Solution
= Solution
\b[Dopant atoms and carrier type.] <Substitutional semiconductor doping> places a small fraction of foreign atoms on host <lattice> sites. In a tetrahedrally bonded group-IV <semiconductor>, a group-V substitution such as <phosphorus> in <silicon> has an extra valence <Electron>. A shallow donor level just below the conduction edge can release that <Electron>, leaving a positively charged donor. In an <n-type semiconductor>, mobile <Electrons> are the majority carriers and the <chemical potential> shifts toward the <conduction band>.
A group-III substitution such as <boron> in <silicon> has one fewer valence <Electron>. An acceptor level just above the valence edge captures an <Electron> from the valence states, leaving a mobile <electron hole> and a negatively charged acceptor. In a <p-type semiconductor>, <electron holes> are the majority carriers and the <chemical potential> shifts toward the <valence band>. The material can remain electrically neutral in both cases: mobile carrier <electric charge> is compensated by fixed ionized impurity <electric charge>. <Dopant freeze-out>, compensation and degeneracy qualify the simplest one-carrier-per-dopant picture.
\b[Making the junction.] Choose p material at $x<0$ and n material at $x>0$. <Electrons> initially diffuse from n to p, and <electron holes> from p to n, recombining near the interface. Negative ionized acceptors are exposed on the p side and positive ionized donors on the n side. They produce an <electric field> directed from n to p. The field opposes further majority-carrier <diffusion> until drift and <diffusion> balance and the equilibrium electrochemical potential is uniform.
The depletion regions are depleted of mobile majority carriers, not of the fixed dopant charges that create the <space charge>. At equilibrium the conduction and valence edges bend together, while the <chemical potential> is flat. The specified $\Delta\phi_0$ is a <electric potential difference>: the corresponding initial chemical-potential <energy> separation is $e\Delta\phi_0$. This convention is required by the quoted units and width formulas. Write $\varepsilon_s=\epsilon\epsilon_0$ for the absolute <permittivity>; the paper's $\epsilon$ is <relative permittivity>.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2013-doping-and-junction.png]
{title=Substitutional donor/acceptor levels, equilibrium band bending, fixed space charge and the junction electric field}
\b[<Poisson equation> and widths.] Let the depleted intervals be $-w_p<x<0$ and $0<x<w_n$. Fully ionized number densities $N_A,N_D$ correspond to <electric charge> densities $-eN_A,+eN_D$, respectively. Thus
$$
\rho(x)=\begin{cases}-eN_A,&-w_p<x<0,\\eN_D,&0<x<w_n,\\0,&\text{in the neutral regions},\end{cases}\qquad \phi''(x)=-\frac{\rho(x)}{\varepsilon_s}.
$$
The factor $e$ converts a dopant concentration to a space-charge density. Requiring zero field in both neutral regions gives <charge neutrality>,
$$
N_Aw_p=N_Dw_n.
$$
With $\mathcal E=-\phi'$, direct integration gives the triangular field
$$
\mathcal E(x)=\begin{cases}-eN_A(x+w_p)/\varepsilon_s,&-w_p\leq x\leq0,\\eN_D(x-w_n)/\varepsilon_s,&0\leq x\leq w_n.\end{cases}
$$
Continuity at the interface is the same neutrality condition. The field is negative throughout the depleted interval. Set $\phi(-w_p)=0$. Its integral gives
$$
\phi(x)=\begin{cases}eN_A(x+w_p)^2/(2\varepsilon_s),&-w_p\leq x\leq0,\\\Delta\phi_0-eN_D(w_n-x)^2/(2\varepsilon_s),&0\leq x\leq w_n.\end{cases}
$$
Potential continuity therefore requires
$$
\Delta\phi_0=\frac{e}{2\varepsilon_s}(N_Aw_p^2+N_Dw_n^2).
$$
Combining it with neutrality yields the <abrupt-junction depletion width and capacitance> results
$$
\boxed{w_n=\sqrt{\frac{2\varepsilon_sN_A\Delta\phi_0}{eN_D(N_A+N_D)}},\qquad w_p=\sqrt{\frac{2\varepsilon_sN_D\Delta\phi_0}{eN_A(N_A+N_D)}}.}
$$
The more lightly doped side depletes farther, because equal and opposite space charges must be exposed.
