How to improve the sensitivity of a piezo assembly?

Question

First and foremost I’d like to warn you that I’m not a native English speaker. Sorry in advance for my mistakes. And also I’m kinda a newbie in terms of electronics; so go easy on me please, I will try to be as understandable as I can.

I have a project to carry out dealing with piezo technology. I have to flash an LED using a piezo film. Here is the assembly put in place and it’s working (fortunately):

Some information regarding the components:

LED: LW3333-R1t2-5K8L

Op amp : LM358N

Q1: NPN transistor BC337-025G

Piezo film: don’t ask me about it, I don’t have any information. I took it from a device. However it generates within the assembly 50mv (for low pressure) to 1V (for high pressure)

I have some questions in regards of the circuit that I don’t quite understand.

1) I used to have a resistor right before the diode as you can see on the following illustration, but I got rid of it because it appears to be irrelevant. Having it or not makes no difference. Why?

2) When I squeeze the piezo (a pulse is generated) then maintain the pressure for some time and then release the piezo generates a positive pulse. Why not a negative pulse? I’d like to get rid of this second pulse that occurs when I maintain the pressure and then release (for brief pressure it doesn’t occur). I’ve added a Schmitt trigger right after the op amp on this purpose but I have the feeling that the sensitivity has reduced somehow. Do you have any suggestion?

3) The resistor of 560kΩ plays the role of a pull down resistor for me. Is it the case? The assembly is working for a resistor of 250 kΩ (and even less but requires high pressure) to 600 kΩ (perhaps a bit more). Why this gap? This resistor has an effect on the sensitivity, I can more or less guess why but would you like to explain it for me?

4) At last the main question: How to improve the sensitivity of the circuit?

Thanks in advance, I'd be truly grateful if you can help me out.

Answer 1

1. At first sight, I would say the \$1k\Omega\$ resistor is used in combination with the \$560k\Omega\$ resistor as a voltage divider. This might sound like it decreases the sensitivity but the voltage output level of a piezo depends highly on the frequency spectrum of a push and the load impedance. The internal impedance of a piezo element was also relatively high in my experience, so for perfect matching the load resistance would have to be high as well. For that reason it is indeed possible that the voltage divider would render a larger voltage at the positive input of the operational amplifier.

2. The piezo does generate a negative pulse, but it is being short circuited by the diode. Depending at which point you measure your signal, it is possible that you measure two positive pulses.

3. If you remove the \$1k\Omega\$ resistor you could say that this is the case. But as I mentioned in answer 1, I would say that these two resistors were intended as a voltage divider. And again: the output voltage of the piezo element is highly dependent on the load resistance. If you choose unsuitable resistors the output voltage can become arbitrarily small. I would suggest you to conduct several measurements where you can estimate which resistor combination is best.

4. With this circuit the sensitivity can mainly be improved by two factors:

   1. Adjusting the directly connected load resistors (now \$560k\Omega\$ and \$1k\Omega\$) for optimal source/load matching.
   2. Adjusting the noninverting amplifier by adjusting the \$68k\Omega\$ and the \$1.2k\Omega\$ resistor using: $$ U_{out} = \Big(1+ \frac{R_2}{R_1}\Big)U_{in}$$ where: $$U_{in} = \frac{560k\Omega}{1k\Omega + 560k\Omega} \cdot U_{piezo}$$, $$R_1 = 1.2k\Omega$$ and $$R_2 = 68k\Omega$$

Moreover I'd say that the schmitt trigger is not really necessary unless the LED has to specifically be switched on at a certain level. Additionally if the task is only to switch on the LED by using a piezo element, you would not need the amplifier, the schmitt trigger etc, you could only implement a simple passive half wave or full wave rectifier and connect the LED to its output. With this setup there is also no need for an external power supply since the piezo element supplies the LED directly.

I hope this helps.

Answer 2

First it would be nice if you had a part number for the piezo.

The next thing I'd would try and do is make a model of the piezo. (So do a search.. maybe a pressure sensitive voltage source with a capacitor in series?) Now your questions. (my answers are my best guess and should be checked against your circuit.)

1.) If we think of the piezo as a voltage source then (ideally) the opamp draws no current and the voltage seen by the opamp is a 1k/560k voltage divider... the 1 k ohm makes no real difference.

2.) Well I'm not very sure about this. The 560 k resistor is slowly discharging the piezo cap. Then when releasing the pressure the cap "recharges again". I'd really like to look at the waveforms on a 'scope.

3.) I wouldn't call the 560k ohm resistor a pull down. It's providing a DC path for the bias current of the opamp. (I don't know the lm358.. looks to be a bjt input... ~20nA of bias current.) Are you running the opamp as a single supply? There will be an output voltage due to the bais current. 20nA*0.5Meg*gain(60)... I get something like 600mV but you should check my math.)

4.) Well first thing I would try is a FET input opamp. I like the opa134, but it's not a single supply.

Answer 3

A piezo is in fact a voltage source with a series capacitance. If you have a resistor as a load then the capacitor will be discharged by the resistor. Suppose you press the piezo in a step function (means press and do not release), if the resistor has a high value (mostly mega ohms) then the output will follow the press and will then go to zero because of the discharge. If the resistor is very low in value (kohms) then the RC circuit works more as a differentiator and will respond more to how fast the press changes than to the force itself. If you have a high resistance as a load and if pressed continuously then the resistor will discharge the piezo and if the press is released (which is a negative press change for the piezo) then the output goes negative and finds the diode as a short circuit, which means the capacitor of the piezo is precharged negative. Best is to remove the diode, keep the series resistor(which is premarily there to protect the electronics when contacting the piezo for the first time with the electronics. The piezo capacitor could hold a high charge when connecting for the first time and the peak discharge current could distroy the opamp. I always used piezobuzzers for sensors, 5 microns of bending can easily generate a couple of voltages when connected to a 22Mega ohm resistor (to ground), we always used 5V cmos logic as an interface. Piezo's have also an output response on temperature changes.