Hello Everyone,

If anyone here can please suggest me best sites for designing Ultrasonic transmiiter/Receiver circuits....

And say If an ultrasonic signal of  (0 to 5v ) @ 40 khz is transmitted and it bounces back to the receiver what could be the probable range of the reflected signal....

Vpool..
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Commented:
Commented:
Airmar Technology Corp has a wide range of transducers and a lot of technical infor available:

According to the spec sheet, the Airmar AT50 transducer (which I have used)  has a maximum range of 10 to 15 meters.

Commented:
By the way, the AT50 runs at 50 kHz.  It is fairly expensive, on the order of \$100.

I have also used surplus 40 kHz transducers from Marlin P Jones.  They are considerably less powerful.  You would be lucky to get 3 to 5 meters range.  It also looks like they are out of stock.

http://www.mpja.com/category/Security_Accessories/40KHz_ULTRASONIC_TRANSDUCERS_12941_UT.asp

The second Airmar link discusses transducer power and range calculations.  A lot depends on how you are using the transducers.  The range numbers I've given are for sonar (ping and listen for the echo).

If you are trying to tranmit information or make a make an intruder alarm (acoustic beam that an intruder would interrupt), your ranges could be 4X larger.

If you are willing and able to do fancy signal processing on the receiver, you can increase the range even more.
Commented:
Airmar has a simple but flexible circuit that will drive all of their transducers:

You could simplify it a little for a singel fixed frequency.

I should correct a mistake:

>> your ranges could be 4X larger.      [That should be 2X not 4X]
Author Commented:
Thanks for all the reply...But still i couldnt find the information regarding the range of reflected ultrasonic waves IF TRANSMITTED WAVE IS (5V@40khZ).....PLEASE HELP
Commented:
Range will depend on the reflector. For example a flat steel wall is a good reflector.
Most transducers need a higher drive voltage than 5V.
For specific applications you may consider parabolic focusing reflectors.
Commented:
>> Most transducers need a higher drive voltage than 5V.

Absolutely correct.  But you can still run a transducer off of a 5 volt power supply and use a tuned resonant circuit
and/or a transformer to boost the drive voltage.

>> vpool

Exactly what are you trying to do?
What are your requirements?  What range do you need?
Do you have particular transducers you want to use?  If so, what are the part numbers?
5 volts doesn't tell us anything.  It would be much more helpful to know the power rating of the transducer.
Are you looking to build a system or buy it?  What is your budget?  Do you have any electronic experience?

Commented:
Here is a data sheet for the Airmar AR41 transducer:

According to the spec sheet you can get a range of 15 meters if you run it at 1800 volts peak-to-peak.
If you run it at 10 volt peak-to peak (that is +5 to -5) your range would be 8.3 cm (15m/180) or 3.25 inches.

Im guessing that is not enough, but I don't know for sure.
So here is a scematic for a ultrasonic transmitter and receiver board from Airmar.

It runs on 12 volts, and uses a transformer to boost the drive to ~200 volts.
You could modify the circuit so that it would run on 5 volts.
Most of the stuff on the left side side of the page is so you can adjust the frequency from 30 to 300 kHz.
You can simplify the circuit considerably if you operate just at 40 kHz.
You could build something like this for ~ \$25 in parts.

With a circuit like this one (running at 5 volts and boosting the drive to 200 volts)
you would get an operating range of 1.66 meters (15m/9) or 65 inches.

If you still need more range, you need more power, which means more drive voltage,
which means a better circuit design.
Commented:
One lat note on Range, Power, and Voltage.

An acoustic beam spreads out as it travels.
So if you increase the range by a factor of two, you have to increase the power by a factor of four to maintain the same power at the receiver.

Electrical power is proportional to V².
So if you increase the voltage on the transducer by a factor of 2, you increase the transmitted power by a factor of four.

So there is a linear relationship between voltage and range.

Since the AR41 has a range of 15 meters at 1800 volts.

At 200 volts, the range is (15 meters) (200v/1800v) = 15m/9 =  1.66 meters.

If you have a data sheet for your transducer, you should be able to do similar calculations.
Author Commented:
I am making a parking sensor.....
The transducer I am using is Au5550 from Jaycar....
At the max I can drive the Transmitter circuit at 12 v.....
Range of parking sensor is....5cm to 1.5 m

Commented:
That is very helpful.  I have found the data sheet:

The sound pressure at 30 cm is: 105 dB  ==>  10^(105/20) * (0.002 ubar)  ==>  35.6 ubar with a 10 volt drive.

