part05c

(Problem 5.9 in part05b is already complete; the solutions below pick up with 5.10.)

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Problem 5.10: A stepped n-side profile

Problem. A pn diode has the doping profile of Fig. P5.10, and we assume $x_n>0$ (the depletion region reaches beyond the junction) for all biases of interest. (a) What is the built-in voltage? Justify. (b) Using the depletion approximation, sketch $\rho (x)$.

Reading the figure. The figure is mostly lost in the OCR; only the labels $N_D-N_A$, $N_D$, $N_D/2$, $x_0$, $-N_A$ survive. I read it as: $N_D-N_A=-N_A$ for $x<0$, then $N_D/2$ for $0<x<x_0$, then $N_D$ for $x>x_0$. If your figure differs, the method below is unchanged.

Solution (a): built-in voltage. Think of $V_{bi}$ as the height difference between two flat “lakes” (the neutral p and n regions) once their Fermi levels line up. Only the lake levels matter, not the terrain between them. So we need only the far-side doping.

Far p-side (neutral): $p_0=N_A,E_i-E_F=kT\ln \frac{N_A}{n_i}$ Far n-side (neutral, $x\gg x_0$): $n_0=N_D,E_F-E_i=kT\ln \frac{N_D}{n_i}$ At equilibrium $E_F$ is constant, so the total bending of $E_i$ is the sum: $q{V}_{bi}=kT\ln \frac{N_A}{n_i}+kT\ln \frac{N_D}{n_i}$ $\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}}$ Justification: the intermediate $N_D/2$ layer is inside the band-bending region, so it changes how the bending is distributed in space but not its total. The result is the same as for a plain step junction with the same outer dopings.

Solution (b): charge density. The depletion approximation gives $\rho =q(N_D-N_A)$ for $-x_p\le x\le x_n$, zero outside: $\rho (x)=\{\begin{array}{llccc}-q{N}_A,& -x_p\le x<0+q{N}_D/2,& 0\le x<x_0+q{N}_D,& x_0\le x\le x_{n}0,& \text{otherwise}\end{array}$ With $x_n>x_0$ (our assumption) all three levels appear. Sketch: a negative block of height $-q{N}_A$ and width $x_p$; then a positive step of height $q{N}_D/2$ and width $x_0$; then a taller positive step of height $q{N}_D$ out to $x_n$.

Neutrality (equal areas, the “debts equal credits” rule): $N_A{x}_p=\frac{N_D}{2}{x}_0+N_D(x_n-x_0)=N_D\left(x_n-\frac{x_0}{2}\right)$ $\Rightarrow x_n=\frac{N_A}{N_D}{x}_p+\frac{x_0}{2}$ Why the assumption $x_n>x_0$ matters: if $x_n<x_0$ the depletion region would end inside the $N_D/2$ layer and the top level $q{N}_D$ would never appear.

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Problem 5.11: The p-i-n diode

Problem. A p-i-n diode has uniform p ($N_A$) and n ($N_D$) regions and a narrow middle region of width $x_i$ with $N_D-N_A=0$. (a) Sketch $\rho$, $\mathcal{E}$, $V$ and the equilibrium band diagram. (b) Built-in voltage? (c) Quantitative expressions for $\rho$, $\mathcal{E}$, $V$ and the depletion widths.

Setup. From Fig. P5.11 (labels lost in OCR except $x_i/2$), place the i-region symmetrically: $-x_i/2\le x\le x_i/2$. Let $x_p$ be the depleted thickness inside the p-region (from $-x_i/2$ leftward) and $x_n$ the depleted thickness inside the n-region (from $x_i/2$ rightward). Assume the i-region is fully depleted, which is true because it is narrow and has no mobile charge to spare.

Solution (a): sketches.

