--- title: "Rhombic antenna working and design" subject: "Propagation and Antenna" chapter: 3 key: "semester_6/propagation_and_antenna:rhombic-working-design" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/rhombic-working-design site: BCT Engineering --- # Rhombic antenna working and design Propagation and Antenna · Chapter 3: Antenna Classification - Asked in: 2071 Bhadra, 5 marks - Video: [Rhombic Antenna (Basics, Structure, Types, Operation, Designing, Radiation & Array) Explained](https://www.youtube.com/watch?v=ub-CA0H5YX8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/rhombic-working-design - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3 ## Question Explain the working principle and design of Rhombic antenna. ## Short answer A rhombic antenna is a travelling-wave antenna of four long horizontal wires in a diamond, fed at one acute end and terminated in about 600–800 Ω. Each leg radiates a conical lobe; the tilt angle makes lobes add along the major axis, giving a unidirectional horizontally polarised beam toward the load. For elevation Δ: H = λ/(4 sin Δ), φ = 90° − Δ, L = 0.371λ/sin²Δ. ## Answer A **rhombic antenna** is a non-resonant (travelling-wave) antenna made of four long horizontal wires arranged as a rhombus, fed at one acute corner and terminated in a matched resistor at the other. ```text B / \ L = leg length feed / \ ===== A < → > C --R≈800Ω-- gnd \ / \ / height H above ground D φ = tilt angle = half of the obtuse angle at B, D ``` #### Working principle - The terminating resistor (≈ 600–800 Ω) absorbs the wave, so each leg carries a **travelling wave** and has a unidirectional conical lobe at an angle θ to the wire, pointing toward the load. - The tilt angle is chosen so that one lobe of each of the four legs points along the major axis A→C. These four lobes add in phase; the other lobes cancel in pairs. - The ground image adds to the direct wave at the wanted elevation angle Δ when the height is right. - The result is a unidirectional, horizontally polarised beam toward the terminated end, tilted up at angle Δ for sky-wave links. #### Design (for elevation angle Δ) The standard design for **maximum field intensity** at angle Δ: 1. Height above ground (image in phase): **H = λ/(4 sin Δ)** 2. Tilt angle: **φ = 90° − Δ** 3. Leg length: **L = 0.371λ / sin² Δ** The **alignment design** keeps H and φ the same but uses L = 0.5λ/sin² Δ, so that the main lobe of each leg points exactly at Δ (gives slightly less field but a cleaner pattern). Example: f = 15 MHz (λ = 20 m), Δ = 20° (sin Δ = 0.342): H = 20/(4 × 0.342) = 14.6 m, φ = 70°, L = 0.371 × 20 / 0.117 = 63.4 m. #### Characteristics - Wide bandwidth (about 2:1); impedance stays near 700–800 Ω. - High gain (≈ 10–15 dB), simple construction, easy to build. - Used for HF (3–30 MHz) long-distance point-to-point and sky-wave communication. - Drawbacks: very large area; about half the power is wasted in the termination; large side lobes. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What makes each leg of a rhombic antenna carry a travelling wave? A. Open far end B. A matched terminating resistor (600-800 Ω) at the far end C. Very short legs D. Vertical mounting Answer: B. The terminating resistance absorbs the wave, so there is no reflection and radiation is unidirectional toward the load. (Key point: Travelling wave operation) 2. For maximum field at elevation angle Δ, what is the height of the rhombic above ground? A. H = λ/(4 sin Δ) B. H = 0.371λ/sin²Δ C. H = λ sin Δ D. H = 90° − Δ Answer: A. The ground image adds in phase when H = λ/(4 sin Δ). (Key point: Height design) 3. For the maximum-output design, which pair is correct? A. φ = Δ, L = 0.5λ/sin Δ B. φ = 90° − Δ, L = 0.371λ/sin²Δ C. φ = 90° + Δ, L = λ/4 D. φ = 90° − Δ, L = 0.5λ/sin²Δ Answer: B. Maximum field design uses φ = 90° − Δ and L = 0.371λ/sin²Δ; 0.5λ/sin²Δ is the alignment design. (Key point: Tilt angle and leg length) 4. At 15 MHz (λ = 20 m) and Δ = 20° (sin Δ = 0.342), what is the height H? A. 14.6 m B. 5.0 m C. 29.2 m D. 63.4 m Answer: A. H = 20/(4 x 0.342) = 14.6 m. (63.4 m is the leg length.) 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