# How can I maximize the amount of energy an antenna is receiving?

Bear in mind that I am very new to electrical design and have only a beginners level circuit course under my belt.

I am trying to build an AM radio and although there are a lot of tutorials out there for this, none of them answer this specific question about designing a receiver. The best resource I have found so far is: http://www.antenna-theory.com/basics/gain.php, but it seems to be talking about sending (not receiving) signals.

I would like to know how I can make/modify a receiver so that I can maximize the amount of energy it will get from the carrier wave.

Is there a very good resource on making receivers that will explain the math and theory? Please point me in the right direction so I can gain some understanding and get through this project.

• Receiving and transmitting antennas are identical in terms of dimensions, so you can use the same calculations to maximise receive efficiency. Sep 28 '15 at 18:55
• RF isn't my focus, but the most important item is 1/4 carrier wavelength. If it's shorter than this, you need to inductively load the antenna, and if it's longer you need to capacitively load the antenna. This shifts the reactive components to cancel each other out and make the reception purely resistive. Look up 'electrical length.' Sep 28 '15 at 19:14
• Are you trying to extract energy from a radio transmission? Sep 28 '15 at 20:13
• @Andyaka Not extract energy, but maximize power received. There was a previous design that failed because the power received wasn't enough to be input into the RF Amp circuit.
– Klik
Sep 28 '15 at 23:19
• What we're the parameters of the transmitter and receiver. It's quite possible that a more specific answer can be provided. Sep 29 '15 at 7:09

By the principle of reciprocity, antennas work the same way in receive and transmit directions. So if you make a directional antenna that's better at transmitting in a particular direction, it's better at receiving in that direction by the same amount.

If your goal is maximizing power received, then a useful concept is effective aperture. This is the size of the metaphorical "net" the antenna uses to capture energy. A bigger net means capturing more energy. If you are familiar at all with the term "aperture" in optics and photography, it is a similar concept.

A somewhat counter-intuitive mathematical fact is that effective aperture ($A_{eff}$) and gain ($G$) are related:

$$G = \frac {4 \pi A_{eff} } { \lambda^2 }$$

$\lambda$ is the wavelength, and people tend to develop a misconception here: that the physics of energy transmission are somehow different depending on frequency. They aren't. All electromagnetic radiation decreases with distance according to the inverse square law, be it AM broadcasts, visible light, or gamma radiation. See Is free space path loss dependent on frequency? and Why is antenna aperture a function of wavelength?

What this equation is telling you is this: as frequency decreases and gain remains constant, the aperture increases. But a half-wavelength dipole for 750 kHz is also physically much larger than a half-wavelength dipole for 2.4 GHz, so although they each have the same gain, it would make sense that the 750 kHz dipole has a larger aperture.

Another counter-intuitive result is that an ideal dipole which is infinitesimally small has about the same gain (and effective aperture) as a half-wavelength dipole: 1.76 dBi compared to 2.15 dBi, respectively.

If that's true, then why don't we use infinitesimally small dipoles everywhere? We could save a lot of space. The reason is that to efficiently couple energy with this tiny antenna, you need some kind of matching network, and those parts introduce losses, degrading the efficiency of the system.

But this is still insightful: if you want to maximize received power, focus on minimizing losses first. With the tiny powers you will capture from an antenna, you probably need it to be as efficient as possible, so design for simplicity since each component adds loss. And match your load to the antenna impedance to make the transfer of energy as efficient as possible.

• Thank you for this answer. You've given me a lot to go off of and continue researching the topic. I had thought that gauge of the wire played a role, but I guess not. Anyway, this will definitely help me construct my radio antenna receiver. Cheers!
– Klik
Sep 28 '15 at 23:23
• The wire gauge plays an enormous role. "Focus on minimizing losses first." Antennas far shorter than 1/2-wv dipoles will involve proportionally higher current, and therefore larger losses to conductor heating. Ideally you'd use superconductors and vacuum capacitors. In other words, get your conductor resistances down way below an ohm, and chose dielectrics for extreme low loss (low heating at high current.) At 750KHz, a small desktop antenna, if composed of ideal materials, will pull in all the EM wave energy within a 50-meter radius. Of course for real materials YYMV! Sep 29 '15 at 0:52