• $\rho$: a negative block $-q{N}_A$ over $-x_i/2-x_p\le x\le -x_i/2$; zero across the i-region; a positive block $+q{N}_D$ over $x_i/2\le x\le x_i/2+x_n$. Think of a parallel-plate capacitor whose plates are the two blocks and whose gap is the i-region.
• $\mathcal{E}$: rises linearly (in magnitude) through the p-block, stays constant across the i-region (no charge, so no slope), then falls linearly back to zero through the n-block. The shape is a trapezoid with a flat top. This is exactly the field of a capacitor: uniform in the gap.
• $V$: parabola (concave up) in the p-block, a straight ramp across the i-region, parabola again (concave down) in the n-block, flat outside.
• Bands: $E_C$, $E_V$ bend with the same shape as $-qV$: curved down-slope in the p-block, a straight, constant-slope section in the i-region, curved in the n-block. $E_F$ is flat; $E_F$ sits near $E_V$ on the far left and near $E_C$ on the far right.

Solution (b): built-in voltage. Same argument as in 5.10: the total bending equals the difference of Fermi-level positions in the neutral outer regions, and the lightly doped i-region only sits in between: $\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}}$

Solution (c): quantitative.

Charge density: $\rho (x)=\{\begin{array}{llcc}-q{N}_A,& -\frac{x_i}{2}-x_p\le x<-\frac{x_i}{2}\text{0},& |x|\le \frac{x_i}{2}\text{+}q{N}_D,& \frac{x_i}{2}<x\le \frac{x_i}{2}+x_n\end{array}$

Neutrality: $q{N}_A{x}_p=q{N}_D{x}_n\Rightarrow x_n=\frac{N_A}{N_D}{x}_p$

Electric field. Integrate Poisson’s equation $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$ from the left edge where $\mathcal{E}=0$:

p-block: $\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}\left(x+\frac{x_i}{2}+x_p\right),-\frac{x_i}{2}-x_p\le x\le -\frac{x_i}{2}$ i-region (constant, equal to the value at the block’s right edge): $\mathcal{E}=-\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}\equiv -{\mathcal{E}}_0,|x|\le \frac{x_i}{2}$ n-block: $\mathcal{E}(x)=-\frac{q{N}_D}{K_S{\epsilon }_0}\left(\frac{x_i}{2}+x_n-x\right),\frac{x_i}{2}\le x\le \frac{x_i}{2}+x_n$

Potential. The drop across each piece is the area under $|\mathcal{E}|$ (triangle, rectangle, triangle): $\Delta V_p=\frac{1}{2}{\mathcal{E}}_0{x}_p=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}$ $\Delta V_i={\mathcal{E}}_0{x}_i=\frac{q{N}_A{x}_p{x}_i}{K_S{\epsilon }_0}$ $\Delta V_n=\frac{1}{2}{\mathcal{E}}_0{x}_n=\frac{q{N}_D{x}_n^2}{2K_S{\epsilon }_0}$ In terms of $V(x)$, with $V=0$ at the left edge: $V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}{\left(x+\frac{x_i}{2}+x_p\right)}^2\text{(p-block)}$ $V(x)=\Delta V_p+{\mathcal{E}}_0\left(x+\frac{x_i}{2}\right)\text{(i-region)}$ n-block, written from the right edge where $V=V_{bi}-V_A$: $V(x)=V_{bi}-V_A-\frac{q{N}_D}{2K_S{\epsilon }_0}{\left(\frac{x_i}{2}+x_n-x\right)}^2,\frac{x_i}{2}\le x\le \frac{x_i}{2}+x_n$

Depletion widths. Require the total drop to equal $V_{bi}-V_A$: $V_{bi}-V_A=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}\left(1+\frac{N_A}{N_D}\right)+\frac{q{N}_A{x}_i{x}_p}{K_S{\epsilon }_0}$ (using $x_n=N_A{x}_p/N_D$ in $\Delta V_n$). This is a quadratic in $x_p$: $\frac{q{N}_A}{2K_S{\epsilon }_0}\left(1+\frac{N_A}{N_D}\right)x_p^2+\frac{q{N}_A{x}_i}{K_S{\epsilon }_0}{x}_p-(V_{bi}-V_A)=0$ Solve with the quadratic formula (positive root): $\boxed{x_p=\frac{1}{1+N_A/N_D}\left[-x_i+\sqrt{x_i^2+\frac{2K_S{\epsilon }_0(V_{bi}-V_A)}{q{N}_A}\left(1+\frac{N_A}{N_D}\right)}\right]}$ $\boxed{x_n=\frac{N_A}{N_D}{x}_p}$ Checks: (i) $x_i\to 0$ gives $x_p=\sqrt{2K_S{\epsilon }_0(V_{bi}-V_A)/(q{N}_A(1+N_A/N_D))}$, the ordinary step-junction result. (ii) For very large $x_i$, $x_p\approx \frac{K_S{\epsilon }_0(V_{bi}-V_A)}{q{N}_A{x}_i}$, which is tiny: the long i-region takes most of the voltage, as in a capacitor with a wide gap.

