I am using the MAX4466 microphone amplifier to calculate the db SPL level of sound. I have obtained the peak to peak amplitude in volts and I know that the microphone sensitivity is -44 db which translates to 0.0063096 V/Pa. 1 Pa is 94 db SPL.

so, say my voltage reading is 1.7 volts ; in db it = 20 log (1.7/0.0063096) = 48.6 db

Now acc. to a formula I found on this thread How to convert Volts in dB SPL

to calculate db SPL, I must add the mic sensitivity + 94 db SPL i.e 48.6 + 94 - 44 = 98.6 db SPL.

I don't completely understand why we do this?

Secondly, this formula is for unity gain. I don't know the gain of my device to subtract from the db SPL level and I can't seem to find this anywhere. Please help!

  • \$\begingroup\$ Tried reading the data sheet? \$\endgroup\$ – PlasmaHH Feb 27 '16 at 9:17
  • \$\begingroup\$ Hi, So from the data sheet I could gather that the gain margin is 20 dB. However, my db SPL calculations still seem off. I'm not sure if gain "margin" is in fact the gain of the op amp. \$\endgroup\$ – Shruti M. Feb 27 '16 at 10:15
  • \$\begingroup\$ "Gain margin" is a measure of how close to unstable the amp is. "Gain" tells you its gain. And if it's an opamp the gain is determined by external components you can see and we can't. Try linking the datasheet and adding your circuit (use the circuit editor) if you need more help. \$\endgroup\$ – Brian Drummond Feb 27 '16 at 11:02
  • \$\begingroup\$ If your microphone has a gain of 6.3 mV/Pa then an output level of 1.7 volts (RMS and not DC) is a massive sound pressure of 270 Pa or 143 dB SPL. I think you are erroneously measuring a DC level of 1.7 volts and believing that this indicates SPL. \$\endgroup\$ – Andy aka Feb 27 '16 at 11:43

The MAX4466 doesn't have a 'default' gain. Gain is set by the ratio of feedback and input resistors connected to the inverting input (IN-). For example if the feedback resistor is 220k and the input resistor is 22k then the gain will be 220/22 = 10, or 20dB. To determine the gain of your circuit you need to know the values of those resistors.

SPL is rms sound pressure, so you must convert your 1.7V peak-to-peak measurement to rms by dividing it by 2.828 (assuming a sine waveform) which works out to 0.6Vrms. dB is a relative measurement, representing a power gain or loss. To convert 0.6V to dB you must first decide on a reference power or voltage level. dBV is voltage gain or loss relative to 1V. 20(log 0.6/1) = -4.4dbV.

If your amplifier gain is (say) 20dB then the mic must have generated -4.4 - 20 = -24.4dBV. The mic has a sensitivity of -44dBV, which means that it generates -44dBV at a sound pressure of 94dBSPL. In your test it generated -24.4dBV which is 19.6dB higher than -44dBV, so the sound pressure level must have been 19.6dB higher than 94dBSPL, ie. 113.6dBSPL.

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  • \$\begingroup\$ Thank you. I should have converted to rms before hand itself. the explanation was helpful. Can you also help me with a good technique to test my calculations? A decent online SPL meter perhaps? \$\endgroup\$ – Shruti M. Feb 29 '16 at 6:16

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