Why we use dBA or any other type of decibel unit instead of basic decibels unit (dB)? Why we have so many types of this unit?


Decibels are a nice way to measure things which come in a very wide dynamic range as it is a logarithmic measurement.

At a certain amount of zeros it just gets hard to understand what's going on and decibels help greatly. Someone telling you a signal is 10000000000 more powerful is just harder to grasp then a number of 100dB.

But as Olin already said, a decibel is not a unit as it is, it is a ratio. But a ratio is often only useful when you know which was the reference given.

So when you talk about 100dB more power - it still makes some orders of magnitude of a difference if the reference was 1mW, 1W or 1µW.

dBA was specifically designed for sound or noise measurement. dBA has a frequency dependent weighting behind it, so not all frequencies are considered to have the same strength or effect on the ear. The usefulness of dBA is quite debated, but that is covered quite well on Wikipedia.

Then there is also the difference between field quantities and power quantities. A value of 100dBV is different to a value of 100dBW. The factor to the reference is different:

For power quantities: $$L_P = 10 \log_{10}\!\left(\frac{P}{P_0}\right)\!~\mathrm{dB}$$

For field quantities: $$L_F = 20 \log_{10} \left(\frac{F}{F_0}\right)\!~\mathrm{dB}$$

So 100dBV turns out to be a factor of 10^5 = 100 000 to 1V. Whereas 100dBW is a factor of 10^10 = 10 000 000 000 to 1W.

So giving the reference is quite critical to handle the ratios correctly.


We don't. Deci-bels are deci-bels, which is a way to express a power ratio.

dBA is just dB with a known fixed value for 0 dB. Instead of expressing the ratio of two arbitrary power levels, dBA is the ratio of some power level relative to the reference level.

For example, dBm is sometimes used in radio electronics. 1 mW was chosen as the 0 dB reference. 10 mW is therefore 10 dBm, 100 mW 20 dBm, etc.

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    \$\begingroup\$ dBA is a bit more involved as it has a frequency dependent weighting behind it. \$\endgroup\$ – Arsenal Oct 11 '15 at 21:51
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    \$\begingroup\$ DbA refers to the ratio of two sound pressure levels, where \$dB = 20log10\frac {P1}{P2}\$, which isn't the same as power levels, which would be described by \$dB = 10log10\frac {W1}{W2}\$. \$\endgroup\$ – EM Fields Oct 11 '15 at 23:36
  • \$\begingroup\$ No, this answer states a severe misunderstanding of dBA. It is not analogous to a singularly referenced unit like dBm. \$\endgroup\$ – Chris Stratton Oct 12 '15 at 5:47
  • \$\begingroup\$ @EMFi: 20Log of a pressure level ratio is pretty much the same as 10Log of a power ratio. The only reason they are not the same is due to some non-linearities of air. dB is always a power ratio, but is often handy for expressing voltage (or pressure level for sound) ratios, so we multiply the log by 20 instead of 10 to compensate. \$\endgroup\$ – Olin Lathrop Oct 12 '15 at 11:31

In the beginning was the Bel, which was considered to be a perceived change in loudness of 2:1 of some sound level. It turned out that the unit was impractically large and cumbersome to use, so it was divided into 10 parts: deciBels, or "dB".

It turned out later that a change in sound pressure level (loudness) of 2:1 was closer to 6dB, and a change in power of 2:1 was equal, consequently, to 3dB.

dB always implies a ratio and, electrically, is defined for voltage as:

$$ dB = 20 log 10 \text{ } \frac {V1}{V2}\text { ,}$$

for current as:

$$ dB = 20 log 10 \text{ } \frac {I1}{I2}\text { ,}$$

and for power as:

$$ dB = 10 log 10 \text{ } \frac {P1}{P2}\text { .}$$

The suffixes you refer to are only meant to indicate what absolute quantity zero db is being referred to.

For example, the "m" in 0dBm refers to 1 milliwatt, so 10 dBm would be 10 milliwatts, 20 dBm would be a 100 milliwatts, and so on.

The converse is also true, but with a sign change, so -20dBm would be 0.01 milliwatts.

A fanciful way to make the point might be - if dollars is power - to use 0dB$ to refer to one dollar, from which 10dB$ would be 10 dollars, 20dB$ would be 100 dollars, -10dB$ would be 10 cents, and -20dB$ would be a penny.

  • \$\begingroup\$ This is not really an answer to the question which was asked. The "A" in dBA is not a single reference unit, but rather a weighting curve. \$\endgroup\$ – Chris Stratton Oct 12 '15 at 5:44
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    \$\begingroup\$ The question was: "Why we use dBA or any other type of decibel unit instead of basic decibels unit (dB)?" So if you take little time to parse his question with some care you may notice that he wasn't asking about dBA specifically, but why suffixes are used at all. I addressed the general nature of the question with a little history, a little science, and a couple of examples to illustrate why suffixes are used. Sorry if you missed the point. \$\endgroup\$ – EM Fields Oct 12 '15 at 6:23
  • \$\begingroup\$ It is important to note that when deriving the voltage and current notation (20 log10 V1/V2 for instance) that this holds strictly true only if R (ref) and R(measurement) are exactly the same (because the dB is a power ratio, so P = V^2/R). I know we use it as a convenient way to express large gains for amplifiers (as an example), but for an op-amp (where R(in) can be >> R(out)) the use of the dB is mathematically incorrect. Many datasheets recognise this and give gains of V/mV. \$\endgroup\$ – Peter Smith Oct 12 '15 at 6:45
  • \$\begingroup\$ The problem with this answer is that you don't understand dBA at all, and in assuming it is something different than it is, have failed to address the question asked as you provide only examples that are unrelated to it. \$\endgroup\$ – Chris Stratton Oct 12 '15 at 16:15
  • \$\begingroup\$ @ChrisStratton: What you seem to be getting your knickers in a bunch about is that you're unable to fathom that the OP's question wasn't about the meaning of dBA, it was about the use of suffixes. Either that or you're trying to make me culpable for your gaffe. In reality, his question has nothing more to do with dBA than to use it as an example of a member of a set of suffixes, and my response to his question, I believe, provided a more than adequate answer. \$\endgroup\$ – EM Fields Oct 12 '15 at 17:06

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