part04_p1

Chapter 3 (continued): Carrier Action

3.4 The Continuity and Diffusion Equations (continued)

3.4.1 Using the Continuity Equations (continued)

When the full continuity equations are used, the appropriate thermal recombination-generation (R-G) relationships (which may or may not be the special-case results of Eqs. 3.34), together with the concentration changes due to “other processes” (light, impact ionization, etc.), are substituted into Eqs. (3.46). One then seeks numerical solutions for $n(x,y,z,t)$ and $p(x,y,z,t)$.

Intuition: Think of the continuity equation as a bank statement for carriers in a tiny box: (change in balance) = (deposits minus withdrawals by current flowing across the walls) + (carriers created minus carriers destroyed inside). Solving it exactly in 3-D is a job for a computer. For pencil-and-paper work we shrink the problem to something we can do by hand, and that is what the next subsection does.

When a closed-form answer is wanted, the continuity equations are used only indirectly. The real starting point of such analyses is a simplified version, the minority carrier diffusion equation.

3.4.2 Minority Carrier Diffusion Equations

The “workhorse” minority carrier diffusion equations come from the continuity equations under seven simplifying assumptions:

1. The system is one-dimensional: all variables depend on at most one coordinate (say $x$).
2. The analysis is restricted to minority carriers.
3. $\mathcal{E}\approx 0$ in the semiconductor (or in the region being analysed).
4. The equilibrium minority carrier concentrations are not functions of position: $n_0\ne n_0(x)$, $p_0\ne p_0(x)$.
5. Low-level injection prevails.
6. Indirect thermal R-G (through R-G centers) is the dominant thermal mechanism.
7. There are no “other processes” except possibly photogeneration.

Derivation (electrons in a $p$-type material). For a 1-D system,

\frac{1}{q}\nabla\cdot\vec J_N \;\to\; \frac1q\frac{\partial J_N}{\partial x}\tag{3.47}

The electron current is drift plus diffusion,

J_N = q\left(\mu_n n\mathscr{E} + D_N\frac{\partial n}{\partial x}\right)\;\approx\; qD_N\frac{\partial n}{\partial x}\tag{3.48}

when $\mathcal{E}\approx 0$ and we care only about minority carriers (assumptions 2 and 3). Why can drift be dropped? $\mathcal{E}$ is small by assumption, and minority concentrations are small too, so the product $n\mathcal{E}$ is doubly tiny. (The same argument fails for majority carriers, whose $n$ is huge.) Since $n_0\ne n_0(x)$ and $n=n_0+\Delta n$,

\frac{\partial n}{\partial x}=\frac{\partial n_0}{\partial x}+\frac{\partial \Delta n}{\partial x}=\frac{\partial \Delta n}{\partial x}\tag{3.49}

Combining (3.47)–(3.49):

\frac1q\nabla\cdot\vec J_N = D_N\frac{\partial^2\Delta n}{\partial x^2}\tag{3.50}

For the remaining terms: R-G through centers plus low-level injection plus the minority-carrier restriction lets us use the special-case expression (3.34),

\left.\frac{\partial n}{\partial t}\right|_{\text{thermal R-G}}=-\frac{\Delta n}{\tau_n}\tag{3.51}

Simplification 7 gives

\left.\frac{\partial n}{\partial t}\right|_{\text{other processes}}=G_L\tag{3.52}

($G_L=0$ if the sample is dark). Finally, the equilibrium concentration never changes in time, so

\frac{\partial n}{\partial t}=\frac{\partial n_0}{\partial t}+\frac{\partial\Delta n}{\partial t}=\frac{\partial\Delta n}{\partial t}\tag{3.53}

Substituting (3.50)–(3.53) into the continuity equation (3.46a), and writing the analogous result for holes:

\boxed{\frac{\partial\Delta n_p}{\partial t}=D_N\frac{\partial^2\Delta n_p}{\partial x^2}-\frac{\Delta n_p}{\tau_n}+G_L}\tag{3.54a}

\boxed{\frac{\partial\Delta p_n}{\partial t}=D_P\frac{\partial^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{\tau_p}+G_L}\tag{3.54b}

These are the minority carrier diffusion equations. The subscripts remind us that they are valid only for minority carriers: electrons in $p$-type material ($\Delta n_p$) and holes in $n$-type material ($\Delta p_n$).

