![]() ![]() ![]() maybe a pressure sensitive voltage source with a capacitor in series?) The next thing I'd would try and do is make a model of the piezo. With this setup there is also no need for an external power supply since the piezo element supplies the LED directly.įirst it would be nice if you had a part number for the piezo. 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. 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. Adjusting the noninverting amplifier by adjusting the \$68k\Omega\$ and the \$1.2k\Omega\$ resistor using: $$ U_$$, $$R_1 = 1.2k\Omega$$ and $$R_2 = 68k\Omega$$.Adjusting the directly connected load resistors (now \$560k\Omega\$ and \$1k\Omega\$) for optimal source/load matching.With this circuit the sensitivity can mainly be improved by two factors: I would suggest you to conduct several measurements where you can estimate which resistor combination is best. If you choose unsuitable resistors the output voltage can become arbitrarily small. And again: the output voltage of the piezo element is highly dependent on the load resistance. But as I mentioned in answer 1, I would say that these two resistors were intended as a voltage divider. If you remove the \$1k\Omega\$ resistor you could say that this is the case. Depending at which point you measure your signal, it is possible that you measure two positive pulses. The piezo does generate a negative pulse, but it is being short circuited by the diode. For that reason it is indeed possible that the voltage divider would render a larger voltage at the positive input of the operational amplifier. 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. 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. ![]() Thanks in advance, I'd be truly grateful if you can help me out.Īt first sight, I would say the \$1k\Omega\$ resistor is used in combination with the \$560k\Omega\$ resistor as a voltage divider. ![]() 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?Ĥ) At last the main question: How to improve the sensitivity of the circuit? 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). Do you have any suggestion?ģ) The resistor of 560kΩ plays the role of a pull down resistor for me. 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. 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). Why?Ģ) 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. I have some questions in regards of the circuit that I don’t quite understand.ġ) 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. However it generates within the assembly 50mv (for low pressure) to 1V (for high pressure) Piezo film: don’t ask me about it, I don’t have any information. Some information regarding the components: Here is the assembly put in place and it’s working (fortunately): I have to flash an LED using a piezo film. I have a project to carry out dealing with piezo technology. 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. First and foremost I’d like to warn you that I’m not a native English speaker. ![]()
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