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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): enter image description here

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? enter image description here

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.

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  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.

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  • \$\begingroup\$ I followed a part of your advice, I got rid of the schmitt trigger. Instead, I've put a capacitor of 0.1µF in parallel before the npn transistor in order to get rid of the second pulse & an issue that occurs when I kept maintaining the pressure on the piezo film (I had such a sensitivity that the LED was still flashing. The brightness was weak tough). And also I've adjusted the 68kΩ and 1.2kΩ resistor to get the sensitivity desired. Actually I've changed the 68kΩ by a potentiometer. It's quite convenient because I can adjust the sensitivity as wished. \$\endgroup\$ – Alibaba Jul 16 '14 at 13:28
  • \$\begingroup\$ Can you explain me if you can why with the capacitor I get rid of the second pulse? I don't really get it \$\endgroup\$ – Alibaba Jul 18 '14 at 11:08
  • \$\begingroup\$ I'm afraid I can't. Capacitors should not behave different for negative pulses as they do with positive pulses. What was your measurement setup? And what was your intention when you added the capacitor? \$\endgroup\$ – David Wright Jul 21 '14 at 6:35
  • \$\begingroup\$ I thought that the negative pulse was filtered by the diode? I added this capacitor because when the pressure was maintained on the piezo film, I had such a sensitivity that the LED was still flashing. The brightness was weak though. So I wanted to discharge this phenomenon through a capacitor. And it's working well. Do you think I should make a new topic and ask what is the role of the capacitor within the assembly? \$\endgroup\$ – Alibaba Jul 21 '14 at 10:15
  • \$\begingroup\$ Yes I thought the negative pulse is short circuited by the diode, that's why I don't understand that you apparently filter the negative pulse with a capacitor. \$\endgroup\$ – David Wright Jul 22 '14 at 11:55
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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.

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  • \$\begingroup\$ Any op amp can run from a single supply. \$\endgroup\$ – Matt Young Jul 15 '14 at 13:52
  • \$\begingroup\$ Sure @MattYoung, but I usually think of a single supply opamp as one whose common mode voltage range goes all the way to the power rails... or at least the negative rail. The opa134 only gets within 1.5 to 2 V. And would not work very well in the above circuit... if it is run as a single supply. \$\endgroup\$ – George Herold Jul 15 '14 at 16:08
  • \$\begingroup\$ @GeorgeHerold You are right on your answer 3.) but I don't get how did you calculate the input biais current? and espcially what are those 0.5Meg? \$\endgroup\$ – Alibaba Jul 18 '14 at 7:42

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