\b[Applied bias.] Define positive $V$ as forward bias, raising the p-side terminal potential relative to the n-side terminal. It lowers the remaining barrier from $\Delta\phi_0$ to $\Delta\phi_0-V$. <Reverse bias> has $V<0$ and raises that barrier. The same Poisson integration applies, with this replacement:
$$
\boxed{w_n(V)=\sqrt{\frac{2\varepsilon_sN_A(\Delta\phi_0-V)}{eN_D(N_A+N_D)}},\quad w_p(V)=\sqrt{\frac{2\varepsilon_sN_D(\Delta\phi_0-V)}{eN_A(N_A+N_D)}}.}
$$
Forward bias narrows the depleted region; <reverse bias> widens it. Under bias there need not be one equilibrium <chemical potential> across the device, so the equilibrium flat-level argument must not be reapplied unchanged. The barrier/width change is illustrated in the next figure.
\b[Differential <capacitance> and <electric charge> sign.] Put $B=\Delta\phi_0-V>0$ and $N_{\rm eff}=N_AN_D/(N_A+N_D)$. The signed p-side fixed <electric charge> per unit area is
$$
Q_p=-eN_Aw_p=-\sqrt{2\varepsilon_seN_{\rm eff}B}.
$$
Differentiating with respect to forward <electric potential difference> gives the positive <capacitance> per unit area
$$
\boxed{C_A=\frac{dQ_p}{dV}=\sqrt{\frac{\varepsilon_seN_AN_D}{2(N_A+N_D)(\Delta\phi_0-V)}}=\frac{\varepsilon_s}{w_n+w_p}.}
$$
The n-side <electric charge> is $Q_n=-Q_p$, so $dQ_n/dV=-C_A$ instead. If $Q$ denotes a positive space-charge magnitude, the positive <capacitance> is $-dQ/dV$ for this forward-voltage convention. This resolves the sign implicit in the paper's positive $dQ/dV$. The total device <capacitance> is $C=\mathcal A C_A$, where $\mathcal A$ is its cross-sectional area.
\b[Finite device and numerical sketch.] For the symmetric device, $N_A=N_D=N$ and each side has thickness $d$. Before full depletion,
$$
w_n=w_p=w=\sqrt{\frac{\varepsilon_s(\Delta\phi_0-V)}{eN}},\qquad C(V)=\frac{\mathcal A}{2}\sqrt{\frac{\varepsilon_seN}{\Delta\phi_0-V}}.
$$
Using $\varepsilon_s=12\epsilon_0=1.0625\times10^{-10}\,\mathrm{F\,m^{-1}}$, the zero-bias width on each side is $4.07\,\mathrm{\mu m}$, less than $d=10\,\mathrm{\mu m}$. The total zero-bias <capacitance> is $\boxed{C(0)\simeq1.30\,\mathrm{pF}}$.
Full depletion occurs when $w=d$, so
$$
\boxed{V_{\rm fd}=\Delta\phi_0-\frac{eNd^2}{\varepsilon_s}=0.5-3.016\simeq-2.52\,\mathrm V.}
$$
At larger <reverse bias> the depleted thickness cannot exceed $2d$. In the ideal fully depleted slab, incremental terminal <electric charge> changes the field uniformly through that fixed thickness, giving the <full-depletion capacitance of a symmetric junction slab>
$$
\boxed{C_{\rm fd}=\frac{\varepsilon_s\mathcal A}{2d}\simeq5.31\times10^{-13}\,\mathrm F=0.531\,\mathrm{pF}.}
$$
Thus the idealized piecewise sketch over the specified range is
$$
C(V)=\begin{cases}0.531\,\mathrm{pF},&-5.0\leq V\leq V_{\rm fd},\\\displaystyle\frac{\mathcal A}{2}\sqrt{\frac{\varepsilon_seN}{0.5\,\mathrm V-V}},&V_{\rm fd}<V<0.5\,\mathrm V.\end{cases}
$$
It is continuous at full depletion, rises through $1.30\,\mathrm{pF}$ at zero bias, and has a formal inverse-square-root divergence as $V\to0.5\,\mathrm V$ from below.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2013-junction-capacitance.png]
{title=Bias-dependent depletion geometry and the numerical capacitance curve, including full depletion}
There are two physical limits to this extension. Once all dopants are exposed, their fixed <space charge> no longer increases, but the terminal electrodes can still acquire incremental <electric charge>; differentiating only the fixed ionized <electric charge> would incorrectly predict zero <capacitance>. Near $V=\Delta\phi_0$, the <depletion approximation> itself fails as mobile carriers and forward injection become important, so the formal infinite endpoint is not a prediction of an actual infinite measured <capacitance>. The fully depleted plateau assumes ideal contacts and no breakdown, injection storage or extra series <capacitance>.