dBm to Watts Calculator

dBm is a power level expressed on a logarithmic scale referenced to one milliwatt, and it exists because RF systems span an absurd dynamic range — a transmitter putting out 100 W and a receiver hearing −110 dBm differ by a factor of 1014. This converter moves a level between dBm, dBW, watts, milliwatts and microwatts, and then works out the RMS and peak-to-peak voltage that level corresponds to in a stated system impedance. Enter a value, pick the unit it is already in, and everything else follows.

Calculator

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Inputs this calculator takes, with typical values
InputWhat to enterExample
Power levelThe number as you have it — a level in decibels or a power in watts, depending on the unit you choose next.30
Unit of the value aboveChoose what the number represents; the converter normalises everything to dBm internally and then expands it.dBm (decibels relative to 1 mW)
System impedance50 Ω for almost all RF and test equipment, 75 Ω for video and broadcast coax, 600 Ω for legacy audio lines. Only the voltage outputs depend on it.50 Ω

It returns

  • Power — The absolute power the level represents.
  • Power in milliwatts
  • Level in dBm
  • Level in dBW
  • RMS voltage into the stated impedance
  • Peak-to-peak voltage (sine wave)

The formula

P=10dBm3010
Vrms=PZ
Vpp=22Vrms

In plain text: P(W) = 10^((dBm − 30)/10) and dBm = 10·log₁₀(P in mW)

  • PAbsolute power (W)
  • dBmPower level referenced to 1 milliwatt (dBm)
  • dBWPower level referenced to 1 watt; always dBm − 30 (dBW)
  • ZSystem characteristic impedance (Ω)
  • V_rmsRMS voltage the power develops across Z (V)

The −30 in the exponent converts the milliwatt reference to watts, because 1 W is 1,000 mW and 10·log₁₀(1000) = 30.

Updated Category Electronics Design & RF Verified against published test cases Reading time 13 min

What dBm actually means

dBm is an absolute power level, not a ratio between two signals. The 'm' fixes the reference at one milliwatt, so 0 dBm is exactly 1 mW, 30 dBm is exactly 1 W, and −30 dBm is exactly 1 µW. That single anchoring is what separates dBm from plain decibels: a gain of 20 dB tells you an amplifier multiplied its input by a hundred, while 20 dBm tells you the signal is 100 mW no matter what came before it.

The scale is logarithmic because RF power spans a range that decimal notation handles badly. A cellular base station transmits around 43 dBm and its handset receives usable signal down to −110 dBm — a ratio of 15,800,000,000,000 to one. On the decibel scale that is a difference of 153, which fits on a chart. Logarithms also turn cascaded multiplication into addition: an amplifier with 20 dB of gain followed by 3 dB of cable loss and a 6 dB pad gives a level of input + 20 − 3 − 6, which you can do in your head. That is the reason the whole RF industry works in dBm rather than watts.

dBW is the same idea with a one-watt reference, and is used mostly in satellite and radar work where levels are large. Because 1 W is 1,000 mW and 10·log₁₀(1000) = 30, dBW is always dBm minus 30. There is no other relationship to remember.

Two things dBm is not. It is not a voltage — converting to volts needs an impedance, which is why the third field on this page exists. And it is not additive with dBm: two 0 dBm signals combined incoherently give 3 dBm, not 0 dBm, because you add the powers (1 mW + 1 mW = 2 mW) and then take the logarithm.

The conversions, in both directions

From dBm to power. P in watts = 10((dBm − 30) ÷ 10). The division by ten reverses the ten in the decibel definition; the subtraction of thirty shifts the reference from a milliwatt to a watt. If you want milliwatts directly, drop the −30: P in mW = 10(dBm ÷ 10).

From power to dBm. dBm = 10·log₁₀(P in mW). If your power is in watts, that becomes 10·log₁₀(P) + 30. Note the ten, not twenty: the factor of twenty appears only when you are taking the logarithm of a voltage or current ratio, because power goes as the square of voltage and the square becomes a factor of two outside the logarithm.

From power to voltage. Power delivered into a resistive load is Vrms² ÷ Z, so Vrms = √(P·Z). This is the step that requires an impedance, and it is where 50 Ω enters everything. The historical reason for 50 Ω is a compromise: air-dielectric coaxial line has minimum loss near 77 Ω and maximum power handling near 30 Ω, and 50 Ω sits between them. Video and broadcast distribution standardised on 75 Ω instead, where the loss minimum matters more than power handling.

