My goal is to figure out a relation between output voltage and input SPL:

$$dB(SPL) = f(V_{out})$$

First, I get the sensitivity in volts from this formula:

$$Sensitivity_{dB(V)} = 20 * log_{10} (Sensitivity_{mV/PA})$$

For example, a microphone's sensitivity is -46dB(V)/Pa.

$$Sensitivity_{mV/PA} = 10^{-46/20} = 5.0119mV/Pa = 5mV/Pa$$

Since 1 Pa = 94dB(SPL), can it be written as 5mV/94dB(SPL)? Can that sensitivity then be rewritten as 53uV/dB(SPL)? So the final equation is this?

$$dB(SPL) = V_{out} / 5.3e^{-5}$$

Something tells me that it doesn't work this way, but I can't figure out where I've gone wrong.

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    \$\begingroup\$ Having tried to make a good bugging microphone (no, not related to Nixon) I can tell you it's not easy. but, they make different microphones (ribbon, condenser, etc.). Condensers are sensitive microphones, in general. \$\endgroup\$ Apr 8, 2016 at 17:00
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    \$\begingroup\$ My question has nothing to do with making my own microphone? \$\endgroup\$
    – andrey
    Apr 8, 2016 at 17:03
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    \$\begingroup\$ I meant bugging device. I stand corrected. but different mics would give different dB gains and sensitivity. how's the weather there? \$\endgroup\$ Apr 8, 2016 at 17:06
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    \$\begingroup\$ Frequency response is the dependent variable, and it's pretty nonlinear. Trying to correlate SPL to a voltage over any frequency range is an exercise in futility. \$\endgroup\$
    – Matt Young
    Apr 8, 2016 at 17:06
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    \$\begingroup\$ @MattYoung Then what is the use for a sensitivity characteristic? \$\endgroup\$
    – andrey
    Apr 8, 2016 at 17:13

1 Answer 1


For example, a microphone's sensitivity is -46dB(V)/Pa

-46 dBV is about 5 mV RMS and bear in mind we are talking about pure sinewaves at 1kHz (mid band). It's 5 mV because \$10^{\frac{-46}{20}}\$ = 5 mV.

This voltage arises from an SPL of 1 Pa RMS (unit of sound or any pressure in newtons per square metre) hence for 2 Pa RMS the output voltage will be 10 mV RMS. For 0.1 Pa the output will be 0.5 mV RMS.


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