part05

Part II: pn Junction Diodes and Chapter 5: pn Junction Electrostatics

Source slice. Pierret, Semiconductor Device Fundamentals, Part II opener (Part IIA: Junction Diodes) and Chapter 5, book pages 195-234. Equation numbers (5.x) follow the book. OCR damage has been repaired using physics; wherever the original text or a figure was unreadable, the assumption made is stated explicitly.

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Part II Opener: Where We Are Going

Part I taught us the materials: how electrons and holes are born, how they drift and diffuse. Part II is where we build machines from them. The machines are built from one magical structure: the pn junction, the place where p-type meets n-type material.

Think of Part II as a family tree.

Family member	Where it appears	What it is
pn junction diode	Part IIA (the “charter member”)	One junction, two terminals
Optoelectronic diodes	End of Part IIA (read-only chapter)	Solar cells, LEDs, photodiodes, lasers
Bipolar junction transistor (BJT)	Part IIB	Two junctions, three terminals
PNPN devices	Part IIB	More than two junctions (thyristors)
Schottky diode	Part IIB (last chapter)	Metal-semiconductor “cousin” of the pn diode

The analysis in Part IIA matters beyond diodes: it establishes concepts and procedures of universal utility for every device in the family.

The four-step recipe of every device analysis

A complete, systematic device analysis is broken into four segments:

1. Electrostatics: the charge density $\rho$, electric field $\mathcal{E}$, and electrostatic potential $V$ inside the device under equilibrium and steady state. (This chapter.)
2. Steady-state (d.c.) response: the I-V characteristic.
3. Small-signal (a.c.) response: how it behaves for small wiggles on top of a d.c. bias.
4. Transient (pulsed) response: how it switches.

Analogy. Before you can predict how traffic flows through a city, you must first know the map: where the hills are, where the roads slope. Electrostatics is the map of potential hills and valleys that carriers must navigate. Chapters 6-8 will use this map to compute the traffic (current).

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Chapter 5: pn Junction Electrostatics

5.1 Preliminaries

5.1.1 Junction Terminology and Idealized Profiles

Thought experiment: making a junction

Take a wafer that is uniformly n-type with donor concentration $N_D$. Now diffuse a p-type dopant (acceptors, $N_A$) in from the surface. Near the surface the acceptor concentration is huge; deeper inside it decays away (Fig. 5.1(a): $N_A$ (diffused) falling with $x$, $N_D$ (original) flat).

• Near the surface where $N_A>N_D$ the semiconductor is p-type.
• Deep inside where $N_D>N_A$ it is n-type.
• The dividing plane is the metallurgical junction, defined by

$N_D-N_A=0.$

Key point. Only the net doping $N_D-N_A$ matters, both for locating the metallurgical junction and for finding the electrostatic variables. (A donor and an acceptor next to each other “cancel” in effect: the electron from the donor falls into the acceptor.) So instead of two superimposed plots of $N_A(x)$ and $N_D(x)$, we plot the single curve $N_D-N_A$ versus $x$ (Fig. 5.1(b)). That plot is the doping profile.

Idealized profiles

Real diffusion and ion-implantation profiles make the mathematics miserable. Luckily only the doping very near the metallurgical junction matters, so surprisingly accurate results come from idealizations (Fig. 5.2):

Idealization	Profile	When appropriate
Step junction	$N_D-N_A=-N_A$ for $x<0$, $+N_D$ for $x>0$ (abrupt jump)	Ion implantation or shallow diffusion into a lightly doped wafer
Linearly graded junction	$N_D-N_A=ax$	Deep diffusion into a moderately to heavily doped wafer

Which is better depends on the slope of the real profile at the metallurgical junction and the background doping of the starting wafer. In most of the book we use the step junction purely to keep the algebra manageable.

5.1.2 Poisson’s Equation

From electromagnetism, the 3-D form is

\nabla\cdot\boldsymbol{\mathcal{E}} = \frac{\rho}{K_S\varepsilon_0} \tag{5.1}

In one dimension, $\mathcal{E}={\mathcal{E}}_x$:

\frac{d\mathcal{E}}{dx} = \frac{\rho}{K_S\varepsilon_0} \tag{5.2}

• $K_S$ is the semiconductor dielectric constant (11.8 for Si) and ${\epsilon }_0=8.85\times 10^{-14}$ F/cm.
• Beware the symbol: earlier, $\rho$ meant resistivity. In the electrostatics context $\rho$ is the charge density in C/cm${}^3$.

With totally ionized dopants,

\rho = q\,(p - n + N_D - N_A) \tag{5.3}

(this is the first part of Eq. (2.23), used earlier to derive charge neutrality). Far from any surface in a uniformly doped semiconductor in equilibrium, $\rho \equiv 0$. Under less restrictive conditions $\rho$ can be nonzero and position dependent.

A handy visual trick. Since $\rho \propto d\mathcal{E}/dx$, you can read the shape of $\rho (x)$ off an $\mathcal{E}$-vs-$x$ plot simply by looking at its slope: flat field means zero charge, steeply rising field means large positive charge, and so on. Conversely, the area under $\rho$ gives the change in $\mathcal{E}$. This slope/area dance between $\rho$, $\mathcal{E}$, $V$ will carry us through the whole chapter.

5.1.3 Qualitative Solution

Before any algebra: what should the answer look like? Experienced physicists always sketch first.

