I would like to know if my reasoning is correct for this situation, as I am unsure.

I have two piezoelectric substrates, on both of them I deposit an active film on one side of each, that will change length under a certain applied mechanism. The first film is, x, in length, the second is, x/2, (half the length of the first film). Both these film will exhibit the same, let's say 100 parts per million, length change. So the overall length change of the first film, x length, will be twice that of the second, x/2, film.

So if I were to measure the piezoelectric voltage produced when these films were activated, would the first with the x length film show a voltage of twice that of the second? I think it's the overall length change here that is more important here for the voltage produced, not the strain value of the active film. This is what I am unsure about as I often seen the strain value noted in the formula for calculating the piezoelectric voltage. But I am thinking in this case the overall length may be an important factor to consider.

If now I take this further and have a third substrate with two areas of the active film deposited, both of, x/2, in length. If I connect these two films in series like a battery/voltage source it would produce double the voltage of an individual film element area, and equal to that of the first, x, length film of substrate piece.

I think there are some complexities I am missing out as I'm trying to describe this as simply as possible. And I am describing this as a strain/stress applied parallel to the surface of the substrate to clarify. I am thinking about this in terms of surface acoustic wave devices, and whether the number of inter digital transducers produces a greater voltage as there is an increased active area. Or if it's just an improved acoustic signal with a longer overall length.

Any help would be appreciated.


1 Answer 1


I don't think the area of the substrate affects the voltage produced. The piezo element is almost like a capacitor and increasing it's size won't change the voltage (for a constant strain value) just the amount of charge stored on it. Doubling the area should double the stored charge and hence double the stored energy. If the voltage were doubled by doubling the area and the charge remained constant, this would increase the stored energy by 4x (because E = 1/2CV^2).

I think you are right that putting two in series would double the voltage in the same way that putting two capacitors in series charged to the same voltage doubles the voltage.

Having a sensor with a larger area is beneficial because all amplifiers have a finite impedance meaning they will have some charge flow into them when connected to the sensor. The larger the charge stored on the device (the larger the capacitance or equivalently area) the less the measured voltage will change because of this leakage current.

I believe that the voltage produced by a piezoelectric element (assuming nothing is connected to it) would only depend on the material, the strain induced, and the thickness of the piezoelectric material.

  • $\begingroup$ thank you for your reply, I was wondering about area as in the case of magneto-electric devices the magneto-electric coefficient is given in terms of (V per cm per Oe) however as yet I do not fully understand why. $\endgroup$
    – Dave
    Jul 25, 2019 at 11:51

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