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Calcrivo

WiFi Coverage

Estimate WiFi access point coverage radius and area based on power and environment.

Inputs

dBm
dBm

Minimum signal a client needs for a usable connection, typically -65 to -75 dBm

dB

Typical interior wall: 3-6dB; concrete/brick: 10-15dB

Estimated Range per AP

75.7m

Coverage Area per AP

17,997

Total Coverage Area (all APs)

17,997

Max Allowed Path Loss Budget

84.0dB

Step by step

  1. Values used

    Number of Access Points = 1; AP Transmit Power = 20 dBm; Frequency Band = 5 GHz (less range, less interference); Client Receiver Sensitivity = -70 dBm; Wall/Obstruction Attenuation = 6 dB

  2. Free-space path loss

    FSPL(dB) = 20·log10(distance_km) + 20·log10(freq_MHz) + 32.44

  3. Maximum range from link budget

    range = 10^((tx_power − rx_sensitivity − wall_loss − 20log10(f) − 32.44) / 20)

  4. Estimated Range per AP

    = 75.7 m

  5. Coverage Area per AP

    = 17,997 m²

  6. Total Coverage Area (all APs)

    = 17,997 m²

  7. Max Allowed Path Loss Budget

    = 84.0 dB

How it works

WiFi range is estimated using the free-space path loss (FSPL) model, which relates signal loss to distance and frequency: FSPL(dB) = 20·log10(distance_km) + 20·log10(freq_MHz) + 32.44. Rearranging for distance, given the maximum path loss the link budget allows (transmit power minus receiver sensitivity minus any wall/obstruction attenuation), yields the theoretical range. Higher frequencies (5/6 GHz) attenuate faster over distance than 2.4 GHz, which is why 2.4 GHz networks reach further but 5/6 GHz offer more capacity and less interference at closer range. Real-world range is usually shorter than this free-space estimate due to additional obstructions, multipath, and interference not captured by this simplified model.

Formulas

Free-space path loss

FSPL(dB) = 20·log10(distance_km) + 20·log10(freq_MHz) + 32.44

d
Distance in kilometers
f
Frequency in MHz

Maximum range from link budget

range = 10^((tx_power − rx_sensitivity − wall_loss − 20log10(f) − 32.44) / 20)

Frequently Asked Questions

Why does 5 GHz have less range than 2.4 GHz at the same power?

Higher frequency signals attenuate faster over distance and are more easily blocked by obstacles, which is a direct consequence of the FSPL formula's 20·log10(freq_MHz) term — doubling frequency from 2.4 to roughly 5 GHz adds about 6.3 dB of extra path loss at the same distance.

Why is real-world WiFi range usually shorter than the free-space estimate?

The free-space model assumes an unobstructed line-of-sight path with no reflections, interference, or absorption. Real environments have walls, furniture, other WiFi networks sharing spectrum, and multipath fading, all of which reduce effective range below the theoretical free-space number.

What receiver sensitivity should I use for a typical client?

Most consumer WiFi clients need roughly -70 dBm or stronger for a solid, high-throughput connection, though they may still connect (at lower speeds) down to around -85 to -90 dBm. Video calls and high-bandwidth applications generally want -65 dBm or better.

How does adding more access points help coverage?

Each additional AP adds its own coverage cell, so total coverage area scales with AP count assuming cells don't need to overlap much — but AP placement, overlap for roaming, and interference between APs on the same channel all matter in real deployments, which this simplified model doesn't capture.

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