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Problem 5.12: From a given $V(x)$ back to the doping

Problem. In equilibrium the depletion-region potential is $V(x)=\frac{V_{bi}}{2}\left[1+\sin \frac{\pi x}{W}\right],-\frac{W}{2}\le x\le \frac{W}{2}$ (the OCR is garbled; I take the sine argument as $\pi x/W$, the only choice that runs from $0$ to $V_{bi}$ across $\pm W/2$). Find (a) $\mathcal{E}(x)$, (b) $\rho (x)$, (c) $N_D-N_A$ under the depletion approximation, with sketches.

This is Poisson’s equation run backwards: instead of doping to field to potential, we differentiate down the staircase.

Check of the given $V$. $V\left(-\frac{W}{2}\right)=\frac{V_{bi}}{2}[1+\sin (-\pi /2)]=0,V\left(\frac{W}{2}\right)=\frac{V_{bi}}{2}[1+1]=V_{bi}✓$

Solution (a): electric field. Use $\mathcal{E}=-dV/dx$: $\mathcal{E}(x)=-\frac{d}{dx}\left\{\frac{V_{bi}}{2}\left[1+\sin \frac{\pi x}{W}\right]\right\}$ $\mathcal{E}(x)=-\frac{V_{bi}}{2}\cdot \frac{\pi }{W}\cos \frac{\pi x}{W}$ $\boxed{\mathcal{E}(x)=-\frac{\pi V_{bi}}{2W}\cos \frac{\pi x}{W}}$ Checks: at $x=\pm W/2$, $\cos (\pm \pi /2)=0$, so $\mathcal{E}=0$ at both depletion edges, as required. At $x=0$, $\mathcal{E}=-\pi V_{bi}/(2W)$, the most negative value. Sketch: an inverted half-cosine arch, zero at both edges, minimum at the center.

Solution (b): charge density. Poisson: $\rho =K_S{\epsilon }_{0}d\mathcal{E}/dx$. $\rho (x)=K_S{\epsilon }_0\frac{d}{dx}\left[-\frac{\pi V_{bi}}{2W}\cos \frac{\pi x}{W}\right]$ $\rho (x)=K_S{\epsilon }_0\cdot \frac{\pi V_{bi}}{2W}\cdot \frac{\pi }{W}\sin \frac{\pi x}{W}$ $\boxed{\rho (x)=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2W^2}\sin \frac{\pi x}{W}}$ Sketch: an odd sine wave, negative for $-W/2<x<0$ (reaching $-{\rho }_{max}$ at $x=-W/2$), zero at $x=0$, positive for $x>0$ (reaching $+{\rho }_{max}$ at $x=W/2$), with ${\rho }_{max}=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2W^2}.$ Net charge: $\int \rho dx=0$ by oddness, so the region is neutral overall, as it must be.

Solution (c): doping profile. The depletion approximation says $\rho =q(N_D-N_A)$ inside the region, so $\boxed{N_D-N_A=\frac{\rho }{q}=\frac{{\pi }^2{K}_S{\epsilon }_0{V}_{bi}}{2q{W}^2}\sin \frac{\pi x}{W},-\frac{W}{2}\le x\le \frac{W}{2}}$ Sketch: the same odd sine shape, p-type on the left, n-type on the right. Near $x=0$, $\sin (\pi x/W)\approx \pi x/W$, so the profile is approximately linearly graded with $a\approx \frac{{\pi }^3{K}_S{\epsilon }_0{V}_{bi}}{2q{W}^3}$ Cross-check: the exact linear-graded result is $a=12K_S{\epsilon }_0{V}_{bi}/(q{W}^3)$. The coefficients ${\pi }^3/2\approx 15.5$ and $12$ are similar, as expected for a smoothed linear profile.

