part05b

5.2.5 Linearly Graded Junctions

Step junctions are the “straight-line” idealization. But real junctions made by a deep diffusion into a moderately or heavily doped wafer look very different: the doping does not jump, it slides smoothly from p-type to n-type. Picture a ramp, not a cliff. The linearly graded junction models that ramp (Fig. 5.13a) as

N_D - N_A = a\,x \tag{5.40}

where $a$ (units ${\text{cm}}^{-4}$) is the grading constant. The metallurgical junction sits at $x=0$, where the net doping is exactly zero.

Treat this subsection as an unnumbered exercise: a chance to see the procedure you learned for the step junction applied to a new profile. A few things make it pleasantly different.

1. The profile is continuous through $x=0$. So there is no need to solve the p-side and n-side separately and then stitch them together at $x=0$. There is one $\rho$, one $\mathcal{E}$, one $V$ for the whole depletion region.
2. The profile is symmetric about $x=0$ (it is an odd function), so the depletion region extends equally far to both sides, $x_n=x_p=W/2$, and all electrostatic variables inherit a symmetry.
3. A complication we choose to ignore. The doping outside the depletion region is nonuniform too, so strictly there is a small residual charge, field, and potential drop out there. We ignore this and take $\rho =0$, $\mathcal{E}=0$, $V=\text{constant}$ outside the depletion region.
4. $V_{bi}$ cannot be taken from the step-junction formula (5.10). It still has to be recomputed, and this is done at the end.

Analogy. In a step junction the two sides are like two countries with a sharp border wall. In a linearly graded junction it is a mountain slope: the “no-man’s-land” is the whole gentle gradient, and the point where it crosses zero is the middle of the pass.

Depletion approximation applied

With (5.40) and the depletion approximation, the charge density is just $q$ times the net doping inside the depletion region:

\rho(x) = \begin{cases} q\,a\,x, & -W/2 \le x \le W/2 \\[2pt] 0, & x \le -W/2 \ \text{and}\ x \ge W/2\end{cases} \tag{5.41}

Because $\rho$ is odd in $x$, the net charge is zero only if the two edges are symmetric:

x_n = x_p = \frac{W}{2} \tag{5.42}

Thought experiment. The positive charge on the right ($qax$ for $x>0$) and the negative charge on the left must cancel exactly (neutrality). Because the ramp is a straight line through the origin, the only way two triangles can have equal area is for them to have equal base. That is the whole argument for (5.42).

Electric field

Poisson’s equation, $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$, with the boundary condition $\mathcal{E}(-W/2)=0$ (integrate from the p-edge):

$\mathcal{E}(x)=\frac{qa}{K_S{\epsilon }_0}{\left[\frac{x^{\prime 2}}{2}\right]}_{-W/2}^x=\frac{qa}{2K_S{\epsilon }_0}\left(x^2-\frac{W^2}{4}\right)$

\boxed{\mathcal{E}(x) = -\frac{qa}{2K_S\varepsilon_0}\left(\frac{W^2}{4} - x^2\right), \qquad -\frac{W}{2}\le x\le\frac{W}{2}} \tag{5.43}

A linear $\rho$ gives a parabolic $\mathcal{E}$. The field is negative everywhere (pointing from n toward p, as always), zero at both edges, and most negative at the center:

$|\mathcal{E}{|}_{max}=|\mathcal{E}(0)|=\frac{qa{W}^2}{8K_S{\epsilon }_0}$

Electrostatic potential

Use $\mathcal{E}=-dV/dx$ and integrate from the p-edge, where we set $V=0$:

$V(x)=-{\int }_{-W/2}^x\mathcal{E}d{x}^{\prime }=\frac{qa}{2K_S{\epsilon }_0}{\int }_{-W/2}^x\left(\frac{W^2}{4}-x^{\prime 2}\right)d{x}^{\prime }$

$V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^2}{4}\left(x+\frac{W}{2}\right)-\frac{1}{3}\left(x^3+\frac{W^3}{8}\right)\right]$

$V(x)=\frac{qa}{2K_S{\epsilon }_0}\left[\frac{W^{2}x}{4}-\frac{x^3}{3}+\frac{W^3}{8}-\frac{W^3}{24}\right]$

\boxed{V(x) = \frac{qa}{6K_S\varepsilon_0}\left(\frac{3W^2}{4}\,x - x^3 + \frac{W^3}{4}\right), \qquad -\frac W2\le x\le\frac W2} \tag{5.44}

A parabolic field gives a cubic potential. Sanity check at the n-edge, $x=W/2$:

$V\left(\frac{W}{2}\right)=\frac{qa}{6K_S{\epsilon }_0}\left(\frac{3W^3}{8}-\frac{W^3}{8}+\frac{W^3}{4}\right)=\frac{qa}{6K_S{\epsilon }_0}\cdot \frac{W^3}{2}=\frac{qa{W}^3}{12K_S{\epsilon }_0}$

Depletion width

The total potential drop across the depletion region must be $V_{bi}-V_A$. Setting $V(W/2)=V_{bi}-V_A$:

$\frac{qa{W}^3}{12K_S{\epsilon }_0}=V_{bi}-V_A$

\boxed{W = \left[\frac{12K_S\varepsilon_0}{qa}\,(V_{bi}-V_A)\right]^{1/3}} \tag{5.45}

Remember this contrast. The linearly graded depletion width varies as the cube root of $V_{bi}-V_A$, whereas the step-junction width varies as the square root. Why? In a step junction $\rho$ is uniform, so exposed charge grows linearly with $W$ and $V\propto W^2$. In a graded junction the charge density is zero at the center and grows outward, so the exposed charge grows like $W^2$ and the potential like $W^3$.

The built-in voltage of a linearly graded junction

Equation (5.10) was derived for a step junction. But the derivation of (5.8) placed no restriction on the profile except nondegeneracy, so we just re-evaluate the carrier concentrations at the two edges of the equilibrium depletion region ($W_0=W{|}_{V_A=0}$). At the edges the net doping magnitude is $a{W}_0/2$, so

n(x_n) = (N_D - N_A)\big|_{x=W_0/2} = \frac{aW_0}{2} \tag{5.46a}

p(-x_p) = -(N_D-N_A)\big|_{x=-W_0/2} = \frac{aW_0}{2} \tag{5.46b}

Substituting into (5.8), $V_{bi}=(kT/q)\ln [n(x_n)p(-x_p)/n_i^2]$:

$V_{bi}=\frac{kT}{q}\ln \left[\frac{(a{W}_0/2)(a{W}_0/2)}{n_i^2}\right]$

\boxed{V_{bi} = \frac{2kT}{q}\ln\!\left(\frac{aW_0}{2n_i}\right)} \tag{5.47}

Now insert $W_0$ from (5.45) with $V_A=0$:

\boxed{V_{bi} = \frac{2kT}{q}\ln\!\left\{\frac{a}{2n_i}\left[\frac{12K_S\varepsilon_0V_{bi}}{qa}\right]^{1/3}\right\}} \tag{5.48}

$V_{bi}$ appears on both sides. Equation (5.48) cannot be solved explicitly and must be iterated numerically: guess $V_{bi}$, compute the right-hand side, feed the result back in, and repeat. Because the dependence on $V_{bi}$ is only through the logarithm of a cube root, the iteration converges in two or three passes. (Problem 5.13 does exactly this.)