From RMS to peak-to-peak. For an unmodulated sine wave, the peak is √2 times the RMS value and the peak-to-peak swing is twice that, so Vpp = 2√2 · Vrms ≈ 2.828 · Vrms. This factor is specific to sine waves. A modulated carrier, an OFDM signal or band-limited noise has a much higher crest factor, so the same average power reaches a considerably larger instantaneous peak — which is why an amplifier that is comfortable with a 30 dBm carrier can clip badly on a 30 dBm modulated signal.

The two shortcuts worth memorising: 3 dB is a factor of two in power (2.00 against the exact 1.995, an error of a quarter of a percent), and 10 dB is a factor of ten. Between them you can do most conversions mentally — 23 dBm is 10 + 10 + 3, so 1 mW × 10 × 10 × 2 = 200 mW.

Worked example: a 5 W transmitter through a lossy feed

A handheld transmitter is rated 5 W into 50 Ω. It feeds an antenna through 30 m of coax with 6.2 dB of loss at the operating frequency, and there is a 0.3 dB connector allowance. What arrives at the antenna, and what voltage appears at the transmitter's output connector?

  1. Convert the transmitter output to dBm. 5 W is 5,000 mW, so the level is 10 · log₁₀(5,000) = 10 × 3.69897 = 36.99 dBm. As a check, 5 W is 30 dBm (1 W) plus 7 dB, and 7 dB is a factor of 5.01 — consistent.
  2. Voltage at the connector. Vrms = √(5 × 50) = √250 = 15.81 V RMS, and the peak-to-peak swing on a scope with a 50 Ω through-termination would be 2√2 × 15.81 = 44.72 V. That is why you never probe a transmitter output with a 10× scope probe rated for a few volts.
  3. Subtract the losses. 36.99 − 6.2 − 0.3 = 30.49 dBm at the antenna. In decibels the losses simply subtract; there is no multiplication anywhere.
  4. Convert back to watts. P = 10((30.49 − 30) ÷ 10) = 100.049 = 1.12 W.

So 6.5 dB of total loss threw away 3.88 W of the original 5 W — 78% of the transmitter's output — while the level fell by less than 7 on the decibel scale. That compression is exactly what makes decibels convenient and simultaneously makes them easy to under-respect. A number that looks like a small subtraction is a large fraction of the power.

If you need to work the gain and loss budget itself rather than a single conversion, the decibel gain calculator handles cascades of amplifiers, attenuators and cable runs.

What the numbers mean in practice

Know the landmarks. 0 dBm is 1 mW and 223.6 mV RMS into 50 Ω — this is the level most bench signal generators default to. 30 dBm is 1 W. A Wi-Fi access point transmits between 15 and 20 dBm; a Bluetooth Low Energy radio around 0 to 4 dBm; a 100 W amateur transceiver 50 dBm; a broadcast FM transmitter 70 dBm or more.

Receiver sensitivity is where negative numbers live. A GPS receiver works with a signal around −130 dBm, which is 10−16 W. The thermal noise floor of a matched resistor at room temperature is about −174 dBm per hertz of bandwidth, so in 1 MHz the noise alone is around −114 dBm. Any claimed sensitivity below the noise floor for the stated bandwidth is either using a narrower bandwidth than you assumed or is wrong.

Voltage figures are only valid at the stated impedance. Change the field from 50 Ω to 75 Ω and the voltage rises by a factor of √1.5 = 1.225 for the same power. The dBm, dBW, watt and milliwatt figures do not move at all, because power is power. This is the single most common misunderstanding in level conversion, and it is why professional audio uses dBu — a voltage reference of 0.7746 V that deliberately has no impedance attached — rather than dBm.

