Can you explain this statement? The diagram you show in post #19 shows an integrator, before the opamp, with a transfer characteristic of:

\[ Vout = -\int_0^tVin\,\,\frac{\mathrm{d}t} {Rin.C} \]

sorry, where do you se this integrator?

 

Can you explain this statement? The diagram you show in post #19 shows an integrator, before the opamp, with a transfer characteristic of:

\[ Vout = -\int_0^tVin\,\,\frac{\mathrm{d}t} {Rin.C} \]

sorry but i can't see this integrator, where have you find it?

 

Charge to voltage converter is the integrator.

 

Charge to voltage converter is the integrator.

this charge to voltage should send the charge generated by the piezo into voltage to go into the opamp

 

It is an integrator. You need to add a resistor across the capacitor in order to discharge the capacitor.

 

this charge to voltage should send the charge generated by the piezo into voltage to go into the opamp

Hmmm, do you have details of this piezo device? All the ones I've used generate a voltage proportional to the stress experienced by the crystal. When we talk about electrical charge, in Coulombs, we usually mean the product of current and time ie Q = i . t, or the charge stored in a capacitor, the product of the capacitance and voltage, Q = V . C

In this context I suspect 'charge' means the time-integrated output of the piezo device (voltage) across a known load resistor, e.g. ∫V(t)/R . dt which is a measure of Impulse (J = F . dt ), hence the integrator component shown.

Can you shed more light on the application?

 

Hmmm, do you have details of this piezo device? All the ones I've used generate a voltage proportional to the stress experienced by the crystal. When we talk about electrical charge, in Coulombs, we usually mean the product of current and time ie Q = i . t, or the charge stored in a capacitor, the product of the capacitance and voltage, Q = V . C

In this context I suspect 'charge' means the time-integrated output of the piezo device (voltage) across a known load resistor, e.g. ∫V(t)/R . dt which is a measure of Impulse (J = F . dt ), hence the integrator component shown.

Can you shed more light on the application?

Thanks for the clarification.


The piezoelectric material used is PVDF piezofilm.


So, as we discussed, by connecting the piezo to this circuit, it should operate independently, without the use of a sort of "charge to voltage". Is that correct?

 

It is an integrator. You need to add a resistor across the capacitor in order to discharge the capacitor.

I've reached out with this in the end in easy eda to give the arduino a dig. The big question is: would it work?

 

1mF for C1 and C2 is overkill for two op amps. 100μF is good enough.

Op amp gain of 500 and 910 is pushing it. I would reduce the gain per stage to about 200-300. You will need to add AC coupling between stages.

 

1mF for C1 and C2 is overkill for two op amps. 100μF is good enough.

Op amp gain of 500 and 910 is pushing it. I would reduce the gain per stage to about 200-300. You will need to add AC coupling between stages.

so i adjust the F for C1 and C2, if I reduce the resistance I have to put another stage in between and it will cost more, Idk, the simulation on LTSpice works, what may be the problem that occurs if I leave it like that?
what do you mean by AC coupling between stage? could you maybe send a drawing? thanks.
the piezo will work like I have done it? and so the ADS?
thank you very much!

 

If voltage gain = 1000, a 10 mV DC offset at the input would become 10 V at the output which will saturate the op amp.

 

If voltage gain = 1000, a 10 mV DC offset at the input would become 10 V at the output which will saturate the op amp.

so, what I have to do? :,)
anyway if it's possible I'll like to give you something for your help

 

If voltage gain = 1000, a 10 mV DC offset at the input would become 10 V at the output which will saturate the op amp.

by my calculations on the piezo, the maximum value of V generated by 300kg is 0,0000024V (g33 is 0,2 v*m/N and i have a piezo of 5cmx5cm with a height of 30micron under compression load (300kg->3000N)). so with this V, after the OpAmp i got a maximum of 3,6V

 

To AC couple op amps, put capacitors between stages.

 

If you need DC coupled op amps, you will have to add DC trim (DC offset adjustment) in one of the stages.

 

If you need DC coupled op amps, you will have to add DC trim (DC offset adjustment) in one of the stages.

so it isn't enough to do like that?

 

You will need a resistor to ground at the non-inverting input of U2, otherwise there will be DC offset present at the non-inverting input.

 

You will need a resistor to ground at the non-inverting input of U2, otherwise there will be DC offset present at the non-inverting input.

if I put a resistor at ground as you have done, the second amp won't work anymore
it will work if I do like this

is now ok?

 

is now ok?

No.
It may work in the simulation but it won't work in a real circuit.
The (+) input of U2 needs a DC path for its bias current as MrChips stated.
Do you not believe that?

To minimize the output offset, you can provide a U2 DC gain of one by connecting a capacitor (Cx) of appropriate size for the lowest AC signal frequency, in series with R6 to ground.
Then the DC gain for offset will be 1 but the AC gain above the R6Cx rolloff is 161.

Your plots are useless to us without knowing where those signals are coming from.
Give the nodes a label (Label Net) on the schematic (e.g. In, Out, etc.) so we can tell.

 

by my calculations on the piezo, the maximum value of V generated by 300kg is 0,0000024V (g33 is 0,2 v*m/N and i have a piezo of 5cmx5cm with a height of 30micron under compression load (300kg->3000N)). so with this V, after the OpAmp i got a maximum of 3,6V

OK, but what are you trying to measure. Variations in that Z force? Or something else? What is the d33 coefficient of your piezo-film?