From the Einstein-relationship discussion (Subsection 3.2.4) we already know that non-uniform doping produces band bending and an internal field. Assume a 1-D step junction in equilibrium.

Building the equilibrium band diagram in three moves (Fig. 5.3)

1. (a) Far from the metallurgical junction the material is just uniform p-type on the left and uniform n-type on the right. Draw each region’s own band diagram: on the p-side $E_F$ sits near $E_v$; on the n-side $E_F$ sits near $E_c$.
2. (b) In equilibrium the Fermi level is a constant, independent of position. So slide the two diagrams vertically until the $E_F$ lines coincide. Now the n-side bands sit lower than the p-side bands.
3. (c) Connect $E_c$, $E_i$, $E_v$ across the middle with a smooth monotonic curve that is flat (zero slope) at both ends of the central region.

The exact shape in the middle is not known yet, only that it is monotonic and joins smoothly. That is Fig. 5.3(c) and Fig. 5.4(a).

A footnote in the text: from the band diagram alone one can only conclude that the magnitude of the field (slope of $E_c$) rises from zero on the p-side, peaks near the metallurgical boundary, and drops to zero on the n-side. The pseudo-linear shape sketched in Fig. 5.4(c) anticipates the quantitative result.

Reading the diagram for $V$, $\mathcal{E}$, $\rho$ (Fig. 5.4)

• Potential: from Subsection 3.1.5 on band bending, $V(x)$ has the same shape as the upside-down $E_c$ (or $E_i$ or $E_v$), because electron energy $=-qV$. Setting $V=0$ on the far p-side, $V$ rises to $V_{bi}$ on the n-side (Fig. 5.4(b)).
• Field: $\mathcal{E}=-dV/dx$, i.e. proportional to the slope of $E_c$ (Fig. 5.4(c)). It is negative (pointing from n to p) and nonzero only in the central region.
• Charge: $\rho =K_S{\epsilon }_{0}d\mathcal{E}/dx$: negative on the p-side (field becoming more negative as $x$ increases), positive on the n-side (Fig. 5.4(d)).

The big question: where does that charge come from?

Perform a thought experiment (Fig. 5.5). Hold a p-block and an n-block apart.

• In the p-block, mobile holes ($+$) exactly balance fixed ionized acceptor ions ($-$). In the n-block, mobile electrons ($-$) exactly balance fixed ionized donor ions ($+$). Both blocks are neutral.
• Now press them together with a perfect connection. There are enormously more holes on the p-side than on the n-side, so holes diffuse to the n-side; electrons diffuse to the p-side. Think of two rooms, one crowded with red balloons, one with blue; open the door and each color spills into the other room.
• But the dopant ions are bolted into the crystal. When the mobile carriers leave the neighborhood of the junction, they leave behind uncovered ions: negative acceptor ions on the p-side, positive donor ions on the n-side.

This is the source of the junction charge. The region around the metallurgical junction where this significant charge exists is the space charge region or depletion region (the latter because the mobile carriers there are greatly reduced, “depleted”).

The charge build-up does not continue forever. The growing field pushes carriers back (drift) while concentration gradients push them forward (diffusion). The process halts when diffusion and drift precisely balance. Individually $J_N=0$ and $J_P=0$ in equilibrium: each carrier type’s drift cancels its own diffusion.

Analogy. Think of a hiker climbing a hill while a crowd pushes from behind. The hill (potential barrier) grows until the push from the crowd (diffusion) is exactly matched by the pull back down the slope (drift).

5.1.4 The Built-in Potential ($V_{bi}$)

The voltage drop across the depletion region in equilibrium is the built-in potential $V_{bi}$. We want a computational formula.

Take a nondegenerately doped pn junction in equilibrium with $x=0$ at the metallurgical boundary, and depletion region edges at $-x_p$ (p-side) and $x_n$ (n-side).

Derivation (Eqs. 5.4-5.10)

Start from the definition of potential:

\mathcal{E} = -\frac{dV}{dx} \tag{5.4}

Integrate across the depletion region:

${\int }_{-x_p}^{x_n}\mathcal{E}dx=-{\int }_{V(-x_p)}^{V(x_n)}dV=-\left[V(x_n)-V(-x_p)\right]=-V_{bi}$

(the book writes this with the sign convention $V_{bi}=V(x_n)-V(-x_p)$, i.e. $-\int \mathcal{E}dx=V_{bi}$):

-\int_{-x_p}^{x_n}\mathcal{E}\,dx = \int_{V(-x_p)}^{V(x_n)} dV = V(x_n) - V(-x_p) = V_{bi} \tag{5.5}

In equilibrium the electron current is zero:

J_N = q\mu_n n\mathcal{E} + qD_N\frac{dn}{dx} = 0 \tag{5.6}

Solve for $\mathcal{E}$ and use the Einstein relation $D_N/{\mu }_n=kT/q$:

\mathcal{E} = -\frac{D_N}{\mu_n}\frac{dn/dx}{n} = -\frac{kT}{q}\frac{dn/dx}{n} \tag{5.7}

Substitute (5.7) into (5.5) and integrate:

V_{bi} = -\int_{-x_p}^{x_n}\mathcal{E}\,dx = \frac{kT}{q}\int_{n(-x_p)}^{n(x_n)}\frac{dn}{n} = \frac{kT}{q}\ln\frac{n(x_n)}{n(-x_p)} \tag{5.8}