Summary table, step versus linearly graded.

Quantity	Step junction	Linearly graded
$\rho (x)$	constant on each side	linear in $x$
$\mathcal{E}(x)$	piecewise linear (triangle)	parabolic
$V(x)$	piecewise parabolic	cubic
$W\propto$	$(V_{bi}-V_A{)}^{1/2}$	$(V_{bi}-V_A{)}^{1/3}$
$V_{bi}$	closed form (5.10)	iterate (5.48)

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5.3 Summary

Take a breath and look back at how far we have come. The electrostatics of the pn junction is the foundation for the operational model of the diode and of every other device containing a pn junction.

What we did, in order.

1. Vocabulary. We defined the metallurgical junction and doping profile, and introduced the idealized step and linearly graded profiles used throughout the analysis.
2. Poisson’s equation, the usual starting point for quantitative solutions of $\rho$, $\mathcal{E}$, $V$.
3. A qualitative solution read directly from the energy band diagram: bands bend, a potential hill forms, and a space charge region appears.
4. The built-in voltage, $V_{bi}=(kT/q)\ln (N_A{N}_D/n_i^2)$ for a step junction, the height of the potential hill.
5. The depletion approximation, the most important and most widely used simplification in device modeling: the transition region is completely empty of mobile carriers and everything beyond it is perfectly neutral.
6. Quantitative solutions for $\rho$, $\mathcal{E}$, $V$ in a step junction at equilibrium, then extended to $V_A\ne 0$ by replacing $V_{bi}\to V_{bi}-V_A$.
7. Effect of bias on the electrostatic variables and the resulting band diagrams: forward bias shrinks the hill and the depletion region, reverse bias grows both.
8. A linearly graded junction, which showed that the same procedure gives parabolic fields and cubic potentials.

What you should now be able to do. Feel (not just compute) what is happening electrostatically inside a pn junction, and, given any other doping profile, set up and carry out the solution: write $\rho$ from the depletion approximation, integrate Poisson once for $\mathcal{E}$, again for $V$, and force the total potential drop to equal $V_{bi}-V_A$.

Encouragement. The ritual ”$\rho \to \mathcal{E}\to V\to$ match $V_{bi}-V_A$” is one of the most reusable ideas in this course. You will meet it again in the MOS capacitor, the MOSFET, and the Schottky diode. Master it here and later chapters will feel like old friends.

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Chapter 5 Problems

A note on the source. The OCR of the problem statements is damaged in a few places (garbled exponents, a missing figure label). Wherever a value or a figure could not be read, the reconstruction I used is stated explicitly at the start of that solution, chosen to be consistent with the problem-information table (difficulty level, topic, and “NA ~ ND” or “NA $\ge$ ND” descriptions). The method is the point; if your textbook shows different numbers, just substitute them into the same chain of steps.

Constants used throughout (Si, 300 K). $kT/q=0.0259$ V, $n_i=10^{10}{\text{cm}}^{-3}$, $K_S=11.8$, ${\epsilon }_0=8.85\times 10^{-14}$ F/cm so $K_S{\epsilon }_0=1.044\times 10^{-12}$ F/cm, $q=1.6\times 10^{-19}$ C, $E_G=1.12$ eV.

Problem information table (from the book).

Problem	After Section	Level	Description
5.1	5.2.5	1	True-or-false quiz
5.2	5.1.4	1	$V_{bi}$ from energy band
5.3	5.1.4	2	Isotype step junction
5.4	5.2.2	2	Step junction, $N_A\sim N_D$
5.5	5.2.2	2	Step junction, $N_A\ge N_D$
5.6	5.2.3	2	Compute using diary
5.7	5.2.4	3	Step junction program
5.8	5.2.5	2	Combination profile
5.9	5.2.5	2	Exponential profile
5.10	5.2.5	2-3	Modified-step profile
5.11	5.2.5	3-4	PIN diode
5.12	5.2.5	2-3	Given $V$, find $\mathcal{E}$, $\rho$, $N_D-N_A$
5.13	5.2.5	3-4	Linearly graded program
5.14	5.2.5	5	Exact solution

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Problem 5.1: True or false

Problem. Decide whether each statement is true or false (and, as a good habit, say why).

Solution. One statement at a time.

(a) The step junction is an idealized doping profile used to model $p^{+}$-n and $n^{+}$-p junctions. $\text{Heavy doping on one side and light on the other, with an abrupt switch, is exactly a step profile.}$ TRUE.

(b) The $\rho$ in Poisson’s equation is the charge density, with units of C/cm${}^3$. $[\rho ]=\frac{\text{C}}{{\text{cm}}^3},\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}\Rightarrow \frac{\text{V/cm}}{\text{cm}}=\frac{{\text{C/cm}}^3}{\text{F/cm}}✓$ TRUE.

(c) The space charge region about the metallurgical junction is due to a pile-up of electrons on the p-side and holes on the n-side. $\text{Actually, the carriers diffuse across and recombine/are swept away, leaving behind uncovered ionized dopants }(N_A^{-},N_D^{+}).$ Mobile carriers are depleted, not piled up. FALSE.

(d) The built-in potential is typically less than the band gap energy converted to volts. $q{V}_{bi}=kT\ln \frac{N_A{N}_D}{n_i^2}<E_G\text{(it approaches }{E}_G\text{ only for very heavy doping)}$ For Si, $V_{bi}\sim 0.6$-$0.9$ V $<1.12$ V. TRUE.

(e) Invoking the depletion approximation makes the charge density inside the depletion region directly proportional to the net doping concentration. $\rho \approx q(N_D-N_A)(p=n=0)$ TRUE.

(f) Ohmic contacts reduce the built-in voltage drop across a junction. $\text{Contact potentials at the two metal-semiconductor contacts exactly cancel the difference, so the total equilibrium voltage around the loop is zero; the voltage across the junction itself is unchanged.}$ FALSE.

(g) In solutions based on the depletion approximation, the magnitude of the electric field reaches a maximum right at the metallurgical boundary. $\mathcal{E}(x)\text{ is a triangle that peaks where }\rho \text{ changes sign, i.e., at }x=0.$ TRUE.

