How to calculate the erfc of a number

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SUMMARY

The discussion focuses on calculating the complementary error function, erfc, particularly in the context of determining junction depth in phosphorus diffusion into a p-type wafer. The user expresses confusion over the absence of an erfc function on their calculator and seeks clarification on how to isolate x from the equation involving erfc. Key insights include the relationship between erfc and the error function (erf), where erfc(x) = 1 - erf(x), and the use of normal distribution functions to compute values. Additionally, methods for finding the inverse, erfc^-1(z), are discussed, including trial-and-error and Newton's Method.

PREREQUISITES
  • Understanding of the complementary error function (erfc) and its properties.
  • Familiarity with the error function (erf) and its relationship to normal distribution.
  • Basic algebra skills for manipulating equations involving erfc.
  • Knowledge of numerical methods such as Newton's Method for solving equations.
NEXT STEPS
  • Research the properties and applications of the complementary error function (erfc).
  • Learn about the error function (erf) and its significance in statistics and probability.
  • Explore numerical methods for solving equations, specifically Newton's Method.
  • Investigate calculators or software that provide built-in functions for erfc and erf, such as Python's SciPy library.
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Students and professionals in fields such as semiconductor physics, materials science, and applied mathematics, particularly those involved in diffusion processes and statistical analysis.

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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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