RF Unit Converter (dBm, dBW, W, dBµV, dBmV, V at 50/75 Ω)

Understanding the RF unit converter — concept and principle

This converter changes radio-frequency signal levels between dBm, dBW, watts, milliwatts, dBµV, dBmV and RMS volts for a given system impedance, usually 50 Ω or 75 Ω. It also shows the peak-to-peak voltage of a sine wave. Radio amateurs, RF engineers, technicians and students use it to read datasheets and test equipment.

dBm is power on a log scale — with the default inputs: 0.707107 V RMS.

How it is calculated

dBm = 10 × log₁₀(P ÷ 1 mW); dBW = dBm − 30
P = V² ÷ R, with V the RMS voltage across a resistive load R
dBµV = 20 × log₁₀(V ÷ 1 µV); dBmV = dBµV − 60
Peak-to-peak voltage of a sine = 2√2 × V(RMS)

The principle

Decibels express ratios on a logarithmic scale, which keeps huge ranges of signal level manageable and turns gains and losses into additions. Power ratios use 10 × log₁₀; voltage ratios use 20 × log₁₀ because power is proportional to the square of voltage. A level in dBm or dBµV is an absolute level relative to a fixed reference (1 mW or 1 µV). Because P = V²/R, the same power produces a different voltage in a 50 Ω and a 75 Ω system: 0 dBm is 107.0 dBµV at 50 Ω and 108.8 dBµV at 75 Ω.

Worked example

Defaults: 10 dBm at 50 Ω.

P = 10^((10 − 30) ÷ 10) = 0.01 W = 10 mW; dBW = -20.
V = √(0.01 × 50) = 0.707107 V RMS.
dBµV = 20 × log₁₀(0.707107 ÷ 10⁻⁶) = 116.99; dBmV = 56.99.
Peak-to-peak = 0.707107 × 2√2 = 2 V.

dBm: 10.

Each point is the calculator re-run with only “Value” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Value	Power (dBm)
2.5	2.5
5	5
7.5	7.5
10	10
12.5	12.5
15	15
20	20

How to read the result

Every 3 dB doubles the power and every 10 dB multiplies it by ten. A signal 6 dB stronger has twice the voltage. When comparing receiver sensitivity or cable loss, make sure both figures use the same impedance and reference.

Common mistakes

Using 10 × log₁₀ for voltage ratios.
Mixing 50 Ω and 75 Ω figures.
Reading peak-to-peak or peak voltage as RMS.
Applying the sine-wave peak factor to modulated or noise signals.

Keep in mind: The conversions assume a purely resistive, matched load. Real systems with mismatch or complex impedance need more detailed analysis.

Frequently asked questions

How many milliwatts is 0 dBm?

1 mW.

What is the difference between dBm and dBW?

dBW is referenced to 1 W, so dBW = dBm − 30.

Why does impedance matter for dBµV?

Voltage depends on impedance for the same power, since P = V²/R.

Sources: ITU-R Recommendation V.574-4 (2000), "Use of the decibel and the neper in telecommunications"

Checked 9 Nov 2026. ITU-R Recommendation V.574-4 (2000), “Use of the decibel and the neper in telecommunications”, §6 (absolute power levels in dBW and dBm with 10 lg; field quantities with 20 lg); dBµV and dBmV as 20 lg of a voltage ratio to 1 µV and 1 mV; P = V²/R for the RMS voltage across a resistive load.

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