Intuition: Read (3.54b) as a story about a crowd of excess holes in a sea of electrons. The crowd (i) spreads out by random jostling (the $D_P{\partial }^2/\partial x^2$ term: large where the profile is sharply curved), (ii) dies off by recombination at rate $1/{\tau }_p$ per carrier, and (iii) is replenished by light at $G_L$. Three effects, three terms, one equation.

3.4.3 Simplifications and Solutions

Real problems often allow further simplifications. They are collected in Table 3.1.

Simplification	Effect on Eq. (3.54)
Steady state	$\partial \Delta n_p/\partial t\to 0$ (and $\partial \Delta p_n/\partial t\to 0$)
No concentration gradient (or no diffusion current)	$D_N{\partial }^2\Delta n_p/\partial x^2\to 0$, $D_P{\partial }^2\Delta p_n/\partial x^2\to 0$
No drift current, or $\mathcal{E}=0$	No further simplification ($\mathcal{E}\approx 0$ was already assumed in the derivation)
No thermal R-G	$\Delta n_p/{\tau }_n\to 0$, $\Delta p_n/{\tau }_p\to 0$
No light	$G_L\to 0$

Frequently used solutions to simplified forms are given in Table 3.2 (written for electrons in $p$-type material; holes in $n$-type are identical with $N\to P$, $n\to p$).

No.	Given	Simplified diffusion equation	Solution
1	Steady state, no light	$0=D_N\frac{d^2\Delta n_p}{d{x}^2}-\frac{\Delta n_p}{{\tau }_n}$	$\Delta n_p(x)=A{e}^{-x/L_N}+B{e}^{x/L_N}$, with $L_N=\sqrt{D_N{\tau }_n}$; $A,B$ are constants fixed by boundary conditions
2	No concentration gradient, no light	$\frac{d\Delta n_p}{dt}=-\frac{\Delta n_p}{{\tau }_n}$	$\Delta n_p(t)=\Delta n_p(0)e^{-t/{\tau }_n}$
3	Steady state, no concentration gradient	$0=-\frac{\Delta n_p}{{\tau }_n}+G_L$	$\Delta n_p=G_L{\tau }_n$
4	Steady state, no R-G, no light	$0=D_N\frac{d^2\Delta n_p}{d{x}^2}$	$\Delta n_p(x)=A+Bx$

Intuition: $L_N=\sqrt{D_N{\tau }_n}$ is the diffusion length: a carrier random-walks for a time ${\tau }_n$ before it dies, and a random walker with diffusivity $D$ wanders about $\sqrt{D\tau }$ from where it started (like a drunkard: after $N$ steps he is about $\sqrt{N}$ steps from the lamppost, not $N$). Solution 1 says a disturbance decays exponentially with this characteristic length; Solution 2 is radioactive-style decay; Solution 3 is a bathtub with a tap ($G_L$) and a drain ($1/{\tau }_n$) at balance; Solution 4 is what you get when nothing can be lost: a straight line.

3.4.4 Problem Solving

All the necessary information is now in hand, but it may not be obvious how to actually use the diffusion equations. Device analyses later in the text provide examples, but a few simple, well-defined problems illustrate the procedure. They also provide a basis for two supplemental concepts introduced in the next section.

Sample Problem No. 1 (photoconductor turn-on)

Problem. A uniformly donor-doped silicon wafer at room temperature is suddenly illuminated at $t=0$. Assume $N_D=10^{14}/{\text{cm}}^3$, ${\tau }_p=10^{-6}$ s, and a light-induced creation of $G_L=10^{17}$ electron-hole pairs per cm${}^3$-s throughout the semiconductor. Determine $\Delta p_n(t)$ for $t>0$.

Step 1 — Review the information. The semiconductor is Si, $T=300$ K, $N_D=10^{14}/{\text{cm}}^3$ everywhere, $G_L=10^{17}/{\text{cm}}^3\text{-s}$ at all points. The statement also implies equilibrium for $t<0$.

Step 2 — Equilibrium state. In Si at room temperature $n_i=10^{10}/{\text{cm}}^3$. Since $N_D\gg n_i$:

$\begin{array}{rl}{n}_0& \approx N_D=10^{14}{\text{cm}}^{-3}\\ p_0& =\frac{n_i^2}{N_D}=\frac{10^{20}}{10^{14}}=10^6{\text{cm}}^{-3}\end{array}$

Doping is uniform, so $n_0$, $p_0$ are the same everywhere.

