# Trying to relate microphone sensitivity and SPL

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 VRMS stated.

How do I transform the Vout 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?

• Leaving this as a comment since it doesn't answer your whole question, but I didn't see any other answers mention it -- I think there's an error in the book's last equation. The -94dB should be +94dB. Big difference! Also, note that their .399 is in mV, not V. That might explain your factor of 1000. (RMS means "root mean square", which is just a type of average. I think it's just a distraction here, not relevant to your issues.) Commented Aug 2, 2022 at 16:36

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.

• Where is 20uPa coming from? Commented Aug 2, 2022 at 3:40
• @DKNguyen It's in the usual standard for the term SPL. I forget where. Probably find it in Wiki. Want me to look it up?
– jonk
Commented Aug 2, 2022 at 3:47
• Nah, I'm not that invested in it right now. lol Commented Aug 2, 2022 at 3:48
• @DKNguyen I added a reference, anyway. ;)
– jonk
Commented Aug 2, 2022 at 4:09
• @G0tBlackOps It isn't clear to me what you want from me. I can see the edit. Anyway, I need to edit my answer so that I include "RMS" when I speak about 1 V. Also, there something called the Overload SPL, which is NOT RMS, but an absolute maximum not to exceed.
– jonk
Commented Aug 2, 2022 at 18:50

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.

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