Decibels, Explained

After reading this you will be able to turn any ratio into decibels and back, keep the 10·log and 20·log rules straight, and convert absolute power between dBm and watts without guessing.

What a decibel actually is

A decibel is not a unit of anything physical. It is a way to write a ratio on a logarithmic scale. When you see a number like 40 dB, it always means "this quantity compared to some reference." The dB compresses huge ranges into small readable numbers.

Here is the hook. A signal that is 1,000,000 times stronger than another is not written as 1000000. It is written as 60 dB. A signal 100 times weaker is -20 dB. The logarithm turns multiplication into addition, so cascading gains and losses becomes simple arithmetic. Amplify by 30 dB, lose 12 dB in a cable, lose 6 dB in a splitter, and the total is 30 - 12 - 6 = 12 dB. No multiplying required.

When to use dB and when not to

Use decibels when you are comparing two power levels or two amplitudes that span many orders of magnitude: antenna gain, amplifier output, filter attenuation, acoustic intensity, optical loss in fiber. The scale matches how these systems behave and how your ears and instruments respond.

Do not use plain dB when you need an absolute value. A statement like "the transmitter puts out 30 dB" is meaningless because there is no reference. For absolute power you need a suffixed unit such as dBm (relative to 1 milliwatt) or dBW (relative to 1 watt). This tool handles dBm.

The suffix tells you the reference. dBm is referenced to 1 mW, dBW to 1 W, dBV to 1 volt. Plain dB is a pure ratio between two quantities of the same kind.

The two formulas and why there are two

For a power ratio, the definition is:

L_{dB} = 10 \log_{10}\!\left(\frac{P_1}{P_2}\right)

Here P_1 is the power you are measuring, P_2 is the reference power, and L_{dB} is the result in decibels. The factor 10 comes from the "deci" prefix: the bel (log of the ratio) was too coarse, so we use tenths of a bel.

For an amplitude quantity such as voltage, current, or sound pressure, the formula uses 20 instead of 10:

L_{dB} = 20 \log_{10}\!\left(\frac{V_1}{V_2}\right)

The reason is physical, not arbitrary. Power in a resistor scales with the square of voltage: P = V^2/R. So if voltage doubles, power quadruples. Put a squared ratio inside the log and the exponent comes out front:

10 \log_{10}\!\left(\frac{V_1^2}{V_2^2}\right) = 10 \cdot 2 \log_{10}\!\left(\frac{V_1}{V_2}\right) = 20 \log_{10}\!\left(\frac{V_1}{V_2}\right)

That is the whole story. The 20·log rule is the 10·log rule applied to a quantity whose square is the power. This is why doubling power is +3 dB but doubling voltage is +6 dB: same physical doubling of power in the second case, just counted through voltage.

To go back from dB to a ratio, invert the log. For power:

\frac{P_1}{P_2} = 10^{L_{dB}/10}

For amplitude, divide by 20 in the exponent instead of 10.

Worked example: 100× power, and 20 dB

Reproducing the demo values

The demo loads a ratio of 100 as a power quantity. Convert it to dB.

Take the ratio: P_1/P_2 = 100.
Take the base-10 log: \log_{10}(100) = 2.
Multiply by 10 for a power quantity: 10 \times 2 = 20 dB.

So a hundredfold power increase is exactly 20 dB. Now reverse it, matching the dB → ratio default of 20 dB:

Divide the dB by 10: 20/10 = 2.
Raise 10 to that power: 10^{2} = 100.

You are back to 100. The two directions are exact inverses.

Now try the same 100 ratio as an amplitude quantity. Use 20·log: 20 \times \log_{10}(100) = 20 \times 2 = 40 dB. The same numeric ratio gives twice the dB because for amplitude the log is scaled by 20.

dBm and watts: the absolute scale

dBm answers "how much power, in absolute terms." The reference is 1 milliwatt, so the formula is the power dB rule with P_2 = 1 mW:

P_{dBm} = 10 \log_{10}\!\left(\frac{P}{1\,\text{mW}}\right)

The demo loads 30 dBm. Convert to watts. First back out the ratio to 1 mW: 10^{30/10} = 10^{3} = 1000. So the power is 1000 mW, which is 1 W. That reproduces the footnote: 30 dBm = 1 W.

Going the other way, the Watts → dBm default is 1 W. Convert: 1 W is 1000 mW, and 10 \log_{10}(1000) = 10 \times 3 = 30 dBm. Consistent.

Common dBm values and their power
dBm	Ratio to 1 mW	Power
-30	0.001	1 µW
0	1	1 mW
10	10	10 mW
20	100	100 mW
30	1000	1 W
60	1,000,000	1 kW

Reading the numbers on a log scale

A few landmarks make dB values readable at a glance. Memorize these for power quantities:

+3 dB ≈ 2× the power (exact value is 10^0.3 = 1.995).
+10 dB = 10× the power, exactly.
-3 dB ≈ half the power. This is the standard "half-power point" used to define bandwidth.
0 dB = equal, the ratio is 1.

For amplitude, halve those dB thresholds' effect: +6 dB doubles voltage, +20 dB is a tenfold voltage increase. The chart below shows how the power ratio climbs with dB.

The power ratio rises by a factor of 10 for every 10 dB. The marker sits at 20 dB, where the ratio is 100.

Common mistakes

The single most frequent error is using 10·log for a voltage or sound-pressure measurement, or 20·log for a power measurement. Check the quantity first. If it is power, energy, or intensity, use 10. If it is voltage, current, pressure, or field strength, use 20.

Two more traps to watch:

Confusing dB with dBm. A cable spec of "-3 dB" is a relative loss (half the power gets through). A receiver spec of "-90 dBm" is an absolute power (about 1 picowatt). They look similar but one is a ratio and one is a level.
Adding ratios instead of dB. Gains multiply as linear ratios but add as dB. Two amplifiers of 100× each give 10000× total, which is 20 dB + 20 dB = 40 dB, not 40× and not 200×.

One more subtlety on the 20·log rule: it assumes the voltages are measured across the same impedance. If the input and output impedances differ, converting a voltage ratio to a true power ratio requires an impedance correction. On a single line with fixed impedance, the 20·log rule is exact.

Related converters on this site

If you need everyday physical units rather than ratios, the Universal Unit Converter handles length, mass, temperature, pressure and energy. For vehicle efficiency figures use the Fuel Economy Converter, and for kitchen quantities the Cooking Measurement Converter maps cups to grams by ingredient.

Frequently asked questions

Why is doubling power +3 dB but doubling voltage +6 dB?

Power scales with the square of voltage. Doubling voltage quadruples power, which is +6 dB of power. Written as amplitude with the 20·log rule, doubling voltage gives 20 \log_{10}(2) \approx 6.02 dB. Doubling power alone gives 10 \log_{10}(2) \approx 3.01 dB.

What does 0 dB mean?

It means the two quantities are equal: the ratio is 1, since \log_{10}(1) = 0. It does not mean "no signal." For dBm, 0 dBm means exactly 1 mW, a real and nonzero power.

Can decibels be negative?

Yes. A negative dB value means the ratio is less than 1, that is, a loss or attenuation. -20 dB of power means the output is one hundredth of the input.

How do I convert dBm to watts by hand?

Compute 10^{(P_{dBm}/10)} to get milliwatts, then divide by 1000 for watts. For 30 dBm: 10^{3} = 1000 mW = 1 W.

Is a decibel the same in acoustics and electronics?

The math is identical. The difference is the reference. Acoustic sound pressure level uses 20·log against 20 µPa, while electrical power uses 10·log against a power reference. Pick the correct factor and reference for your field.

Try it yourself in the tool