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Admittance for a piezoelectric transducer in Comsol 3.5a
Posted 30 nov. 2011, 13:22 UTC−5 Low-Frequency Electromagnetics, MEMS & Nanotechnology, Piezoelectric Devices Version 3.5a 1 Reply
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Hello,
I am performing a frequency response analysis on a 2D piezoelectric transducer in Comsol 3.5a, and I would like to plot the magnitude and phase of the admittance (relative to voltage). Following the "composite piezoelectric transducer" tutorial, the magnitude of the current is given by integrating -imag(nJ) over the boundary of interest; dividing that quantity by V gives the admittance. I am a bit confused about this definition - why is the imaginary part of current density integrated over the boundary instead of the real part?
It was posted elsewhere that the phase(A) = atan2(imag(A), real(A))*180/pi. Intuitively, to get the phase of current, I would think that "A" would be the integration of nJ over the boundary of interest. Is this correct? Also, what is this phase relative to?
In summary, my questions are:
1. Why is the magnitude of current = integrating the imaginary part of nJ?
2. How would I plot the phase of the admittance relative to the voltage?
Any help would be appreciated. Thanks.
I am performing a frequency response analysis on a 2D piezoelectric transducer in Comsol 3.5a, and I would like to plot the magnitude and phase of the admittance (relative to voltage). Following the "composite piezoelectric transducer" tutorial, the magnitude of the current is given by integrating -imag(nJ) over the boundary of interest; dividing that quantity by V gives the admittance. I am a bit confused about this definition - why is the imaginary part of current density integrated over the boundary instead of the real part?
It was posted elsewhere that the phase(A) = atan2(imag(A), real(A))*180/pi. Intuitively, to get the phase of current, I would think that "A" would be the integration of nJ over the boundary of interest. Is this correct? Also, what is this phase relative to?
In summary, my questions are:
1. Why is the magnitude of current = integrating the imaginary part of nJ?
2. How would I plot the phase of the admittance relative to the voltage?
Any help would be appreciated. Thanks.
1 Reply Last Post 7 déc. 2011, 19:53 UTC−5
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Posted:
1 decade ago
In the "composite piezoelectric transducer" tutorial, the goal is to calculate the susceptance which is the imaginary part of the admittance. Therefore, the surface integral in the tutorial only includes the imaginary part.
You are correct that "A" should be the integral of nJ over the boundary of interest which is a complex quantity.
Best
Ye
You are correct that "A" should be the integral of nJ over the boundary of interest which is a complex quantity.
Best
Ye