Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.
Impedance Simulation - Solution at angle (phase) setting
Posted Aug 4, 2016, 6:54 a.m. EDT Low-Frequency Electromagnetics Version 5.2 2 Replies
Please login with a confirmed email address before reporting spam
Hi,
I'm evaluationg the impedance of a 2D structure which is similar to a parallel capacitor. Between the two electrodes a material with a certain conductivity and permittivity is placed. One Electrode has a fixed potential of 1V and the other one has ground potential.
To evaluate the impedance a line integration with the expression ec.nJ is used. This leads to a negative imaginary part. The real part is also negative, which is certainly wrong (the structure is passive --> no negative resistance).
By searching for similar problems in this forum, the setting "Solution at angle (phase)" was mentioned. So if I compute the results for an angle of 180° instead of 0° I get the expected signs (+ for Re and + for Im) if I interprete the result by the admittance (Y = 1/R + j * omega * C).
By having a look at the 2D surface plot of the Electrical Potential the result is -1V to 0V across the geometry for 180°. So it looks like a cosine shifted by 180°, which leads to a negative cosine, is used as excitation source. But that would be the same as using a negative potential of -1 V. Which again leads to the wrong results ( Y = Re(ec.nJ)/V + Im(ec.nJ)/V ).
If a negative sine (+90°) is used I get a negative imaginary part and a positive real part. For a more complex structures without prior knowledge of the imaginary part (capacitive or inductive) it becomes tricky ;-).
Is there any recommendation wich phase angle should be used for impedance analyses?
BR
I'm evaluationg the impedance of a 2D structure which is similar to a parallel capacitor. Between the two electrodes a material with a certain conductivity and permittivity is placed. One Electrode has a fixed potential of 1V and the other one has ground potential.
To evaluate the impedance a line integration with the expression ec.nJ is used. This leads to a negative imaginary part. The real part is also negative, which is certainly wrong (the structure is passive --> no negative resistance).
By searching for similar problems in this forum, the setting "Solution at angle (phase)" was mentioned. So if I compute the results for an angle of 180° instead of 0° I get the expected signs (+ for Re and + for Im) if I interprete the result by the admittance (Y = 1/R + j * omega * C).
By having a look at the 2D surface plot of the Electrical Potential the result is -1V to 0V across the geometry for 180°. So it looks like a cosine shifted by 180°, which leads to a negative cosine, is used as excitation source. But that would be the same as using a negative potential of -1 V. Which again leads to the wrong results ( Y = Re(ec.nJ)/V + Im(ec.nJ)/V ).
If a negative sine (+90°) is used I get a negative imaginary part and a positive real part. For a more complex structures without prior knowledge of the imaginary part (capacitive or inductive) it becomes tricky ;-).
Is there any recommendation wich phase angle should be used for impedance analyses?
BR
2 Replies Last Post Aug 31, 2016, 3:41 a.m. EDT
Please login with a confirmed email address before reporting spam
Posted:
8 years ago
I have the same situation here, waiting for someone's help.
Mingyu
Mingyu
Please login with a confirmed email address before reporting spam
Posted:
8 years ago
Hi,
by having a look on the orientation of the normal vector you can see that the vector points outwards of the electrode (towards your excitation source). Thus It seems you have to multiply your excitation voltage with minus one, when calculating your impedance.
For a capacitor you will get:
Z = V_exc * (-1)/(-Re - j * Im) = +Re -j *Im
Which leads to a positive real part and a imaginary part of capacitive nature.
BR
by having a look on the orientation of the normal vector you can see that the vector points outwards of the electrode (towards your excitation source). Thus It seems you have to multiply your excitation voltage with minus one, when calculating your impedance.
For a capacitor you will get:
Z = V_exc * (-1)/(-Re - j * Im) = +Re -j *Im
Which leads to a positive real part and a imaginary part of capacitive nature.
BR