Trying to relate microphone sensitivity and SPL

Question

In this:

I don't understand equation 3. How does 20·log₁₀(...) - 94 dB = SPL?

If I did the example and solved the equation for 70 dB, I would get something around:

10^(70+94)/20 · 6.3 · 10^-3 = 998483 V

The result doesn't seem to make any sense and has nothing to do with the 0.399 V_RMS stated.

How do I transform the V_out value computed from that equation to RMS?

Edit:

So using the things you guys told me I put the equation in MATLAB to do a sketch. This is what I got:

Very similar with the curve from the first image, what differs is the y axis. In the first image the y axis is in Vrms while the one i made in matlab is in V, if I multiply the y values with 1000 I seem to get the same result.

What is RMS in this context?

Answer 1

A microphone specification of \$-44\:\text{dBV}\$ implies a reference of \$1\:\text{Pa}\$ or \$\approx 94\:\text{dB SPL}\$ and the V is with reference to \$1\:\text{V}_{_\text{RMS}}\$. not peak. The spec is usually first linearized (taken out of log domain.) To do that, just compute \$10^{^\frac{-44\:\text{dB}}{20}}\:\frac{\text{V}_{_\text{RMS}}}{\text{Pa}}\approx 6.31\:\text{m}\frac{\text{V}_{_\text{RMS}}}{\text{Pa}}\$. (You can see now how the dB and V get separated on the left side and also how the Pa gets added.)

Also \$70\:\text{dB SPL}\$, linearized, is just \$10^{^\frac{70}{20}}\cdot 20\:\mu\text{Pa}\approx 63.25\:\text{mPa}\$. The \$20\:\mu\text{Pa}\$ comes from how dB SPL is defined.

If you multiply these you get about \$\left[10^{^\frac{-44}{20}}\:\frac{\text{V}}{\text{Pa}}\right]\cdot \left[10^{^\frac{70}{20}}\cdot 20\:\mu\text{Pa}\right]\approx 399.05\:\mu\text{V}_{_\text{RMS}}\$.

(Clearly, you could also have just computed \$10^{^\frac{-44+70}{20}}\cdot 20\:\mu\text{V}_{_\text{RMS}}\$ to get the same answer.)

A POM-3535P-R microphone has \$-35\:\text{dBV}\pm 4\:\text{dBV}\$. This means anywhere from \$710\:\mu\text{V}_{_\text{RMS}}\$ to \$1780\:\mu\text{V}_{_\text{RMS}}\$ given \$70\:\text{dB SPL}\$. Note that for any given microphone of this type, it can be a factor of more than 2. So you also need to account for part variations in any design. Not just some average value.

Answer 2

This is the best I got:

Ignoring as much of what they said as possible and doing it the way I think it should be done:

If your microphone has a sensitivty of 6.3mV/Pa and your output voltage is 0.399mV, what is your SPL?

Find how many Pascals your output reprsents: \$P=0.399\text{mV}\div6.3\frac{mV}{Pa}=0.06\overline3\text{Pa}\$

Now we just want to convert that to SPL which is in dB. We know the reference pressure \$P_0 = 1 \text{Pa}\$. So just put everything in Equation 1.

\$SPL=20log\frac{P}{P_0}=20log\frac{0.06\overline3\text{Pa}}{1 \text{Pa}} = -23.97\text{dB}\$

Which doesn't match their result.

But if, for some unknown reason, I add 94dB to that then I end up with their 70dB. Which means that their -94dB should be +94dB. However, I do not know why they are adding 94dB.

If you change the -94dB to +94dB in their Equation 3, then if you put \$Vout =0.399\text{mV}\$ and the \$Sens_{mV/Pa} = 6.3\text{mV/Pa}\$ you get their result.

Answer 3

Work backwards. If you gave an SPL of 94dB you would have 6.3mV from the mic.

0.399mV is 24dB less than 6.3mV - in voltage terms, use

20 log (0.399/6.3)

So you need an SPL of 94-24 = 70dB to get 0.399mV out