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I have some questions regarding my assembly and especially on the role of the capacitor within the circuit.

enter image description here

Before I added the capacitor I encountered 2 issues:

  • when the piezo was squeezed (a pulse is generated) then the pressure was maintained for some time and then released, the piezo generated a positive pulse that I wished to get rid of.
  • 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.

Adding the capacitor fixed these issues but I can’t explain why for the 1st one or rather I don’t know why. Of course, why I added the capacitor was for the 2nd point, I wanted to discharge this phenomenon through a capacitor. And it's working well.

And also I don’t get why since I’ve added the capacitor, the voltage seen on the oscillator is different. It’s pretty low now around 20mV while without the capacitor I have an input voltage of 20 mV to 800 mV depending on the pressure applied on the piezo film and an output voltage (right after the op amp) around 2.5V-2.8V.

Thanks in advance ;)

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  • \$\begingroup\$ By the voltage on the "oscillator" do you mean the opamp? Did you look up a model for the Piezo? (A voltage source with a capacitor in series looked the simplest.) So the source capacitance is in series with the 0.1 uf.. and that's a voltage divider. \$\endgroup\$
    – George Herold
    Jul 21, 2014 at 12:38
  • \$\begingroup\$ I mean I try to read on the oscillator the input voltage before the op amp and also before the npn transistor. With the capacitor in serie that doesn't solve my two issues that I spoke about. And sorry I don't have access to the model of the piezo. \$\endgroup\$
    – Alibaba
    Jul 21, 2014 at 14:01
  • \$\begingroup\$ Hi Alibaba, In a previous post I suggested that you go on the web and find a model for your piezo. (So the electrical model of some thing might be a voltage source, or current source and then some other passive components, R's C's and/ or L's.) One model I found of a piezo said it looked like a voltage source with a capacitor in series. (Does that make any sense to you?) Once you have a model of your signal source it may be a lot easier to figure out how to make it work they way you'd like... or you'll see why it can't work and adjust you approach. \$\endgroup\$
    – George Herold
    Jul 21, 2014 at 14:40
  • \$\begingroup\$ There is no oscillator in your diagram only an amplifier \$\endgroup\$
    – Andy aka
    Jul 21, 2014 at 16:22
  • \$\begingroup\$ @Andyaka Was I supposed to show the oscillator on the diagram? It won't be part of the assembly at the end that's why I didn't see the point to put it there. I'm using it to check that my calculations match with the experiment. \$\endgroup\$
    – Alibaba
    Jul 22, 2014 at 8:21

1 Answer 1

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You are low-pass filtering the piezo-sensor's voltage by adding a capacitor. Since your fluctuations (pressure applied) most probably lies in the high region, low-pass filtering has gotten rid of it.

the piezo generated a positive pulse that I wished to get rid of

Are you sure that you want to get rid of +ve pulse at the input to the opamp? You need to reverse the polarity of the diode then.

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  • \$\begingroup\$ What I needed to get rid of was the 2nd positive pulse that occured when I released the pressure. I still wish to keep the pulse that occurs when I apply a pressure on the piezo. \$\endgroup\$
    – Alibaba
    Jul 21, 2014 at 13:52
  • \$\begingroup\$ OK. By the addition of a capacitor, the response has gone sluggish (slow). Thus, you do not see the flicker in LED that was caused by your tiny muscular vibrations in the fingers. Try playing with different capacitors if you are not happy with the current performance. \$\endgroup\$
    – nvd
    Jul 21, 2014 at 18:41
  • \$\begingroup\$ ignore please... \$\endgroup\$
    – George Herold
    Jul 21, 2014 at 20:30
  • \$\begingroup\$ @nvd I have no way to find the part number of the piezo element, sorry. Thank you for your explaination about my second point and how about the 1st one? Someone answered me this before his post get deleted (dunno why) : the release is quick , so the negative voltage pulse generated thru the diode causes a positive voltage to be created by a positive ramping current when it reaches the -0.65V clamp on the release spike, resulting in another +ve voltage when let go. The 0.1 cap acts as a huge voltage divider and AC noise filter. But I still don't understand why the capacitor got rid of this :( \$\endgroup\$
    – Alibaba
    Jul 22, 2014 at 8:30
  • \$\begingroup\$ Please explain the observation 1 in detail especially the spikes' polarities before and after the application of pressure. It is not so clear now. \$\endgroup\$
    – nvd
    Jul 22, 2014 at 12:54

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