RF exposure around small and macro cells
How macro sites and small cells differ for RF safety: geometry, near/far-field implications, boundary setting, and defensible measurement workflows.
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ContinueA lower-power antenna can give a larger field at a nearby accessible point than a higher-power antenna far away. The comparison depends on direction, distance, access and operating state. “Macro” and “small cell” describe deployments; they do not supply an exposure ranking.
Use near and far fields and field and power-density quantities. For a survey, you will also need measurement uncertainty. The free-space calculations below are explicit screening models, not site compliance determinations.
Find the places people can occupy
Begin with accessible sidewalks, balconies, rooftops, ladders and maintenance positions. State which people can enter each location and under what operating conditions. A rooftop is not automatically a controlled-exposure area because people there are employed: the access and training conditions must actually apply.
An antenna high above pedestrians can direct its main lobe towards an occupied region some distance away. A low-mounted node can place accessible locations close to its aperture. Record heights, antenna directions and three-dimensional distances, rather than measuring only the nearest point on a map.
Use downtilt to find candidate locations
For a straight beam-centre direction on level ground, the vertical drop is h-hₚ and the horizontal distance is d = (h-hₚ)/tan θ. Here h is antenna height, hₚ is the observation height and θ is downtilt below horizontal. This locates one geometric candidate, not the field maximum: beam width, sidelobes, obstructions and reflections remain relevant.
Worked example 1
Locate a macro antenna beam at pedestrian height
Problem
A quick geometry check helps you avoid sampling only “closest point”.
Assume level ground, an antenna centre at h = 25 m and a straight beam-centre direction downtilted by θ = 6° below horizontal. Approximate where it intersects pedestrian height hₚ = 1.5 m.
Interpret what the estimated intersection means for a public-access sampling plan.
Show full solution
Identify the height difference and downtilt
Method
Use antenna height, pedestrian height and the downtilt angle.
Reason
The beam-centre intersection follows from the vertical height difference and beam direction.
Working
Answer
The vertical drop is h-hₚ = 25-1.5 = 23.5 m. The downtilt is 6° below horizontal; use degrees when evaluating its tangent.
Estimate the horizontal intersection distance
Method
Divide the height difference by the tangent of the downtilt angle.
Reason
The right-triangle geometry relates the vertical drop to horizontal distance.
Working
Answer
d ≈ (h-hₚ)/(tan θ) = (25-1.5)/(tan 6°) ≈ 23.5/0.105 ≈ 224 mInterpret the sampling implication
Working
Answer
Interpretation: the main lobe can intersect occupied space hundreds of meters away. That doesn’t mean the maximum is exactly at 224 m (pattern and clutter matter), but it strongly suggests you should sample beyond “under the tower” if your goal is maximum public-access assessment.
Include direction before comparing powers
Equivalent isotropically radiated power (EIRP) expresses radiated power in a specified direction as an equivalent isotropic source. If P₀ is boresight EIRP and g is the normalised power-pattern factor in the direction of the point, directional EIRP is P₀g. Do not multiply a directional EIRP by the antenna gain a second time.
For a free-space travelling-wave model in a justified far field, S = P₀g/(4π r²). Use slant distance r in metres; g is dimensionless and S is in W/m². Doubling distance at fixed directional EIRP quarters the modelled power density. That scaling is conditional: it does not model a near field or indoor interference.
Guided practice 2
Compare two accessible points
Problem
Two 3.50 GHz sources are assessed separately in free space, at points already justified as far-field-like. Their EIRPs describe the same operating and averaging state. The macro source has boresight EIRP 2000 W; its point is 100 m away with normalised power-pattern factor 0.100. The small-cell source has boresight EIRP 20.0 W; its point is 10.0 m away with factor 0.250.
Find the modelled incident power density at each point in mW/m² and the small-cell-to-macro ratio. Explain why boresight power alone gives the wrong ranking, and why these two calculations cannot establish a whole site’s maximum.
Make the decisions, then compare your working
Hints
Hint 1
Use the direction of each point
Directional EIRPs are 2000(0.100) W and 20.0(0.250) W. The given values already include boresight antenna gain.
