# How do half-wave RF rectifiers manage efficiencies above 50%?-A theory

First off, this exact question was directly asked here EDAboard and I think it was hinted at in atleast two other posts on this exchange programmingEnthusiast, frick, by Andy AKA, who mentioned a 100% conversion efficiency for ideal diodes. Efficiencies over 50% are obtained from multiple rectenna papers that designed RF rectifiers using half-wave topologies, such as in the paper by McSpadden et.al. The referenced paper presents some equations for the half-wave rectifier efficiency but its honestly hard to see why the efficiency is over 50%. It also mentions and distinguishes overall (no reflected power) and conversion (with reflected power) efficiencies though they only ultimately only differ by 1%.

Based on the aforementioned posts, I think the reasons given are that 1)Providing a DC path separate from the RF path gives higher efficiencies, 2)Ideal diodes have 100% conversion efficiency, 3)Reflections within the signal path leading to further rectification of harmonics or fundamental.

The thing I dont understand with reasons 1) and 2) is how anything over 50% power can be achieved when only half the input power is being rectified over one period, regardless of diode efficiency or DC path. What I'm thinking that makes makes more sense to me is related to reason 3) which is that the ground connected diode, reflects the polarity of the negative half of the input waveform (as RF shorts do) passing through it to a positive voltage. You then effectively get full-wave rectification once you add in the original positive half-waveform that didnt originally pass through the diode.

I'm having a tough time understanding that EDAboard post and some of the other answers posted here so could someone please elaborate on this?

Edit: To clarify the image, the low pass filter serves as a matching network between the dipole and diode impedance, and has a secondary purpose of reflecting generated harmonics. The diode is a schottky diode (MA40150-119) with a 0.4V forward voltage. The DC pass filter used in the paper is a capacitor used to short the RF signal and it also is spaced apart from the diode to cancel the diode capacitance.

Edit2: Paper that calculates rectification efficiencies for ideal diode with half-wave topology. Achieves 93% efficiency.[https://ieeexplore.ieee.org/document/9055534]

• I must be missing something. How would a dipole antenna with an LPF inserted ever generate enough voltage to forward bias the diode? Commented Apr 22 at 20:08
• Its a schottky diode, so its got a really low forward voltage. The LPF is mainly to match the antenna to the diode, and also reflect any generated harmonics for re-rectification. Commented Apr 22 at 20:10