(h) If one has a $p^{+}$-n step junction, with $N_A\gg N_D$, it follows that $x_p\le x_n$. $N_A{x}_p=N_D{x}_n\Rightarrow x_p=\frac{N_D}{N_A}{x}_n\le x_n$ TRUE.

(i) The potential hill between the n-side and the p-side of a junction increases with forward biasing. $\text{Hill height}=V_{bi}-V_A\text{ decreases for }{V}_A>0.$ FALSE.

(j) The depletion width in a linearly graded junction varies as the square root of $V_{bi}-V_A$. (The OCR of this item is garbled; this is the statement it matches, consistent with the chapter’s emphasis.) $W={\left[\frac{12K_S{\epsilon }_0}{qa}(V_{bi}-V_A)\right]}^{1/3}$ It varies as the cube root. FALSE.

Answer key. (a) T, (b) T, (c) F, (d) T, (e) T, (f) F, (g) T, (h) T, (i) F, (j) F.

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Problem 5.2: $V_{bi}$ from the energy band diagram

Problem. (OCR-damaged; reconstruction.) A pn junction has its Fermi level at $E_F=E_v+2kT$ on the p-side and $E_F=E_c-kT/4$ on the n-side. (a) Draw the equilibrium energy band diagram. (b) Determine the built-in voltage, symbolically and numerically.

Assumption note. The garbled text reads ”$E_F=E-2kT$ on the p-side and $E_F=E-/4$ on the n-side.” I interpret it as above, with the reference levels being the band edges on each side. Note that a Fermi level only $kT/4$ below $E_c$ means the n-side is just on the edge of degeneracy, so we are using the energy-band route (5.12)-type argument, not the doping formula (5.10).

Solution (a): the diagram. Equilibrium means one flat $E_F$ through the whole structure. Draw:

1. A horizontal line for $E_F$.
2. Far on the p-side, put $E_v$ a distance $2kT$ below $E_F$ ($E_F$ sits $2kT$ above $E_v$), and $E_c=E_v+E_G$ above it.
3. Far on the n-side, put $E_c$ a distance $kT/4$ above $E_F$, and $E_v=E_c-E_G$ below.
4. Join the p-side bands to the n-side bands by a smooth S-curve across the depletion region. The n-side bands lie lower in energy than the p-side bands.

Solution (b): the built-in voltage. The total band bending equals the difference between the band-edge energies far from the junction. Everything is measured from the common Fermi level $E_F$.

$E_c^{p\text{-side}}-E_F=(E_v+E_G)-E_F=E_G-2kT$

$E_c^{n\text{-side}}-E_F=\frac{kT}{4}$

The electron potential energy drop (bending) is the difference of the conduction-band edge on the two sides:

$q{V}_{bi}=E_c^{p\text{-side}}-E_c^{n\text{-side}}=\left(E_G-2kT\right)-\frac{kT}{4}$

$\boxed{V_{bi}=\frac{E_G}{q}-\frac{9kT}{4q}}$

Numerically, for Si ($E_G/q=1.12$ V, $kT/q=0.0259$ V):

$V_{bi}=1.12-2.25(0.0259)=1.12-0.0583=1.062\text{V}$

Check. $V_{bi}<E_G/q$, as it must be (consistent with Problem 5.1d).

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Problem 5.3: The $p_1$-$p_2$ “isotype” step junction

Problem. Consider the p${}_1$-p${}_2$ isotype step junction of Fig. P5.3: $N_A(x)=N_{A1}$ for $x<0$ and $N_{A2}$ for $x>0$. (a) Draw the equilibrium band diagram taking the doping nondegenerate and $N_{A1}>N_{A2}$. (b) Derive an expression for the built-in voltage. (c) Sketch potential, field and charge density. (d) Briefly describe the depletion approximation. (e) Can it be used here?

Solution (a): band diagram. Both sides are p-type, so $E_F$ lies below $E_i$ on both. The more heavily doped side ($p_1$) has $E_F$ closer to $E_v$ than the lighter side ($p_2$). With one flat $E_F$, the $p_1$ bands must therefore sit higher in energy than the $p_2$ bands (we need $E_F-E_v$ to be small on side 1). So the bands bend downward as we go from $p_1$ (left) to $p_2$ (right), by an amount equal to the difference of the Fermi-level positions. No step in the picture: it is a smooth transition.

Solution (b): built-in voltage. Use the equilibrium Fermi-level-from-$E_i$ relation, $E_i-E_F=kT\ln (p_0/n_i)$, on each side with $p_0=N_A$:

${\left(E_i-E_F\right)}_1=kT\ln \frac{N_{A1}}{n_i},{\left(E_i-E_F\right)}_2=kT\ln \frac{N_{A2}}{n_i}$

The band bending is the difference of $E_i$ between the sides (Fermi level is flat):

$q{V}_{bi}=E_{i,1}-E_{i,2}={\left(E_i-E_F\right)}_1-{\left(E_i-E_F\right)}_2=kT\ln \frac{N_{A1}}{n_i}-kT\ln \frac{N_{A2}}{n_i}$

$\boxed{V_{bi}=\frac{kT}{q}\ln \frac{N_{A1}}{N_{A2}}}$

Alternative check: at equilibrium the hole drift and diffusion currents cancel, so the hole populations at the two far edges are Boltzmann-related, $p_1/p_2=N_{A1}/N_{A2}=e^{q{V}_{bi}/kT}$, which gives the same result.

Solution (c): sketches.

Charge density. Holes diffuse from the heavily doped $p_1$ side (many holes) to the lightly doped $p_2$ side. Therefore:

• On the $p_1$ side near $x=0$: holes left behind, uncovered acceptors, so $\rho <0$ (a thin layer of negative charge, depleted of holes).
• On the $p_2$ side near $x=0$: extra holes arrive on top of the fixed $N_{A2}^{-}$, so $\rho >0$ (a layer of positive charge, hole accumulation).

So $\rho (x)$ is a “dipole”: negative lump left of $x=0$, positive lump right of $x=0$, total zero.

Electric field. From $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$, $\mathcal{E}$ starts at zero, goes negative through the negative-$\rho$ layer, reaches its most negative value near $x=0$, and returns to zero through the positive-$\rho$ layer. This is a single negative bump.

Potential. $\mathcal{E}=-dV/dx$ with $\mathcal{E}<0$ means $V$ rises monotonically from $0$ (far left) to $V_{bi}$ (far right) in an S-shape.

Solution (d): the depletion approximation. Assume that (i) the transition region has carrier concentrations so small that the mobile-carrier charge is negligible, $p\approx n\approx 0$, so $\rho =q(N_D-N_A)$ inside; (ii) outside, the material is exactly neutral, $\rho =0$; and (iii) the boundary between the two is abrupt.

