# Tag Info

38

An active buzzer will generate a tone using an internal oscillator, so all that is needed is a DC voltage. A passive buzzer requires an AC signal to make a sound. It is like an electromagnetic speaker, where a changing input signal produces the sound, rather than producing a tone automatically. To identify them, if you apply a DC voltage to them and it ...

17

A contact microphone is basically an accelerometer. It does not require any air for its operation.

9

When trying to detect a signal in noise, there are two filter stages to use. 1) Signal to noise improvement - removal of out-of-band noise The first uses conventional filter concepts. The signal and the noise in question are at different frequencies, so you can separate them with a filter. You have to know the bandwidth of the signal, and design a filter ...

6

Add a decoupling cap (100nF) across the AVR's power pins, with leads as short as possible to the ciontroller; For the zener diode use a voltage close, preferably slightly lower to the battery voltage. The trigger threshold of the input is ½ Vcc, so you definitely want the zener voltage higher than that; Did you actually check the sound output volume level ...

6

10 Vpp is NOT the same thing as 10 Vrms. In order to get the full output from the transducer at 10 Vrms, you're going to have to drive it with a 20 Vpp square wave, or a 28.3 Vpp sine wave. With a single supply and an H-bridge driver, you will need a 12 to 15 V power supply in order to get there. An alternative would be to build a relatively high-currrent ...

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

if it is mounted on the same surface with the same vibrations, it in theory should measure nothing. When sound waves propagate through solid material, it's not just the whole rigid object shaking back and forth: It's waves of strain. Those strain waves will pass through a contact microphone that is attached to some small area of the surface. It will work ...

6

It looks like aliasing of some higher frequency signal (perhaps mains frequency). Compare your sample rate to mains frequency and other relatively high frequency noise sources.

6

Because you have ground as the negative supply rail JP1 can only go positive which means that JP2 wants to go negative but it can't because there is no negative supply rail so JP2 is locked at 0V. Add a negative supply rail to the op amps or bias everything up to half the supply voltage. At the moment JP1 is half wave rectified because it cannot swing ...

6

First, you should start by drawing the schematic in the "standard" way: from left to right. The first item should be the power supply and then the rest of the components. Also try to isolate the circuits in small functional parts. I mean, one can suppose that the Wemos and the 7805 will work as expected. So I would draw the schematic for your questions ...

5

Using almost any article posted on the Internet, about sensing a piezo sensor digitally will provide the frequency of the vibration. This has the huge advantage that a digital pin can be sensed about 1000 times faster than the ADC. A problem with using the ADC to sample the actual signal is the Arduino ADC is so slow that the the samples my suffer aliasing (...

5

Got a brilliant idea from one of the EE's at work... iPhone wall adapter as AC/DC converter. Plug tongs fit inside the clips that connected the previous buzzer. USB cable cut and wired to Piezo. The iPhone adapter outputs DC, drops the voltage to 5V which the Piezo can handle, and it work like a charm. My wife can't understand how I did it. I'm a champ (for ...

4

You got the non-inverting amplifier wrong. See for example this image from Wikipedia: Note that the feedback goes to the negative input of the op-amp.

4

I don't see how a serious transient that would affect a relatively slow op-amp is being caused directly by that switch, nor do I see any problem the way you're using it in general. Do you have a resistor directly across the piezo (not from the non inverting input to ground)? If not, then leakage may be causing the voltage to rise across the piezo while it ...

4

I don't have the reputation to post this a comment so I've had to 'answer'. Is the ADC Vdd really connected as shown in the schematic? If so it doesn't have a current path from the power rail and so doesn't have power. It looks as though C3 and C4 are intended to provide decoupling. Capacitors performing this function should be connected between the ...

4

You are demanding too much from your power supply. Op-amps aren't magic. All they do is to connect the output more or less strongly to the voltage rails depending on the strength of the input voltage. So, to get 120V out of an opamp, you have to power it off of at least 120V. But you CANNOT operate an opamp on that voltage. Most are only rated up to ...

4

This type of thing can be done with event discrimination. An event discriminator accepts signals with the proper amplitude and frequency content. To accomplish this in a simple micro-controller of moderate capability (say 16MHz or better): Forget the A2D, and apply your piezo signal to an analog comparator. The comparator's output then connects to an ...

4

Why isn't there a diode between the LC circuit and the supply rail? Wouldn't the coil put some pretty nasty voltage spikes onto the supply rail, too? Because the piezoelectric device is a (nonlinear) capacitor from the electrical point of view, therefore prevents the $V_\mathrm{CE}$ of $T_1$ to rise toward infinity. Precisely, during $T_1$ turn-off ...

4

Add another Schottky diode from the non-inverting input to the positive power supply rail. The Schottky diode will be reverse-biased in normal operation, but when the power is off it will conduct any spikes to the positive rail where the bypass capacitors will kill it off, and even if the user pumps it the zener will prevent the supply rail from going too ...

4

As this is a fixed frequency application, tuning out the capacitance with an inductor would improve things dramatically. There's a choice to either do this in parallel, and drive the resulting tank at 200Vpp. Or it could be driven in series, with the benefit that if the transducer has a reasonable Q, the L and C together will achieve a significant voltage ...

4

Not too complicated. Quite neat in fact: Firstly: Q5 is an inverting amplifier stage implemented in PNP; Then Q6, is another inverting amplifier stage, but implemented in NPN; these two together, give a high gain from the signal already amplified in the LF411. But MJE340 / 350s are 300-volt-capable transistors, so form a "final amplification" ...

4

Aside from obvious questions as to whether your piezo is the same as whatever some random site on the internet might have used, and whether yours is being hit as hard, there is really no way to measure the output signal with a multimeter. The pulse will be brief. If you are seeing 1V on a multimeter it may well be railing at +3-4V (the most the op-amp can ...

3

You can separate the mic audio from the earphone audio fairly easily. But I don't think that a piezo pickup is going to work into your phone very well (or at all) without an active buffer. The phone is expecting the mic to have a source impedance down around 1k and your piezo pickup wants to work into something with an input impedance in the megohms range (...

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

As already mentioned, some bias on the input would be an idea if you are only using a single supply - half the supply voltage supplied through resistors with a parallel value of > 10x the piezo impedance. You also need the bias voltage at the base of the Rf/Rg resistors then. Due to the high impedances involved, you will have to be careful with layout, wire ...

3

The 1 Mohm bias resistor is far too low - it will load the bass signals and attenuate them much more than mid range or the treble. This happens because the piezo has capacitance and this capacitance is in series with the signal being produced. It might only be a few hundred pF and, for instance, 1nF and an external 1 Mohm bias resistor will form a high pass ...

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

You can multiplex a single ADC, with throughput reduced, of course. So if you have a 1MSPS ADC that settles in 1usec (probably available on-chip with some micros) you could sample 100 inputs at 5kHz. Sampling at 5kHz with a realistic analog anti-aliasing filter on each input you might be able to capture 1-1.5kHz fairly well (the theoretical limit is 2.5kHz)...

3

Yes, this is the simplest way to do it. C1 and C2 are needed for supplying current to the transformer as it switches, as its supply is decoupled from the main +5V by a 100Ω resistor. This is done to avoid ripple coming from the transducer circuit leak into the main +5V. The center-tapped transformer is needed to keep it simple, as you cannot deliver a DC ...

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