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Problem 5.13: Linearly graded junction by computer

Problem. For a nondegenerately doped Si linearly graded junction at $T=300$ K: (a) plot $\mathcal{E}(x)$ and $V(x)$ for a given gradient $a$ (iterating Eq. 5.48 for $V_{bi}$); (b) overlay $V_A=V_{A0}/2^n$, $n=0$ to 3; (c) also print $a$, $V_A$, $V_{bi}$, $W$, $\mathcal{E}$ at $x=0$, $V$ at $x=0$; (d) vary $a$ from $10^{18}$ to $10^{23}$ cm${}^{-4}$; (e) compare with Problem 5.7.

What is not recoverable. This is a programming task, so there is no figure data to read. The OCR gives no numerical $V_{A0}$ or specific $a$, so I choose $V_{A0}=-4$ V (reverse bias) and $a=10^{20}$ cm${}^{-4}$ for illustration. Eq. 5.48 is quoted from the standard linear-graded theory because its text did not survive.

Equations (depletion approximation, junction at $x=0$, $N_D-N_A=ax$).

Neutrality makes the region symmetric, $x_n=x_p=W/2$: $\rho (x)=qax,|x|\le W/2$ $\mathcal{E}(x)=-\frac{qa}{2K_S{\epsilon }_0}\left[{\left(\frac{W}{2}\right)}^2-x^2\right]$ $V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^{2}x}{4}-\frac{x^3}{3}+\frac{W^3}{12}\right]$ Check: $V(W/2)=\frac{qa}{2K_S{\epsilon }_0}\frac{W^3}{6}=\frac{qa{W}^3}{12K_S{\epsilon }_0}$, which must equal $V_{bi}-V_A$: $W={\left[\frac{12K_S{\epsilon }_0(V_{bi}-V_A)}{qa}\right]}^{1/3}$ Peak field and midpoint potential: $\mathcal{E}(0)=-\frac{qa{W}^2}{8K_S{\epsilon }_0},V(0)=\frac{V_{bi}-V_A}{2}$ $V_{bi}$ is not known in closed form, because the doping at the depletion edges, $N=aW/2$, depends on $W$, which depends on $V_{bi}$. Eq. 5.48 (as I recall it) reads $V_{bi}=\frac{2kT}{q}\ln \frac{a{W}_{eq}}{2n_i},W_{eq}={\left[\frac{12K_S{\epsilon }_0{V}_{bi}}{qa}\right]}^{1/3}$

Algorithm for $V_{bi}$ (fixed-point iteration).

1. Guess $V_{bi}^{(0)}=0.6$ V.
2. Compute $W^{(k)}$ from the $W$ formula with $V_A=0$ and $V_{bi}^{(k)}$.
3. Update $V_{bi}^{(k+1)}=\frac{2kT}{q}\ln \frac{a{W}^{(k)}}{2n_i}$.
4. Stop when $|V_{bi}^{(k+1)}-V_{bi}^{(k)}|<10^{-6}$ V.

It converges fast because $V_{bi}$ depends on $W$ only logarithmically and $W$ on $V_{bi}$ only as a cube root.

Sample MATLAB.

q = 1.602e-19;  e0 = 8.854e-14;  Ks = 11.8;   % C, F/cm
ni = 1e10;  kT = 0.0259;                       % cm^-3, V
a = 1e20;   VA0 = -4;                          % cm^-4, V
 
Vbi = 0.6;                                     % iterate Eq. 5.48
for k = 1:100
    Weq  = (12*Ks*e0*Vbi/(q*a))^(1/3);
    Vnew = 2*kT*log(a*Weq/(2*ni));
    if abs(Vnew-Vbi) < 1e-6, Vbi = Vnew; break, end
    Vbi = Vnew;
end
 
figure; subplot(2,1,1); hold on; subplot(2,1,2); hold on;
for n = 0:3
    VA = VA0/2^n;
    W  = (12*Ks*e0*(Vbi-VA)/(q*a))^(1/3);      % cm
    x  = linspace(-W/2, W/2, 400);
    E  = -q*a/(2*Ks*e0)*((W/2)^2 - x.^2);      % V/cm
    V  = q*a/(2*Ks*e0)*(W^2*x/4 - x.^3/3 + W^3/12);
    subplot(2,1,1); plot(x*1e4, E);
    subplot(2,1,2); plot(x*1e4, V);
    fprintf('a=%g VA=%g Vbi=%.4f W=%.4g E0=%.4g V0=%.4f\n', ...
        a, VA, Vbi, W, -q*a*W^2/(8*Ks*e0), (Vbi-VA)/2);
end
subplot(2,1,1); xlabel('x (um)'); ylabel('E (V/cm)');
subplot(2,1,2); xlabel('x (um)'); ylabel('V (V)');

