15

Parallel capacitor plates of 25 mm by 25 mm seperated by 4mm of glass with a relative permittivity of 4 would give a coupling capacitance of about 5 pF. That capacitance is in series with an antenna signal and at 2.5 GHz, would act as a blocking impedance of about 13 ohms so, it's feasible it could be used without disrupting the VSWR too much. k is the ...


12

A potentiometer gives output (resistance) in response to a physical input (position). So that puts it pretty squarely in the ‘sensor’ category. And they all go to 11... At the risk of dissecting the live frog, from here: https://en.m.wikipedia.org/wiki/Up_to_eleven


11

I never fully understood how or why it worked Well, it's not magic ;-) Actually it can either be done magnetically using coupled inductors. This is like a transformer without a magnetic core. Wireless charging as used in some mobile phones uses the same principle. Basically, one coil creates a magnetic field from an electrical signal which is then picked ...


7

You need to run it at 2.40 kHz get maximum sound pressure level (SPL). The transducer is mounted in a ported acoustic cavity which has a definite resonant frequency. Here is a typical response curve for a 2kHz transducer: You'll still get a fair bit of noise at other frequencies (you can even use these things as horrible little speakers), but not maximum. ...


6

820kHz and 2.5MHz (=3*820kHz) are eigenfrequencies of the mechanical system. You will be able to drive the transducer at any other frequency, but the damping will be greater by orders of magnitude, so it won't be usable. Some ultrasound transducers can oscillate at both longitudinal and transverse mode. It is possible that that 820kHz and 2.5MHz are the ...


6

An idea once I had, but never tried, is to treat an ultrasonic transducer like a high-power crystal oscillator. Your typical crystal oscillator circuit looks like this: simulate this circuit – Schematic created using CircuitLab Simplistically, the crystal (along with the 2 capacitors) provides a 180 degree phase shift at its resonant frequency and ...


6

Such transducers work best with higher voltages, like 9V or more. A straightforward solution would be to generate that voltage using e.g. a charge pump, then use the NPN transistor to amplify the MCU signal to that voltage. Depending on the project, you may already have that voltage available. Another common option is to drive a small step-up transformer ...


6

An electrical sensor is a device that converts a specific physical parameter to an electrical signal. A potentiometer as a level sensor, for example, would fit into this category. An electrical actuator is an electromechanical device that converts an electrical signal to a mechanical movement. A solenoid actuator would be one such example.


5

Try these linear amplifiers made by Apex. They are designed specifically for ultrasound aplications.


4

I'm going to take a stab at this though I don't know if it applies to sonar. If you have a signal received on a transducer there will be both signal and noise. If you have two transducers (in an array) it can be presumed that the same signal is received on both but the noise on each is likely gaussian and not coherent. The upshot of this is that when you ...


4

First, generally your transducer is usually in close proximity to your receiving unit (or if they are one and the same). Thus, any change in resonance is likely to occur on both elements at the same time, so it won't affect their send/receive performance. Secondly, the speed of sound in air depends on density and thus temperature, so you may want to take ...


4

This is a pressure transducer that would typically consist of a resistive bridge powered from a single voltage that outputs a small differential signal (perhaps 50mV full scale). If the bridge is energized with (say) 4.0V, then the two outputs would be around 2V and the difference would change from +50mV to -50mV depending on the pressure differential (this ...


4

A similar question piqued my own interest when I was building a Time Difference of Flight system. There is a difference in the impedance of the components at the exact frequency for which it was designed. Low impedance on the transmitter allows it to be driven at high power, high impedance on the receiver turns the received signals into a high voltage. I ...


3

It's likely a flex sensor, or force sensitive resistor. These are long devices that change resistance according to the amount of flexure (flex sensor) or force applied (force sensor) along the tape. If this is the case, you can measure the resistance, and should find that it changes as you bend or put pressure on the tape. Breathing and movements aren't ...


3

I'm not aware of a standard symbol for a pressure sensor. Usually, it's drawn as an IC. It's a good idea to indicate in the schematic that this is a sensor, and therefore something special. You can make a text block next to the symbol saying what the sensor is sensing. If you know what kind of sensing element the sensor is based on, you can draw an icon ...


3

Assuming you are using multiple Arduino outputs to drive the ladder, the ladder output is attempting to drive a current into the LM358. This ensures that the - input will always be larger than ground, and the op amp output will always be zero, or whatever your - supply is (I'm assuming you're powering the LM358 from ground and +5). Your amplifier is ...


