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I am really confused at this moment. I have the following microphone: http://www.farnell.com/datasheets/1660938.pdf

What I want to do is make a actual dB meter with this sensor. So I would net to know how much the voltage changes for example when the sound changes with one 1dB? How do I calculate that with the above datasheet? It has something to do with the sensitivity but what and how?

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You assume the microphone is linear.

This means that when the sound level input to the microphone changes by 1dB, the output voltage of the microphone changes by 1dB.

Some old (very old) pointer type AC reading meters actually have a dB scale, but I've not seen any digital meters, at least down the hobby end, that offer a conversion from volts to dB.

To make a dB reading, you take a reference at one level, make your measurement at another level, then take the ratio. For a voltage ratio, the dB difference is 20*log10(ratio).

Conventionally, 0dBA is the threshold of hearing, not easy to judge.

So you have two issues. One, to get a meter that reads dB ratio, or use an Arduino or something that can compute and display dB ratio. Two, to establish your sound reference level.

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  • \$\begingroup\$ And that reference point is the sensitivity? or do I actually have to measure that? \$\endgroup\$ – user3892683 Nov 1 '15 at 18:27
  • \$\begingroup\$ That's the tricky bit. It's easy for an amateur to calculate dB(ref), that is the dB ratio with respect to some ref. However, without a calibrated microphone, or access to a calibrated sound source, you won't have an absolute value for ref. You could look up the rated SPL of your speakers, or the rated value for a 747 taking off 2000ft above you, but without an absolute reference, you will not be measuring dBA, just dB(some_arbitrary_reference). Ideally, you will know someone who knows someone in a calibration lab, or an audiometer's, buy them a beer, and use their kit for a moment. \$\endgroup\$ – Neil_UK Nov 1 '15 at 20:05
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From the data sheet it says this: -

Sensitivity (at F = 1KHz, 0dB = 1V/Pa): -41 ± 3dB

This means with 1 Pa RMS of sound (newtons per square metre or 94 dB sound pressure level) you will get -41 dB voltage out - that's approximately 8.9 mV RMS.

Plus or minus 3 dB and only at 1kHz - there's no guarantee iwhat it will be at any other frequency because all they say is 50 to 16,000Hz with no spec about how flat that may be.

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  • \$\begingroup\$ How do you get from -41 dB to 8.9mV? \$\endgroup\$ – user3892683 Nov 1 '15 at 21:29
  • \$\begingroup\$ -41 divided by 20 = -2.05. Then take the antilog = 8.9mV. The ratio of voltages in dB terms is 20log\$_{10}\$(V1/V2) so I just did the reverse. 0 dBV = 1V RMS. \$\endgroup\$ – Andy aka Nov 1 '15 at 21:52
  • \$\begingroup\$ Ok, and if I want to know the decibels for 5.4 mV -> 20log(0.0054/0.0089)=-4.3 dB , correct? \$\endgroup\$ – user3892683 Nov 2 '15 at 6:30
  • \$\begingroup\$ When assuming that it is linear \$\endgroup\$ – user3892683 Nov 2 '15 at 7:30
  • \$\begingroup\$ 5.4mV is -45.3 dBV. Think about it... 1V = 0 dBV, 0.1V is -20 dBV and 0.01V (10mV) is -40 dBV. 5.4mV in dB form is 20*log(5.4E-3). \$\endgroup\$ – Andy aka Nov 2 '15 at 8:20

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