Stereo dB to wattage

[knight_rider1977]

I just finished watching blog #49 covering Decibels.  Using certain formulas has always been a struggle for me even though I have been working with computers since 1986 and started programming a Commodore 64 in the 3rd grade.

The information somewhat makes sense but I would like to understand an application to better understand the relation between dB and wattage.

I had the stereo in my truck in a sound competition a couple years ago.  The sound meter registered 113 dB.  What/How would I go about converting that to wattage?  Would this use a variation on the P=10*log(p1/p2) or am I totally out of the ball park on this one.

Thanks,

[wemme]

Sound is another kettle of fish.
0db is the point of human hearing. So 112dB is that many dB from that point.
Keep in mind audio has different weightings for example dBa and dBc.
I think it is dBa is weighted to the frequency response of the human ear.
Where as dBc is flat at the lower frequency.
Regards
Bart

[Sebastian]

The dB you are talking about is the sound pressure level. The dB in the formula is for calculating the ratio between two power levels.
The sound pressure level dB is also a ratio, I think in reference to some constant level.
But you can't calculate the sound pressure with just the power you are putting into your speaker, or vice versa. I am not an expert with audio stuff, but I think there should be some relationship between power and sound pressure in the datasheet of the speaker.
In conclusion, dB is a way of describing a ratio. Your 113dB tells you how loud the sound is. The formula P=10*log(p1/p2) is used to calculate the ratio between to power levels, which is not useful in this case.
There is of course a relationship between the sound pressure produced by the speaker and the power you put into that speaker. But it depends on multiple factors and you can not convert this just like this. You need additional information.

[cyr]

There is no direct conversion between sound pressure (what a sound meter measures) and sound system wattage, the only way you could do that conversion is by knowing the efficiency of that particular sound/speaker system.

Sound pressure is often said to be simply "X decibels", but like for any other measurement that is a ratio. In the case of sound, it's the ratio of the measured sound pressure to a reference level that is defined as 20 micro pascal (approximately the threshold of human hearing). More correctly it should be called dBSPL (Sound Pressure Level).

So 113 dB on your sound pressure meter means it's almost 450000 times louder than the quietest sound you can hear (or could, before you attended the competition...).

[knight_rider1977]

Thanks for the reply's.  Good information. :-)  I will do more research on the sound aspects and see what I can find.  So the formula would be more suited to say determining the input and output of say an audio amplifier or a wifi router to boost the output signal to increase signal strenght.  For example a netgear router that comes with 2 6dB gain antenna and I replaced them with 9dB gain antenna.  Using the formula would help determine the additional power needed to increase the signal strength or how much additional power is used to boost the signal.  Would that understanding be correct?

Thanks,

[barry14]

As an addition to the previous answer, sound level meters generally use either A or C weightings.  Both of these are specified frequency responses.  A weighting is designed to approximate human hearing for low level sounds and falls off significantly for both high and low frequencies.. C weighting has a much more uniform response.

[ConKbot]

And as other have said for sound its relavtive to 20uPa, in practice, you will see dBa, or dBc, both pretty common "weighting" standards for measuring SPL, and correlating it "more correctly" to how the human ears respond. http://en.wikipedia.org/wiki/A-weighting  As far as in a competition, assuming bass, even C weighting would be considerably rolled off, so unweighted response is probably used, so dBspl would be more appropriate, as cyr said.

As far as watts to dBspl  it depends on everything.  Your bare driver will have an efficiency rating in dB/watt at 1 watt, but that with a given set of conditions, either the driver in free air, infinite baffle, or a specified enclosure volume, at a specified frequency, and at a specified distance (usually 1 meter)  When you put it in a vehicle, with a different enclosure, lots more wattage, add reflections, and a variety of frequencies, it all changes.  Estimating dBspl from watts input, would be better served by someone with experience and a good gut feel for that particular application, rather than calculation. Could it be done with a sufficiently detailed model, but it wouldn't be trivial.

[ConKbot]

Thanks for the reply's.  Good information. :-)  I will do more research on the sound aspects and see what I can find.  So the formula would be more suited to say determining the input and output of say an audio amplifier or a wifi router to boost the output signal to increase signal strenght.  For example a netgear router that comes with 2 6dB gain antenna and I replaced them with 9dB gain antenna.  Using the formula would help determine the additional power needed to increase the signal strength or how much additional power is used to boost the signal.  Would that understanding be correct?

Thanks,

Audio amplifiers are kind of messy for the calculations because the impedance changes too, usually 10k+ on the input to 8,4,2, or less ohms on the output.  An amplifier taking 1Vrms (consumer line level is lower than this, but just for the example) and putting 1 Vrms out, would be taking 100 microwatts into 10k ohms, and putting out half a watt into 2 ohms. Obviously half a watt isnt crap, so they would provide voltage gain too, but your math would be different, and would have to accommodate the impedance change.

