 117 dB. There is another assumption here; the antennas used are undirectional and, in a lot of practical situations you can use somewhat directional antennas.

Amended answer due to math mistake

So you might be able to get a loss of about 113 dB using a dipole antenna and this means you need to transmit -110 dBm + 113 dB = 3 dBm or 2 mW. This is feasible from a small battery and is certainly feasible with a module that is 5 cm x 5 cm.

However, at VHF frequencies, you get no reflections from the ionosphere so you need a line-of-site communication and this means mounting transmit antenna and receive antenna very high up. For instance, using THIS calculator you would need 150 metres height at both ends to get line of sight.

  • \$\begingroup\$ Thanks for your answer. 150 metres or even 4 meters high is not possible sadly (the device could be in the street for example). Should I look for higher or lower frequency then? \$\endgroup\$ – Basj Sep 15 '17 at 20:06
  • \$\begingroup\$ I think you have all the relevant formulas to calculate this yourself. The link loss formula tells you that you'll get less attenuation at lower frequencies but the antenna gets bigger proportional to frequency dropping so that will become a problem if it isn't already. The trouble is you won't get the distance without dropping down to a few MHz maximum and even then weather conditions will mean you won't get bounce from the ionosphere. Rethink your objectives. I've even left a link to a calculator for line of sight versus height so, get cracking and adjust the objectives. \$\endgroup\$ – Andy aka Sep 15 '17 at 20:17

You will not get 100km with VHF (the english term for UKW). At these frequencies the wave almost completely travels line of sight. Hence the curvature of earth will block your signal at these distances.

A CR2032 will last 10 years, no problem. It should be even able to power a micro-controller to do some measurements from time to time. But this means that you have to keep all leakage current in check. I.e. you have to keep the average current consumption at less than 2µA. And that does not yet account for self-discharge. Even if your circuit itself, uses less than 500nA on average (and that's not trivial to achieve), and we assume that you can transmit whatever you want within 1s once per day that will still leave you with at most 100mA available to do the transmission (still not accounting for self-discharge and not for ESR of the battery or decreased capacitance due to high output current).

At these low energy levels, I do not see how it should be possible to output enough RF energy to cover more than a few 100m or maybe 1-2km. And even that would require a sophisticated communication protocol that makes use of all kind of tricks to save power on the transmitter side.

On lower distances, i.e. within a few 10m, this can be achieved. There are BTLE and sub-GHz low power radio chipsets that are meant for this kind of applications. Though I am not sure whether they actually get to 10y on a CR2032, as I've never done long term measurements with those.

The PCB size sounds ok. I've done full ARM9+wifi on 2x2cm boards. With a few tricks you can get even smaller than that.

  • \$\begingroup\$ Thanks for your detailed answer. Let's lower the specs: 20 km range and only 1-year battery. Does this seem possible? If so, what kind of emitting chip would you use? \$\endgroup\$ – Basj Sep 15 '17 at 19:51
  • \$\begingroup\$ Also I'm open to change from VHF to something else: would a higher, or lower frequency help to have more distance / range? \$\endgroup\$ – Basj Sep 15 '17 at 19:57
  • \$\begingroup\$ 20km sounds more likely. But it's still not easy. The problem is your limited power that you can draw at any time due to the use of a coin cell. If you could get of something close to 1W for a few ms, than this might actually work. Another problem is that you need a very well designed protocol, because the receiver has to know exactly when the transmitter is going to send, the precise frequency etc pp so it will be able to find it in the noise. Or you can just throw a lot of computational power at it. \$\endgroup\$ – Attila Kinali Sep 15 '17 at 22:52
  • \$\begingroup\$ Just to make sure you understand what the problem is: Operating a radio system from a single coin cell is a non-trivial problem and a lot of engineering has to go into it to make it work under these severe power constraints. BTLE and other low power systems work, because companies spend a few years to come up with suitable IC designs that do everything on chip to save as much power as possible. But the power source still limits the range these systems have. You are asking now for several magnitudes more range. This means you have to optimize the whole system, from power source to receiver. \$\endgroup\$ – Attila Kinali Sep 15 '17 at 23:12

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.