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I am currently working on a microcontroller based project to measure the blood flow velocity using ultrasound doppler.

I have purchased two Spengler 8 MHz continuous mode probes in order to measure the doppler shift.

I have disassembled one of the probes to test it. It is formed of an 8 MHz crystal, which outputs a 2 V ppk 8 MHz sine wave (as viewed on the oscilloscope.) It also has an MC1496 modulator/demodulator chip to remove the carrier frequency followed by an LM324 chip for amplification. The final probe output are two wires, sig1 and sig2.

I have tested the outputs on the oscilloscope and compared them to the RX and TX signals of the sensor after breaking the probe open.

My experience in analog circuit design is very limited, especially in piezoelectric crystal signal conditioning and fetal doppler.

I have exported the signals to Matlab with the hope of detecting the blood velocity using FFT. I performed AM demodulation as well.

I am sure it is based on AM because when I hit the probe head the sig1 amplitude changes and rises, but I am not understanding how the CW doppler probe circuit works.

Based on what I found, this is a common and standard circuit used in probes and devices to detect fetal heartbeat.

How does it work? I do not see any shift in frequency in the heartbeat range using FFT. I have finally found the probe circuit, my outputs of interest are output2 and output3, from the final stage it seems they are using output3: enter image description here (This circuit is taken from https://patents.google.com/patent/CN203970415U/en) I have attached the simulation outputs:

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chaosmind
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  • You would need to draw a reverse engineered schematics of the probe. It's not likely that a differential output would produce 2.7Vpp on one output and 133mV on the other. – Marko Buršič Apr 09 '21 at 21:14
  • I wasn't getting any much signal on the RX of the crystal eventhough the TX clearly shows a sine wave being sent. I did not place it on my hand just putted it facing an obstacle. but when I hit on it with my finger i can see the pulses in the signals – chaosmind Apr 09 '21 at 21:17
  • The probe had 4 wires out which were the VCC, GND and two others I am supposing they are for data. Is it possible that the two other wires be Signal and its own ground and not differential? there was also an 401 inverter chip inside. – chaosmind Apr 09 '21 at 21:20
  • An interesting challenge. Unless you have a very cooperative volunteer, maybe you should first build what's called a "phantom" to simulate the moving fluid in the body and give you a consistent signal to measure. Then you'll be able to do testing without holding the probe against your own body. – Mark Leavitt Apr 09 '21 at 21:36
  • But first i need to design the analog circuit to have a clean signal for the microcontroller. this is the part I am stuck on – chaosmind Apr 10 '21 at 00:01
  • Aren't you making an assumption as to how the MC1496 is being used? Why do you think it's being used detect an AM modulated signal? – SteveSh Apr 25 '21 at 13:05
  • It is used to suppress the carrier and mostly remove it. the FFT showing sig1 and sig2 should confirm the very low carrier power. I am also supposing that the DC offset was added in order to pass the signal to the ADC. but using sig1 and sig2 I do not understand their relationship and how to get the doppler from them. since these are the two output signals from the probe – chaosmind Apr 25 '21 at 13:12
  • I think this is the circuit patent : https://patents.google.com/patent/CN203970415U/en It list the same chips used inside the probe and the LM328 in the handheld doppler device. I will need to see if it say what does its two output signals represent – chaosmind Apr 25 '21 at 15:17

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I did not place it on my hand just putted it facing an obstacle.

Ultrasonic probes require proper coupling to the target. They do not work through air!

That means using ultrasound probe gel and a target that has ultrasonic impedance similar to body tissues. Otherwise there'll be reflections - like in a mismatched antenna in a radio - and the sensor won't "hear" any useful signal to work with.

The output of the probe is demodulated Doppler shift. There's no need to do FFT or anything else on it. Feed it to a an amplified speaker and listen to it first. Use the favorite artery on your body (and gel!) to hear it first. Then you'll know what conditions the probe works in - just based on the audio quality. Then you can also look at the signal.

But first I need to design the analog circuit to have a clean signal for the microcontroller

Listen to it first. Really. And have speaker output available whenever you're feeding it to the MCU, so you'll know that there's any signal to begin with.

And again, just to be clear: These probes do essentially nothing unless they are coupled to tissues via gel, or at least to water with something moving in it. As a phantom you can use a magnetic stirrer bar in a beaker with water. The probe has to be angled a little bit at least, instead of just looking straight down at the stirrer, to get better signal/noise ratio.

Kuba hasn't forgotten Monica
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