Solution (e): can we use it here? No. The depletion approximation assumes the space charge is made of uncovered dopant ions with the mobile carriers swept away. In the $p_2$ half, the space charge is excess holes (accumulation), which are mobile carriers, so the "$p\approx 0$" assumption is violated. On the $p_1$ side there is a partial hole depletion, but the hole concentration there never falls near zero, because both sides are p-type and the potential hill is only $\frac{kT}{q}\ln (N_{A1}/N_{A2})$, typically a few $kT/q$. A full solution requires Poisson’s equation with the Boltzmann hole distribution (the type of “exact” solution in Problem 5.14).

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Problem 5.4: Step junction at equilibrium, $N_A\sim N_D$

Problem. A Si step junction at room temperature, in equilibrium, has $N_A$ (p-side) and $N_D$ (n-side) doping. Compute (a) $V_{bi}$; (b) $x_n$, $x_p$ and $W$; (c) $\mathcal{E}$ at $x=0$; (d) $V$ at $x=0$; (e) sketch $\rho$, $\mathcal{E}$, $V$.

Assumption note. The OCR gave ”$N_A=2\times 10^1/{\text{cm}}^3$ and $N_D=10^1/{\text{cm}}^3$” with the last digit of each exponent lost. The problem table says $N_A\sim N_D$, so I use $N_A=2\times 10^{15}{\text{cm}}^{-3}$ and $N_D=10^{15}{\text{cm}}^{-3}$.

Solution (a): built-in voltage.

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}\text{[eq. (5.10)]}$

$V_{bi}=0.0259\ln \frac{(2\times 10^{15})(10^{15})}{(10^{10}{)}^2}$

$V_{bi}=0.0259\ln \left(2\times 10^{10}\right)\text{[since }2\times 10^{30}/10^{20}\text{]}$

$\ln \left(2\times 10^{10}\right)=\ln 2+10\ln 10=0.693+23.026=23.719$

$\boxed{V_{bi}=0.0259\times 23.719=0.614\text{V}}$

Solution (b): depletion widths. First the total width from (5.31):

$W={\left[\frac{2K_S{\epsilon }_0}{q}{V}_{bi}\frac{N_A+N_D}{N_A{N}_D}\right]}^{1/2}$

The doping factor:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{3\times 10^{15}}{2\times 10^{30}}=1.5\times 10^{-15}{\text{cm}}^3$

The prefactor:

$\frac{2K_S{\epsilon }_0{V}_{bi}}{q}=\frac{2(1.044\times 10^{-12})(0.614)}{1.6\times 10^{-19}}=\frac{1.282\times 10^{-12}}{1.6\times 10^{-19}}=8.02\times 10^6{\text{cm}}^{-1}$

Multiply and take the root:

$W={\left[(8.02\times 10^6)(1.5\times 10^{-15})\right]}^{1/2}={\left[1.203\times 10^{-8}\right]}^{1/2}$

$\boxed{W=1.10\times 10^{-4}\text{cm}=1.10\mu \text{m}}$

Split using neutrality $N_A{x}_p=N_D{x}_n$ with $x_n+x_p=W$:

$x_n=W\frac{N_A}{N_A+N_D}=1.097\times 10^{-4}\times \frac{2}{3}$

$\boxed{x_n=7.31\times 10^{-5}\text{cm}=0.731\mu \text{m}}$

$x_p=W-x_n=3.66\times 10^{-5}\text{cm}\Rightarrow \boxed{x_p=0.366\mu \text{m}}$

Check: $N_A{x}_p=2\times 10^{15}(3.66\times 10^{-5})=7.31\times 10^{10}$ and $N_D{x}_n=10^{15}(7.31\times 10^{-5})=7.31\times 10^{10}$. These match.

Solution (c): the field at $x=0$. The field at the metallurgical junction is the peak field, from the n-side expression (5.21) at $x=0$:

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(10^{15})(7.31\times 10^{-5})}{1.044\times 10^{-12}}$

$\mathcal{E}(0)=-\frac{1.170\times 10^{-8}}{1.044\times 10^{-12}}$

$\boxed{\mathcal{E}(0)=-1.12\times 10^4\text{V/cm}}$

(The p-side formula gives the same, $-q{N}_A{x}_p/K_S{\epsilon }_0$, as it must by continuity.)

Solution (d): the potential at $x=0$. From (5.26) at $x=0$ (p-side, with $V=0$ at the p-edge):

$V(0)=\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=\frac{(1.6\times 10^{-19})(2\times 10^{15})}{2(1.044\times 10^{-12})}{\left(3.66\times 10^{-5}\right)}^2$

$V(0)=\frac{3.2\times 10^{-4}}{2.088\times 10^{-12}}\left(1.339\times 10^{-9}\right)=(1.532\times 10^8)(1.339\times 10^{-9})$

$\boxed{V(0)=0.205\text{V}}$

Check. The potential drop on each side is in the ratio of the doping fractions: $V(0)=V_{bi}{N}_D/(N_A+N_D)=0.614/3=0.205$ V.

Solution (e): sketches.

• $\rho (x)$: $-q{N}_A=-3.2\times 10^{-4}$ C/cm${}^3$ for $-0.366\le x\le 0$ $\mu$m; $+q{N}_D=+1.6\times 10^{-4}$ C/cm${}^3$ for $0\le x\le 0.731$ $\mu$m; zero elsewhere. The negative block is taller and narrower (twice as tall, half as wide) than the positive block, so the two areas are equal.
• $\mathcal{E}(x)$: zero outside; a straight line falling from 0 at $x=-0.366\mu$m to $-1.12\times 10^4$ V/cm at $x=0$, then a (shallower-slope) straight line rising back to 0 at $x=0.731\mu$m. A skewed triangle with the sharp peak at $x=0$.
• $V(x)$: 0 for $x<-0.366\mu$m; a parabola rising to $0.205$ V at $x=0$ (zero slope at the p-edge); another parabola (concave down) continuing to $0.614$ V at $x=0.731\mu$m, then constant. Slope at $x=0$ is continuous.

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Problem 5.5: Step junction with $N_A\gg N_D$

Problem. Repeat Problem 5.4 taking a larger p-side doping, and compare.

Assumption note. The OCR shows "$N_A=10^7/{\text{cm}}^3$" (an exponent digit lost). The table says $N_A\ge N_D$ and the n-side stays as before, so I use $N_A=10^{17}{\text{cm}}^{-3}$ and $N_D=10^{15}{\text{cm}}^{-3}$.

Solution (a): built-in voltage.

