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ptc-17936981-Visitor
1-Visitor
February 26, 2022
Solved

Help me to learn Mathcad

  • February 26, 2022
  • 2 replies
  • 25240 views

Hello,

 

I need some help in recalling and learning skills in order to do some research in material science.

This implies using Mathcad as a tool for interpreting electrical impedance data of semiconductor diodes.

I have been away from academic life for 10 years so I do not understand easily those mathcad files I used then.

 

Can you explain me how to integrate a function found on the basis of fitting?

 

Kristjan Laes

Best answer by AlanStevens

Do you want something more like this:

Int.jpg

 

Alan

2 replies

W
Werner_E25-Diamond I
25-Diamond I
February 28, 2022

One possible way is to make the Integral function dependent on the index k

Werner_E_0-1646024771094.png

But your function Z.s returns far too high values for I > -2*10^-6 and so the integral will fail.

 

Note thats it a big difference if you call Z.s(I) vectorized or not vectorized.

I used "on error" to show where Z.s(I) fails:

Werner_E_1-1646025601424.png

 

P
ptc-17936981-VisitorAuthor
1-Visitor
February 28, 2022

Is this due to the bad fit of the data?

 

Z.s(I) looks almost like a constant number, not big variations. In more simpler calculations it is even taken as a constant as a formula for Ohms law V=R*I.

 

 

 

 

A
AlanStevens19-TanzaniteAnswer
19-Tanzanite
February 28, 2022

Do you want something more like this:

Int.jpg

 

Alan

P
ptc-17936981-VisitorAuthor
1-Visitor
February 28, 2022

Yes, this I wanted. Except Io there must be 0. 

A
AlanStevens19-Tanzanite
19-Tanzanite
February 28, 2022
@ptc-1793698 wrote:

Yes, this I wanted. Except Io there must be 0. 

Why?  If you aren't interested in values of I that are less than zero it would be best to remove them (and the corresponding values of Z) from the start.

 

Incidentally, in my solution I integrated logZs. If you actually want the integral of the values of Zs,  you will need to set  Zs = 10^(logZs) first.

 

Alan

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