Do not average decibels. The mean of 0 dBm and 20 dBm is not 10 dBm. Convert both to power (1 mW and 100 mW), average those (50.5 mW), then convert back (17.0 dBm). Decibels are logarithms, and logarithms do not commute with addition.

dBm reference table with 50 Ω voltages

Every 10 dB is a factor of ten in power and √10 ≈ 3.162 in voltage. Voltages assume a 50 Ω resistive load and a sine wave.
LevelPowerMilliwattsV RMS at 50 ΩTypical example
−90 dBm1 pW0.000000001 mW7.07 µVWeak cellular signal
−60 dBm1 nW0.000001 mW223.6 µVStrong Wi-Fi at the edge of a room
−30 dBm1 µW0.001 mW7.07 mVSpectrum analyser mid-scale
−10 dBm100 µW0.1 mW70.71 mVMixer LO drive, low side
0 dBm1 mW1 mW223.6 mVSignal generator default
10 dBm10 mW10 mW707.1 mVBluetooth Class 1 transmitter
20 dBm100 mW100 mW2.236 VWi-Fi access point
30 dBm1 W1,000 mW7.071 VHandheld radio, low power
40 dBm10 W10,000 mW22.36 VMobile VHF transceiver
50 dBm100 W100,000 mW70.71 VAmateur HF transceiver
60 dBm1,000 W1,000,000 mW223.6 VLegal-limit amplifier

Voltages come straight from √(P × 50). Examples are indicative of where equipment in each class typically operates, not specifications.

Conversion mistakes and how to avoid them

  • Using 20·log instead of 10·log for power. The factor of twenty belongs to voltage and current ratios only. Applying it to power doubles every decibel figure you produce.
  • Adding dBm to dBm. Adding levels is only meaningful when one of them is a gain or loss in plain dB. Two signals at 0 dBm combine to 3 dBm, and only if they are uncorrelated; two coherent in-phase signals combine to 6 dBm.
  • Quoting a voltage without an impedance. −10 dBm is 70.71 mV in 50 Ω and 86.6 mV in 75 Ω. State the impedance or the number means nothing.
  • Assuming peak-to-peak is 2.828 × RMS for a modulated signal. That ratio is a property of the sine wave. A modulated or noise-like waveform has a much higher crest factor, so its peak voltage exceeds what this page reports for the same average power.
  • Confusing dBm with dBmV or dBµV. Cable television uses dBmV, referenced to one millivolt across 75 Ω; RF measurement often uses dBµV. They are voltage references, not power references, and converting between them requires the impedance.
  • Feeding a spectrum analyser too much power. Most analysers have a maximum safe input around +30 dBm with attenuation engaged and far less at the mixer. Compute the level before connecting anything to a transmitter.
  • Reading 0 dBm as 'no signal'. Zero on a logarithmic scale is the reference point, not the absence of a quantity. Absence of signal is minus infinity.

Why decibels use ten for power and twenty for voltage

The bel was defined as the base-ten logarithm of a power ratio, and the decibel is a tenth of one, so a power ratio in dB is 10·log₁₀(P₂/P₁). Because power in a resistive load is proportional to the square of voltage, a voltage ratio expressed in the same decibels becomes 10·log₁₀((V₂/V₁)²) = 20·log₁₀(V₂/V₁). The twenty is not a separate convention; it is the ten with the square pulled out of the logarithm. This also means the identity only holds when both voltages appear across the same impedance — comparing a voltage before and after a transformer using 20·log gives a number that is not a power ratio.

The other decibel units you will meet

dBW uses a one-watt reference and is standard in satellite link budgets and radar. dBW = dBm − 30, always. dBk occasionally appears with a kilowatt reference, and equals dBW − 30.

dBi and dBd describe antenna gain rather than power level, referenced to an isotropic radiator and to a half-wave dipole respectively; dBi = dBd + 2.15. Multiplying transmitter power by antenna gain gives effective radiated power, which is why a 20 dBm transmitter into a 12 dBi antenna produces 32 dBm EIRP — the figure most regulators actually limit. The antenna length calculator covers the physical side of that.

dBc is a ratio relative to the carrier, used for spurious emissions and phase noise. dBFS is relative to a digital converter's full scale and is always negative or zero. dBu and dBV are audio voltage references — 0.7746 V and 1 V respectively — and carry no impedance, which is the correct modern convention for audio because line inputs are high-impedance bridging loads rather than matched terminations.

For circuit-level work where you are converting between power, voltage, current and resistance rather than between decibel scales, the electrical power calculator and the Ohm's law calculator handle the underlying relations. And when the load is not purely resistive — a real antenna at the edge of its band, say — the power delivered depends on the complex impedance, which the RLC impedance calculator covers.

