How to calculate the erfc of a number

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The discussion centers on calculating the complementary error function (erfc) for a homework problem related to phosphorus diffusion in a p-type wafer. The original poster is confused about how to compute erfc, as their calculator lacks this function and they are unsure how to manipulate it algebraically. It is clarified that erfc can be expressed in terms of the error function (erf), and many calculators have a normal distribution function that can be used instead. While there are no closed-form solutions for erfc or its inverse, numerical methods such as trial-and-error or Newton's Method can be employed to find values. Understanding these properties and alternative methods is essential for solving the given problem.
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So the HW is actually about calculating the junction depth of phosphorus diffusion into a p-type wafer however that is not the problem I am having. The problem I am having is that the book tells me to calculate the erfc(#). However, my calculator does not seem to have an erfc function. In the book the do algebra with the function, for instance it says:

(1.1x10^30)erfc(x/(2sqrt(DT)) = (3x10^16) Solving for x yeilds: x = 2sqrt(DT)erfc^-1(.000273)

What am I missing here? How are they pulling x out of the erfc function? I have never even heard of an erfc function I have no idea what the properties of the function are. For instance if it were exp(x) then I can find x by taking the ln(exp(x)) but how do I do that with erfc?
 
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vysero said:
So the HW is actually about calculating the junction depth of phosphorus diffusion into a p-type wafer however that is not the problem I am having. The problem I am having is that the book tells me to calculate the erfc(#). However, my calculator does not seem to have an erfc function. In the book they do algebra with the function, for instance it says:

(1.1x10^30)erfc(x/(2sqrt(DT)) = (3x10^16) Solving for x yeilds: x = 2sqrt(DT)erfc^-1(.000273)

What am I missing here? How are they pulling x out of the erfc function? I have never even heard of an erfc function I have no idea what the properties of the function are. For instance if it were exp(x) then I can find x by taking the ln(exp(x)) but how do I do that with erfc?
The Wikipedia page on the Error Function may be helpful to you.
 
vysero said:
So the HW is actually about calculating the junction depth of phosphorus diffusion into a p-type wafer however that is not the problem I am having. The problem I am having is that the book tells me to calculate the erfc(#). However, my calculator does not seem to have an erfc function. In the book the do algebra with the function, for instance it says:

(1.1x10^30)erfc(x/(2sqrt(DT)) = (3x10^16) Solving for x yeilds: x = 2sqrt(DT)erfc^-1(.000273)

What am I missing here? How are they pulling x out of the erfc function? I have never even heard of an erfc function I have no idea what the properties of the function are. For instance if it were exp(x) then I can find x by taking the ln(exp(x)) but how do I do that with erfc?

You can get ##\text{erfc}(x)## in terms of the so-called erf-function: ##\text{erfc}(x) = 1 - \text{erf}(x)##. Most calculators lack an "erf" button, but many of them have a "normal distribution" button, giving the cumulative distribution (CDF) of the standard normal distribution. If ##\Phi(x)## is the normal CDF we have
$$\Phi(x) = \frac{1}{2} + \frac{1}{2} \text{erf} \left( \frac{x}{\sqrt{2}} \right)$$
Thus,
$$ 1-\Phi(\sqrt{2} y) = \frac{1}{2} - \frac{1}{2} \text{erf}(y)= \frac{1}{2} \text{erfc}(y)$$
There are no exact, closed-form formulas for erfc or ##\Phi##, but many fast and accurate algorithms are available to compute numerical values reliably, so getting ##\Phi(x)## by pressing a button is really no different from getting ##\sin(x)## by pressing a button.

Note, however, if you want ##\text{erfc}^{-1}(z)## you need to solve the equation ##\text{erfc}(x) = z##. You can do that fairly quickly by trial-and-error methods, or by plotting, etc. You could also use fancy techniques like Newton's Method.
 

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