As far as the router antenna example, youre correct. When youre in the main beam, you would get double the irradiance (watts per square meter)  So a 1"x1" patch antenna recieving 1 microwatt before (-30 dBm) would recieve 2 microwatts (-27 dBm) afterwards.  But the same power is going into the coax feed of the antenna, so it concentrates it into a smaller beam, so "antenna gain" is a measure of the directivity. Its unit is dBi, or decibel isotropic, comparing it to a isotropic radiator, which emits power evenly in all directions, in a sphere. Its really is all about the unit after dB which does all the explaining, though the context can provide the unit, for casual conversation, though it can be a bit sloppy at times.  I.e. "I was getting 0dB out of the 4db antenna and my SNR was 25dB"  would make sense to a ham radio operator, and every unit there is different.

[radioFlash]

As others have mentioned, speakers have an efficiency specification that's expressed in dB per Watt at 1 meter. A typical value might be 87 dB/W, which means that 1 watt of input will produce 87 dB SPL at 1 meter. At 10 W, the speaker would produce 97 dB. 100 W would produce 107 dB. Note the exponential rise in power--10X power for a 10 dB increase in output. Note that the actual environment the speaker is in will have an effect on it's actual SPL--so if you put that speaker in a closet, it will of course produce higher SPL than if it were in free air.

[Delta]

Car amp manufacturers would probably answer with "20,000 watts".

[knight_rider1977]

Thanks for the input!!  Giving my brain lots to process and research.

[Richard Crowley]

Remember that deciBels is a RELATIVE measurement. It only expresses a RATIO between two values.  You seem to be asking about SPL (sound pressure level) where, as was already mentioned, the "reference" of "0dB" is the theoretical threshold of human hearing.

Certainly we can express specific values of voltage, power (watts), etc. in "dB".  But to be accurate, the term "dB" cannot stand alone.  The proper use of deciBels is to state the reference.  For example "dBv" is referenced to 1 volt.  And "dBm" is referenced to 1 milliwatt.
Ref: http://en.wikipedia.org/wiki/Decibel#Suffixes_and_reference_values

There is no direct correlation between dB SPL and watts.  You can probably produce 113 dB SPL with 5 watts inside your refrigerator. But it might take 5 million watts to produce 113 dB SPL in just the narrow end of the Grand Canyon.

Remember also that DOUBLING the SPL is only 3 dB higher because it uses a logarithmic scale.

[TimFox]

As with voltage or current, the sound pressure or displacement is an amplitude, and the actual acoustic power is proportional to the square of the pressure, measured by a pressure-sensitive sensor.

[Electro Fan]

As mentioned above, a couple key considerations are 1) remembering that the acoustics of a room will impact the sound from the speakers, including frequency response (including amplitude levels such as +/- Xdb), generally measured at various frequencies from 20-20kHz, and 2) speaker sensitivity.

Often speaker sensitivity is measured as described in the link below (note this talks about a non-reverberant environment - even more generally accepted is the notion of measuring at 1 meter with an input of 2.83 volts RMS):

http://www.audioholics.com/loudspeaker-design/loudspeaker-sensitivity
This is usually specified for a loudspeaker in a non-reverberant environment, in dB SPL and referenced to 1 meter on the reference axis with an input of 2.83 volts RMS, typically at one or more specified frequencies (often 300, 400, 500, 600Hz or the average of these). Sensitivity should always be accompanied by an impedance specification.

Speaker design can greatly influence how sensitive (or in laymen's terms, how efficient) a given speaker is - some speakers will require considerably more power than others to achieve the same output level.  If you study speaker spec sheets it is possible to find speakers that range from over 100 db sensitivity to under 90 db sensitivity, for example:

http://www.klipsch.com/cornwall-ii-floorstanding-speaker/details (see specs tab)
SENSITIVITY:  101dB @ 1watt/1meter 

http://www.soundlab-speakers.com/millenium-series.html (M-3PX)
Sensitivity: 89 dB/1W/1m

So, wattage doesn't directly convert to decibels. Some amount of wattage, generally measured per channel (2 channels for stereo) over 20-20 kHz RMS will in turn produce a certain number decibels for a speaker design of a certain sensitivity.

[flynwill]

And to put those number in perspective if you can make those soundlab speakers loud enough to keep you happy with a 100W/channel amp, you can get the Klipsch just as loud with a 6.3 Watt/channel amp...

[CatalinaWOW]

All of the caveats mentioned above apply, but you can get some idea for your system the old fashioned way.  Measure the power going into your sound system.  Watts = amps*volts.   It still will be a very rough number, as the actual sound power delivered will depend on several efficiencies, and the sound pressure measured will include reflections etc.

The number will be very accurate for the power coming out of your cars electrical system, and will explain why you might have needed to upgrade alternators, batteries etc.

[IconicPCB]

A very important aspect may have been drowned in the cacophony of the responses.

Electrofan has pointed to the UNITS of measurement  of sound pressure level as so many dB SPL/1 meter distance from source.

Sound pressure waves radiate outwards and as the distance increases the SPL value drops. Generally SPL level is assumed ( in SI system ) to be dBSPL at 1 meter.

Normal SPL between two humans conversing in a relaxed tone is said to be approximately 74dBSPL@1m. The roar of a jet engine is at the point of death since Your body will start to bleed from many an orifice and death will ensue.