Notice that nothing here required a step junction, only nondegenerate doping. Now specialize to a nondegenerate step junction:

n(x_n) = N_D \tag{5.9a}

n(-x_p) = \frac{n_i^2}{p(-x_p)} = \frac{n_i^2}{N_A} \tag{5.9b}

(on the p-side, $p=N_A$ and $n=n_i^2/N_A$). Therefore

\boxed{V_{bi} = \frac{kT}{q}\ln\!\left(\frac{N_AN_D}{n_i^2}\right)} \tag{5.10}

Sample computation. $N_A=N_D=10^{15}$/cm${}^3$, Si at 300 K ($n_i=10^{10}$/cm${}^3$, $kT/q=0.0259$ V):

$V_{bi}=0.0259\ln \left(\frac{10^{15}\cdot 10^{15}}{10^{20}}\right)=0.0259\ln (10^{10})=0.0259(23.03)\approx 0.6\text{V}$

This is typical. In nondegenerately doped diodes $V_{bi}<E_G/q$. Nondegenerately doped Ge, Si, GaAs diodes at room temperature have $V_{bi}$ below 0.66 V, 1.12 V, 1.42 V respectively.

Alternative derivation from the band diagram

From Figs. 5.4(a) and (b), $q{V}_{bi}$ is the total band bending. Since $E_i$ (the reference for electron energy) drops by $q{V}_{bi}$ from the p-side to the n-side:

V_{bi} = V(x_n) - V(-x_p) \tag{5.11a}

qV_{bi} = (E_i)_{p\text{-side}} - (E_i)_{n\text{-side}} \tag{5.11b}

Since $E_F$ is constant, add and subtract $E_F$ on the right:

$q{V}_{bi}=(E_i-E_F{)}_{p\text{-side}}+(E_F-E_i{)}_{n\text{-side}}$

or

V_{bi} = \frac{1}{q}\left[(E_i - E_F)_{p\text{-side}} + (E_F - E_i)_{n\text{-side}}\right] \tag{5.12}

In a nondegenerate diode each term is less than $E_G/2$, so $q{V}_{bi}<E_G$, which explains the bound above. For a nondegenerate step junction,

(E_i - E_F)_{p\text{-side}} = kT\ln(N_A/n_i) \tag{5.13a}

(E_F - E_i)_{n\text{-side}} = kT\ln(N_D/n_i) \tag{5.13b}

Inserting (5.13) into (5.12):

$V_{bi}=\frac{kT}{q}\left[\ln \frac{N_A}{n_i}+\ln \frac{N_D}{n_i}\right]=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2},$

which is (5.10). Important: (5.13), hence (5.10), need nondegeneracy, but (5.12) has no doping restrictions.

Common confusion. “If $V_{bi}\approx 0.6$ V exists inside the diode, why can’t I measure it with a voltmeter or light a bulb?” Because every metal-semiconductor contact has its own built-in potential, and these exactly cancel $V_{bi}$ around the closed loop (a diode in equilibrium cannot deliver free energy; the Fermi level is flat everywhere).

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Exercise 5.1: $V_{bi}$ for $p^{+}$-n and $n^{+}$-p junctions (computer-based)

Problem. Most real diodes are heavily doped on one side. In computing $V_{bi}$ of $p^{+}$-n and $n^{+}$-p step junctions, it is common practice to assume that the Fermi level on the heavily doped side is at the band edge: $E_F=E_v$ in a $p^{+}$ material and $E_F=E_c$ in an $n^{+}$ material. With this assumption, compute and plot $V_{bi}$ versus the doping ($N_A$ or $N_D$) on the lightly doped side of Si $p^{+}$-n and $n^{+}$-p step junctions at 300 K, for $10^{14}/{\text{cm}}^3\le N_A$ or $N_D\le 10^{17}/{\text{cm}}^3$.

Solution logic.

• On the $p^{+}$ side, $E_F=E_v$ so $(E_i-E_F{)}_{p\text{-side}}=E_i-E_v=E_G/2$ (midgap).
• On the lightly doped n-side, $(E_F-E_i{)}_{n\text{-side}}=kT\ln (N_D/n_i)$.
• Use (5.12), valid for arbitrary doping:

$V_{bi}=\frac{1}{q}\left[\frac{E_G}{2}+kT\ln \frac{N_D}{n_i}\right]=\frac{E_G}{2q}+\frac{kT}{q}\ln \frac{N_D}{n_i}$

For $n^{+}$-p junctions replace $N_D$ by $N_A$ (same computation by symmetry).

Hand check. At $N_D=10^{14}$: $V_{bi}=0.56+0.0259\ln (10^4)=0.56+0.2385=0.80$ V. At $N_D=10^{17}$: $0.56+0.0259\ln (10^7)=0.56+0.417=0.98$ V. This matches the plot’s range (0.8 to 0.98 V, gently rising, almost a straight line on the semilog axis).

MATLAB. logspace(14,17) gives 50 log-spaced doping values; the formula is vectorized with .* and ./; semilogx gives a logarithmic horizontal axis.

