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Sound Pressure Level - Ultrasonic Help!! Maths Stuff

Discussion in 'Electronic Basics' started by roots_of_culture, Jun 9, 2006.

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  1. 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
     
  2. Phil Allison

    Phil Allison Guest

    "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
     
  3. Guest

    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
     
  4. Phil Allison

    Phil Allison Guest


    ** You missed the *peak* bit.

    106.7mV x 1.414 = 150.1 mV

    Comparator windows are DC voltages.





    ........ Phil
     
  5. Guest

    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
     
  6. Phil Allison

    Phil Allison Guest

    <>


    ** Go away - troll.
     
  7. Guest

    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.. (",)
     
  8. Guest

    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
     
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