dB · dBm Calculator
Convert power ratios, dB, dBm and voltage levels.
P = 1 mW · 10^(dBm/10) · Vrms = √(P·Z)
This tool handles the three decibel jobs that come up at the bench: converting power or voltage ratios to dB and back, converting absolute dBm to watts, and telling you the RMS voltage a given dBm level produces across 50, 75 or 600 Ω. Reference points: ×2 power = 3.01 dB, and 0 dBm = 1 mW = 0.2236 Vrms at 50 Ω.
How it works
The decibel compresses multiplicative chains into additions: dB = 10·log₁₀(P₂/P₁) for power, and since power goes as voltage squared, dB = 20·log₁₀(V₂/V₁) for voltage at equal impedance. dBm anchors the scale to an absolute reference of 1 mW — P = 1 mW·10^(dBm/10) — and the voltage across an impedance follows from V = √(P·R). Cascade gains and losses by simply summing their dB values.
Worked example
0 dBm = 1 mW → √(0.001 × 50) = 0.2236 Vrms @ 50 Ω
30 dBm = 1 W · −30 dBm = 1 µW
Parameters
| Parameter | Value | Notes |
|---|---|---|
| Power ratio | dB = 10·log₁₀(P₂/P₁) | |
| Voltage ratio | dB = 20·log₁₀(V₂/V₁) | equal impedances |
| dBm | ref 1 mW | absolute level |
| Vrms | √(P·R) | 50 / 75 / 600 Ω presets |
| Rules of thumb | +3 dB ≈ ×2 P · +20 dB = ×10 V |
Python implementation
import math
db_power = lambda ratio: 10 * math.log10(ratio)
db_voltage = lambda ratio: 20 * math.log10(ratio)
dbm_to_w = lambda dbm: 1e-3 * 10 ** (dbm / 10)
vrms = lambda watts, ohms: math.sqrt(watts * ohms)
assert round(db_power(2), 4) == 3.0103
assert dbm_to_w(0) == 1e-3
assert round(vrms(dbm_to_w(0), 50), 4) == 0.2236FAQ
▸What is the difference between dB and dBm?
dB is a ratio — dimensionless, always relative to something. dBm is an absolute power level: decibels relative to exactly 1 mW. '−10 dB' says a signal lost 90 % of its power; '−10 dBm' says it is 0.1 mW, full stop.
▸Why 10·log for power but 20·log for voltage?
Power is proportional to voltage squared (P = V²/R), and the square becomes a factor of two outside the logarithm. Both formulas describe the same power ratio when the impedance is equal on both sides — doubling voltage is +6 dB precisely because it quadruples power (+6 dB).
▸How does 0 dBm become 0.224 Vrms at 50 Ω?
0 dBm is 1 mW by definition. Across an impedance R the RMS voltage is √(P·R) = √(0.001 × 50) ≈ 0.2236 V. The same 1 mW is 0.775 Vrms at 600 Ω — which is exactly where the audio industry's dBu reference comes from.
▸What are the handy mental-math values?
+3 dB ≈ ×2 power, +10 dB = ×10 power, +6 dB ≈ ×2 voltage, +20 dB = ×10 voltage. Chain them: +13 dB is 3+10, so ×20 in power. The calculator confirms the exact figures (3.01 dB, not 3).
▸Is my data uploaded?
No. All math runs locally in your browser.