1
\$\begingroup\$

So I'm considering different antenna shapes to most effectively detect electric field component of far-field VLF waves. I saw some papers that used this method but couldn't find any broad explanation on its electrical design.

I know for a transmitting antenna the impedance of the antenna should be purely resistive(resonant frequency). However, I don't know if that's the case for also a receiving dipole antenna. If it is, then can dipole antennas be used for detection of waves at VLF since it would be practically impossible to build a half-wave dipole at such low frequencies?

So should a receiving dipole antenna also have resistive impedance or is there a way around this?

\$\endgroup\$
  • \$\begingroup\$ Some VLF receivers can utilize dipole loops around ferrite rods, like in the old transistor radios, which detect the B field enstead of the E field. \$\endgroup\$ – SDsolar Jun 30 '17 at 21:55
1
\$\begingroup\$

Maintaining the correct impedance for a transmitting antenna is much more important than for a receiving antenna. Simple reason 1: avoidance of power turned directly to heat due to a mismatch. Simple reason 2: reflection of power back to the transmitting power amp that might cause the transistors to fail.

A quarter wave monopole can be made "short" and its impedance will look both capacitive and resistive: -

enter image description here

So, at a quarter wavelength the reactive impedance (magenta colour) is zero and the radiation resistance (blue) is about 37 ohms. If you shortened the antenna to 0.05 wavelengths, the radiation resistance is maybe an ohm or two in series with a capacitive reactance of 1000 ohms.

This makes it difficult to drive as a transmit antenna but still usable as a receive antenna. You can tune out the capacitance with an inductive reactance of 1000 ohms but you get a very tight bandwidth and this may not be suitable as a receive antenna. It all depends on what bandwidth the transmission has that you are interested in.

Alternatively, if the transmission you are trying to receive is a proper EM wave then you can make a loop receiver and just pick-off the magnetic part of the EM wave. If you look at long-wave radio receivers, they use a ferrite rod and loop antenna and this can be quite compact. It makes a useless transmit antenna but a very practical receive antenna.

\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer. Yes, I know about magnetic loop antennas and have designed one but I am trying to get the electric field component in addition to the magnetic field component I'll get from the loop antenna. Also can you expand more on why I would get a tight bandwidth with smaller resistance(1-2 Ohm). What are the factors that limit the bandwidth of a receiving dipole antenna? \$\endgroup\$ – Starior Mar 28 '17 at 14:06
  • \$\begingroup\$ With a shorter antenna you inevitably get capacitive reactance and this can be cancelled by inductive reactance but this works OK over a limited bandwidth because you are using electrical components to resonate an antenna that is not resonant because it is short. But it all depends on the input impedance you are feeding. If high then the extra 1000 ohms cap reactance may not have a serious effect. \$\endgroup\$ – Andy aka Mar 28 '17 at 14:11
  • \$\begingroup\$ Re the receive coil, if you can calculate the H portion of the EM wave then the E portion is \$120\pi\$ higher because \$120\pi\$ is the impedance of free space. I.e. you don't need to measure both. \$\endgroup\$ – Andy aka Mar 28 '17 at 14:13
  • 1
    \$\begingroup\$ @Starior Andy's short monopole can work very nicely with a FET source-follower buffer right at its base (use NO coax). Impulse noise is worse than magnetic-loop antennas. And your monopole antenna/buffer should be outside, in clear space. \$\endgroup\$ – glen_geek Mar 28 '17 at 14:49
  • \$\begingroup\$ @glen_geek Thanks, that's actually a really good idea. Can you estimate how much would that solution widen the bandwidth of the receiver if the frequency I want to operate the antenna at is 1-30 kHz? \$\endgroup\$ – Starior Mar 29 '17 at 3:29

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.