$V_{bi}=0.0259\ln \frac{(10^{17})(10^{15})}{10^{20}}=0.0259\ln \left(10^{12}\right)$

$\ln 10^{12}=12(2.3026)=27.631$

$\boxed{V_{bi}=0.0259\times 27.631=0.716\text{V}}$

Solution (b): widths. Doping factor:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{1.01\times 10^{17}}{10^{32}}=1.01\times 10^{-15}{\text{cm}}^3$

Prefactor:

$\frac{2K_S{\epsilon }_0{V}_{bi}}{q}=\frac{2(1.044\times 10^{-12})(0.716)}{1.6\times 10^{-19}}=\frac{1.495\times 10^{-12}}{1.6\times 10^{-19}}=9.34\times 10^6{\text{cm}}^{-1}$

$W={\left[(9.34\times 10^6)(1.01\times 10^{-15})\right]}^{1/2}={\left[9.43\times 10^{-9}\right]}^{1/2}$

$\boxed{W=9.71\times 10^{-5}\text{cm}=0.971\mu \text{m}}$

$x_n=W\frac{N_A}{N_A+N_D}=9.71\times 10^{-5}\times \frac{10^{17}}{1.01\times 10^{17}}=9.62\times 10^{-5}\text{cm}$

$\boxed{x_n=0.962\mu \text{m}}$

$x_p=\frac{N_D}{N_A}{x}_n=10^{-2}(9.62\times 10^{-5})\Rightarrow \boxed{x_p=9.6\times 10^{-7}\text{cm}=9.6\text{nm}}$

Solution (c): field at $x=0$.

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(10^{15})(9.62\times 10^{-5})}{1.044\times 10^{-12}}=-\frac{1.539\times 10^{-8}}{1.044\times 10^{-12}}$

$\boxed{\mathcal{E}(0)=-1.47\times 10^4\text{V/cm}}$

Solution (d): potential at $x=0$.

$V(0)=V_{bi}\frac{N_D}{N_A+N_D}=0.716\times \frac{10^{15}}{1.01\times 10^{17}}=0.716\times 9.9\times 10^{-3}$

$\boxed{V(0)=7.1\times 10^{-3}\text{V}\approx 7\text{mV}}$

(Direct check: $\frac{q{N}_A}{2K_S{\epsilon }_0}{x}_p^2=\frac{1.6\times 10^{-2}}{2.088\times 10^{-12}}(9.2\times 10^{-13})=7.1\times 10^{-3}$ V. Same.)

Solution (e): sketches. The same shapes as in 5.4, but extremely lopsided: $\rho$ has a very tall, very thin negative spike ($-q{N}_A=-1.6\times 10^{-2}$ C/cm${}^3$, 10 nm wide) next to a low, wide positive block ($+1.6\times 10^{-4}$ C/cm${}^3$, 0.96 $\mu$m wide). $\mathcal{E}$ is a triangle with an almost vertical left side. $V$ is essentially flat at 0 until $x\approx 0$, then climbs almost the whole $0.716$ V on the n-side.

Comparison with Problem 5.4.

Quantity	5.4 ($N_A=2\times 10^{15}$)	5.5 ($N_A=10^{17}$)
$V_{bi}$ (V)	0.614	0.716
$W$ ($\mu$m)	1.10	0.971
$x_n$ ($\mu$m)	0.731	0.962
$x_p$ ($\mu$m)	0.366	0.0096
$	\mathcal{E}(0)	$ (V/cm)
$V(0)$ (V)	0.205	0.007

Making one side 50 times more heavily doped raised $V_{bi}$ only slightly (a logarithm), nearly eliminated the p-side depletion region, and pushed almost the entire junction into the lightly doped n-side. The lightly doped side controls everything: $W\approx x_n\approx \sqrt{2K_S{\epsilon }_0{V}_{bi}/(q{N}_D)}$.

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Problem 5.6: Computation using the MATLAB diary function

Problem. A Si step junction at room temperature has instructor-supplied $N_A$, $N_D$, $V_A$. Working in the Command window, with the MATLAB diary function recording the session, compute (a) $V_{bi}$; (b) $x_n$, $x_p$, $W$; (c) $\mathcal{E}$ at $x=0$; (d) $V$ at $x=0$.

Assumption note. The three values are supplied by the instructor. As a worked illustration I use $N_A=10^{16}{\text{cm}}^{-3}$, $N_D=10^{15}{\text{cm}}^{-3}$, $V_A=-2$ V (reverse bias), and verify the numbers by hand. Replace these three lines for your own assignment.

Solution: the session. The sequence of Command-window entries (each entry is a step; explanations in comments):

diary prob56.txt          % start recording the session
kT = 0.0259; q = 1.6e-19; ni = 1e10;
KSe0 = 11.8*8.85e-14;     % K_S * eps_0 (F/cm)
NA = 1e16; ND = 1e15; VA = -2;
Vbi = kT*log(NA*ND/ni^2)                      % (a) eq. (5.10)
W   = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND)) % eq. (5.38)
xn  = W*NA/(NA+ND)                            % eq. (5.37)
xp  = W*ND/(NA+ND)                            % eq. (5.34)
E0  = -q*ND*xn/KSe0                           % (c) eq. (5.35) at x = 0
V0  = q*NA*xp^2/(2*KSe0)                      % (d) eq. (5.33) at x = 0
diary off                 % stop recording

Hand check of each output.

(a) $V_{bi}=0.0259\ln \frac{(10^{16})(10^{15})}{10^{20}}=0.0259\ln \left(10^{11}\right)=0.0259(25.328)=0.656\text{V}$

(b) $V_{bi}-V_A=0.656+2=2.656\text{V}$

$W={\left[\frac{2(1.044\times 10^{-12})(2.656)}{1.6\times 10^{-19}}\cdot \frac{1.1\times 10^{16}}{10^{31}}\right]}^{1/2}={\left[(3.466\times 10^7)(1.1\times 10^{-15})\right]}^{1/2}={\left[3.81\times 10^{-8}\right]}^{1/2}$

$W=1.95\times 10^{-4}\text{cm}=1.95\mu \text{m}$

$x_n=W\frac{N_A}{N_A+N_D}=1.95\times 10^{-4}\times \frac{10}{11}=1.78\times 10^{-4}\text{cm}=1.78\mu \text{m}$

$x_p=W-x_n=1.8\times 10^{-5}\text{cm}=0.178\mu \text{m}$

(c) $\mathcal{E}(0)=-\frac{(1.6\times 10^{-19})(10^{15})(1.78\times 10^{-4})}{1.044\times 10^{-12}}=-2.72\times 10^4\text{V/cm}$

(d) $V(0)=\frac{(1.6\times 10^{-19})(10^{16})}{2(1.044\times 10^{-12})}(1.78\times 10^{-5}{)}^2=(7.66\times 10^8)(3.17\times 10^{-10})=0.243\text{V}$

(Check: $(V_{bi}-V_A)N_D/(N_A+N_D)=2.656/11=0.241$ V, agreeing to rounding.)

Why diary? It leaves a text log you can hand in: every command and every numeric answer, so the grader can follow your steps. Reverse bias ($V_A<0$) widens $W$ (here 1.95 $\mu$m compared with 1.2 $\mu$m at equilibrium) and raises the peak field, as Section 5.2.4 predicted.

