Convert between dBm and power in mW or W, as used in RF, communications and audio, and see the voltage that power produces across a load. You can also check what a dB value means as a power ratio and as a voltage ratio.
Your inputs are kept in this page's URL, so a bookmark or a shared link opens the same calculation. Favorites and recently used calculators are stored only in this browser (localStorage).
Voltages assume a sine wave delivering this power into the load resistance. Editing either dBm or mW updates the other.
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
dBm expresses power relative to 1 mW: P[dBm] = 10·log10(P/1 mW). To convert back, P[mW] = 10^(dBm/10) and P[W] = 10^(dBm/10)/1000. dBW uses 1 W as the reference, so it equals dBm − 30. dBµV is a voltage unit referenced to 1 µV, equal to 20·log10(V/1 µV). The default of 0 dBm is 1 mW, or −30 dBW. Taking a ratio and then a logarithm lets you handle powers that span many orders of magnitude with simple addition and subtraction.
Once the power is known, the RMS voltage of a sine wave across a load R is Vrms = √(P·R). With the defaults of 0 dBm (1 mW) and R = 50 Ω, Vrms = √(0.001 × 50) ≈ 223.6 mV, the peak-to-peak voltage is 2√2 times that, about 632.5 mV, and the level is about 106.99 dBµV. In a 50 Ω system, “dBµV ≈ dBm + 107” is a handy shortcut. 50 Ω is the characteristic impedance used as the standard in much RF equipment and coaxial cable. Video and similar equipment also use 75 Ω, where the same 0 dBm corresponds to about 273.9 mV.
A dB value is not a unit but a ratio. For power, dB = 10·log10(P2/P1). Because power is proportional to the square of voltage (P = V²/R), a voltage ratio across the same resistance becomes dB = 20·log10(V2/V1). The default of 3 dB is a power ratio of about 1.995 and a voltage ratio of about 1.413. Remembering that 3 dB is roughly twice the power, 6 dB roughly twice the voltage (four times the power), 10 dB exactly ten times the power and 20 dB ten times the voltage (100 times the power) makes mental estimates easy.
Useful reference points: −30 dBm is 1 µW, 0 dBm is 1 mW, 10 dBm is 10 mW, 20 dBm is 100 mW, 30 dBm is 1 W and 40 dBm is 10 W. Adding a gain or loss in dB to a power in dBm gives the power in dBm after that stage. Adding two dBm values together, however, has no meaning; to combine the power of two signals, convert them to mW first and then add. Two uncorrelated 0 dBm signals, for example, combine to about 3 dBm. Converting to a voltage only works for a sine wave into a known load resistance. With the 600 Ω long used in audio, 0 dBm corresponds to about 0.775 V.
Practice analog circuit questions
Answer: B. 100 mW
dBm expresses power relative to 1 mW as P[dBm] = 10·log₁₀(P/1 mW). Conversely, P = 10^(20/10) mW = 10² mW = 100 mW. Power rises tenfold every 10 dB and roughly doubles every 3 dB, so 0 dBm is 1 mW, 10 dBm is 10 mW and 23 dBm is about 200 mW. Working in dBm lets gains and losses (dB) be handled by addition and subtraction.
Use P[W] = 10^(dBm/10)/1000. 10 dBm is 0.01 W, 20 dBm is 0.1 W, 30 dBm is 1 W and 40 dBm is 10 W, so every 10 dB is a factor of ten in power. To go from watts to dBm, use dBm = 10·log10(P[W]) + 30.
dB expresses a ratio between two quantities and has no absolute size on its own. dBm is a ratio to 1 mW, so it represents an absolute power. Amplifier gain and cable loss are given in dB, while transmit power and received levels are given in dBm. The two combine: a 10 dBm signal through a 20 dB amplifier becomes 30 dBm (1 W), as long as the amplifier is not saturated.
dB is defined as 10·log10 of a power ratio. Since power is proportional to voltage squared, across the same resistance 10·log10(V2²/V1²) = 20·log10(V2/V1). So doubling the voltage is about 6 dB, and ten times the voltage is 20 dB. When comparing voltages at two points with different resistance (impedance), the 20·log value no longer equals the power ratio in dB.