dBm to Watts Calculator (dBm, dBW, dBu and Volts)
Convert dBm to watts and back, with dBW, milliwatts and dBu, plus the RMS voltage and current that power gives across 50, 75, 600 Ω or any impedance.
dBm to Watts Calculator
Background.
A router's datasheet lists its transmit power as 20 dBm. The phone two rooms away reports the signal at −65 dBm, and the radio inside it is rated to work down to −90 dBm. In watts those figures run from 100 milliwatts to one picowatt, a span of a hundred billion. On the decibel scale they are 110 apart, and that compression is the whole reason radio engineers work in dBm.
dBm is an absolute power level: ten times the base-10 logarithm of a power measured in milliwatts. The m names the reference, one milliwatt, and the reference is what lets the number stand for an actual power. ITU-R Recommendation V.574-5, which sets out how the decibel is used in telecommunications, writes the level as dBm when the reference is a milliwatt and dBW when it is a watt. A bare dB has no reference at all; it describes a gain or a loss. That is why an amplifier's 20 dB can be added to a level in dBm, while adding two dBm figures together means nothing.
Three anchors cover most of the scale. 0 dBm is 1 mW, +30 dBm is 1 W, −30 dBm is 1 µW, and each further 30 dB is another factor of a thousand.
Enter a level in dBm or dBW, a power in watts or milliwatts, or an audio voltage level in dBu, and every other form comes back, along with the RMS voltage and current that power produces across the impedance you set. That last step is the one you cannot do in your head. The same milliwatt is about 0.2236 V across 50 Ω and about 0.7746 V across 600 Ω.
What is dbm to watts calculator?
An absolute power level compares a power with a fixed reference and states the ratio in decibels: L = 10 · lg(P ÷ P₀). With P₀ = 1 mW the result is in dBm; with P₀ = 1 W it is in dBW, and because a watt is a thousand milliwatts, the dBW figure is always exactly 30 lower. Satellite and broadcast link budgets tend to use dBW, where powers are large, while receivers, test equipment and radio modules use dBm. dBu belongs to a neighbouring family: it is a voltage level, 20 · lg(V ÷ 0.775 V), and lines up with the power level in dBm only in a 600 Ω circuit, the resistance in which 0.775 V dissipates about a milliwatt. ITU-R V.574-5 §6.5 gives the bridge as Lu = Lp + 10 · lg(R ÷ 600).
How to use this calculator.
- Under "What have you got?" choose the form your number is already in: dBm, dBW, watts, milliwatts or dBu. The list works in both directions, so watts to dBm is simply the watts option. Only the field for the chosen option is read.
- Enter the value. Negative dBm and dBW figures are normal and just mean the power is below the reference; watts and milliwatts must be above zero, because zero power has no logarithm.
- Set System impedance to what your equipment actually uses before reading volts or amps: 50 Ω for RF test gear and most radio coax, 75 Ω for TV antenna feeds and video, 600 Ω for legacy line-level audio, 4 to 8 Ω for a loudspeaker. In dBu mode it also sets the power.
- Read the summary for the power in a unit that suits its size and the thousand-fold band it falls in, then the individual lines for each unit.
- To include an amplifier, an antenna gain or a cable loss, add or subtract its dB figure from the level first and convert the result afterwards.
The formula.
From a level to a power: P(mW) = 10^(L ÷ 10), and dividing by 1,000 gives watts. A dBW entry goes straight to watts, P(W) = 10^(L ÷ 10). In the other direction, L(dBm) = 10 · lg(P in mW) and L(dBW) = L(dBm) − 30, since 10 · lg 1,000 is exactly 30.
Voltage and current come from P = V² ÷ Z: V = √(P × Z) and I = √(P ÷ Z), both RMS, with P in watts and Z in ohms. Multiplying V by I gives P back, a quick check on any result. A dBu entry runs the other way round: the voltage comes first, V = 0.775 × 10^(L ÷ 20), and the power follows from V² ÷ Z, so in this mode alone the impedance changes the power.
The 0.775 V reference is the rounded value ITU-R V.574-5 records as generally adopted. The voltage that really dissipates 1 mW in 600 Ω is √0.6, about 0.7746 V, so even at 600 Ω a signal's dBu figure is about 0.0045 lower than its dBm figure: 0 dBu works out to 0.0045 dBm. The calculator keeps 0.775 V because that is the figure audio equipment is calibrated to.
