Depletion width of linearly doped PN-junction

Mr_Allod
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Homework Statement
Derive expressions for the electric field distribution and potential for a PN Junction with linear doping gradient then use this to calculate the depletion width as a function of the bias voltage ##V_bi##
Relevant Equations
##x<0## (P-Region) doping concentration: ##N_A = -Sx##
##x>0## (N-Region) the doping concentration is: ##N_D = Kx##
Electric Field in P-Region: ##E_P=\frac {qS}{2\epsilon}(x^2 - x_p^2)##
Electric Field in N-Region: ##E_N=\frac {qK}{2\epsilon}(x^2 - x_n^2)##
Potential in P-Region: ##V = V_p + \frac {qM_A}{\epsilon} \left( \frac {x_p^2x}{2} - \frac {x^3}{6} + \frac {x_p^3}{3} \right)##
Potential in N-Region: ##V = V_n + \frac {qM_A}{\epsilon} \left( \frac {x_n^2x}{2} - \frac {x^3}{6} - \frac {x_p^3}{3} \right)##
Continuity at x = 0:
$$Sx_p^2 = Kx_n^2$$
$$V_{bi} = V_n - V_p = \frac {q}{3\epsilon} \left( Sx_p^3 + Kx_n^3\right)$$
Hello there, I have derived the expressions for electric field and potential to be the ones above, then for continuity at ##x = 0## I set the electric fields and potentials to be equal to yield the expressions:
$$Sx_p^2 = Kx_n^2$$
$$V_{bi} = V_n - V_p = \frac {q}{3\epsilon} \left( Sx_p^3 + Kx_n^3\right)$$

I am now stuck on how to find an expression for the depletion region width, which in our class we have defined as ##|x_p|+x_n##, where ##-x_p## and ##x_n## are the edges of the depletion layer in the P and N regions respectively. I just can't seem to find a way to isolate them without ending up with one as a function of the other, which is not terribly useful so I would really appreciate some help with this.
 
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The solution to this problem is to substitute the equation for the biasing voltage into the equation for electric field:$$\frac{q}{3\epsilon} \left( Sx_p^3 + Kx_n^3\right) = Sx_p^2 - Kx_n^2$$Rearranging this equation gives us:$$Kx_n^3 + (K-S)x_n^2 - \frac{q}{3\epsilon} = 0$$This is a cubic equation in terms of ##x_n##, which can be solved using the Cardano formula. Once we have the solution for ##x_n##, we can then solve for ##x_p## by substituting it into the equation for electric field and solving for ##x_p##.
 
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