2026-11-05
Many inquiries start from the model: "quote me 1/2-inch cable". But 1/2 inch is only an outer diameter; what really decides the choice is how many dB the link can afford. Working backwards, three steps land you on a specific size.
Step 1 — Compute the allowable loss
Allowable cable loss = TX power − RX sensitivity − antenna gain − all non-cable losses − design margin.
Example: RRU output 43 dBm, receiver sensitivity −100 dBm, antenna gain 17 dBi, connectors and devices 3 dB combined, margin 3 dB. The cable may lose 43 − (−100) − 17 − 3 − 3 = 74 dB.
Step 2 — Take the figure at the top band
The same cable loses far more at 2600 MHz than at 900 MHz. Always check the budget at the highest operating frequency; lower bands then take care of themselves.
Step 3 — Work back through the dB/100 m curve
Divide the allowable loss by the run length to get dB per 100 m, then read the attenuation curves to find the smallest diameter that fits.
Example: 60 m run, 74 dB budget, 12.3 dB per 100 m allowed. Curves at 2600 MHz: 1/2-inch corrugated copper tube ≈ 10.7 dB/100 m — fits; 3/8-inch ≈ 16 dB/100 m — does not. So 60 m at 2600 MHz needs at least 1/2 inch. Stretch the run to 120 m and the allowance drops to 6.2 dB/100 m, which calls for 7/8 inch (≈ 6.1 dB/100 m).
Three common traps
- Sizing from the 900 MHz curve, then running out of budget at 2.6 GHz.
- Treating 0.1–0.3 dB per joint as rounding error: twenty joints is 2–6 dB.
- Ignoring the extra loss from a bad VSWR — at 1.5 the return loss is about 14 dB, and the margin is gone.
Send us the run length, band and loss budget and our engineers will size it from measured curves and offer two options.