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For a linearly graded p-n junction, the signed p-side depletion charge per area is $Q_p=-e\alpha w^2/8$. Differentiating with respect to positive p-side voltage $V$, using the linearly graded junction depletion width, gives the positive differential junction capacitance per areaThus $C_A^{-3}$ is linear in $V$. If stored charge denotes the positive n-side magnitude $Q_n=-Q_p$, its derivative with respect to $V$ is negative; the positive capacitance is instead $-d{Q}_n/dV=d{Q}_n/d{V}_R$ for reverse-voltage magnitude $V_R=-V$. Neither definition uses the zero net charge of the whole depletion layer.

A direct band gap has its conduction minimum and valence maximum at the same crystal momentum. A photon can then connect the band-edge states with negligible momentum change. An indirect band gap requires an additional phonon to supply or remove momentum, making radiative recombination much less probable. Direct-gap materials such as gallium arsenide or gallium nitride are therefore usually better for a light-emitting diode than silicon.

For the graded junction, let $q=|e|>0$, ${\epsilon }_s={\epsilon }_0{\epsilon }_r$, and use electrostatic potential $\phi$ with $E_x=-{\phi }^{\prime }$. Fully ionized dopants have charge density $\rho =q(N_D-N_A)=q\alpha x$, so Poisson's equation is ${\phi }^{\prime \prime }=-q\alpha x/{\epsilon }_s$. The paper's displayed plus sign corresponds to using the negative potential $V=-\phi$, or a signed electron charge in that formula. The positive barrier magnitude and charge magnitude must be used consistently in the width below.

The intended compensation point is $x=0$, where $N_D=N_A$ and the net doping changes sign. A locally neutral model with comparable electron and hole mobilities has its lowest carrier conductivity near there. The depletion approximation neglects mobile-carrier conductivity throughout the layer, so it does not by itself locate an exact conductivity minimum; unequal mobilities or asymmetric carrier profiles can shift it.

Let the depletion edges be $x=\pm d$. With zero exterior field, integrating $E_x^{\prime }=\rho /{\epsilon }_s$ and imposing $E_x(\pm d)=0$ gives the linearly graded p-n junction fieldIts integral gives $\phi (x)=\phi (0)+q\alpha (d^{2}x-x^3/3)/(2{\epsilon }_s)$, so the electric potential increases toward the n-type side while the electron band edges $E_c,E_v$ bend downward together. Their separation is the local band gap, and the equilibrium electrochemical potential is flat. The built-in potential arises when carrier diffusion exposes ionized dopants until the opposing field balances further diffusion.

Finally the positive barrier isWith total width $w=2d$, this yields the linearly graded junction depletion widthHere $\alpha >0$ labels p-to-n progression with increasing $x$; otherwise use its magnitude in the width formula.

The electric potential difference is the negative integral of the field:ConsequentlyThe linearly graded junction depletion width has this cube-root dependence because of the linearly growing space charge: the charge on one side scales as $w^2$, the field as $w^2$, and its potential integral as $w^3$. The formula assumes $V_b-V_{\mathrm{ext}}>0$ and that the depletion approximation remains applicable.