# Discussion Forum

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.

## Piezoresistive Coefficent of n-type Silicon- sigma0 function of N or nd?

The below post is related to an archived discussion

Hi!

I am trying to simulate resistivity change in n-type doped Si given some mechanical stress. My Comsol results for resistivity change are 10E-5 smaller than my theoretical math.

Can anyone please help me understand where the "sigma0" analytic function for n-type piezoresistive Silicon conductivity comes from?

I don't understand the sigma0 equation in terms of 'N'. Is 'N' number density which is equal to dopant concentration or total number of particles? When I define number density nd= dopant conc[1/m^3] my Comsol resistivity change is 10e-5 times smaller than my hand calculations. So I am guessing my definition of nd[1/m3] should not be dopant conc.

This is the comsol eq: (Ne_const0.1400/sqrt(1+N/(N/350 +3e22)))

Thank you so much!

2 Replies Last Post May 1, 2020, 9:33 AM EDT

Posted: 9 months ago

Hi, Lodhi Based on the application "piezoresistive pressure sensor", I want to use n-type SiC to replace Si. Same as you, I cannot understand the expression of conductivity of sigma0(nd[m^3]). Is it a default variable of comsol? And the piezoresistive coupling matrix is: {-102.2e-11[1/Pa]/sigma0(nd[m^3])[S/m], 53.4e-11[1/Pa]/sigma0(nd[m^3])[S/m], -102.2e-11[1/Pa]/sigma0(nd[m^3])[S/m], 53.4e-11[1/Pa]/sigma0(nd[m^3])[S/m], 53.4e-11[1/Pa]/sigma0(nd[m^3])[S/m], -102.2e-11[1/Pa]/sigma0(nd[m^3])[S/m], 0, 0, 0, -13.6e-11[1/Pa]/sigma0(nd[m^3])[S/m], 0, 0, 0, 0, -13.6e-11[1/Pa]/sigma0(nd[m^3])[S/m], 0, 0, 0, 0, 0, -13.6e-11[1/Pa]/sigma0(nd[m^3])[S/m]}.