Ohmsums

dBm to watts converter

Type in any field: dBm, dBW, W, mW or RMS voltage. The others follow as you type. Voltage is across the impedance you choose (50 Ω by default).

Indicative calculation; verify before use.

dBm
W
mW
dBW
V rms
Ω

Power
–
Voltage across 50 Ω
–
dBW
–
Peak voltage (sine)
–
Peak-to-peak voltage (sine)
–
RMS current
–
dBµV (across the load)
–

Common power levels

Tap a dBm value to load it. Voltage column uses the impedance above.

dBmPowerdBWV rms, 50 ΩTypical example

How it works

P [W] = 10^((dBm − 30) / 10) dBm = 10 × log10(P / 1 mW) dBW = dBm − 30 V rms = √(P × R) I rms = √(P / R) V peak = √2 × V rms V p-p = 2 × V peak (sine wave) dBµV = 20 × log10(V rms / 1 µV)

dBm is power relative to one milliwatt on a logarithmic scale: +3 dB doubles the power, +10 dB is ten times the power, and 0 dBm is exactly 1 mW. dBW is the same thing relative to one watt, so it is always 30 lower than the dBm figure.

A power level alone has no voltage. The voltage depends on the resistance it is delivered into: 0 dBm is 0.224 V rms in a 50 Ω RF system, 0.274 V in 75 Ω (TV and video) and 0.775 V in 600 Ω (classic audio, the 0 dBu reference). The peak and peak-to-peak values assume a pure sine wave; for modulated signals they describe the carrier, not the envelope peaks.

This converts a power that is actually delivered into a matched resistive load. It does not include cable loss, antenna gain (dBi/ERP/EIRP) or mismatch; use the SWR calculator for the last one.

Worked example: 5 W is 5000 mW, so 10 × log10(5000) = 36.99 dBm, or 6.99 dBW. Across 50 Ω: V rms = √(5 × 50) = 15.81 V, V peak = √2 × 15.81 = 22.36 V, V p-p = 44.72 V, and I rms = √(5 / 50) = 0.316 A.

Sources (checked 2026-10-31):