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I would like to emit as much RF energy in the frequency range of 433.9 to 434.5 MHz into surrounding fresh water with a 100mW transceiver.

The antenna is in the middle of a 4mx8mx2m fresh-water tank connected to a VA by a 4m thin coax cable.

So I took a simple half-wavelength dipole sleeve-antenna design. I calculated the 433 MHz half-wavelength in fresh water as round about 19mm, but when I compare several antennas of the same type with different pole length, the one that has a peak in the reflected power plot at 433 MHz has a pole length of 80mm and for antenna variants with shorter poles the peak moves to the higher frequencies.

  • So I ask myself now why is my resonance antenna length longer than my calculated value?
  • Could it be that the RF field will be disturbed by the air around the test tank and if how would this be related to the length?
  • If I would like to emit as much RF energy as possible into the water, what do I have to respect in my antenna design?
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    \$\begingroup\$ can you add a photo of your submerged antenna? \$\endgroup\$ Feb 8, 2017 at 14:26
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    \$\begingroup\$ Can you show you working out for antenna lengths \$\endgroup\$ Feb 8, 2017 at 14:27
  • \$\begingroup\$ Take care to build a balanced antenna - you can do better than a sleeve dipole. If there are significant currents on the coax, they might run out of the water and radiate happily from the dry portion of the coax. Check for currents by watching the SWR graph while running your hand along the cable. \$\endgroup\$
    – tomnexus
    Feb 8, 2017 at 18:45
  • \$\begingroup\$ Out of curiosity, which country allows 100mW on the 433 band? I haven't heard of that before. \$\endgroup\$
    – Lundin
    Feb 9, 2017 at 11:02
  • \$\begingroup\$ I will check the cable. Is the maximum emitted power regulated for water, and if not, would 1.5 meter water not absorb most of the energy? \$\endgroup\$
    – AndRe
    Feb 9, 2017 at 21:01

1 Answer 1

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why is my resonance antenna length longer than my calculated value?

I calculate wavelength at 76 mm for sea water with a conductivity of 4 S/m using this formula: -

\$\lambda=1000\sqrt{\dfrac{10}{f\cdot \sigma}}\$ = 76 mm

Taken from this document.

Could it be that the RF field will be disturbed by the air around the test tank and if how would this be related to the length?

No, because the attenuation at 434 MHz will be about 721 dB per metre. I used the same document as reference with the formula being: -

Attenuation/metre = \$0.0173\sqrt{f\cdot\sigma}\$ = 721 dB/m

This means, that given the size of your tank, the air outside the tank would be quite invisible due to the massive attenuation factor.

If I would like to emit as much RF energy as possible into the water, what do I have to respect in my antenna design?

100 mW should be OK but this is very specialist and I've never been one for respecting antennas too much. Just make sure you know the conductivity of the water and don't use it to try and ward-off sharks because the range at 433 MHz is tiny.

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  • \$\begingroup\$ Assuming I have a conductivity of 0.19 S/m (tap water with sweat). I would get a half Wave length of 17.4 cm (the length of my actual antenna) and Attenuation per meter of 156.9 dB/m... If I assume a transceiver power of 100mW (20dBm)(neglecting cable loss) and a measurable power of the receiver as -60dBm. What would be the maximum measurable distance of my two antennas? \$\endgroup\$
    – AndRe
    Feb 9, 2017 at 20:34
  • \$\begingroup\$ Well it's a complex shaped bath and the edges might begin to play a role but if they don't then what do you think the answer is. I'm not at a pc at mo. \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 20:38
  • \$\begingroup\$ If I do the calculation from the document (above) right... then(20dBm+60dBm)/156.9dB/m = 0.51m (?) \$\endgroup\$
    – AndRe
    Feb 9, 2017 at 20:52
  • \$\begingroup\$ You also need to bear in mind that if your receive antenna is outside the tank there is also a refraction loss. This is covered in the document I linked. You also need to think about the aperture of your receive antenna and imagine how much of the transmitted power it can "net" if there were no water losses. This means that because the transmit power is sprayed in all directions, the Rex antenna only nets a small fraction. \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 21:10
  • \$\begingroup\$ Rex was a tablet spelling mistake. I should have said Rx \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 21:16

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