Depletion width of linearly doped PN-junction

[Mr_Allod]

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.

 

[Nate810]

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 $x_p$.