Having read through this post I tried an experiment.

I connected a Extech sound level calibrator to a Type 4189 microphone. The output sensitivity of the microphone is

50 mV/Pa,-26 dB ± 1.5 dB re 1 V/Pa

This microphone is connected to a Type 2671 preamp. Using a BNC 50 ohm cable the preamp is connected to a signal conditioning amplifier through a 50 Ohm BNC-BNC cable. The output from the amplified is connected to a standard oscilloscope.

The calibrator generates a consistent 1 kHz, 94dB SPL sound. The calibrator was recently calibrated, so the output is assumed to be the standard.

The gain at the signal conditioning amplifier is 100mV/unit.

The Vpp measured at the oscilloscope was 16.2 Vpp. Converting that to Vrms,

Vrms = 16.2/2.828 = 5.72843

Calculating for dBrms = 20*log10(5.72843/0.05012) = 41.1605

Since the gain at the signal conditioner is 100 mV/unit. the dB value is -20 dB.

For dB SPL = 41.1606+94-26-20 = 89.1605.

Would I be wrong in expecting this to have amounted to 94 dB SPL instead? If so, why the discrepancy ? what am I missing?


Forget decibels for a moment. There is 1 Pa sound. Microphone output for 1 Pa sound is 50mVrms. Since the preamp does not have gain, and oscilloscope output is about 5Vrms, so there must be about 100x gain somewhere in the systen, so this must happen in the signal conditioner. I don't know what 100mV/unit means in this case so my assumption is that is it means 100x, or +40dB. If you work this backwards, from 16.2Vpp you get 5.7Vrms, so without 100x gain this is 57mVrms from mic. This is within the given tolerance for a 1 Pa sound, which nominally is 50mVrms. So 50mVrms is 1 Pa and thus 94dB SPL, and 57mVrms is 1.14 Pa and thus 95dB SPL.

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  • \$\begingroup\$ What you said totally makes sense. But i got confused with that earlier post because the Math just didn't add up in that post. \$\endgroup\$ – whoknowsmerida Feb 20 '19 at 1:16

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