% Vbi computation (p+/n and n+/p junctions)
% Constants
EG = 1.12;      % band gap, eV
kT = 0.0259;    % thermal energy, eV
ni = 1.0e10;    % intrinsic concentration, cm^-3
% Computation
ND = logspace(14,17);
Vbi = EG/2 + kT.*log(ND./ni);
% Plotting
close
semilogx(ND,Vbi); grid
axis([1.0e14 1.0e17 0.75 1])
xlabel('NA or ND (cm^-3)'); ylabel('Vbi (volts)')
text(1e16,0.8,'Si, 300K')
text(1e16,0.78,'p+/n and n+/p diodes')

Result (Fig. E5.1). A slowly rising, nearly straight line on a semilog plot from about 0.80 V at $10^{14}$ to about 0.98 V at $10^{17}$. Intuition: each decade of doping on the light side moves $E_F$ by $kT\ln 10=59.6$ meV, so $V_{bi}$ rises by about 60 mV per decade.

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5.1.5 The Depletion Approximation

Now we need to actually solve Poisson’s equation. Rewrite it:

\frac{d\mathcal{E}}{dx} = \frac{\rho}{K_S\varepsilon_0} = \frac{q}{K_S\varepsilon_0}\left(p - n + N_D - N_A\right) \tag{5.14}

The doping profile $N_D-N_A$ is known. But $p(x)$ and $n(x)$ in the region where $\rho \ne 0$ are not known in advance; they depend on the potential, which is what we are solving for (see the band diagram, Fig. 5.4(a)). This is a chicken-and-egg problem. Exact closed-form solutions exist in certain cases (they are complicated; see Problem 5.14), but the depletion approximation sidesteps the issue and is “far and away the most important and most widely encountered of the simplifying approximations used in the modeling of devices.”

Statement of the approximation (Fig. 5.6)

1. In a region $-x_p\le x\le x_n$ straddling the metallurgical junction, the carrier concentrations are assumed negligible compared to the net doping: $p,n\ll |N_D-N_A|$.
2. Outside that region, the charge density is taken to be identically zero (quasineutral).

Within the depletion region this gives $\rho =q(N_D-N_A)$, and Poisson’s equation becomes

\frac{d\mathcal{E}}{dx} = \frac{q(N_D - N_A)}{K_S\varepsilon_0}, \quad -x_p\le x\le x_n \tag{5.15a}

\frac{d\mathcal{E}}{dx} = 0, \quad x\le -x_p \ \text{and}\ x \ge x_n \tag{5.15b}

Analogy. The real charge distribution has soft, fuzzy edges: carrier numbers drop gradually. The approximation replaces the fuzzy edge by a razor-sharp cliff: “Inside this box, no carriers. Outside it, perfect neutrality.” Fig. 5.6(b) shows the exact $\rho$ (smooth, rounded) and the depletion-approximation $\rho$ (sharp rectangles) for a step junction. The picture is crude but it captures the area (total charge) and location of the real charge, which is what matters for $\mathcal{E}$ and $V$.

Everything about $\rho$ is now specified except the unknown edges $-x_p$, $x_n$. Moreover, inside the depletion region $\rho$ has precisely the same functional form as $N_D-N_A$ (Fig. 5.7 shows two moderately complicated examples): to get $\rho (x)$ just copy the doping profile shape (times $q$) inside the box.

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5.2 Quantitative Electrostatic Relationships

Now the main event: quantitative relationships. We emphasize the step junction, and give every derivation in detail, partly to get the formulas but also to establish procedures that apply to any doping profile. The plan: set up the system, solve the step junction at $V_A=0$, extend to $V_A\ne 0$, and finally treat the linearly graded junction.

5.2.1 Assumptions and Definitions

The diode (Fig. 5.8):

• All variables depend only on $x$, the coordinate normal to the surface: the device is one-dimensional (edge effects at lateral ends are ignored).
• $x=0$ sits at the metallurgical boundary.
• The end contacts are ohmic: by definition a negligible portion of an applied voltage appears across an ohmic contact.
• $V_A$ denotes the applied voltage (subscript A distinguishes it from internal junction voltages). It is positive when the p-terminal is positive relative to the n-terminal.
• To start, $V_A=0$ (equilibrium).

5.2.2 Step Junction with $V_A=0$

Take $N_A>N_D$ for illustration (Fig. 5.9(a)).

Solution for $\rho$

By the depletion approximation:

\rho = \begin{cases} -qN_A, & -x_p\le x\le 0 \\ +qN_D, & 0\le x\le x_n \\ 0, & x\le -x_p \text{ and } x\ge x_n \end{cases} \tag{5.16a-c}

($x_n$, $x_p$ unknown for now.)

Solution for $\mathcal{E}$

Poisson’s equation gives

\frac{d\mathcal{E}}{dx} = \begin{cases} -qN_A/(K_S\varepsilon_0), & -x_p\le x\le 0 \\ +qN_D/(K_S\varepsilon_0), & 0\le x\le x_n \\ 0, & x\le -x_p \text{ and } x\ge x_n\end{cases} \tag{5.17a-c}

Boundary conditions: $\mathcal{E}=0$ far from the junction, thus $\mathcal{E}=0$ everywhere outside the depletion region; and since $\mathcal{E}$ can’t jump, $\mathcal{E}(-x_p)=0$ and $\mathcal{E}(x_n)=0$.

p-side. Separate variables and integrate from the edge $-x_p$ to a point $x$:

\int_0^{\mathcal{E}(x)} d\mathcal{E}' = -\int_{-x_p}^{x}\frac{qN_A}{K_S\varepsilon_0}\,dx' \tag{5.18}

\mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}\left[x' \right]_{-x_p}^{x} \;\Rightarrow\; \mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}(x + x_p), \quad -x_p\le x\le 0 \tag{5.19}

n-side. Integrate from $x$ to the edge $x_n$:

\int_{\mathcal{E}(x)}^{0} d\mathcal{E}' = \int_{x}^{x_n}\frac{qN_D}{K_S\varepsilon_0}\,dx' \tag{5.20}

-\mathcal{E}(x) = \frac{qN_D}{K_S\varepsilon_0}(x_n - x) \;\Rightarrow\; \mathcal{E}(x) = -\frac{qN_D}{K_S\varepsilon_0}(x_n - x), \quad 0\le x\le x_n \tag{5.21}

So the field is always negative inside the depletion region, zero outside, and linear in position with a “peak” (largest magnitude) at $x=0$. It looks like a tent/triangle (Fig. 5.9(c)). This agrees with the qualitative sketch of Fig. 5.4(c).

