How do I calculate the number of elements required for a phased array to achieve a given beamwidth?
Array Sizing
The number of elements is driven by two independent requirements: beamwidth (which sets the total aperture size) and element spacing (which prevents grating lobes). For half-wavelength spacing: N = aperture_length / (λ/2) = 2 × aperture_length/λ. For a 1° beamwidth at broadside: aperture_length ≈ 50λ, requiring N ≈ 100 elements per dimension (10,000 for a pencil beam).
| Parameter | Low Gain | Medium Gain | High Gain |
|---|---|---|---|
| Gain Range | 2-6 dBi | 6-15 dBi | 15-45 dBi |
| Beamwidth | 60-360° | 15-60° | 1-15° |
| Typical Types | Dipole, monopole, patch | Yagi, helical, horn | Parabolic, array, Cassegrain |
| Bandwidth | Narrow to wide | Moderate | Narrow to moderate |
| Complexity | Low | Medium | High |
Frequently Asked Questions
What if I need more gain but can't add elements?
Increase the element gain (use horn or stacked patch elements instead of simple patches), improve the aperture efficiency (optimize the amplitude taper), or increase the operating frequency (gain scales as f²). But the beamwidth narrows with higher gain, which may not be desired.
How does this scale at mmWave?
At mmWave, the small element spacing (2.5 mm at 60 GHz) allows thousands of elements in a compact area. A 1024-element array at 60 GHz fits in a 80mm × 80mm area and provides about 33 dBi gain with a 3° beamwidth. This is the approach used in 5G mmWave base stations.
Does element count affect cost linearly?
Approximately yes. Each element requires a T/R module (phase shifter, amplifier, receiver), and the T/R module cost dominates the array cost. For military AESA: $100-1000 per element. For commercial 5G: $1-10 per element (with integrated RFIC packages that contain multiple channels).