Step 3 — Qualitative analysis. Before $t=0$, $\Delta p_n=0$. After $t=0$ the light adds pairs and $\Delta p_n$ starts to grow. But a growing $\Delta p_n$ raises the recombination rate (proportional to $\Delta p_n$), so more and more excess holes vanish per second. Eventually recombination balances light generation and a steady state results. Hence we expect $\Delta p_n(t)$ to start at zero, rise at a decreasing rate, and level off. At steady state $G_L=\Delta p_n(\infty )/{\tau }_p$, so $\Delta p_n{|}_{\max }=G_L{\tau }_p$ (if low-level injection holds).

Intuition: A bathtub filling from a tap while draining through a hole: the deeper the water, the faster it drains. The level rises quickly at first, then flattens at the height where inflow equals outflow.

Step 4 — Quantitative analysis. Check the assumptions behind the diffusion equation: only the minority-carrier concentration matters; equilibrium concentrations don’t depend on position; indirect R-G dominates in Si; the only “other process” is photogeneration; uniform photogeneration makes the perturbation position-independent, so there is no gradient and $\mathcal{E}=0$; and low-level injection must be verified at the end. Start from

\frac{\partial\Delta p_n}{\partial t}=D_P\frac{\partial^2\Delta p_n}{\partial x^2}-\frac{\Delta p_n}{\tau_p}+G_L\tag{3.55}

with initial condition

\Delta p_n(t)\big|_{t=0}=0\tag{3.56}

Because $\Delta p_n$ is not a function of position, the diffusion term vanishes and

\frac{d\Delta p_n}{dt}+\frac{\Delta p_n}{\tau_p}=G_L\tag{3.57}

The general solution is the particular (steady) solution plus a homogeneous decaying exponential:

\Delta p_n(t)=G_L\tau_p+Ae^{-t/\tau_p}\tag{3.58}

The initial condition gives

\begin{aligned}0&=G_L\tau_p+A\\ A&=-G_L\tau_p\end{aligned}\tag{3.59}

so

\boxed{\Delta p_n(t)=G_L\tau_p\left(1-e^{-t/\tau_p}\right)}\tag{3.60}

Numbers:

$\begin{array}{rl}{G}_L{\tau }_p& =(10^{17}{\text{cm}}^{-3}{\text{s}}^{-1})(10^{-6}\text{s})=10^{11}{\text{cm}}^{-3}\\ \Delta p_n(t)& =10^{11}\left(1-e^{-t/(10^{-6}\text{s})}\right){\text{cm}}^{-3}\end{array}$

Low-level check: $\Delta p_{\max }=10^{11}\ll n_0=10^{14}$, so low-level injection holds. At $t={\tau }_p$, $\Delta p_n=0.632\times 10^{11}$; at $t=3{\tau }_p$, $0.95\times 10^{11}{\text{cm}}^{-3}$.

Step 5 — Examine the solution. “Failing to examine the mathematical solution to a problem is like growing vegetables and then failing to eat the produce.” $G_L{\tau }_p$ has units of a concentration (number/cm${}^3$), so the result is dimensionally correct. Fig. 3.25 plots the result: $\Delta p_n$ rises from zero at $t=0$, the slope is initially $G_L$ and steadily decreases, and it saturates at $G_L{\tau }_p$ (within a few percent after roughly $3$–$5{\tau }_p$), as predicted qualitatively.

Epilogue — link to photoconductive decay (Subsection 3.3.4). The stroboscope light in that measurement can be modeled by the pulse train of Fig. 3.26(a) (a square pulse of height $G_{L0}$ that turns on and off). Eq. (3.60) approximately describes the build-up during a pulse. But the stroboscope pulse lasts $t_{on}\approx 1\mu$s while ${\tau }_p\lesssim 150\mu$s; with $t_{on}/{\tau }_p\ll 1$ we see only the very first portion of the Fig. 3.25 transient before the light is switched off and the sample decays back to equilibrium.

For the light-off part, the derivation parallels the light-on one except that $\Delta p_n$ at the start of the decay equals the value at the end of the light-on transient, and $G_L=0$ in Eqs. (3.55) and (3.57). The solution of (3.57) with $G_L=0$ is Solution No. 2 of Table 3.2:

\Delta p_n(t)=\Delta p_n(0)\,e^{-t/\tau_p}\tag{3.61}

with $t=0$ reset to the start of the light-off transient. This is identical to Eq. (3.37) in the measurement analysis. Fig. 3.26(b) sketches the combined light-on/light-off solution: a rising exponential toward $G_L{\tau }_p$ during the pulse, then a falling exponential afterwards.