The receiving sensitivity is  -74 dB  ==>  0.2 mV at 1 ubar

So if you put a receiver 30 cm from a transmitter your would see a voltage of

( 35.6 ubar) x ( 0.2 mV/ubar) = 7.1 mV  with a 10 volt drive.
=============================================================
Assume you build a receiver circuit with a gain A= 1000 and a detector threshold of 0.1 volts.

With a target range of  30 cm, the sound has to travel out and back for a total of 60 cm
Your signal would be 7.1 volts/4   =  1.78 volts  ==>  OK

For a target at 1.5 meters ( 5 x 30 cm)
Your signal would be 1.78 volts/25 = 0.071 volts  ==> which is a little too small.
=============================================================

But don't despair.  You can increse the gain, reduce the dector threshold, or drive the signal just a little harder.

The best thing to do is drive harder.  And it is pretty easy.

Note that capacitance of the transducer is 1800 pF.    That gives a reactance of 2*PI*40kHz*1800pf = - 2200 ohms
To make a resonant circuit with the transducer you need an inductor of 8.8 mH.

If you drive the series combination of the inductor and transducer with a 5 volt square wave you will boost the voltage on the transducer by a the Q of the resonant circuit (typically 10 to 20).

Good luck.

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Author Commented:

Its a very useful information...

I have got my circuit

http://geocities.com/naymesh/circuit.html

could you please review it  ...

Transmitter will be driven at 12v...

Cheers...

Vipul
Commented:
Hard to read part numbers, but if LM393 is used, that is a comparator and not an opamp.
You should use some type of bandpass filtering for the receiver or it may be swamped by noise.
The +pin of the output amp is not well biased.

The drive is poor. Use a discrete fet and a stepup.
Commented:
public is correct, although I think you are off to a pretty good start.

The CD4069 is a poor driver for this sort of thing. The max output current is ~6 mA.
You could try something like the MIC4428, which can sink and source 100 mA.

If the MIC doesn't give you enough power, you should look at switching power supply designs.
I have used transformers from PICO in the past.  They have sample circuit designs on line.
I have a circuit somewhere.  I will try to post it tomorrow.

Your amplifier only has a gain of 100, this is probably not going to be enough.
I think you need a gain of at least 1000.
Three stages of A=10 are probably better than two stages of A=33.

You are relying on the transducer to filter out noise.  This may be okay.
But it would be much better practice to configure each of gain stages as a Sallen-Key bandpass filter.

Your detector is just a diode and capacitor.  It doesn't start to work until the signal get up to a few tenths of a volt.
You can make a precision full wave rectifier with an op amp and two diodes that will let you work with smaller signals.

You do want to use a comparator after the detector.

The detector capacitor needs some sort of bleed resistor.  RC should be about the same as the ping pulse length.

Certainly build a seperate transmitter and receiver to start.  But it is possible to use a single transducer for both functions.  Look at the Airmar circuit to see how they did it if you are interested.

Author Commented:

Hey Public,

How to calculate the exact value of bleeder resistor connected to C=01uf..,,

Commented:
How long is your transmit pulse?

If your transmit pulse is 1 ms then your detector time constant should be something like 0.5 ms.

RC = 5 x 10^-4    ==>    R = (5 x 10^-4) / (1 x 10^-7) = 5 x 10^3 = 5K
Commented:
Here is the PICO Transformer website:             http://www.picoelectronics.com/plugin/pe50_51.htm

And here is one of my Piezo Driver Circuits:      http://www.geocities.com/d_glitch/Piezo_Driver.pdf

With the MIC4428 I mentioned earlier, you can get 100 mA drive at 12 volts.  But this is still only 1.2 watts.

With the PICO transformers, you can get up to 40 watts.  The capacitors in parallel with the transducer allow you to increase the Q of the tank circuit.  Note that you have to turn both FETs off between pulses because the transformer is a DC short circuit.

Commented:
>http://www.geocities.com/d_glitch/Piezo_Driver.pdf

the transformer must be driven with a 50% duty cycle. A flip-flop in front of the driver is recommended.
Q of the transducer should be controlled, otherwise the pulse would decay very slowly.
Author Commented:
Thanks a lot d glitch......public...
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