Derivation of width of depletion layer in the pn-junction

AI Thread Summary
The discussion focuses on the derivation of the depletion layer width (W) in a pn-junction, referencing Robert F. Pierret's "SEMICONDUCTOR DEVICE FUNDAMENTALS." The user expresses confusion over the derivation provided, particularly regarding the equations for the lengths of the n-doped (x_N) and p-doped (x_P) regions. They question why the expected formula for W differs from the one found in various sources, noting the mathematical principle that the sum of square roots does not equal the square root of the sum. The conversation highlights the complexities of semiconductor physics and the importance of understanding the underlying equations. Clarification on this derivation is sought to resolve the confusion.
tcsv018
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Hello,

I read a derivation for the width of the depletion region W in "SEMICONDUCTOR DEVICE FUNDAMENTALS" by Robert F. Pierret in which at one point it says:
http://imageshack.com/i/ipbgKsK9p
Here again for better readability:

x_N = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_A}{N_D\cdot(N_A+N_D)}V_{bi}}
x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D}{N_A\cdot(N_A+N_D)}V_{bi}}
W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D+N_A}{N_A\cdot N_D}V_{bi}}

Which is confusing to me as I would expect the same containing:
W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D^2+N_A^2}{N_A\cdot N_D\cdot(N_A+N_D)}V_{bi}}This same outcome though is found on various places in the internet.

Does anyone know what I am missing?

Kind regards,

Name

Symbols:
x_N length of n-doped region
x_P length of p-doped region
N_Adensity of acceptors
N_Ddensity of donors
V_{bi} builtin potential

All the others are known constants
 
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##\sqrt{x}+\sqrt{y} \neq \sqrt{x+y}##

Ignoring the common prefactors,
$$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
$$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
$$W = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(\sqrt{N_A^2} + \sqrt{N_D^2}\right) = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(N_A + N_D\right) \\= \sqrt{\frac{(N_A+N_D)^2}{N_D N_A (N_A+N_D)}} = \sqrt{\frac{N_A+N_D}{N_D N_A}}$$
 
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