Ultrasonic Receiver Circuit at 25kHz

[duffy]

Where should I connect my ultrasonic receiver? Connect at the pin 2 from U1?

It may work if you connect it directly to pin 2 of U1, but I would suggest using the connection in the schematic. C1 provides DC blocking, this is sometimes important with piezo elements (your transducer is a piezo element) because of strain related to temperature change and mounting stress. The 10k "R2" (which should be labeled R1) sets the gain of the amplifier with the other R2 (the 1M pot). Without that, you will always have infinite gain because the ratio will be:

[LATEX] \frac{R2}{R1} &= \frac{R2}{0} &= \infty [/LATEX]

In practice, it is really only around 100,000, because the transducer has some impedance and no amplifier really has infinite gain. But you do not want gain this high, it will amplify too much noise. With R1 = 10k and R2 = 1M you only have a maximum gain of 100:

[LATEX] \frac{R2}{R1} &= \frac{1000000}{10000} &= 100 [/LATEX]

So a 50 millivolt signal will give you a 5V output which will be accepted by the comparator (which is set for +4.5V), but a 10 millivolt signal will only give you a 1V output and will not be detected.

 

[killer_fighting]

It may work if you connect it directly to pin 2 of U1, but I would suggest using the connection in the schematic. C1 provides DC blocking, this is sometimes important with piezo elements (your transducer is a piezo element) because of strain related to temperature change and mounting stress. The 10k "R2" (which should be labeled R1) sets the gain of the amplifier with the other R2 (the 1M pot). Without that, you will always have infinite gain because the ratio will be:

[LATEX] \frac{R2}{R1} &= \frac{R2}{0} &= \infty [/LATEX]

In practice, it is really only around 100,000, because the transducer has some impedance and no amplifier really has infinite gain. But you do not want gain this high, it will amplify too much noise. With R1 = 10k and R2 = 1M you only have a gain of 100:

[LATEX] \frac{R2}{R1} &= \frac{1000000}{10000} &= 100 [/LATEX]

Please refer to the circuit diagram from ecelab again. If I want to know whether I receive the 25 kHz, what should I connect at pin 8 (NE567, the decoded output)? Is it I have to connect an oscillator?

 

[duffy]

You do not have to connect to an oscillator. The NE567 has an oscillator inside. Refer to the block diagram at the bottom of page 1 of the spec sheet.

Do you wish to connect the output of your circuit to an input port pin of a microcontroller? If so, you must connect a 10K resistor to the +Vcc that your microprocessor operates at (for instance: +5V). Look at Figure 1 on page 5 of the spec sheet. This is how the resistor must be connected. Your schematic from ecelab is wrong. Your processor port pin will be connected directly to pin 8. It will go low (0V) when 25khz is detected. It will go high (+5V) when 25khz is not detected. You can use a voltmeter to verify this. Do you have an oscilloscope?

 

[killer_fighting]

You do not have to connect to an oscillator. The NE567 has an oscillator inside. Refer to the block diagram at the bottom of page 1 of the spec sheet.

Do you wish to connect the output of your circuit to an input port pin of a microcontroller? If so, you must connect a 10K resistor to the +Vcc that your microprocessor operates at (for instance: +5V). Look at Figure 1 on page 5 of the spec sheet. This is how the resistor must be connected. Your schematic from ecelab is wrong. Your processor port pin will be connected directly to pin 8. It will go low (0V) when 25khz is detected. It will go high (+5V) when 25khz is not detected. You can use a voltmeter to verify this. Do you have an oscilloscope?

Yes, I wish to connect the decoded output from NE567 to the input port pin of a microcontroller. And, yes again, I had an oscilloscope. Does NE567 (Philip Semiconductor) and LM567 (National Instrument) are the same?

 

[duffy]

Yes, they are the same, so is the NJM567.

With the oscilloscope, you can view the 567 oscillator at pin 5. If you have C5 and R8 correct, you will always see 25khz from the internal oscillator.

You can also trace the signal. At pin 6 (not marked) of U1 you will see the amplifier signal - sine waves. At pin 6 of U2 you will see the comparator signal - square waves. At pin 8 of U3, if the 10k resistor is correctly connected to microprocessor +V, you will see the signal go low when 25khz is present, and high when it is not.

 

[killer_fighting]

Yes, they are the same, so is the NJM567.

With the oscilloscope, you can view the 567 oscillator at pin 5. If you have C5 and R8 correct, you will always see 25khz from the internal oscillator.

You can also trace the signal. At pin 6 (not marked) of U1 you will see the amplifier signal - sine waves. At pin 6 of U2 you will see the comparator signal - square waves. At pin 8 of U3, if the 10k resistor is correctly connected to microprocessor +V, you will see the signal go low when 25khz is present, and high when it is not.

Sure, I will order the LM567 from farnell tomorrow. Initially, I actually ignored the U3 as I tot it might not important. But now only I realise that is the main core for my whole ultrasonic receiver system. After I complete the ultrasonic receiver circuit, I will have to write some programming. For your information, I am using PIC Programming, CCS compiler and the PIC that I use is PIC16F684A.

Just now you said the circuit diagram from ecelab is wrong, I will have to draw the circuit diagram again. Do you need me upload the updated circuit diagram to this forum? So that we can share to other members if they need.

 

[duffy]

No, the errors were explained. Good luck with your circuit!

 

[killer_fighting]

No, the errors were explained. Good luck with your circuit!

Thanks, duffy! Does this forum provides help for PIC Programming? Just curious and want to ask.

 

[duffy]

Yes, there are many experts on PIC programming here.

 

[audioguru]

At 25kHz, the maximum gain of a lousy old 741 opamp is only about 30 and its maximum output swing will be about only plus and minus 5V.

 

[Sceadwian]

As no range requirements or specifics were given for gain AG, this 'lousy' solution may work just fine. The distance may not be what is imagined, but this is what learning is about. =)