Continuity at $x=0$. The field is continuous across a boundary unless a sheet of charge lies on it, and there is none. Equating (5.19) and (5.21) at $x=0$:

-\frac{qN_A}{K_S\varepsilon_0}x_p = -\frac{qN_D}{K_S\varepsilon_0}x_n \;\Rightarrow\; \boxed{N_Ax_p = N_Dx_n} \tag{5.22}

Gauss’ law view. $\mathcal{E}=0$ outside means the total charge in the depletion region is zero: $-q{N}_A{x}_p\cdot (\text{area})+q{N}_D{x}_n\cdot (\text{area})=0$, i.e. the rectangular areas in Fig. 5.9(b) are equal. So (5.22) is just charge balance. A see-saw: the more heavily doped side has the shorter arm.

Solution for $V$

Since $\mathcal{E}=-dV/dx$, with (5.19), (5.21):

\frac{dV}{dx} = \begin{cases}\dfrac{qN_A}{K_S\varepsilon_0}(x + x_p), & -x_p\le x\le 0 \\[2mm] \dfrac{qN_D}{K_S\varepsilon_0}(x_n - x), & 0\le x\le x_n\end{cases} \tag{5.23a,b}

Take the arbitrary reference $V(-x_p)=0$; the total drop across the depletion region in equilibrium is $V_{bi}$:

V(-x_p) = 0, \qquad V(x_n) = V_{bi} \tag{5.24a,b}

p-side. Integrate from $-x_p$ to $x$:

\int_0^{V(x)}dV' = \frac{qN_A}{K_S\varepsilon_0}\int_{-x_p}^{x}(x'+x_p)\,dx' \tag{5.25}

V(x) = \frac{qN_A}{K_S\varepsilon_0}\left[\frac{(x'+x_p)^2}{2}\right]_{-x_p}^{x} = \frac{qN_A}{2K_S\varepsilon_0}(x+x_p)^2, \quad -x_p\le x\le 0 \tag{5.26}

n-side. Integrate from $x$ to $x_n$:

\int_{V(x)}^{V_{bi}}dV' = \frac{qN_D}{K_S\varepsilon_0}\int_x^{x_n}(x_n - x')\,dx' \tag{5.27}

$V_{bi}-V(x)=\frac{q{N}_D}{K_S{\epsilon }_0}{\left[-\frac{(x_n-x^{\prime }{)}^2}{2}\right]}_x^{x_n}=\frac{q{N}_D}{2K_S{\epsilon }_0}(x_n-x{)}^2$

V(x) = V_{bi} - \frac{qN_D}{2K_S\varepsilon_0}(x_n - x)^2, \quad 0\le x\le x_n \tag{5.28}

$V(x)$ is quadratic: it rises with an upward-opening (concave-up, “smile”) curvature on the p-side and an upside-down (concave-down) curvature on the n-side, giving the S-shaped hill of Fig. 5.9(d). Continuity of $V$ at $x=0$ is justified because there is no dipole layer (closely spaced sheets of plus and minus charge) at the boundary. Equating (5.26) and (5.28) at $x=0$:

\frac{qN_A}{2K_S\varepsilon_0}x_p^2 = V_{bi} - \frac{qN_D}{2K_S\varepsilon_0}x_n^2 \tag{5.29}

Solving for $x_n$ and $x_p$

The only unknowns in (5.22) and (5.29) are $x_n$, $x_p$. Rewrite (5.29) as

$V_{bi}=\frac{q}{2K_S{\epsilon }_0}\left(N_A{x}_p^2+N_D{x}_n^2\right)$

Eliminate $x_p$ with (5.22), $x_p=(N_D/N_A)x_n$:

$V_{bi}=\frac{q}{2K_S{\epsilon }_0}\left(\frac{N_D^2}{N_A}{x}_n^2+N_D{x}_n^2\right)=\frac{q{N}_D}{2K_S{\epsilon }_0}\frac{N_D+N_A}{N_A}{x}_n^2$

Solve for $x_n$, and do the same for $x_p$ (using $x_n=(N_A/N_D)x_p$):

\boxed{x_n = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\,\frac{N_A}{N_D(N_A+N_D)}\right]^{1/2}} \tag{5.30a}

\boxed{x_p = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\,\frac{N_D}{N_A(N_A+N_D)}\right]^{1/2}} \tag{5.30b}

The total depletion width $W$:

$W=x_n+x_p=x_n\left(1+\frac{N_D}{N_A}\right)=x_n\frac{N_A+N_D}{N_A}$

\boxed{W = \left[\frac{2K_S\varepsilon_0}{q}\,V_{bi}\left(\frac{N_A+N_D}{N_AN_D}\right)\right]^{1/2}} \tag{5.31}

$W$ is what you usually need in practical device computations. Another quick result: peak field, from (5.21) at $x=0$:

$|\mathcal{E}{|}_{max}=|\mathcal{E}(0)|=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{q{N}_A{x}_p}{K_S{\epsilon }_0}$

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Exercise 5.2: How big are $W$ and $|\mathcal{E}{|}_{max}$?