Wrapping the body in for a = logspace(18,23,6) handles part (d).

Expected numbers (hand check, $a=10^{20}$ cm${}^{-4}$, $V_A=0$). $W^{(0)}={\left[\frac{12(11.8)(8.854\times 10^{-14})(0.6)}{(1.6\times 10^{-19})(10^{20})}\right]}^{1/3}\approx 7.8\times 10^{-5}\text{ cm}$ $V_{bi}^{(1)}=2(0.0259)\ln \frac{10^{20}(7.8\times 10^{-5})}{2\times 10^{10}}\approx 0.667\text{ V}$ A second pass gives $W\approx 8.2\times 10^{-5}$ cm (0.82 $\mu$m) and $V_{bi}\approx 0.67$ V, converged in 3 to 4 passes. Then $\mathcal{E}(0)\approx -1.3\times 10^4$ V/cm and $V(0)=V_{bi}/2\approx 0.33$ V.

Part (b) behavior. As $V_A$ goes $-4,-2,-1,-0.5$ V, $W\propto (V_{bi}-V_A{)}^{1/3}$ shrinks (roughly 1.7 $\mu$m down to 0.9 $\mu$m), and the peak field scales as $(V_{bi}-V_A{)}^{2/3}$. The $\mathcal{E}$ curves are nested inverted parabolas; the $V$ curves are nested S-shaped cubics, each rising from 0 to its own $V_{bi}-V_A$.

Part (d), range of $a$. $W\propto a^{-1/3}$, $|\mathcal{E}{|}_{max}\propto a{W}^2\propto a^{1/3}$, and $V_{bi}$ changes only logarithmically (roughly 0.5 V at $10^{18}$ to 0.9 V at $10^{23}$). Over five decades of $a$ the width drops by about $46\times$: a steep gradient gives a thin, high-field junction.

Part (e), comparison with the step junction of 5.7. The step junction has a triangular field (linear in $x$) and a parabolic $V$; the graded junction has a parabolic field and a cubic $V$. The graded junction’s width grows as $(V_{bi}-V_A{)}^{1/3}$ rather than ${}^{1/2}$, so it is less sensitive to bias, and its field is spread more evenly.

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Problem 5.14: The “exact” equilibrium step-junction solution

Problem. Without the depletion approximation, the equilibrium step junction can be solved in closed form using the $F$-function. (a) Plot $\mathcal{E}(x)$ and $V(x)$ for a sample doping and compare with Problem 5.7(a); (b) also plot $\rho /q$ and compare with the depletion approximation.

What is not recoverable. The OCR scrambled the formulas and the sample doping ($N_A$ prints as a garbled “$10^1$/cm${}^3$”; $N_D=2\times 10^{14}$ cm${}^{-3}$). I state the equations in the standard form consistent with the surviving fragments and use $N_A=10^{16}$ cm${}^{-3}$ only as a placeholder, not as the book’s value.

Setup. Define $U=qV/kT$, a length scale, and the Fermi levels: $L_I=\sqrt{\frac{K_S{\epsilon }_{0}kT}{2q^2{n}_i}},U_{FP}=\ln \frac{N_A}{n_i},U_{FN}=-\ln \frac{N_D}{n_i},U_{BI}=U_{FP}-U_{FN}=\frac{q{V}_{bi}}{kT}$ The $F$-function comes from integrating Poisson’s equation once with Boltzmann carrier statistics: $F(U,U_F)={\left[e^{U_F}\left(e^{-U}+U-1\right)+e^{-U_F}\left(e^U-U-1\right)\right]}^{1/2}$ A second integral gives position, and $F$ also gives the field:

