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2 years ago in Electrical Engineering By Sumitra R
How do aperture efficiency and ohmic (conductor/dielectric) losses quantitatively combine to set the upper bound on an antenna’s overall radiation efficiency?
I'm analyzing a high-gain reflector antenna. My simulation shows high radiation efficiency, but I want to understand the fundamental limits. I know radiation efficiency (η_rad) = P_rad / P_accepted. Is it correct to think of this as: η_rad = (Aperture Efficiency) * (1 - Ohmic Loss Fraction)? Or are these factors interrelated in a more complex way? For instance, does a highly tapered illumination (high aperture efficiency) reduce spillover but also increase current density and thus ohmic loss in the feed? How do I budget for these effects in a preliminary design?
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By Martha Whaley Answered 1 year ago
They are independent multiplicative factors in a well-designed system. Think of it as a cascade: Total Efficiency (η_total) = η_illumination * η_spillover * η_ohmic. Aperture Efficiency (η_aperture) itself combines illumination efficiency (how uniform the phase/amplitude is) and spillover efficiency (how much feed energy misses the reflector). These are about power distribution. Ohmic loss (η_ohmic) is about power dissipation in metals and dielectrics. They are largely independent; a perfect illumination doesn't change the conductivity of aluminum. However, your intuition is correct: a highly tapered feed pattern (for good η_spillover) often requires a larger, more complex feed horn, which may have slightly higher ohmic loss due to more surface area. In practice, you design the feed for optimal η_aperture first, then select low-loss materials and large enough conductors to keep η_ohmic > 95%. The product gives you your realistic upper bound.
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