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I was browsing through some OFCOM Frequency Allocation Tables for use on a small project I'm doing. I require the use of some unlicensed ISM bands in the UK, and found that the lowest unlicensed band has a limit on the transmit power/power density of "42 dBµA/m at 10 m"

The document in question is: http://stakeholders.ofcom.org.uk/binaries/spectrum/spectrum-policy-area/spectrum-management/research-guidelines-tech-info/interface-requirements/IR_2030-june2014.pdf and the table in question is on page 17.

1) What does this mean?

2) Can this limit be converted into something a little more intuitive, like an effective radiated power in Watts?

Thanks!

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  • \$\begingroup\$ This document applies this measurement only bands below 27 MHz which may affect the answer. Transmit power will be MUCH lower at say 13.56 MHz than the value given by Andy for 433 MHz. \$\endgroup\$ – Brian Drummond Aug 20 '15 at 13:00
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This answer has been edited to correct a formula and update to a more exact answer that doesn't require thinking about the antennas (credit to @tomnexus for jarring my brain in the right direction).

A H field of 42 dB µA/m = 126 µA/m (in real numbers) and, given the impedance of free space is about 377 ohms, you can take the µA/m, square it and multiply by 377 ohms to get a power of 6 µW at 10 metres. Basically \$I^2 R\$. But this is power per sq metre because E and H fields are volts per metre and amps per metre.

Now imagine the transmitter emits all the power in a spherical pattern so that at any distance, the total power passing thru the surface of a sphere is always the same. At 10m radius, the area of a sphere is \$4 \pi r^2\$ = 1257 sq metres. This means the transmit power is 1257 times greater than the 6 µW (per metre) mentioned above.

Therefore the power emitted is 7.5 mW.

Again this assumes an isotropic TX antenna (emits equally in all directions). An antenna with gain (such as a dipole with a gain of about 2dB over the isotropic antenna) cannot transmit 7.5mW but 2 dB less.

Specifying a H field magnitude as the legislative limit - there can be no argument about that because it is the (E or H) field that causes disruption to other equipment and specifying a TX power doesn't stop someone using a high gain antenna and causing localized (more directional) problems.

I've also taken a look at the table mentioned in the question. The 42 dB µA/m specified is for the 6.8 MHz band - this has a wavelength of 44 metres and therefore at a 10m distance it would be totally wrong to specify power because a coherent EM wave only begins to form (typically) at one wavelength from a TX antenna.

Note, that at 27 MHz both 10 mW e.r.p. and 42 dB µA/m are specified because 27 MHz has a wavelength of 11m and the EM wave has pretty much "formed" at 10m. At higher frequencies only e.r.p. is specified for reasons given above.

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  • \$\begingroup\$ How have you converted dB µA/m into just µA/m? What formula are you using? Also, how did the number of "13 wavelengths squared" appear when talking about the aperture of a half dipole antenna? \$\endgroup\$ – Airlangga Gunawan Aug 20 '15 at 12:34
  • \$\begingroup\$ Nice answer. Would be useful to know more about the "magic number" for aperture - either a link, or a comparison with the aperture of a 1/4 wave whip for example. \$\endgroup\$ – Brian Drummond Aug 20 '15 at 12:40
  • \$\begingroup\$ I have made a mistake on one formula so bear with me while I edit it. \$\endgroup\$ – Andy aka Aug 20 '15 at 12:40
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    \$\begingroup\$ @BrianDrummond I was going to try and reconcile the 10 mW and 42 dB µA/m H field at 27 MHz but me brain is fried now. I need a drink LOL. \$\endgroup\$ – Andy aka Aug 20 '15 at 13:23
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    \$\begingroup\$ Why do you need to involve a dipole? EiRP is simply (power density at x m) * (area of x m sphere). Or, (Power density) = (EiRP) * (Area of sphere). No frequency dependence, as long as you're in the far field. So 6 uW/m2 @ 10m = EiRP of 7 mW. Whether this is 4 mW into a dipole, or 0.7 mW into a yagi, either will touch the legal limit. \$\endgroup\$ – tomnexus Aug 20 '15 at 14:07
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It is the field strength at a certain distance. It refers to the amount of power your device is allowed to transmit in that frequency band. You could do calculations and convert it to power.

In practice you're allowed to transmit 20 dBm (= 100 milli Watt) from a device. Look at the specifications for all devices (Bluetooth, WIFI, Zigbee etc. they are all limited to 20 dBm.

With a special antenna you could focus all that energy at one point and get more than the 42 dBuA/m at 10m so you see how this limitation is impractical, it depends on so many variables and conditions. Therefore the 20 dBm.

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  • \$\begingroup\$ Hi, what calculations would I need to do in order to turn dBµA/m at X distance into a power? \$\endgroup\$ – Airlangga Gunawan Aug 20 '15 at 12:17
  • \$\begingroup\$ Sorry, too lazy for that ;-) Luckily Andy is not so lazy so see above :-) \$\endgroup\$ – Bimpelrekkie Aug 20 '15 at 12:48
  • \$\begingroup\$ He might not be using the 2.4 GHz ISM band, hence the lower limit. And, in many countries, they limit both the power and the erp, hence the gain of the antenna is also limited. \$\endgroup\$ – tomnexus Aug 20 '15 at 14:12

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