Problem. Si step junction at 300 K with $N_A=10^{17}$/cm${}^3$ and $N_D=10^{14}$/cm${}^3$. Compute $W$ and $|\mathcal{E}{|}_{max}$ at equilibrium.

Solution.

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2}=0.0259\ln \frac{(10^{17})(10^{14})}{10^{20}}=0.0259\ln (10^{11})=0.656\text{V}$

Use $K_S{\epsilon }_0=11.8(8.85\times 10^{-14})=1.044\times 10^{-12}$ F/cm. By (5.30a), since $N_A\gg N_D$, $N_A/(N_A+N_D)\approx 1$:

$x_n={\left[\frac{2(11.8)(8.85\times 10^{-14})(0.656)}{(1.6\times 10^{-19})(10^{14})}\right]}^{1/2}=2.93\times 10^{-4}\text{cm}=2.93\mu \text{m}$

$x_p=\frac{N_D}{N_A}{x}_n=10^{-3}{x}_n=2.93\times 10^{-7}\text{cm}$

$W=x_n+x_p\approx x_n=2.93\mu \text{m}$

$|\mathcal{E}{|}_{max}=\frac{q{N}_D{x}_n}{K_S{\epsilon }_0}=\frac{(1.6\times 10^{-19})(10^{14})(2.93\times 10^{-4})}{(11.8)(8.85\times 10^{-14})}=4.49\times 10^3\text{V/cm}$

(One could have used (5.31) directly. The purpose of computing $x_n$, $x_p$ separately is to make a point.)

Lesson. In an asymmetrically doped junction ($N_A\gg N_D$ or $N_D\gg N_A$) the depletion region lies almost entirely on the lightly doped side. Here the p-side depletion is 0.3 nm worth of … precisely, 2.93 nm versus 2.93 micrometers on the n-side. Intuition: the lightly doped side has few dopant ions per unit length, so it must expose a long stretch to supply the same total charge.

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5.2.3 Step Junction with $V_A\ne 0$

To be practical we must handle $V_A\ne 0$. We could repeat everything, but there is a shortcut.

Apply $V_A$ to the diode terminals (Fig. 5.10). The voltage must drop somewhere inside:

• A negligible portion drops across the ohmic contacts (by definition).
• Under low-level injection (reasonable currents) the resistive drop across the quasineutral p- and n-regions between the contacts and the depletion edges is also negligible.
• Therefore all of $V_A$ is dropped across the depletion region.

When $V_A>0$ (forward bias, p-terminal positive) the potential of the n-side is lowered relative to the p-side, reducing the hill. When $V_A<0$ (reverse bias), the n-side potential rises relative to the p-side, increasing the hill. So the voltage drop across the depletion region becomes

$V(x_n)-V(-x_p)=V_{bi}-V_A$

The only change in the mathematical problem is one boundary condition (5.24b). Hence the $V_A\ne 0$ solution follows from the $V_A=0$ solution by replacing every explicit $V_{bi}$ with $V_{bi}-V_A$. Summary:

For $-x_p\le x\le 0$:

\mathcal{E}(x) = -\frac{qN_A}{K_S\varepsilon_0}(x_p + x) \tag{5.32}

V(x) = \frac{qN_A}{2K_S\varepsilon_0}(x_p + x)^2 \tag{5.33}

x_p = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_D}{N_A(N_A+N_D)}\right]^{1/2} \tag{5.34}

For $0\le x\le x_n$:

\mathcal{E}(x) = -\frac{qN_D}{K_S\varepsilon_0}(x_n - x) \tag{5.35}

V(x) = V_{bi} - V_A - \frac{qN_D}{2K_S\varepsilon_0}(x_n - x)^2 \tag{5.36}

x_n = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_A}{N_D(N_A+N_D)}\right]^{1/2} \tag{5.37}

and

\boxed{W = \left[\frac{2K_S\varepsilon_0}{q}(V_{bi}-V_A)\frac{N_A+N_D}{N_AN_D}\right]^{1/2}} \tag{5.38}

To avoid an imaginary square root, $V_A\le V_{bi}$ is obviously required in (5.34), (5.37), (5.38). The formulation actually fails before this: as $V_A\to V_{bi}$ a large current flows, and the voltage drops across the quasineutral regions can no longer be neglected.

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Exercise 5.3: $W$ versus doping at several biases (computer-based)

Problem. Construct a log-log plot of the depletion width $W$ versus the impurity concentration ($N_A$ or $N_D$) on the lightly doped side of Si $p^{+}$-n and $n^{+}$-p step junctions at 300 K. Include curves for $V_A=0.5$ V, 0 V, $-10$ V over $10^{14}\le N_A$ or $N_D\le 10^{17}$/cm${}^3$.

Solution logic.

1. $V_{bi}$ comes from the Exercise 5.1 relation, $V_{bi}=E_G/2q+(kT/q)\ln (N_B/n_i)$, where $N_B$ is the light-side doping.
2. For a strongly one-sided junction the doping factor in (5.38) simplifies:

$\frac{N_A+N_D}{N_A{N}_D}=\frac{1}{N_D}+\frac{1}{N_A}\approx \frac{1}{N_B}$

(the heavily doped term is negligible).

