Learning Forum

[tade]

hello.

why in this example, the voltage in the boundary condition is more than the applied one?

https://optics.ansys.com/hc/en-us/articles/360042760633-Metal-oxide-semiconductor-MOS-capacitor

[Amrita Pati]

Hi Tade,

I am not sure about this. It might have been just an oversight. I will talk to the development team and get back to you. Thank you.

Regards,
Amrita

[tade]

thanks a million. The answer is so necessary for me. Please answer me as soon as possible.

[Amrita Pati]

Hi Tade,

I reached out to the Development team but I haven't heard back yet. I will get back to you as soon as I can.

Regards,
Amrita

[tade]

Dear Amrita, thank you for your response. The answer you will provide is crucial to my thesis. I had to pause my simulations due to unexpected results on the monitor. I am eagerly awaiting your reply.

[Amrita Pati]

Hi Tade,

I understand. Sorry that it is taking a while. I will remind the team again.

Regards,
Amrita

[Amrita Pati]

Hi Tade,

I apologize that it took me so long to get back to you. But I heard back from the R&D team recently. The response is as follows: "The base grid is defined with dV = 0.25 and then we just add one more voltage point next to every base grid point at the distance of 0.025. So essentially the voltage grid dV = 0.025 and 0.225 alternately. The reason to have small dV = 0.025 next to every point is that we need to calculate a numerical derivative C = dQ/dV on line 41. The numerical derivative is more accurate if dV is smaller.".

Regards,
Amrita