Watt to dB Converter
Convert any power level — watts, milliwatts, microwatts down to femtowatts — to dBm and dBW. Uses dBW = 10·log₁₀(P) and dBm = dBW + 30. Includes Vrms across 50 Ω, zone classification, and a transmit/receive reference table.
Input
Result
Common Power → dB Reference
| Power | dBm | dBW | Vrms @ 50 Ω |
|---|---|---|---|
| 1 MW | +90 dBm | +60 dBW | ~7.07 kV |
| 1 kW | +60 dBm | +30 dBW | ~223.6 V |
| 100 W | +50 dBm | +20 dBW | ~70.7 V |
| 10 W | +40 dBm | +10 dBW | ~22.4 V |
| 1 W | +30 dBm | 0 dBW | ~7.07 V |
| 100 mW | +20 dBm | −10 dBW | ~2.24 V |
| 10 mW | +10 dBm | −20 dBW | ~707 mV |
| 1 mW | 0 dBm | −30 dBW | ~224 mV |
| 100 µW | −10 dBm | −40 dBW | ~70.7 mV |
| 1 µW | −30 dBm | −60 dBW | ~7.07 mV |
| 1 nW | −60 dBm | −90 dBW | ~223.6 µV |
| 1 pW | −90 dBm | −120 dBW | ~7.07 µV |
| 1 fW | −120 dBm | −150 dBW | ~223.6 nV |
| 0 | −∞ dBm | −∞ dBW | 0 V |
How to Convert Watts to dBm: Worked Example
To convert 250 mW to dBm and dBW, apply the formula dBm = 10 · log₁₀(P_mW) directly: 10 · log₁₀(250) = 10 · 2.3979 = +23.98 dBm. To get dBW, either use 10 · log₁₀(0.250) = 10 · (−0.6021) = −6.02 dBW, or simply subtract 30 from the dBm result: 23.98 − 30 = −6.02 dBW. Both routes give the same answer. The Vrms across 50 Ω is √(0.250 × 50) = √12.5 ≈ 3.54 Vrms.
Two common reference values are worth memorizing: 1 mW = 0 dBm (by definition) and 1 W = +30 dBm = 0 dBW. From those anchors you can estimate any power mentally using the rule that each factor of 10 adds 10 dB and each doubling adds ~3 dB. For example, 250 mW is 2 × 125 mW, and 100 mW = +20 dBm, so 200 mW ≈ +23 dBm and 250 mW ≈ +24 dBm — close to the exact +23.98 dBm above. If you need to go the other way, use the dB to Watt converter to reverse the calculation. For link-budget work where transmit power, cable loss, and antenna gain combine logarithmically, the dB Addition Calculator handles the arithmetic.
About Power → dB Conversion
Both dBm and dBW are absolute power units — they specify what 0 dB means. 0 dBm = 1 milliwatt, 0 dBW = 1 watt. Because 1 W is exactly 1000 mW (which is 30 dB), the two scales are offset by 30: dBm = dBW + 30. The conversion from linear power uses the ÷10 form: dB = 10·log₁₀(P / P_ref) -- so converting a watt to decibels (dBW) or milliwatts to dBm always uses this power-ratio formula, not the amplitude (÷20) form.
Why the ÷10 (not ÷20)?
Power-based dB scales use ÷10. Amplitude-based scales (dBV, dB SPL, dBu, dBFS) use ÷20. The factor-of-2 difference comes from P ∝ A² — doubling the amplitude quadruples the power, so the same "+6 dB amplitude" only adds 3 dB of power. dBm and dBW are explicitly power references, so they always use ÷10. This distinction is critical in link budget calculations: an RF power calculator must use the ÷10 rule when working with transmit power (dBm), cable and connector losses (dB), antenna gain (dBi), free-space path loss (dB), and receiver sensitivity (dBm) all adding and subtracting on the same logarithmic scale. If you need to go the other direction — dBm back to watts — use the dB to Watt converter.
What is "Vrms across 50 Ω"?
RF and test-equipment systems are nearly all standardized on a 50-ohm characteristic impedance. Given a power, the equivalent open-circuit RMS voltage across a 50 Ω load is V = √(P × 50). So 1 mW = 0 dBm = ~224 mVrms into 50 Ω. This is the connection between dBm (used by RF folks) and oscilloscope readings (used by everyone). For 75 Ω (cable TV) or 600 Ω (legacy audio) systems, replace 50 with the appropriate impedance. The voltage figure also helps when estimating signal-to-noise ratio (SNR): if you know both the signal power and the noise power in dBm, their difference in dB is the SNR directly.
Why does P = 0 give −∞ dB?
log₁₀(0) is mathematically undefined — the limit as P → 0⁺ is negative infinity. So absolute zero power is represented as "negative infinity decibels". In practice every system has a thermal noise floor (≈ −174 dBm/Hz at room temperature, derived from Johnson–Nyquist noise: kTB where k is Boltzmann's constant). A real receiver adds its own noise on top of thermal noise, quantified as noise figure (NF) in dB. Below the effective noise floor — thermal noise plus noise figure plus bandwidth — signals can't be reliably detected, regardless of how sensitive the receiver is.
Frequently Asked Questions
What is 1 watt in dBm?
How do I convert milliwatts to dBm quickly?
Why are RF receiver sensitivities expressed in negative dBm?
Is +3 dBm really "double the power" of 0 dBm?
How does Vrms change with impedance?
V = √(P × R). So 1 W into 50 Ω = ~7.07 V, but 1 W into 600 Ω = ~24.5 V, and 1 W into 75 Ω = ~8.66 V. RF: usually 50 Ω. Antenna/cable systems: 75 Ω. Pro audio (legacy): 600 Ω. Modern audio: ~10 kΩ (high-impedance bridging). This tool shows the 50 Ω value because that's the most common RF reference.Why doesn't this tool accept negative power?
What is the difference between dBm and dBu?
How do I calculate the EIRP of a transmitter in dBm?
EIRP (dBm) = P_tx (dBm) + G_antenna (dBi) − cable_loss (dB). For example, a 100 mW (+20 dBm) transmitter with a 3 dBi antenna and 1 dB of cable loss gives EIRP = 20 + 3 − 1 = +22 dBm (≈ 158 mW). US FCC Part 15 caps most unlicensed 2.4 GHz devices at +30 dBm EIRP. Use this converter to find the dBm value for any transmit power before adding antenna gain.