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Problem 5.7: A step-junction program

Problem. For a nondegenerately doped Si pn step junction at $T=300$ K: (a) compute and present coordinated plots of $\mathcal{E}(x)$ and $V(x)$ using, as a sample computation, $N_A$ (value lost in OCR), $N_D=2\times 10^{14}{\text{cm}}^{-3}$ and $V_A=-20$ V; (b) modify the program to show several $V_A$ values, $V_A=V_{A0}/2^n$, $n=0,\dots ,3$; (c) add an output list of parameters; (d) explore how the plots change with relative doping; (e) check against Problems 5.4 and 5.5.

Assumption note. The OCR lost the exponent of $N_A$. I use $N_A=10^{16}{\text{cm}}^{-3}$ for the sample run.

The plan (before any code)

Everything is already derived. The program only has to evaluate the closed-form results piece by piece:

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2},W={\left[\frac{2K_S{\epsilon }_0}{q}(V_{bi}-V_A)\frac{N_A+N_D}{N_A{N}_D}\right]}^{1/2}$

$x_n=W\frac{N_A}{N_A+N_D},x_p=W\frac{N_D}{N_A+N_D}$

$\mathcal{E}(x)=\{\begin{array}{ll}-\frac{q{N}_A}{K_S{\epsilon }_0}(x_p+x),& -x_p\le x<0\\ -\frac{q{N}_D}{K_S{\epsilon }_0}(x_n-x),& 0\le x\le x_n\\ 0& \text{otherwise}\end{array}$

$V(x)=\{\begin{array}{ll}0,& x<-x_p\\ \frac{q{N}_A}{2K_S{\epsilon }_0}(x_p+x{)}^2,& -x_p\le x<0\\ V_{bi}-V_A-\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2,& 0\le x\le x_n\\ V_{bi}-V_A,& x>x_n\end{array}$

The suggested plot ranges: $x\in [-x_{max},x_{max}]$ with $x_{max}=2.5\max (x_n,x_p)$; $\mathcal{E}\in [{\mathcal{E}}_{min},0]$ with ${\mathcal{E}}_{min}=1.1\mathcal{E}(0)$; and $V\in [0,V_{max}]$ with $V_{max}=1.1(V_{bi}-V_A)$ (the OCR of this suggestion is garbled; this is the reading consistent with its intent, a 10 percent headroom). subplot(2,1,1) and subplot(2,1,2) stack the two graphs so the $x$ axes line up, which is what “coordinated plots” means.

(a) Single-bias program

% Problem 5.7(a): E(x) and V(x) of a Si step junction
q = 1.6e-19;  KSe0 = 11.8*8.85e-14;  kT = 0.0259;  ni = 1e10;
NA = 1e16;  ND = 2e14;  VA = -20;            % sample values
 
Vbi = kT*log(NA*ND/ni^2);
W   = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND));
xn  = W*NA/(NA+ND);   xp = W*ND/(NA+ND);
 
E0   = -q*ND*xn/KSe0;                        % field at x = 0 (negative)
Emin = 1.1*E0;   Vmax = 1.1*(Vbi-VA);
xmax = 2.5*max(xn,xp);
x = linspace(-xmax, xmax, 2000);
 
E = zeros(size(x));  V = zeros(size(x));
ip = (x >= -xp & x < 0);   in = (x >= 0 & x <= xn);   ir = (x > xn);
E(ip) = -q*NA*(xp + x(ip))/KSe0;
E(in) = -q*ND*(xn - x(in))/KSe0;
V(ip) =  q*NA*(xp + x(ip)).^2/(2*KSe0);
V(in) = (Vbi-VA) - q*ND*(xn - x(in)).^2/(2*KSe0);
V(ir) = Vbi - VA;
 
subplot(2,1,1); plot(x,E); axis([-xmax xmax Emin 0]);
ylabel('E (V/cm)'); grid on
subplot(2,1,2); plot(x,V); axis([-xmax xmax 0 Vmax]);
ylabel('V (V)'); xlabel('x (cm)'); grid on

Expected result for the sample inputs (verified by hand):

$V_{bi}=0.0259\ln \frac{(10^{16})(2\times 10^{14})}{10^{20}}=0.0259\ln (2\times 10^{10})=0.614\text{V},V_{bi}-V_A=20.614\text{V}$

$W={\left[\frac{2(1.044\times 10^{-12})(20.614)}{1.6\times 10^{-19}}\cdot \frac{1.02\times 10^{16}}{2\times 10^{30}}\right]}^{1/2}={\left[(2.69\times 10^8)(5.1\times 10^{-15})\right]}^{1/2}=1.17\times 10^{-3}\text{cm}=11.7\mu \text{m}$

$x_n=W\frac{10^{16}}{1.02\times 10^{16}}=11.5\mu \text{m},x_p=W-x_n=0.23\mu \text{m}$

$\mathcal{E}(0)=-\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=-\frac{(1.6\times 10^{-19})(2\times 10^{14})(1.148\times 10^{-3})}{1.044\times 10^{-12}}=-3.5\times 10^4\text{V/cm}$

$V(0)=(V_{bi}-V_A)\frac{N_D}{N_A+N_D}=20.614\times 0.0196=0.40\text{V}$

So the plot shows a sharp triangle, mostly on the n-side, with the potential nearly flat at 0 for $x<0$ and climbing to 20.6 V across the n-side depletion region. The tiny p-side depletion layer (0.23 $\mu$m) is a thin sliver at the left of the triangle.

(b) Several biases at once

Loop over $n=\mathrm{0..3}$ with $V_A=V_{A0}/2^n$ and use hold on so all curves share the axes. Use the largest depletion width (the first, $n=0$, which is the most reverse-biased) to choose the axis limits.