3. So $W\approx {\left[\frac{2K_S{\epsilon }_0}{q{N}_B}(V_{bi}-V_A)\right]}^{1/2}$.

Hand check at $N_B=10^{15}$: $V_{bi}=0.56+0.0259\ln (10^5)=0.858$ V.

• $V_A=0$: $W=\sqrt{(1.305\times 10^7)(0.858)/10^{15}}=1.06\mu$m.
• $V_A=0.5$ V: $(V_{bi}-V_A)=0.358$, $W=0.68\mu$m.
• $V_A=-10$ V: $(V_{bi}-V_A)=10.86$, $W=3.77\mu$m.

(Here $2K_S{\epsilon }_0/q=1.305\times 10^7$ cm${}^{-1}\cdot$V${}^{-1}\cdot$cm${}^{-2}$ units, i.e. $W^2=(1.305\times 10^7)(V)/N_B$ in cm${}^2$.)

MATLAB logic. Build vectors of doping (logspace), a three-element bias vector, and a $3\times 50$ matrix W whose rows are the three bias curves. Multiply by $10^4$ to convert cm to micrometers. loglog(NB,W,'-') plots all three rows at once. (The OCR rendering of VA=[0.5 0 -10] and of the e0 variable name was garbled; restored below.)

% This program calculates and plots the depletion width vs impurity
% concentration in Silicon p+/n and n+/p step junctions at 300K.
% Three plots are generated corresponding to VA=0.5V, 0.0V and -10V
% The Vbi relationship employed is Vbi=(EG/2q)+(kT/q)ln(NB/ni)
% where NB is the impurity concentration on the lightly doped side.
 
% Constants and Parameters
T  = 300;        % Temperature in Kelvin
k  = 8.617e-5;   % Boltzmann constant (eV/K)
e0 = 8.85e-14;   % permittivity of free space (F/cm)
q  = 1.602e-19;  % charge on an electron (coul)
KS = 11.8;       % dielectric constant of Si at 300K
ni = 1e10;       % intrinsic carrier conc. in Silicon at 300K (cm^-3)
EG = 1.12;       % band gap of Silicon (eV)
 
% Choose variable values
NB = logspace(14,17);   % doping ranges from 1e14 to 1e17
VA = [0.5 0 -10];       % VA values set
 
% Depletion width calculation
Vbi = EG/2 + k*T.*log(NB./ni);
W(1,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(1))./NB);
W(2,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(2))./NB);
W(3,:) = 1.0e4*sqrt(2*KS*e0/q.*(Vbi-VA(3))./NB);
 
% Plot
close
loglog(NB,W,'-'); grid
axis([1.0e14 1.0e17 1.0e-1 1.0e1])
xlabel('NA or ND (cm^-3)')
ylabel('W (micrometers)')
set(gca,'DefaultTextUnits','normalized')
text(.38,.26,'VA=0.5V')
text(.38,.50,'VA=0')
text(.38,.76,'VA=-10V')
text(.77,.82,'Si, 300K')
text(.77,.79,'p+/n and n+/p')
set(gca,'DefaultTextUnits','data')

Result (Fig. E5.3). Three nearly parallel straight lines of slope about $-1/2$ on log-log axes (because $W\propto N_B^{-1/2}$; the slight curvature comes from the weak $\ln$ dependence of $V_{bi}$). The reverse-bias curve ($-10$ V) lies highest, forward bias ($0.5$ V) lowest. Depletion widths span roughly 0.1 to 10 micrometers over this range.

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5.2.4 Examination and Extrapolation of Results

How do the electrostatic variables respond to bias?

• Depletion width. From (5.34) and (5.37), $x_p$ and $x_n$ shrink under forward bias ($V_A>0$) and grow under reverse bias ($V_A<0$), consistent with Exercise 5.3.
• Field. From (5.32), (5.35), smaller $x_p$, $x_n$ under forward bias lower $|\mathcal{E}|$ everywhere in the depletion region; reverse bias gives larger $x_p$, $x_n$ and a larger $|\mathcal{E}|$. Physically: less space charge means weaker field, more space charge means stronger field.
• Potential. From (5.33), (5.36) the potential decreases everywhere for $V_A>0$ and increases everywhere for $V_A<0$. The potential hill shrinks in both height and width under forward bias, and grows taller and wider under reverse bias (Fig. 5.11).

Analogy. Reverse bias is like heightening a dam: more water (carriers) are held back and the dam wall must be wider and thicker (more exposed charge). Forward bias is like lowering the dam gate.

Energy band diagrams under bias (Fig. 5.12)

Run the Subsection 3.1.5 procedure in reverse: take the “upside-down” potential plots of Fig. 5.11(d) and modify the equilibrium band diagram: smaller width and smaller hill for forward bias, larger width and larger hill for reverse bias.

Comments on the $V_A\ne 0$ diagrams:

1. $E_F$ is omitted from the depletion region, because the device is no longer in equilibrium and one level cannot describe both carrier populations there. The levels labeled $E_{Fp}$ and $E_{Fn}$ on the quasineutral sides are actually majority-carrier quasi-Fermi levels; the deviation from equilibrium in the neutral regions is small (especially far from the junction), so we keep calling them $E_F$.
2. By inspection of the diagrams,

E_{Fp} - E_{Fn} = -qV_A \tag{5.39}

(For $V_A>0$, $E_{Fn}$ lies above $E_{Fp}$ by $q{V}_A$.)

