How to calculate “dB” ?

Decibel values can be calculated from any power measurements that use a common linear scale (e.g. Watts). In all cases they are derived from the ratio between two measurements, and in all cases they are calculated by finding the logarithm of the ratio, and multiplying it by ten*.

The decibel can also be used to compare pressure measurements (e.g. Volts, or sound pressure), and is calculated in the same way, save that the logarithm is instead multiplied by twenty. This reflects the fact that any power value corresponds to the square of a pressure value.

Power is calculated from the square of the voltage (P = V2/R). Where P is the power in Watts, V is the voltage in Volts, and R is the resistance in Ohms).

The following examples show how to work out what value are “n” decibels larger or smaller than a reference value.

For power (e.g. watts)

• Multiply your reference value by 10^(n/10).

For example, if your system is rated at 100 watts, a 20 decibel increase represents 100 x 10^(20/10) watts = 100 x 10^2 watts = 100 x 100 watts = 10,000 watts!

Negative values work in exactly the same way, so that if your system is rated at 100 watts, a reduction of 20 decibels (−20 dB) will produce 100 x 10^−(20/10) watts = 100 x 10^−2 watts = 100 x 0.01 watts = 1 watt. At 20 decibels below full output, your 100 watt system is only running at 1 watt!

For pressure (e.g. volts, sound pressure)

• Multiply your reference value by 10^(n/20).

For example, if your microphone produces 2.6 millivolts, an increase of 60 decibels will produce 2.6 x 10^(60/20) millivolts = 2.6 x 10^3 millivolts = 2.6 x 1,000 millivolts = 2,600 millivolts = 2.6 volts.

Where power and pressure combine, the values remain consistent.

For example, an amplifier produces an output of 20 Volts. From this, we can calculate the power generated into a 4Ω load as (20^2)/4 Watts = 400/4 Watts = 100 Watts.

If we double the output voltage (40 Volts) we find that the power generated is now (40^2)/4 Watts = 1,600/4 watts = 400 Watts.

Using decibels we find that doubling the output voltage is an increase of 6dB (20 x log 2 = 20 x 0.3 = 6). However, we have quadrupled the power generated. Yet using decibels we find that the increase is still the same: 6dB (10 x log 4 = 10 x 0.6 = 6).

Some fixed reference “dB” scales:

Examples:

dBm

Reference value (0dB) = 1 milliwatt.

For example, 10 dBm = 1 x 10^(10/10) milliwatts = 1 x 10^1 milliwatts = 1 x 10 milliwatts = 10 milliwatts.

dBSPL

Reference value (0dB) = 20 μPa (micropascals) is the threshold of hearing.

dBu

Reference value (0dB) = 0.775 volts (775 millivolts). The ‘u’ stands for ‘unloaded’.

For example, +4 dBu = 0.775 x 10^(4/20) volts = 0.775 x 10^0.2 volts = 0.775 x 1.58 volts = 1.23 volts.

dBV

Reference value (0dB) = 1 volt.

For example, −10 dBV = 1 x 10^−(10/20) volts = 1 x 10^−0.5 volts = 1 x 0.316 volts = 0.316 volts.

Comparable “dB” values:

Note: Analogue level of +24dBu (12.5Vrms) is where clipping starts. The level +4dBu is same as -20dBFS or 0VU.

For a better working knowledge, try to memorize some of the following common values (dotted values).

Observations:

10dB = 10 x intensity = 10phons = 2 x loudness

Intensity is an objective comparison of sound power per unit area. But the ear responds in a non-linear way to that sound intensity. Loudness is the strength of the ear’s perception of the sound, and by the rule of thumb, loudness is only doubled with a 10-fold increase in intensity.

So, a 10dB (10 x intensity) increase in sound intensity will cause a 10phons (2 x loudness) increase in loudness.

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