% Problem 5.7(b): multiple bias values on the same axes
q = 1.6e-19;  KSe0 = 11.8*8.85e-14;  kT = 0.0259;  ni = 1e10;
NA = 1e16;  ND = 2e14;  VA0 = -20;
Vbi = kT*log(NA*ND/ni^2);
 
W0   = sqrt(2*KSe0/q*(Vbi-VA0)*(NA+ND)/(NA*ND));       % widest case
xmax = 2.5*W0*max(NA,ND)/(NA+ND);                      % 2.5*max(xn,xp)
x    = linspace(-xmax, xmax, 2000);
Emin = 1.1*(-q*ND*(W0*NA/(NA+ND))/KSe0);   Vmax = 1.1*(Vbi-VA0);
 
for n = 0:3
    VA = VA0/2^n;
    W  = sqrt(2*KSe0/q*(Vbi-VA)*(NA+ND)/(NA*ND));
    xn = W*NA/(NA+ND);  xp = W*ND/(NA+ND);
    E = zeros(size(x));  V = zeros(size(x));
    ip = (x >= -xp & x < 0);  in = (x >= 0 & x <= xn);  ir = (x > xn);
    E(ip) = -q*NA*(xp + x(ip))/KSe0;
    E(in) = -q*ND*(xn - x(in))/KSe0;
    V(ip) =  q*NA*(xp + x(ip)).^2/(2*KSe0);
    V(in) = (Vbi-VA) - q*ND*(xn - x(in)).^2/(2*KSe0);
    V(ir) = Vbi - VA;
    subplot(2,1,1); plot(x,E); hold on
    subplot(2,1,2); plot(x,V); hold on
end
subplot(2,1,1); axis([-xmax xmax Emin 0]); ylabel('E (V/cm)'); grid on
subplot(2,1,2); axis([-xmax xmax 0 Vmax]); ylabel('V (V)');
xlabel('x (cm)'); grid on

With $V_A=-20,-10,-5,-2.5$ V you will see a nest of triangles (field) and a nest of S-curves (potential): lower reverse bias gives narrower, shorter curves, with each $V$-curve plateauing at its own $V_{bi}-V_A$. Because $W\propto (V_{bi}-V_A{)}^{1/2}$, halving the bias shrinks $W$ by roughly $1/\sqrt{2}$.

(c) Adding an output list

Insert inside the loop, after computing the numbers (and before the plotting calls for that bias value):

    E0 = -q*ND*xn/KSe0;   V0 = q*NA*xp^2/(2*KSe0);
    fprintf('NA=%9.2e ND=%9.2e VA=%7.3f Vbi=%6.4f\n', NA, ND, VA, Vbi);
    fprintf('  xn=%9.3e xp=%9.3e W=%9.3e cm\n', xn, xp, W);
    fprintf('  E(0)=%10.3e V/cm   V(0)=%8.4f V\n\n', E0, V0);

The list contains exactly what is requested: $N_A,N_D,V_A,V_{bi},x_n,x_p,W$, $\mathcal{E}$ at $x=0$, and $V$ at $x=0$. For the $n=0$ pass you should see $V_{bi}=0.6143$, $W=1.171\times 10^{-3}$ cm, $\mathcal{E}(0)=-3.5\times 10^4$ V/cm, $V(0)=0.404$ V, matching the hand numbers above.

(d) Effect of the relative doping

Run the program for the listed $(N_A,N_D)$ pairs (including equal doping and ratios of 10, 100 and 1000 in both directions). What you will observe:

Case	Where is the depletion region?	Field triangle	Potential
$N_A=N_D$	Equal on both sides, $x_n=x_p=W/2$	Symmetric triangle	Symmetric S-curve, $V(0)=(V_{bi}-V_A)/2$
$N_A$ somewhat $>N_D$	More on the n-side	Skewed, steeper slope on the p-side	Most of the drop on the n-side
$N_A\ge 100N_D$	Essentially all on the n-side	Right triangle, vertical edge at $x\approx 0$	$V$ stays $\approx 0$ for $x<0$ and then rises on the n-side only; $V(0)\approx 0$
$N_D\ge 100N_A$	Essentially all on the p-side	Right triangle, vertical edge at $x\approx 0$ on the right	$V$ rises on the p-side only; $V(0)\approx V_{bi}-V_A$

Description for the extreme cases. When one side is $\ge 100$ times more heavily doped, the junction is one-sided: the lightly doped side supplies all the depletion width, nearly the full $V_{bi}-V_A$ is dropped across it, and the heavily doped side behaves like a metal contact. The width reduces to the single-side formula $W\approx \sqrt{2K_S{\epsilon }_0(V_{bi}-V_A)/(q{N}_{light})}$, and the peak field is ${\mathcal{E}}_{max}=2(V_{bi}-V_A)/W$ (twice the average, because the field is a triangle).

(e) Check against Problems 5.4 and 5.5

Set $(N_A,N_D,V_A)=(2\times 10^{15},10^{15},0)$ and $(10^{17},10^{15},0)$. The output list should reproduce the hand values of Problems 5.4 and 5.5:

Quantity	Problem 5.4	Problem 5.5
$V_{bi}$	0.614 V	0.716 V
$W$	1.10 $\mu$m	0.971 $\mu$m
$x_n$, $x_p$	0.731, 0.366 $\mu$m	0.962, 0.0096 $\mu$m
$\mathcal{E}(0)$	$-1.12\times 10^4$ V/cm	$-1.47\times 10^4$ V/cm
$V(0)$	0.205 V	0.0071 V

Agreement to three significant figures is expected. A mismatch usually means a mistaken exponent in NA/ND, or a forgotten .^ or .* in the vector expressions.

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Problem 5.8: Combination profile

Problem. The doping around the metallurgical junction of a special diode is pictured in Fig. P5.8 (a combination of a step and a linear profile). Sketch, using the depletion approximation, the charge density, electric field and electrostatic potential. Scale and label relevant lengths, and add a few words of explanation.

Assumption note. The figure did not survive OCR; only the axis labels $N_D-N_A$, $x$ and $-N_A$ remain. I read it as: uniform p-type doping $N_D-N_A=-N_A$ for $x<0$, and a linear ramp $N_D-N_A=ax$ for $x>0$, with the metallurgical junction at $x=0$. This is the natural “step on the p-side, graded on the n-side” combination. If your figure differs, apply the same three-step method to its pieces.

Solution: step 1, charge density. The depletion approximation sets $\rho =q(N_D-N_A)$ inside the depletion region $-x_p\le x\le x_n$:

$\rho (x)=\{\begin{array}{ll}-q{N}_A,& -x_p\le x<0\\ +qax,& 0\le x\le x_n\\ 0,& \text{elsewhere}\end{array}$

Sketch: a rectangular negative block of height $-q{N}_A$ and width $x_p$ on the left, and a triangular ramp rising from $0$ at $x=0$ to $+qa{x}_n$ at $x=x_n$ on the right. Note the discontinuity at $x=0$ (the block ends at $-q{N}_A$, the ramp begins at 0).

Neutrality. Equal areas under the two pieces:

$q{N}_A{x}_p=\frac{1}{2}qa{x}_n^2\Rightarrow N_A{x}_p=\frac{a{x}_n^2}{2}$

If $N_A$ is large, $x_p\ll x_n$: the p-side layer is thin, the n-side layer is wide. Label $x_p$ (short) and $x_n$ (long) on the axis.