The voltage and current assume a purely resistive load. A reactive load carries current that delivers no power, which these formulas do not model.
A worked example.
A receiver datasheet quotes a sensitivity of −70 dBm into 50 Ω. Undoing the logarithm, 10^(−70 ÷ 10) = 10^−7 mW, which is 1 × 10⁻¹⁰ W, or 100 picowatts. The summary therefore reads "−70 dBm is 100 pW" and places it between one picowatt and one nanowatt. On the watt-referenced scale it is −70 − 30 = −100 dBW. Across 50 Ω the voltage is √(1 × 10⁻¹⁰ × 50) = √(5 × 10⁻⁹) ≈ 7.071 × 10⁻⁵ V, about 70.71 microvolts RMS, and the current is √(1 × 10⁻¹⁰ ÷ 50) ≈ 1.414 × 10⁻⁶ A, about 1.414 microamps. Multiplying the two returns 1 × 10⁻¹⁰ W. As an audio voltage level the same signal is −80.7963 dBu: 10 · lg(50 ÷ 600) = −10.7918 dB below the dBm figure, plus the 0.0045 dB reference offset. Seventy microvolts disappears into the noise of an ordinary oscilloscope, which is why receiver sensitivity is measured on a spectrum analyser and quoted in dBm rather than in volts.
Frequently asked questions.
How many watts is 0 dBm?
What happens if I combine two signals given in dBm?
Why do I need an impedance to get volts from dBm?
Is dBu the same as dBm?
Why does my Wi-Fi signal show as a negative number?
Is it correct to write dBm at all?
Does the summary tell me whether my signal is good?
References& sources.
- [1]Recommendation ITU-R V.574-5 (08/2015), "Use of the decibel and the neper in telecommunications", Annex 1. Read directly from the ITU PDF on 2026-07-29. Primary source for this page: §6.1 defines absolute power level and states that the symbol dBW is used when the reference power is one watt and dBm when it is one milliwatt; §4.2 records that "common use may give rise to simplified symbols, such as dBm instead of dB(mW)"; §6 sets out the general level notation and the rule that where no number is shown the number 1 is understood; §6.5 defines the absolute voltage level dBu against "a reference voltage with an r.m.s. of 0.775 volt … which corresponds to a 1 milliwatt power dissipated in a resistance of 600 ohms" and gives Lu = Lp + 10 lg(R/600).
- [2]NIST Special Publication 811, 2008 edition, "Guide for the Use of the International System of Units (SI)", §8.7 "Logarithmic quantities and units: level, neper, bel". Read directly from the NIST PDF on 2026-07-29. The independent check on the level definition — it gives LP = lg(P/P0) B = 10 lg(P/P0) dB and states that "when reporting values of LF and LP, one must always give the reference level" — and the source of the notational conflict recorded on this page: Note 3 states that "the rules of Ref. [5: IEC 60027-3] preclude, for example, the use of the symbol dBm to indicate a reference level of power of 1 mW", which is the opposite of ITU-R V.574-5 §4.2.
- [3]IEC 60027-3, "Letter symbols to be used in electrical technology — Part 3: Logarithmic and related quantities, and their units". The standard behind both sources above: ITU-R V.574-5 names it in considering clause (b), and NIST SP 811 §8.7 states that its section is based on it and that its rules are what preclude the dBm symbol. Access: gated — sold by the IEC and not opened for this page; every statement attributed to it here is quoted through NIST SP 811, which was read in full.
- [4]Recommendation ITU-R V.574-5 §1.1 and §1.2. Source of the distinction this page draws between a level and a ratio: §1.1 defines the bel as the decimal logarithm of a power ratio, and §1.2 warns that use of the decibel across unequal impedances "is not appropriate unless adequate information is given concerning the impedances involved" — the same condition that makes dBu and dBm diverge outside 600 ohms. Retrieved 2026-07-29.
How this page was produced
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- Daniel, Quanta Calculator Researcher
- Primary sources
- 4 cited below
- Method
- P(mW) = 10^(L/10) · L(dBm) = 10·lg(P mW) · L(dBW) = L(dBm) − 30 · V = √(P·Z) · L(dBu) = 20·lg(V / 0.775)
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
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