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Tool 01 · Available

RF Unit Converter

Translate absolute power and sinusoidal voltage without losing the impedance or RMS convention.

Typical inputs
Absolute power or sinusoidal voltage, input unit, and real resistance; or a dimensionless dB ratio.
Calculated output
W, dBm, dBW, RMS/peak/peak-to-peak voltage, voltage levels, and linear ratios.
Model status
Release A · documented and numerically tested
Engineering question

What does this level mean in watts, dBm, and volts at a stated resistance?

Conversion workbench

Absolute power and sinusoidal voltage remain separate from dimensionless gain/loss ratios. Resistance is used only when deriving voltage and power from one another.

Absolute level0.000000 dBm1 mW in 50 Ω; voltage assumes a sine wave
Power
1 mW
dBW
-30.000000 dBW
RMS voltage
223.607 mVrms
Peak voltage
316.228 mVpk
Peak-to-peak
632.456 mVpp
Voltage levels
-13.0103 dBV
46.9897 dBmV · 106.9897 dBµV

Calculation path

  1. P[dBm]=10log⁡10(P1mW)P[\mathrm{dBm}] = 10\log _{10}(\frac{P}{1 \mathrm{mW}})
  2. Vrms=PRVpk=2 VrmsVpp=2Vpk\begin{aligned}V_{\mathrm{rms}}&=\sqrt{PR}\\V_{\mathrm{pk}}&=\sqrt2\,V_{\mathrm{rms}}\\V_{\mathrm{pp}}&=2V_{\mathrm{pk}}\end{aligned}
  3. Negative dBm still represents positive physical power below 1 mW.
Model

What this calculator is doing

One input is resolved in linear SI units before every other representation is derived. Voltage-to-power results use the entered real resistance and an RMS sine-wave convention.

Decision

How to read the result

Use absolute results for instrument levels and ratings. Use ratio mode only for relative gain or loss; a bare dB value is not a power level.

Equations & conventions

  • P[dBm]=10log⁡10(P1mW)P[\mathrm{dBm}] = 10 \log _{10}(\frac{P}{1 \mathrm{mW}})
  • P=Vrms2RP = \frac{V_{\mathrm{rms}}^{2}}{R}
  • Vpk=2VrmsV_{\mathrm{pk}} = \sqrt{2} V_{\mathrm{rms}}
  • P is average power in watts; P[dBm] is referenced to 1 mW.
  • R is a positive real resistance in ohms.
  • Vrms, Vpk, and Vpp use a sinusoidal waveform convention.

Independently checked example

Checked example: 0 dBm is exactly 1 mW. In 50 Ω that is 223.6068 mVrms, 316.2278 mVpk, and 632.4555 mVpp.

Common mistakes

  • Treating dB and dBm as interchangeable.
  • Reading negative dBm as negative physical power rather than a positive power below 1 mW.
  • Converting dBm to voltage without stating impedance.
  • Assuming the Vpp/Vrms relationship is identical for every waveform.

Assumptions, validity & omissions

  • Voltage results describe a sine wave.
  • The entered resistance is positive and purely real.
  • A dB ratio is kept separate from an absolute level.

Sources & model provenance

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