Sound Pressure Level - Ultrasonic Help!! Maths Stuff

Hi all,
Im using an ultrasonic transducer(refer link) and reciever to form a
parking sensor..
http://www.jaycar.co.nz/products_uploaded/AU5550.pdf
i am driving the transmitter at 10V rms transmitting a pulse of 40 KHz
using a TC4428 driver - so the output will be at an amplitude to twice
the input voltage..
hecne 28 Vpk-pk

I am trying to calculate the range and intensity of my signal.
i am confused with the calculation of the recieving sensitvity
would it be 1v/1u bar
or would it be 0.2 mV /1u bar??

please exlplain

according to my calculations the SPL @ 30 cm is 35.6u bar


then the output at a distance of 30 cm the reciever will be 35.6 u bar
x 0.2mV/u bar = 7.1 mV
is this correct??

if to calculate the output at a distance of 60 cm( travels 30 cm and
back) : 7.1 mV / 4
( output changes as a function of distance^2)
= 1.78mV



if so how will i calculate the output at 2 m??


i have a gain of 441 and a comparator detector window of 150mv, what
will be the maximum distance that i can detect?? how will i calculate
this??
this is so that i can adjust my gain to be able to extend the distance
to 2m..

i have spent several hours over this... might just be thick.. please
help!


Nick

 

"roots_of_culture"

** The latter is correct.

The transducer has an output 74 dB below a reference level of 1 volt when
the test condition is a SPL of 1 uB.

1 uB = 74 dB SPL.

** Correct.

** Looks good to me.

** You are making an error regarding power and SPL.

When SPL is halved, sound power drops by 4 times.

The *voltage* from the transducer is merely inversely proportional to
distance.

So, at 2 m the voltage level will be 0.3 /2 = 0.15 times.

** Lemme see.

7.1 mV times 441 = 3.13 volts rms

3.13 x 1.41 = 4.41 volts peak.

4.41 / 0.15 = 29.4


So, the distance where the peak signal level matches your 150mV window
threshold

= 29.4 x 0.3 = 8.8 m.



BTW,

Again of 441 times at 40kHz requires an (op) amplifier with at least 20 MHz
gain-bandwidth product.

Or are you using two stages ?




....... Phil

 

Firstly thank you,
that helped me get clear on the concept and calculations to a much
greater degree

Might be being thick again.. but...

the 0.3 u are referring to is the 30 cm mark specified on the
datasheet, yes?

is this to say : the output at a distance of 3 meters ( travels 30 cm
and back) :
= 7.1 mV x .1
==> 0.71 mV
correct??

i understand how u went through this calculation
But this will not be detected by my comparator window.... if u reverese
the calculation

@ 8.8 m : = 0.3/8.8 ==>0.03 times ?
output = 7.1 mV x 0.03 = 0.242 mV

on gain of 441 = 0.242 x 441 = 106.7 mV


Nick

 

** You missed the *peak* bit.

106.7mV x 1.414 = 150.1 mV

Comparator windows are DC voltages.





........ Phil

 

Ur solution got me thinking to calculate max distance:
min voltage to trip the comparator is 150mV

taking away the gain from this = 150mV/441
==> 0.34mV at reciever

Hence 7.1mv x (0.3/ d) = 0.34mV
==>(0.3/d) = 0.05
==>d = 6.3 m??

is this correct???

.......Nick

 

<>


** Go away - troll.

 

lol... last question my main man!!

sori failed to mention
i am rectifying the signal before it reaches the comparator. with a
Schottckly and rc filter combination.
min voltage to trip the comparator is 150mV x 1.414 = 212.16mV

taking away the gain from this = 212.16mV/441
==> 0.34mV at reciever


Hence 7.1mv x (0.3/ d) = 0.48mV
==>(0.3/d) = 0.067
==>d = 4.43 m

and since i am rectifyin the signal will the range halve as im cutting
the signal in half??

ever grateful
Nick.. (",)

 

i would like to thank Phil for his posts,
i now have the theoritical values and am checking the results of the
circuits as compared to the former.

And the results are quite close , this is considering the bode plot for
my circuit( the actual gain) based on test results of applying various
inputs to my circuit.

thanks once again.. (",)
Nick