Key terms

dBm
Power level in decibels referenced to one milliwatt. An absolute quantity: 0 dBm is always 1 mW regardless of impedance.
dBW
Power level referenced to one watt. Numerically dBm minus 30, because a watt is a thousand milliwatts.
Characteristic impedance
The ratio of voltage to current in a transmission line carrying a travelling wave. 50 Ω for RF and test equipment, 75 Ω for video and broadcast distribution.
Crest factor
The ratio of peak to RMS amplitude. Exactly √2 for a sine wave, higher for modulated carriers and noise, which is why peak voltage cannot be inferred from average power alone.
Noise floor
The thermal noise power in a given bandwidth, about −174 dBm per hertz at room temperature. It bounds how small a signal any receiver can usefully detect.

Frequently asked questions

How many watts is 30 dBm?

Exactly 1 watt. The dBm scale is referenced to one milliwatt, and 30 dB is a factor of 1,000, so 30 dBm is 1,000 mW. The same landmark works in both directions: 0 dBm is 1 mW, 60 dBm is 1,000 W, and −30 dBm is 1 µW. Every 10 dB step multiplies or divides the power by ten.

What is 0 dBm in volts?

223.6 mV RMS into 50 Ω, or 273.9 mV RMS into 75 Ω. The voltage depends entirely on the impedance because V = √(P × Z), and 0 dBm is 1 mW in both cases. Peak-to-peak on a sine wave is 2√2 times the RMS figure, so 632.5 mV peak-to-peak at 50 Ω.

Is dBm the same as dB?

No. Plain dB is a ratio between two quantities and carries no absolute meaning — 3 dB of gain doubles whatever went in. dBm is an absolute power level with a fixed one-milliwatt reference. You add dB to dBm and get dBm; you subtract dBm from dBm and get dB. Adding two dBm figures together is meaningless.

How do I convert dBm to dBW?

Subtract 30. A watt is a thousand milliwatts and 10·log₁₀(1000) is exactly 30, so the two scales differ by a constant offset with no other adjustment. 43 dBm is 13 dBW; −20 dBW is 10 dBm. The relationship holds at every level and does not depend on impedance.

Why is 50 ohms the standard RF impedance?

It is a compromise between two competing optima in air-dielectric coaxial line: minimum attenuation occurs near 77 Ω and maximum power handling near 30 Ω. Fifty ohms splits the difference and became the manufacturing standard for test equipment and radio hardware. Video and broadcast distribution chose 75 Ω instead, because low loss over long runs mattered more than power handling.

Do two 0 dBm signals combine to 0 dBm?

No. Convert to power first: 1 mW plus 1 mW is 2 mW, which is 3 dBm. That is the answer for uncorrelated signals such as two independent noise sources. Two coherent signals in phase add in voltage rather than power, giving four times the power and 6 dBm; two coherent signals in antiphase cancel entirely. The decibel scale never adds directly.

Can I convert dBm to volts without knowing the impedance?

No, and this is the most common error in level work. Power fixes the product of voltage and current, not either one alone; only the impedance splits them. The same 0 dBm is 223.6 mV in 50 Ω, 273.9 mV in 75 Ω and 774.6 mV in 600 Ω. If a datasheet quotes a voltage for a dBm level without stating the impedance, assume 50 Ω for RF and 600 Ω for legacy audio, and verify.

What does a negative dBm mean?

A power below one milliwatt. −3 dBm is half a milliwatt, −20 dBm is 10 µW, and −100 dBm is 0.1 picowatts. Negative levels are entirely normal on the receive side of any radio link; a mobile phone routinely works with signals between −70 and −110 dBm. Nothing about the sign implies a problem.

How accurate is the '3 dB equals double' rule?

Very. A factor of two in power is 10·log₁₀(2) = 3.0103 dB, so treating 3 dB as exactly double introduces an error of about 0.24 percent in power. Over a long chain the errors accumulate: ten successive 3 dB steps give 30 dB, which is a factor of 1,000 rather than the 1,024 that ten doublings really produce — a 2.4 percent difference. For mental arithmetic that is fine; for a specification, use the exact conversion.

References