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Why does the differential operator d(f,x) always give a zero result?
Posted 19.01.2016, 22:49 GMT-5 Parameters, Variables, & Functions Version 5.2 2 Replies
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I was using the semiconductor module to do device simulation. After the distribution of electrical potential V had been computed, a contour of V could be plotted. But if I typed d(V,x) into the expression box, wishing to get the x component of the gradient of V, I always got zero althrough the domain. What was wrong with it? Is there any special point I need to pay attentian?
THANKS
THANKS
2 Replies Last Post 20.01.2016, 14:59 GMT-5
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Posted:
9 years ago
Hi
which physics are you using exactly (I do not have the semiconductor module, so I cannot check), what are you dependent variable ? If it's V then Vx = dV/dx and (Vx,Vy,Vz) are the electric potential in [V/m] these are already defined internally in COMSOL.
If you are solving for another variable than V and V is probably derived from the other variables, then it could be that d(V,x) is not defined.
In ES physics, solving for V, plotting both Vx and d(V,x) work OK
Check the doc carefully and check for each physics which are the dependent variable one are really solving for. and take a look "under the hood" by turning on the Equation view in the preferences to see the internal variables of COMSOL
--
Good luck
Ivar
which physics are you using exactly (I do not have the semiconductor module, so I cannot check), what are you dependent variable ? If it's V then Vx = dV/dx and (Vx,Vy,Vz) are the electric potential in [V/m] these are already defined internally in COMSOL.
If you are solving for another variable than V and V is probably derived from the other variables, then it could be that d(V,x) is not defined.
In ES physics, solving for V, plotting both Vx and d(V,x) work OK
Check the doc carefully and check for each physics which are the dependent variable one are really solving for. and take a look "under the hood" by turning on the Equation view in the preferences to see the internal variables of COMSOL
--
Good luck
Ivar
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Posted:
9 years ago
I am assuming that you are using the finite volume formulation in the Semiconductor Module, where shape functions are constant, hence the zero derivatives.
Jeff
Jeff