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I'm attempting to convert a voltage signal originating from a microphone to decibels. However, I do not have the "microphone sensitivity" in terms of dBV/Pa which is used in this previous conversion. Instead, I have the microphone conversion factor as sensitivity.

Given the details below, how would I go about converting this voltage to decibels?

Approximate sensitivity: 250mV/Pa

Pre-amp gain setting: 12 dB

Approximate input-referred self-noise Level (bandwidth: 30-50 kHz): 18 dB SPL

Link to microphone

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  • \$\begingroup\$ how are you planning on calibrating the decibel readings? \$\endgroup\$ – jsotola Feb 5 '18 at 21:11
  • \$\begingroup\$ I wasn't planning on calibrating the microphone myself, although I could. My crude understanding was that if I had the microphone's sensitivity rating in terms of the voltage output for a given pressure on the microphone, I could get back to a decibel reading. From previous posts, this seem like it should be the case, but am I wrong about this? \$\endgroup\$ – E. Camus Feb 5 '18 at 21:22
  • \$\begingroup\$ Remember that decibels are a differential reading, though some have assigned ref points such as 130 dB for a jet engine, etc. \$\endgroup\$ – Sparky256 Feb 5 '18 at 21:33
  • \$\begingroup\$ @E.Camus, are you aware that 250 mV is equal to -12 dBV? \$\endgroup\$ – davidmneedham Feb 5 '18 at 21:38
  • \$\begingroup\$ @davidmneedham I assume you are getting -12 dBV from 20*log10(0.250), where 0.250 is the sensitivity in terms of volts per pascal? If so, how do I use this to calculate a reference voltage so I can use the somewhat standard decibel calculation, 20 * log10 (V/Vref)? \$\endgroup\$ – E. Camus Feb 5 '18 at 21:53
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250mV/Pa is an RMS voltage of 250 mV for an RMS sound pressure of 1 pascal and this is usually specified at 1 kHz for microphones. 1 Pa is a sound pressure level of 94 dB. See this calculator.

So 250 mV RMS is what you get for an SPL of 94 dB. To convert 0.25 V to dBV you take the log then multiply by 20 to give you -12 dBV. If your pre-amp has 12 dB of gain then your microphone is producing -24 dBV for 1 Pa.

The noise level you specified (18 dB SPL) is 76 dB lower than 1 Pa but I'm unsure what relevance it has to your basic question.

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  • \$\begingroup\$ Based on the other packages in the specifications, it looks like the sensitivity before the pre-amp is -12 dBV. So the gain after the pre-amp is 0 dBV with a setting of 12 dB. That does seem like high-sensitivity to me. Based on this page, it's possible the sensitivity is measured at 50 kHz. \$\endgroup\$ – davidmneedham Feb 6 '18 at 16:22
  • \$\begingroup\$ You could easily be correct but the general conversion between the units is the main focus for this answer. \$\endgroup\$ – Andy aka Feb 6 '18 at 16:31
  • \$\begingroup\$ @Andyaka Thanks for the detailed response. Just to make sure I'm performing the right final step in the conversion, if I measure 500 mV as an analog voltage after the preamp, this should be be what in terms of decibels given the microphone sensitivity? My thought is to use the conversion factor of 250 mV/Pa, but now my units are in terms of decibel volts/pascal, not volts/pascal (as in the microphone specification), making the conversion units not cancel out. Am I missing a step? \$\endgroup\$ – E. Camus Feb 6 '18 at 20:10
  • \$\begingroup\$ 500 mV converted to dBV i.e. decibels with respect to 1 volt, is -6 dBV. That conversion has nothing to do with where it is measured or what amplifier gain produced it. You are thinking too hard or in the wrong direction.... 1 volt equals 0 dBV. Half it to get -6 dBV and quarter it to get another reduction of 6 dB i.e. -12 dBV. \$\endgroup\$ – Andy aka Feb 6 '18 at 20:20
  • \$\begingroup\$ @E.Camus , if 250 mV represents 1 Pa (94 dB SPL), 500 mV represents 2 Pa (100 dB SPL). Like Andy said, half of a voltage or pressure level is -6 dB. Double is +6 dB. \$\endgroup\$ – davidmneedham Feb 6 '18 at 20:26

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