integration of exponentials (Gaussian distribution) (1 Viewer)

[amateurguy]

Hello!

Can someone please help me integrate the following Gaussian function (the
function represents an impurity distribution profile in semiconductor
processing):

where Rs is the sheet resistance we're ultimately after, and its a function
of the impurity concentration N(x) from 0 to x

Rs = [ integral(q * u * N(x) dx) ]^-1

the integration limits are from 0 to x and everything in the square brackets
is raised to a power of -1.

q and u are constants (electron charge and electron mobility)

N(x) = No * exp ( -( x/(2*sqrt(Dt)) )^2 ) // the Gaussian function to be
integrated

No and Dt are constants and the exponential term is the negative square of
x/(2*sqrt(Dt))

Thank in advance for any and all help!

Mike

 

[jyu]

Hello!

Can someone please help me integrate the following Gaussian function (the
function represents an impurity distribution profile in semiconductor
processing):

where Rs is the sheet resistance we're ultimately after, and its a function
of the impurity concentration N(x) from 0 to x

Rs = [ integral(q * u * N(x) dx) ]^-1

the integration limits are from 0 to x and everything in the square brackets
is raised to a power of -1.

q and u are constants (electron charge and electron mobility)

N(x) = No * exp ( -( x/(2*sqrt(Dt)) )^2 ) // the Gaussian function to be
integrated

No and Dt are constants and the exponential term is the negative square of
x/(2*sqrt(Dt))

Thank in advance for any and all help!

Mike

 

[amateurguy]

Hi There,

Thank you for the response.

This is the closest I could find:

<IMG class=tex alt="\int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)" src="http://upload.wikimedia.org/math/1/8/a/18a83a3854b2ea8758e38ce615ee33a4.png">


But I can't see how I could use this? Can you explain?

Thanks in advance. (its very long since I have worked with integrals)

 

[Raginsheep]

a is just constant, change your a from the table of standard integrals to another pronumeral like b to stop yourself getting things mixed up and then let b = 1/a.

 

[jyu]

Hi There,

Thank you for the response.

This is the closest I could find:

<IMG class=tex alt="\int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)" src="http://upload.wikimedia.org/math/1/8/a/18a83a3854b2ea8758e38ce615ee33a4.png">


But I can't see how I could use this? Can you explain?

Thanks in advance. (its very long since I have worked with integrals)

You look up the table of indefinite integrals and then work out definite integral (0 to x).

 

[jyu]

Hi There,

Thank you for the response.

This is the closest I could find:

<IMG class=tex alt="\int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)" src="http://upload.wikimedia.org/math/1/8/a/18a83a3854b2ea8758e38ce615ee33a4.png">


But I can't see how I could use this? Can you explain?

Thanks in advance. (its very long since I have worked with integrals)

This is not in the table. Sorry.

 

[Forbidden.]

Wow, that's in the regular Maths course ?

 

[PC]

No. First year stats probably.

 

[amateurguy]

Hi There,

Thanks again for the replies!




Now allow me to embarass myself even more here. How do I incorproate the summation part from 0 to infinity?

I am looking at the last (right most) expression provided. I would envision "a" to be "4Dt" and my particular integration limits are 5e-4 and 1e-4. The part I am still confused on, is if I should be expecting a discreet rational function which I can just plug into the integrated solution then multiply by q and n (constants in the original Rs expression) and take the inverse to get my answer? I plug run this Gaussian expression through Matlab using integration limits of 5e-4 and 1e-4, I got something like:

x/y * erf ( n / m ) * sqrt(pi)

where x, y, n, and m are very large numbers. (sorry I don't have the exact solved expression in front of me as I write this). And this expression does yield the correct answer. So my last question (hopefully) is how I take the summation expression provided above by your posts and get this discrete erf function based on the supplied integration limits that allows me to simply plug in numbers into the expression one time only to get an answer.

I do thank everyone for there patience with me on this!

Mike