Step 2, electric field. Integrate $d\mathcal{E}/dx=\rho /(K_S{\epsilon }_0)$ from each edge where $\mathcal{E}=0$:

p-side: $\mathcal{E}(x)=-\frac{q{N}_A}{K_S{\epsilon }_0}(x+x_p),-x_p\le x\le 0$

n-side (integrating leftward from $x_n$): $\mathcal{E}(x)=-\frac{qa}{2K_S{\epsilon }_0}\left(x_n^2-x^2\right),0\le x\le x_n$

Continuity at $x=0$ (this reproduces the neutrality condition):

$-\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}=-\frac{qa{x}_n^2}{2K_S{\epsilon }_0}\Rightarrow N_A{x}_p=\frac{a{x}_n^2}{2}✓$

Sketch: a straight line descending from 0 at $x=-x_p$ to its most negative value $\mathcal{E}(0)$ at $x=0$ (steep, because it covers the short distance $x_p$), then a parabola rising from $\mathcal{E}(0)$ back to zero at $x_n$ with zero slope at $x=0$ (since $\rho (0^{+})=0$ the field is momentarily flat there). The peak magnitude is at $x=0$.

Step 3, electrostatic potential. Integrate $V=-\int \mathcal{E}dx$, with $V=0$ at the p-edge:

p-side: $V(x)=\frac{q{N}_A}{2K_S{\epsilon }_0}(x+x_p{)}^2,-x_p\le x\le 0,V(0)=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}$

n-side: $V(x)=V(0)+\frac{qa}{2K_S{\epsilon }_0}\left(x_n^{2}x-\frac{x^3}{3}\right),0\le x\le x_n$

The total drop (to fix $x_n$ with the condition $V(x_n)=V_{bi}-V_A$):

$V_{bi}-V_A=\frac{q{N}_A{x}_p^2}{2K_S{\epsilon }_0}+\frac{qa{x}_n^3}{3K_S{\epsilon }_0}$

Sketch: a parabola (concave up) from 0 at $-x_p$ up to the small value $V(0)$ at $x=0$, then a cubic-like curve continuing to rise (still concave up at first, then concave down) until it levels off at $V_{bi}-V_A$ at $x=x_n$; flat outside. The slope $dV/dx=-\mathcal{E}$ is maximum at $x=0$, which is the steepest point of the S-curve. Most of the potential drop occurs on the wide, lightly doped n-side.

Words to forestall misinterpretation. (i) The depletion approximation is assumed, so the sketches are idealized: the real $\rho$ and $\mathcal{E}$ have rounded corners. (ii) $\rho$ is discontinuous at $x=0$; $\mathcal{E}$ is continuous; $V$ and its slope are continuous. (iii) The lengths $x_p\ll x_n$ are not to scale with the potential axis.

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Problem 5.9: The exponential profile

Problem. A pn diode has $N_D-N_A=N_0\left[1-e^{-\alpha x}\right]$, with $N_0$ and $\alpha$ constants (Fig. P5.9). (a) State the depletion approximation concisely. (b) Sketch the charge density. (c) Establish an expression for $\mathcal{E}(x)$ in the depletion region.

Solution (a): the depletion approximation, concisely. Within the (abruptly bounded) depletion region $-x_p\le x\le x_n$, the mobile carrier densities are negligible, $p\approx n\approx 0$, so $\rho =q(N_D-N_A)$; outside the region the semiconductor is perfectly neutral, $\rho =0$ and $\mathcal{E}=0$.

Solution (b): sketch of $\rho$. Apply (a) to the given profile:

$\rho (x)=q{N}_0\left[1-e^{-\alpha x}\right],-x_p\le x\le x_n$

Plot by reading the limits:

• At $x=0$: $\rho =0$. This is the metallurgical junction.
• For $x>0$: $e^{-\alpha x}$ decays, so $\rho$ rises from 0 and saturates at $+q{N}_0$. The n-side density flattens at $q{N}_0$ (a curve like a charging capacitor).
• For $x<0$: $e^{-\alpha x}=e^{\alpha |x|}$ grows, so $\rho =-q{N}_0\left(e^{\alpha |x|}-1\right)$ becomes more and more negative with $|x|$, with an exponentially growing magnitude. The profile therefore represents very heavily doped p-type material far on the left.

So the sketch: zero at the origin, a negative curve plunging downward on the left (steepening as $-x_p$ is approached), a positive curve flattening out toward $+q{N}_0$ on the right, and zero outside the dashed edges $-x_p$ and $x_n$. Neutrality demands that the (large, narrow) negative area balance the positive area, so $x_p$ is considerably smaller than $x_n$.

Solution (c): the electric field. Poisson’s equation:

$\frac{d\mathcal{E}}{dx}=\frac{\rho }{K_S{\epsilon }_0}=\frac{q{N}_0}{K_S{\epsilon }_0}\left[1-e^{-\alpha x}\right]$

Integrate from the p-edge $x=-x_p$, where $\mathcal{E}=0$, to an arbitrary $x$:

${\int }_0^{\mathcal{E}(x)}d{\mathcal{E}}^{\prime }=\frac{q{N}_0}{K_S{\epsilon }_0}{\int }_{-x_p}^x\left[1-e^{-\alpha x^{\prime }}\right]d{x}^{\prime }$

Evaluate the integral, using $\int e^{-\alpha x^{\prime }}d{x}^{\prime }=-e^{-\alpha x^{\prime }}/\alpha$:

${\int }_{-x_p}^x\left[1-e^{-\alpha x^{\prime }}\right]d{x}^{\prime }=[x^{\prime }+\frac{1}{\alpha }{e}^{-\alpha x^{\prime }}{]}_{-x_p}^x=(x+x_p)+\frac{1}{\alpha }\left(e^{-\alpha x}-e^{\alpha x_p}\right)$

Therefore:

$\boxed{\mathcal{E}(x)=\frac{q{N}_0}{K_S{\epsilon }_0}\left[(x+x_p)+\frac{e^{-\alpha x}-e^{\alpha x_p}}{\alpha }\right],-x_p\le x\le x_n}$

Boundary condition at the n-edge. The field must also vanish at $x=x_n$, which gives the relation between $x_n$ and $x_p$ (the neutrality condition for this profile):

$x_n+x_p+\frac{e^{-\alpha x_n}-e^{\alpha x_p}}{\alpha }=0$

Sanity checks. (i) At $x=-x_p$: $\mathcal{E}=\frac{q{N}_0}{K_S{\epsilon }_0}\left[0+\frac{e^{\alpha x_p}-e^{\alpha x_p}}{\alpha }\right]=0$ ✓. (ii) The field is negative in between (since $\rho <0$ at first, then the field is pulled back to zero by the positive $\rho$); its slope $d\mathcal{E}/dx\propto 1-e^{-\alpha x}$ is zero at $x=0$, so the most negative field is at the metallurgical junction, as always for a pn junction.

(As the problem’s note says, completing the solution for $V(x)$, $x_n$, $x_p$ and $V_{bi}$ requires a considerable amount of algebra; the method is the same, integrate once more and force $V(x_n)-V(-x_p)=V_{bi}-V_A$.)

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