Continuing The Discussion
In the June issue of Microwaves & RF, I presented my request to clarify Fig.1 in the article, "Antenna Assists MW Power Transmission". The author failed to respond and explain it, and I still do not understand it. In describing the figure, he wrote that the DC load was next to the detector diode while in a rectenna, the diode is loading the antenna. The DC load can be located farther away, and it affects rectenna efficiency as well as matching. I found no clear explanation to those problems.
I also wanted clarification as to why the rectenna efficiency "exceeded 88 percent." A single-phase rectifier operates (diode opens) for only one-half of the time. In principle, its efficiency cannot be higher than 50 percent. A single-phase rectifier cannot have more than 50 percent efficiency at 60 Hz, so I don't understand how Kumar can make it 88 percent at RF.
I am also surprised by his statement, "This rectenna has nothing to do with the propagation loss." I then wonder how he fed RF power into his rectenna. He failed to describe that he had to use a transmitter with another antenna and to send RF power over some distance (hence with a propagation loss) to his rectenna.
I did study several reports by NASA and JAXA, and I was surprised how those scientists defined efficiency of microwave power transmission from DC to DC. I failed to find a clear definition. The first problem is that DC is converted to RF or microwave power in an electron tube (magnetron, klystron, or similar) or in a solid-state device. To my knowledge, any such device has an efficiency of less than 50 percent. Next, the RF or microwave power must be radiated by an antenna; a good microwave antenna rarely has efficiency better than 50 percent. Then, the radiated power is transmitted over a distance to a remote rectenna. Thus, this power is heavily attenuated by propagation loss: from the geostationary orbit, at 2.45 GHz, this loss is almost 200 dB (from the moon, much more). Therefore, only one part in 10E20 of the transmitted power reaches the rectenna.
So in the end, the overall system efficiency is the product of individual efficiencies. Even with Kumar's miraculous (and I think impossible) 88 percent, the overall efficiency will be: Ktot~ 0.5 x 0.5 x 0.88 = 0.22, without the propagation loss. By recognizing it, there will be no efficiency at all.
Best regards,
Jiri Polivka
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