Hint 2
Compare like operating states
Apply S = P₀g/(4π r²) to each, convert W/m² to mW/m², then form a dimensionless ratio.
Show solution step by step
Determine directional powers
Method
Multiply each boresight EIRP by its power-pattern factor.
Reason
The observer is not necessarily on the strongest beam direction.
Working
Answer
The directional EIRPs are 200 W for the macro point and 5.00 W for the small-cell point. These are not the transmitters’ conducted powers.
Calculate the two incident densities
Method
Divide each directional EIRP by the corresponding spherical area.
Reason
The stipulated far-field free-space conditions make this screening model applicable.
Working
Answer
At the macro point, 200/[4π(100)²] W/m² equals 1.59 mW/m². At the small-cell point, 5.00/[4π(10.0)²] W/m² equals 3.98 mW/m². Retain guard digits before comparing.
Interpret the ranking
Method
Form the small-cell-to-macro ratio.
Reason
A hundredfold difference in boresight EIRP is modified by direction and by the squared-distance ratio.
Working
Answer
The ratio is 2.50. The lower-boresight-power source gives the larger result at these particular points. Other accessible locations, co-located sources, operating states and reflections remain unassessed. Compare the model and scope yourself; the checker verifies the two densities and their ratio only.
Reassess a newly accessible position
Check your understanding 1 An access route changes
A changed access route lets a worker approach to 0.500 m from the same small-cell aperture. Its largest active dimension is 0.300 m at 3.50 GHz. A report multiplies the 10.0 m screening result by (10.0/0.500)² and treats it as the assessment for the new route.
Evaluate that argument and propose the next assessment and control decisions. Use c = 3.00 × 10⁸ m/s.
Show answer
The wavelength is 0.0857 m and the aperture estimate is 2D²/λ = 2.10 m. The new point lies inside it. The 10 m far-field result therefore cannot simply be extrapolated with inverse-square scaling to establish the new exposure.
Map the route and credible operating states, use a validated method suitable for its field region and the required metric, and include other sources and uncertainty. The earlier access assumption has changed: keep access within the established assessed boundary until the revised assessment and required controls are in place. Employment alone does not establish a controlled-exposure classification. Compare your answer for the failed model condition, the new accessible region and an enforceable interim boundary.
Turn candidate points into a survey
Patterns and reflections can create maxima and minima. Search the accessible region across relevant heights and operating states, then make the required time and spatial averages at defined locations. Keep the detector settings, frequency coverage, source configuration, search resolution and uncertainty with the result. IEC 62232:2025 addresses base-station RF assessment methods; use the applicable edition and procedure rather than treating the free-space formula as a complete method.
Review the geometry decisions
Check your understanding 2
Why can “closest distance to the antenna” be a poor proxy for maximum exposure near macro sites?
Show answer
Macro antennas are typically downtilted and mounted high; the main lobe often intersects occupied space at an offset distance. “Closest point” may not be accessible or may be off-beam.
Check your understanding 3
What makes small cell assessments operationally different?
Show answer
Small cells are often mounted lower, so accessible points can be closer to the aperture and may overlap near-field distances. Indoor/street reflections can also dominate spatial variability.
Check your understanding 4
A report calls one sidewalk reading “the site’s maximum.” What information is needed to justify that description?
Show answer
State which accessible region and operating states were assessed, how maxima were searched across positions and heights, the required averaging and the uncertainty. One sidewalk reading establishes only that sampled result; it cannot establish a site maximum without a justified search or bound.
Check your understanding 5
On level ground, an antenna’s beam centre is at h = 20 m and tilted 5° below horizontal. At what horizontal distance does it reach pedestrian height hₚ = 1.5 m, using d = (h-hₚ)/tan θ?
Show answer
d = 18.5 m/tan 5° ≈ 211 m. This is a beam-centre geometry estimate, not proof of the exposure maximum; pattern, sidelobes, reflections and access still determine the sampling plan.
State what the comparison covers
A beam-centre intersection identifies where to investigate; it does not identify a guaranteed maximum. A point calculation supports its stated geometry and operating state. When access, antenna configuration or co-located sources change, reconsider both the model and the controls.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.
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