I'm trying to compare 2 acoustic sensors

In summary, the conversation discusses comparing two acoustic emission sensors with different sensitivity specifications in terms of dB ref 1V/μPa and dB ref 1V/μbar. The use of dB to compare things is confusing and the negative sign may indicate a smaller voltage than the reference. However, the specified sensitivities seem to be off by an order of magnitude and may be a misprint. The conversation also mentions using the "20 log rule" and trying to figure out the expected voltage output for a given input of pressure.
  • #1
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Hi all,

Through my tears, I'm trying to compare two acoustic emission sensors. The first sensor has a specification that says its sensitivity is −150 dBre 1V/μPa. The other sensor says its sensitivity is -70 dB ref 1V/μbar. (Note the different units.)
Wikipedia tells me that 1 Pa = 10-5 bar. So I'm guessing 0.1Pa = 1μbar .

Okay, so it's been a long, long time since I used dB for anything. Frankly, using dB to compare things has never made any sense to me. And with that negative sign thrown in there, I'm really confused. Are they telling me that the sensor provides negative voltage output, or is it a negative sign that gets plugged into the "20 log rule". (When I plug that negative into the 20 log rule, I get insanely small numbers. ) :uhh:

For "20 log rule", I'm looking at this on Wikipedia: en.wikipedia.org/wiki/Decibel

If I ignore the negative and just plug and chug, I get something like this for the first sensor: Pressure needed to get 1 Volt = 10150/20 = 31.6 Pa

and this for the second sensor: Pressure needed to get 1 Volt = 1070/20 = 3162 μbar = 316 Pa

They seem to be off by an order of magnitude, but ... I noticed other sensors, which happen to have integral pre-amps, have a sensitivity rated at -24 dB ref 1V/μbar. So what to do? Anybody understand how this sort of specification is used? I'm trying to figure out how much voltage I can expect the sensors to output vs. some input of pressure. I expect the output to be in the millivolt range. but... :confused:

many thanks,
Mark
 
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  • #2
The log bit is OK.

If you take the log of 3 / 1 you get 0.477
If you take the log of 1 / 3 you get -0.477

So, all the negative sign is doing is telling you that the voltage involved is smaller than the reference assuming you are using 20 log (voltage / reference voltage)

However the data seems pretty odd.
A pascal is already a very small unit. One Pascal is like 100 ml of water spread out over a square meter. Not much pressure.
A micropascal is extremely small.
To have a pressure that is 150 dB less than a micropascal gives crazy results. Could be a misprint.
 
  • #3


Hello Mark,

It seems like you're trying to compare two acoustic emission sensors, one with a sensitivity of −150 dBre 1V/μPa and the other with a sensitivity of -70 dB ref 1V/μbar. First, let's clarify the units being used. The first sensor's sensitivity is measured in dBre 1V/μPa, which means it is referencing the sound pressure level (SPL) to 1 volt per microPascal (μPa). The second sensor's sensitivity is measured in dB ref 1V/μbar, which means it is referencing the SPL to 1 volt per microbar (μbar). As you correctly noted, 1 Pa = 10-5 bar, so 1 μbar = 0.1 μPa.

Now, let's address the negative sign in the sensitivity values. In this context, the negative sign indicates a ratio or a decrease in voltage. For example, a sensitivity of -3 dB means the output voltage is 50% of the input voltage. In your case, the first sensor's sensitivity of −150 dBre 1V/μPa means that the output voltage will be 10^-15 times the input voltage. Similarly, the second sensor's sensitivity of -70 dB ref 1V/μbar means the output voltage will be 10^-7 times the input voltage.

To compare the sensitivity of these two sensors, we can use the following formula:

Sensitivity (dB) = 20 log (Vout/Vin)

Where Vout is the output voltage and Vin is the input voltage. Using this formula, we can calculate the input pressure needed to produce an output voltage of 1 volt for each sensor.

For the first sensor:

Sensitivity (dB) = -150 dB
20 log (Vout/1V) = -150 dB
Vout = 1V x 10^-150/20 = 10^-7.5 = 0.000000316 volts

To convert this voltage to pressure, we can use the formula:

SPL (dB) = 20 log (P/Pref)

Where SPL is the sound pressure level, P is the pressure we want to find, and Pref is the reference pressure. In this case, the reference pressure is 1 μPa.

SPL (dB) = -150 dB
20 log (P/1 μPa) = -150 dB
 

1. How do the acoustic sensors differ in terms of functionality?

The two acoustic sensors may differ in terms of their sensitivity to sound, frequency range, and ability to filter out background noise. One sensor may be better suited for detecting low frequency sounds, while the other may be more sensitive to high frequency sounds.

2. What is the accuracy and precision of each acoustic sensor?

The accuracy and precision of an acoustic sensor refers to its ability to measure sound waves correctly and consistently. This can depend on factors such as the sensor's design, calibration, and signal processing algorithms. It is important to consider both accuracy and precision when comparing sensors.

3. How do the acoustic sensors differ in terms of cost?

The cost of an acoustic sensor can vary depending on factors such as its design, materials used, and manufacturing process. Higher-priced sensors may offer more advanced features and better performance, but it is important to consider the specific needs and budget of your project when comparing costs.

4. What are the potential applications for each acoustic sensor?

The potential applications for an acoustic sensor can vary depending on its capabilities and sensitivity. Some common applications include noise monitoring, sound detection, and acoustic imaging. It is important to choose a sensor that is suitable for your specific application.

5. How do the acoustic sensors differ in terms of size and portability?

The size and portability of an acoustic sensor can be important factors to consider depending on the intended use. Some sensors may be smaller and more lightweight, making them more suitable for portable or remote monitoring applications. Others may be larger and more stationary, but may offer better performance and accuracy.

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