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.

  • 1
    \$\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\$ – Tim Spriggs Apr 8 '16 at 17:00
  • 1
    \$\begingroup\$ My question has nothing to do with making my own microphone? \$\endgroup\$ – andrey Apr 8 '16 at 17:03
  • 1
    \$\begingroup\$ I meant bugging device. I stand corrected. but different mics would give different dB gains and sensitivity. how's the weather there? \$\endgroup\$ – Tim Spriggs Apr 8 '16 at 17:06
  • 1
    \$\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 '16 at 17:06
  • 1
    \$\begingroup\$ @MattYoung Then what is the use for a sensitivity characteristic? \$\endgroup\$ – andrey Apr 8 '16 at 17:13

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.