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Resistance calculation across the electrodes

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Trying to simulate a metaloxide based gas sesnor and I need to calculate the changein resistance across the interdigitated electrodes placed beneath the sensing layer. The problem is that the resistance calculations show the value for the entire structure rather than across the electrodes only. Moreover the gas that was flown  over the sensor wasn't reacting with the oxide. Used the chemistry, surface reaction modules for that but didn't  get the desired results. Thanks in advance!


1 Reply Last Post 2018年1月12日 GMT-5 16:07
Robert Koslover Certified Consultant

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Posted: 7 years ago 2018年1月12日 GMT-5 16:07
Updated: 7 years ago 2018年1月12日 GMT-5 16:19

I don't really know the details of your problem, but it you ever want to know the resistance of a structure or sub-structure in a problem where currents and potentials are well-defined quantities, then you can probably find it pretty easily by: (1) probing the potential (i.e., the voltages) at the appropriate boundaries of the structure (the places the current goes to/from or in/out), and (2) integrating the volume current density across a cross-sectional plane passing through the structure of interest to you. For example, you could define a workplane that crosses the struture, and put a rectangle, circle, or whatever shape you like in that workplane to set up a bound to contain the integration region of interest to you. I.e., you should set it up so that the cross-section only includes the path(s) for which you want to find the associated resistance. Anyway, this will thus provide you the voltage difference V (from V1 and V2, from which you define V = V2-V1) and the current I (integrated J over the aforementioned cross-section), respectively. From that, you can then immediately find R = V/I, in accordance with Ohm's law. Be sure to use fine enough meshes and be careful with the math, if you are using complex quantities. Hope that helps.

I should add that if the vector current density pattern is geometrically messy and/or you can't even envision/figure out the current paths clearly, then you will probably need to isolate the actual geometry and material (aka, the "resistor") of specific interest to you first, impose voltages on its "ends" (wherever you decide them to be) and then solve for the current, from which (as before) you can compute the resistance using R=V/I.

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I don't really know the details of your problem, but it you ever want to know the resistance of a structure or sub-structure in a problem where currents and potentials are well-defined quantities, then you can probably find it pretty easily by: (1) probing the potential (i.e., the voltages) at the appropriate boundaries of the structure (the places the current goes to/from or in/out), and (2) integrating the volume current density across a cross-sectional plane passing through the structure of interest to you. For example, you could define a workplane that crosses the struture, and put a rectangle, circle, or whatever shape you like in that workplane to set up a bound to contain the integration region of interest to you. I.e., you should set it up so that the cross-section only includes the path(s) for which you want to find the associated resistance. Anyway, this will thus provide you the voltage difference V (from V1 and V2, from which you define V = V2-V1) and the current I (integrated J over the aforementioned cross-section), respectively. From that, you can then immediately find R = V/I, in accordance with Ohm's law. Be sure to use fine enough meshes and be careful with the math, if you are using complex quantities. Hope that helps. I should add that if the vector current density pattern is geometrically messy and/or you can't even envision/figure out the current paths clearly, then you will probably need to isolate the actual geometry and material (aka, the "resistor") of specific interest to you first, impose voltages on its "ends" (wherever you decide them to be) and then solve for the current, from which (as before) you can compute the resistance using R=V/I.

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