Decibel Converter — dBm, dBW, watts, dBV and volts

Convert between dB, dBm, dBV, watts, volts and plain ratios.

Example: 20 dBm is 20 dBm, 100 mW, or 2.236 V RMS into 50 Ω.

Inputs

Value

Unit

System impedance(Ω)

50 Ω for RF, 75 Ω for video and coax TV, 600 Ω for legacy audio.

Result

dBm

20 dBm

dBW: -10 dBW
Power: 100 mW
Voltage (RMS): 2.236 V
dBV: 6.99 dBV
Peak-to-peak: 6.325 V
Current: 44.72 mA

Power uses 10 × log₁₀ and voltage uses 20 × log₁₀. They give the same answer in dB for the same physical change — the factor of two compensates for power going as the square of voltage.
Voltage and power only relate through impedance. These figures assume 50 Ω; at a different impedance the same power corresponds to a different voltage.

Common reference points

dBm	Power	Volts RMS into 50 Ω	Power ratio vs 0 dBm
-30 dBm	1 µW	7.071 mV	×0.001
-20 dBm	10 µW	22.36 mV	×0.01
-10 dBm	100 µW	70.71 mV	×0.1
0 dBm	1 mW	223.6 mV	×1
10 dBm	10 mW	707.1 mV	×10
20 dBm	100 mW	2.236 V	×100
30 dBm	1 W	7.071 V	×1000
40 dBm	10 W	22.36 V	×10000

What a given dB change means

dB	Power ratio	Voltage ratio
1 dB	×1.259	×1.122
3 dB	×1.995	×1.413
6 dB	×3.981	×1.995
10 dB	×10	×3.162
20 dB	×100	×10
30 dB	×1000	×31.623

Formula

dBm = 10 × log₁₀(P/1 mW)    dBV = 20 × log₁₀(V/1 V)    P = V² / Z

Worked example

20 dBm is 20 dBm, 100 mW, or 2.236 V RMS into 50 Ω.

P = 10^(dBm/10) / 1000
P = 10^(20/10) mW
100 mW
V = √(P × Z)
V = √(100 mW × 50 Ω)
2.236 V
This is the step that needs an impedance.

Frequently asked questions

How many watts is 20 dBm?

100 mW. dBm is referenced to one milliwatt, so 20 dBm is 10²⁰ᐟ¹⁰ = 100 times 1 mW. 30 dBm is 1 W, and 0 dBm is exactly 1 mW.

Why is it sometimes 10 log and sometimes 20 log?

Because power goes as the square of voltage. Using 10 × log for power and 20 × log for voltage makes the same physical change come out as the same number of dB either way. Mixing them up gives you an answer that is out by a factor of two in dB.

What does 3 dB mean?

Double the power, or 1.414 times the voltage. It is why the filter cutoff is defined at −3 dB: that is the half-power point. 6 dB is double the voltage, and 10 dB is ten times the power.

Why do I need to specify an impedance?

Because dBm measures power and dBV measures voltage, and converting between them needs P = V²/Z. Without an impedance the question has no answer. RF work assumes 50 Ω, video 75 Ω, and older audio 600 Ω.

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