Visualization. Equation (5.39) says the terminals give you direct handles on the two ends of the equilibrium Fermi level. Forward bias: grab the n-side and lift it up by $q{V}_A$ while holding the p-side fixed, which shrinks the hill from $q{V}_{bi}$ to $q(V_{bi}-V_A)$. Reverse bias: pull the n-side down.

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Exercise 5.4: Equilibrium Energy Band Diagram Generator (computer-based)

Problem. Given a quantitative expression for the electrostatic potential, a fully dimensioned band diagram can be drawn. Run the “Diagram Generator” program, which prompts for $N_A$ and $N_D$. Try combinations with $N_A>N_D$, $N_A=N_D$, $N_A<N_D$. Asymmetric junctions give “one-sided” diagrams unlike typical textbooks. Consider modifying the program to draw an arbitrary applied bias.

Program logic.

1. $V_{bi}=(kT/q)\ln (N_A{N}_D/n_i^2)$ and $x_n$, $x_p$ from (5.30) are computed.
2. An $x$ array from $-3.5\mu$m to $+3.5\mu$m is built.
3. $V(x)$ is assembled piecewise: $0$ for $x<-x_p$; (5.26) for $-x_p\le x<0$; (5.28) for $0\le x\le x_n$; $V_{bi}$ for $x>x_n$. In MATLAB the logical masks (x>=0 & x<=xN) multiply each piece, turning it on only in its range.
4. Band edges are $E_i=-V+V_{MAX}/2$ (energy in eV, since $E=-qV$ and $q$ cancels), $E_c=E_i+E_G/2$, $E_v=E_i-E_G/2$, and the Fermi level is placed $kT\ln (N_A/n_i)$ below $E_i$ at the far p-side.

(The OCR damaged the line defining the n-side potential; it has been reconstructed from (5.28), and an x>xN term has been added so that $V$ stays at $V_{bi}$ beyond the depletion edge.)

% Equilibrium Energy Band Diagram Generator
% (Si, 300K, nondegenerately doped step junction)
% Constants
T  = 300;        % Temperature in Kelvin
k  = 8.617e-5;   % Boltzmann constant (eV/K)
e0 = 8.85e-14;   % permittivity of free space (F/cm)
q  = 1.602e-19;  % charge on an electron (coul)
KS = 11.8;       % Dielectric constant of Si
ni = 1.0e10;     % intrinsic carrier conc. in Silicon at 300K (cm^-3)
EG = 1.12;       % Silicon band gap (eV)
% Control constants
xleft  = -3.5e-4;  % Leftmost x position (cm)
xright = -xleft;   % Rightmost x position (cm)
NA = input('Please enter p-side doping (cm^-3), NA=');
ND = input('Please enter n-side doping (cm^-3), ND=');
% Computations
Vbi = k*T*log((NA*ND)/ni^2);
xN = sqrt(2*KS*e0/q*NA*Vbi/(ND*(NA+ND)));   % Depletion width n-side
xP = sqrt(2*KS*e0/q*ND*Vbi/(NA*(NA+ND)));   % Depletion width p-side
x = linspace(xleft,xright,200);
Vx1 = (Vbi - 0.5*q*ND.*(xN-x).^2/(KS*e0)).*(x>=0 & x<=xN) + Vbi*(x>xN);
Vx2 = 0.5*q*NA.*(xP+x).^2/(KS*e0).*(x>=-xP & x<0);
Vx = Vx1 + Vx2;                              % V as a function of x
VMAX = 3;                                    % Maximum Plot Voltage
EF = Vx(1) + VMAX/2 - k*T*log(NA/ni);        % Fermi level
% Plot Diagram
close
plot(x,-Vx+EG/2+VMAX/2);                     % Ec
axis([xleft xright 0 VMAX]);
axis('off'); hold on
plot(x,-Vx-EG/2+VMAX/2);                     % Ev
plot(x,-Vx+VMAX/2,'w:');                     % Ei (dotted)
plot([xleft xright],[EF EF],'w--');          % EF
plot([0 0],[0.15*VMAX 0.5*VMAX],'w-');       % metallurgical junction marker
text(xleft*1.08,(-Vx(1)+EG/2+VMAX/2-.05),'Ec');
text(xright*1.02,(-Vx(200)+EG/2+VMAX/2-.05),'Ec');
text(xleft*1.08,(-Vx(1)-EG/2+VMAX/2-.05),'Ev');
text(xright*1.02,(-Vx(200)-EG/2+VMAX/2-.05),'Ev');
text(xleft*1.08,(-Vx(1)+VMAX/2-.05),'Ei');
text(xright*1.02,EF-.05,'EF');
set(gca,'DefaultTextUnits','normalized')
text(.18,0,'p-side'); text(.47,0,'x=0'); text(.75,0,'n-side');
set(gca,'DefaultTextUnits','data')
hold off

What to expect. For $N_A=N_D$ you get a symmetric S-shaped band bending. For $N_A\gg N_D$ nearly all the bending occurs on the n-side (the p-side is nearly flat, a “one-sided” diagram, with total drop $q{V}_{bi}$). To include bias, replace $V_{bi}$ by $V_{bi}-V_A$ in $x_n$, $x_p$, $V(x)$, and draw the n-side bands raised by $q{V}_A$ relative to the p-side, as in (5.39).

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