3

It all depends on how you want to drive the transducer i.e. the application: - It appears that you can drive at resonance or anti-resonance which indeed does make it pretty similar to how you would use a crystal in an oscillator. The graph above is taken from this interesting website. I can't determine from your question what application you have but, from ...


3

Load cell creep is a problem, and generally is most pronounced early in the life of the cell as the adhesive or weld binding the element to the material under strain relaxes or moves. This will vary depending on the bonding technique and the temperature cycling that the load cell experiences. It is most prominent in the offset; the gain generally remains ...


3

Looks like it's EMI noise coupling into your measurement from your power supply. Make sure that the power cord for the scope is as far away from your probe as possible. To confirm it is noise from your power supply check the frequency by measuring the peak to peak time difference and calculating 1/(t1-t2). Another hug help could be reducing the ground ...


3

A current signal as opposed to a voltage signal isn't so much to reduce environmental noise pickup, but to eliminate errors caused by ground offsets and the series resistance of the cable. Current loop can be lower noise, depending on your noise sources, but there are also scenarios where a voltage signal can be lower noise. There are other ways to ...


3

Your CM ground has an output impedance of 5~10 Ohms with the sensor being ~1K but load current for the MEMs accelerometer "may" be fairly constant. (verify) Otherwise Load regulation of supply may affect noise floor, if important. CMRR depends on frequency of interference and impedance ratio of load to source for CM attenuation and balanced cable and Diff ...


3

The R-C network functions as a low-pass filter that reduces the high-frequency noise (both mechanical and electrical) in the output signal. This noise, if not attenuated, would create aliasing issues if you're gioing to sample the data.


3

If total BW is defined as the 0.707 amplitude threshold ( -3dB) and on the lower half BW averages to -35kHz at 965 KHz, we might approximate it as a BW of 70kHz with a Q= 1000/70= 14 or inversely 1/Q=7.1% Given a 10% 90% rise time,T = 0.35/f(-3dB) we expect a burst envelope rise time of 0.35/70kHz = 5 us or approximately 5 cycles at center f. The center f ...


3

On the board, I see a "220V" input connector, rectifiers, and no smoothing caps. Thus I'm going to assume this runs directly from rectified mains. There doesn't seem to be any voltage regulation, and there are only two active devices. Also the board has "110V" and "220V" versions, which means it can't adjust its power output for lower voltage. So... if you ...


3

US transducers use an array of piezo-electric ceramics mechanically tuned to desired frequencies. You can usually buy what you're looking for, or order custom parts, at places like https://piezotechnologies.com/ Note that the handle you're looking at is probably many individual crystals in an array, not a single element. The US device often sends out a ...


3

Am I using this instrumentation amplifier in a wrong way ? Quite possibly (from your schematic). As with any (maybe most) Instrumentation Amplifiers you need some DC bleed resistors down to 0 volts (mid rail) to remove bias currents because the inputs don't like to float: - Bottom right is the picture I would reference in your particular case. Try 1 Mohm ...


3

An anti aliasing filter should be always present. It would be better to sample at high sample rate, use a FIR filter to eliminate the environment noise 50/60Hz, thus the anti-alias filter can be a small RC with low TAU value. Sampling at 64sps would imply large RC filter and it is very close to 60Hz, so most probably you will pick lots of noise/garbage. See ...


2

I think the Piezo Systems EPA-104-115 fits all your criteria except for the low-cost criteria. It costs $2,639. The AA Lab Systems A-301HS may also fit and is probably as cheap as you'll find. I saw one on ebay for $975. Searching for piezo driver or piezo linear amplifier didn't turn up anything more affordable in my search, but feel free to double ...


2

It certainly looks feasible, but first make sure you NEED to do it. The 24 bit ADC gives you an awful lot of dynamic range that you don't need, and plenty of bits to spare. You often use them to sample small signals without worrying about preamplification. So, work out your noise budget, figure out how many bits you need, and do what you need to get them. ...


2

It is clearly meant to represent a current transformer. The data sheet specifies that the transducers can measure current directly or with a current transformer. The advantage of the current transformer is that it provides isolation between the current source and the current transducer.


Only top voted, non community-wiki answers of a minimum length are eligible