4
\$\begingroup\$

I always try to tune my pcb antennas in length, so they operate in resonance, by cutting the antenna with a scalpel and then matching the tuned antenna to the transmitter.

But recently an RF engineer told me, that you could sometimes get the same radiated power, by matching the impedance without trimming the antenna to resonance.

I have measured an untuned IFA to have impedance of Zin = 7.7+j90 @ 865MHz.

Should I try to tune this antenna to resonance, or will it (in practice) work just fine to match the untuned antenna to my transmitter ?

\$\endgroup\$
5
  • \$\begingroup\$ usually you'd do both – build a resonant antenna of the right length, and have an impedance matcher between whatever generates/consumes the signal and the antenna \$\endgroup\$ Feb 28 '17 at 11:04
  • \$\begingroup\$ I would say that you always get the highest radiated power as long as the antenna is matched to the power amplifier. The idea is that the power has to go into the antenna and not reflect back to the PA. That is achieved when power is matched. Usually the length of the antenna is determined by the frequency at which you want to transmit. \$\endgroup\$ Feb 28 '17 at 11:12
  • \$\begingroup\$ If you can match an untuned antenna, then that means the matching network is storing the reactive energy needed to drive the antenna harder to achieve the same radiated power. \$\endgroup\$
    – Neil_UK
    Feb 28 '17 at 11:42
  • \$\begingroup\$ OK, that makes sense, but does the radiation resistance not depend on the antenna tuning. I just figured that at some point, most of the power into the antenna would be converted to heat. \$\endgroup\$
    – JakobJ
    Feb 28 '17 at 11:54
  • \$\begingroup\$ most of the power into the antenna would be converted to heat Assuming an antenna made of an ideal conductor, it cannot convert the energy into heat. In practice the power which the antenna cannot radiate is mostly send back to the PA. Which has a certain real output impedance, which converts the energy into heat. So the PA gets hot, not the antenna ! Ever tried to run a high-power PA without an antenna or a dummy load (i.e. maximum mismatch) ? It might destroy itself :-( \$\endgroup\$ Feb 28 '17 at 12:22
2
\$\begingroup\$

I have measured an untuned IFA to have impedance of Zin = 7.7+j90 @ 865MHz

The 7.7 ohms "real resistance" measured represents both the radiation resistance AND the antenna losses so, if you can adjust the antenna length to give you a higher value of overall resistance (and therefore a higher figure for radiation resistance) you will get better antenna efficiency and more easily match your output driver (50 ohms maybe) to your electrical radiation resistance.

You should also find (that with a simple dipole or quarter wave monopole) that the reactive component becomes smaller and therefore you more easily can dump power into the antenna. Here's what a monopole impedance looks like: -

enter image description here

At a quarter wavelength (optimum for a monopole) the reactive impedance becomes zero ohms and the resistive part (for an ideal monopole) becomes about 37 ohms. If the antenna is significantly shortened the real resistance rapidly approaches zero so, if there are real antenna losses (as always) any power pushed into the antenna is mainly wasted on heating the antenna and a significantly smaller percentage of power gets radiated.

\$\endgroup\$
4
  • \$\begingroup\$ I read that the impedance of the quaterwave monopole is 37+j21 ohm. So if I understand you correctly, I get the same radiated power, if I impedance match to 37+j21 ohm, or choose to shorten the antenna a bit until the reactive impedance is zero ohms (resonance), and impedance match to the shorted antenna. ? \$\endgroup\$
    – JakobJ
    Mar 1 '17 at 8:22
  • \$\begingroup\$ A theoretical quarter wave monopole has no complex impedance - see the graph in my answer. This may or may not be true of a practical antenna. Regards matching, this is needed if the cable length is a significant part of a wavelength. \$\endgroup\$
    – Andy aka
    Mar 1 '17 at 9:23
  • \$\begingroup\$ Found where I read the stuff about the impedance: monopole impedance I also read somewhere that you would trim the antenna until the imaginary part is zero, but if matched probably, is the trimming necessary ? \$\endgroup\$
    – JakobJ
    Mar 1 '17 at 9:46
  • 1
    \$\begingroup\$ Ah OK, yes I agree. Regards trimming, you can counter the complex impedance with an inductor or capacitor for sure but, if the antenna is too short (normally the case) the electrical radiation resistance drops whilst the antenna losses remain largely the same so you are on a downward spiral but, again, only if the antenna is particularly short (as per my answer). \$\endgroup\$
    – Andy aka
    Mar 1 '17 at 9:55
1
\$\begingroup\$

In practice, I always cut or design the antenna for resonance, and then match impedance if it matters all that much. http://chemandy.com/calculators/return-loss-and-mismatch-calculator.htm

A 2-1 mismatch results in a pretty trivial power loss.

In some cases, RF transceiver chips will radiate high levels of harmonics and cut their output power if the mismatch is large enough.

\$\endgroup\$

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.