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?

Please login with a confirmed email address before reporting spam

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 01/05/2020, 09:33 GMT-4

Please login with a confirmed email address before reporting spam

Posted: 10 months ago 18/04/2020, 21:54 GMT-4

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]}.

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]}.

Please login with a confirmed email address before reporting spam

Posted: 10 months ago 01/05/2020, 09:33 GMT-4

Dear Lukang,

Sorry for seeing this late.

I found out that sigma0 is conductivity of the material which is defined as sigma0=(charge_mobility x q x n). Mobility is the function of doping density (charge density 'n' assumed to be equal to doping density) (see the screenshot attached).

My understanding is that the sigma0 terms in the piezoresistive coef. help adjust the sensitivity of the piezoresistor according to doping. Comsol has the sigma0 term to adjust piezoresistance of the resistor according to its doping... so doping goes up. conductivity increases and piezoresistance decreases. Hope that helps!

Dear Lukang, Sorry for seeing this late. I found out that sigma0 is conductivity of the material which is defined as sigma0=(charge_mobility x q x n). Mobility is the function of doping density (charge density 'n' assumed to be equal to doping density) (see the screenshot attached). My understanding is that the sigma0 terms in the piezoresistive coef. help adjust the sensitivity of the piezoresistor according to doping. Comsol has the sigma0 term to adjust piezoresistance of the resistor according to its doping... so doping goes up. conductivity increases and piezoresistance decreases. Hope that helps!

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.