p-side ($x\le 0$): $x=L_I{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime },U_{FP})},\mathcal{E}=-\frac{kT}{q{L}_I}F(U,U_{FP})$ n-side ($x\ge 0$): $x=L_I{\int }_{U_0}^U\frac{d{U}^{\prime }}{F(U^{\prime }-U_{BI},U_{FN})},\mathcal{E}=-\frac{kT}{q{L}_I}F(U-U_{BI},U_{FN})$ Charge density (p-side), exact Boltzmann form; the n-side is the same with $U_{FN}$ and $U-U_{BI}$: $\frac{\rho }{q}=n_i\left(e^{U_{FP}-U}-e^{U-U_{FP}}+e^{-U_{FP}}-e^{U_{FP}}\right)$ In plain terms: far from the junction $F\to 0$ (the field dies away), and $\rho$ counts ionized dopants plus mobile carriers exactly, so it has smooth tails instead of sharp edges.

Matching at $x=0$. The field is continuous there, so $F(U_0,U_{FP})=F(U_0-U_{BI},U_{FN})$ This is one transcendental equation for $U_0$; solve it with fzero starting at $U_0=U_{BI}/2$.

Sample MATLAB.

q=1.602e-19; e0=8.854e-14; Ks=11.8; ni=1e10; kT=0.0259;  % kT in volts
NA=1e16; ND=2e14;                       % NA is a placeholder
UFP=log(NA/ni);  UFN=-log(ND/ni);  UB=UFP-UFN;
LI=sqrt(Ks*e0*kT/(2*q*ni));             % cm
F=@(U,UF) sqrt( exp(UF).*(exp(-U)+U-1) + exp(-UF).*(exp(U)-U-1) );
 
U0 = fzero(@(U) F(U,UFP)-F(U-UB,UFN), UB/2);
 
Up = linspace(0.1, U0, 300);            % p-side, avoid F=0 at U=0
xp = arrayfun(@(u) LI*integral(@(s) 1./F(s,UFP), U0, u), Up);
Ep = -(kT/LI)*F(Up,UFP);
 
Un = linspace(U0, UB-0.1, 300);         % n-side
xn = arrayfun(@(u) LI*integral(@(s) 1./F(s-UB,UFN), U0, u), Un);
En = -(kT/LI)*F(Un-UB,UFN);
 
x = [xp xn];  E = [Ep En];  V = kT*[Up Un];
subplot(2,1,1); plot(x*1e4,E); ylabel('E (V/cm)');
subplot(2,1,2); plot(x*1e4,V); ylabel('V (V)'); xlabel('x (um)');
 
rp = ni*(exp(UFP-Up)-exp(Up-UFP)+exp(-UFP)-exp(UFP));   % rho/q, p-side
rn = ni*(exp(UFN-(Un-UB))-exp((Un-UB)-UFN)+exp(-UFN)-exp(UFN));

The integrand blows up where $F\to 0$ (at $U=0$ and $U=U_{BI}$), hence the instruction to start at $0.1$ rather than $0$. The far tails are cut off, which is fine for plotting.

Expected behavior and comparison with 5.7(a).

• $\mathcal{E}(x)$: a smooth, rounded triangular spike with its minimum at $x=0$ and exponential tails on both sides, instead of the abrupt zero at $-x_p$ and $x_n$ of the depletion approximation.
• $V(x)$: a smooth S-curve from $0$ to $V_{bi}$, the same total rise as in 5.7(a), with the corners rounded.
• Asymmetry: with $N_A\gg N_D$ nearly all the depletion layer lies on the n-side, $x_n\gg x_p$, in both solutions. Peak field and $V_{bi}=\frac{kT}{q}\ln (N_A{N}_D/n_i^2)$ agree to within a few percent.
• $\rho /q$ (part b): the depletion approximation draws two rectangles ($-N_A$ and $+N_D$) with vertical edges. The exact $\rho /q$ approaches those heights but rounds off over a few Debye lengths, $L_D=\sqrt{K_S{\epsilon }_{0}kT/(q^{2}N)}$, at each edge. The depletion approximation is excellent when the junction is many Debye lengths wide ($V_{bi}\gg kT/q$) and worst near the edges and on the lightly doped side.