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I'm studying Solid State Electronics and at one point my book says: $$\dfrac{\text{d}x_n}{\text{d}V_a}= \dfrac{1}{N_d} \left(\dfrac{\varepsilon_s}{2q(\frac{1}{N_a}+\frac{1}{N_d})(\phi_i -V_a)}\right)^{1/2}$$ where $$ \left\{\begin{align} x_d &= x_n \left( 1+ \frac{N_d}{N_a} \right) \\ x_d &=\sqrt{ \frac{2\varepsilon}{q}\left(\frac{1}{N_a}+\frac{1}{N_d}\right)(\phi_i -V_a)} \end{align}\right. $$ So I tried calculating that derivative by "hand". I actually get the result of the book only, there's a minus sign that doesn't tally with what the books says. Is it me or is the book mistaken?

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Is the second $N_a$ supposed to be $N_d$ in the $x_d$ equation? –  Amzoti Jun 10 '13 at 12:47
    
yes, sorry for the glitch. –  WobblyWindows Jun 10 '13 at 15:39
    
Where does the $s$ in epsilon come from in your first equation since it is not written like that in $x_d$? Also, are all variables constants, except $V_a$? Maybe you can add your work so we can look at those details. Also, if the negative is correct, is there any phase or magnitude data for this SSE circuit (or whatever it is)? Regards –  Amzoti Jun 10 '13 at 15:50
    
What is your derivative supposed to represent? It's entirely possible that your issue is conceptual; remember that the electric field is the negative derivative of the voltage. I ran into similar issues with my textbook when I took a solid state course. –  AHatThrowaway Jun 10 '13 at 15:58
    
You are right, there is a minus sign, unless there are some specific physical meaning in it. –  Shuhao Cao Jun 10 '13 at 16:00

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