In particular, this device: http://www.hexamite.com/hx40str.htm

Electrical Specifications HX40STR
Parameters              Values      Units   
Operating Frequency     40          kHz   
Input Voltage           42 (max)    Volts (p-p)  
Output (SPL) @20V       112         db
Receive (Sensitivity)   -60         db/V/Ubar
Impedance               300         ohm
Beam Angle              ±12         degrees
Bandwidth               1           kHz
Settling Time           5           mS
Temperature             -40 to 100  °C

Assuming the scenario that two trancievers are used, T1 as transmitter, T2 as receiver, 125mm apart. (Assuming no attenuation over distance.)

Output SPL at 20V => 112 db.

Converting pressure to uBar (because sensitivity is in uBar):
P = pressure in uBar

112 = 20 log (P/0.2e-3)

P = 0.2e-3 * 10 ^ (112/20)

P = 79.621 uBar

Input Sensitivity => -60 db V/uBar

S = Sensitivity in V/uBar

-60 = 20 log (S/1)

S = 10 ^ (-60/20)

S = 1e-3 V/uBar

Final Voltage V = P * S

V = 79.621 uBar * 1e-3 V/uBar V = 0.0796721 V

However, looking at the graph "G3, Amplitude vs Excitation"


"Above: Continuous exitation voltage oscillating at 40Khz is applied to T1. The T2 resulting amplitude is plotted above."

We see that the output voltage at 20V is 0.6V. This also excludes the attenuation over the space of 125mm, which I've calculated to decrease the output V to 0.079 V to ~0.075 V.

What am I missing in my generated voltage calculation? Why is there a discrepancy?


1 Answer 1


I don't know for sure, but one possibility might be that the quoted SPL measurement has a condition attached, such as a measurement distance. It ought to, and a common distance for SPL measurements is 1 metre.

In which case the voltage measured at only 125mm distance could be expected to be greater.

If acoustic power varies as the inverse square of distance, then voltage will vary as the inverse of distance, and this would lead to 8x the expected voltage at 1/8 distance.

  • \$\begingroup\$ I think you're onto something with this. \$\endgroup\$
    – stanri
    Dec 9, 2013 at 17:43

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