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

Posted 2020年4月6日 GMT-4 19:10 Electromagnetics, MEMS & Nanotechnology, Structural & Acoustics Version 5.4 2 Replies

Isha Electrical Engineering/MEMS PhD student
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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!

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isha3
2 Replies Last Post 2020年5月1日 GMT-4 09:33
Lukang Wang
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Posted: 5 years ago 2020年4月18日 GMT-4 21:54

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]}.
Isha Electrical Engineering/MEMS PhD student
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Posted: 5 years ago 2020年5月1日 GMT-4 09:33

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!

-------------------
isha3
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!

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