Sample Problem No. 2 (steady-state diffusion from an illuminated end)

Problem. As pictured in Fig. 3.27(a), the $x=0$ end of a uniformly doped semi-infinite bar of silicon with $N_D=10^{14}/{\text{cm}}^3$ is illuminated so as to create $\Delta p_{n0}=10^{10}/{\text{cm}}^3$ excess holes at $x=0$. The wavelength of the illumination is such that no light penetrates into the interior ($x>0$) of the bar. Determine $\Delta p_n(x)$.

Solution — Step 1: information. The semiconductor is again uniformly doped Si with $N_D=10^{14}/{\text{cm}}^3$. Steady state is inferred because we are asked for $\Delta p_n(x)$ and not $\Delta p_n(x,t)$. At $x=0$, $\Delta p_n(0)=\Delta p_{n0}=10^{10}/{\text{cm}}^3$, and $\Delta p_n\to 0$ as $x\to \infty$ (a consequence of the semi-infinite bar: a perturbation created by non-penetrating light cannot reach infinity). Non-penetrating light means $G_L=0$ for $x>0$. The temperature is not stated, so assume $T=300$ K.

Step 2: equilibrium. With the light off, the bar reverts to the same equilibrium as Sample Problem 1: $n_0=10^{14}/{\text{cm}}^3$, $p_0=10^6/{\text{cm}}^3$, uniform throughout.

Step 3: qualitative. Light first creates excess carriers right at $x=0$. With more carriers there than elsewhere, diffusion spreads the excess into the bar. But excess holes inside the bar enhance recombination, and a minority hole lives only about ${\tau }_p$ on average, so fewer and fewer survive the deeper we go. In steady state we expect $\Delta p_n(x)$ to fall monotonically from $\Delta p_{n0}$ at $x=0$ to $0$ as $x\to \infty$.

Intuition: Picture a pile of sand poured at the edge of a beach where the sand slowly dissolves in water. The sand creeps inland (diffusion) but keeps dissolving (recombination); the pile fades out exponentially.

Step 4: quantitative. The system is 1-D; only minority holes matter; equilibrium concentrations are position-independent; indirect R-G dominates; no other processes for $x>0$; low-level injection is obvious ($10^{10}\ll 10^{14}$). The one worry is whether $\mathcal{E}\approx 0$. With the light on, a nonuniform hole distribution (positive charge) appears near $x=0$, but the pile-up is tiny (at most $\Delta p_{n0}$), so the field is small. Moreover, in problems of this type the majority electrons redistribute to partly cancel the minority charge. Experience thus justifies $\mathcal{E}\approx 0$ and the use of the minority-carrier diffusion equation.

Under steady state with $G_L=0$ for $x>0$, the hole equation reduces to

D_P\frac{d^2\Delta p_n}{dx^2}-\frac{\Delta p_n}{\tau_p}=0\qquad (x>0)\tag{3.62}

with boundary conditions

\Delta p_n\big|_{x=0^+}=\Delta p_{n0}\tag{3.63}

\Delta p_n\big|_{x\to\infty}=0\tag{3.64}

This is Solution No. 1 of Table 3.2:

\Delta p_n(x)=Ae^{-x/L_P}+Be^{x/L_P},\qquad L_P=\sqrt{D_P\tau_p}\tag{3.65, 3.66}

Because $e^{x/L_P}\to \infty$ as $x\to \infty$, the only way to satisfy (3.64) is $B=0$. Then (3.63) gives

A=\Delta p_{n0}\tag{3.67}

so

\boxed{\Delta p_n(x)=\Delta p_{n0}\,e^{-x/L_P}}\tag{3.68}

Numerically, with $\Delta p_{n0}=10^{10}{\text{cm}}^{-3}$ and (using the same ${\tau }_p=10^{-6}$ s and ${\mu }_p\approx 458{\text{cm}}^2/\text{V-s}$ as in Section 3.5.1) $L_P\approx 3.44\times 10^{-3}$ cm $=34.4\mu$m:

$\Delta p_n(x)=10^{10}\exp \left(-\frac{x}{3.44\times 10^{-3}\text{cm}}\right){\text{cm}}^{-3}$

Step 5: examine. Eq. (3.68) is plotted in Fig. 3.27(b): a monotonically decreasing exponential from $\Delta p_{n0}$ at $x=0$ toward zero, with characteristic decay length $L_P$, in agreement with the qualitative argument.

3.5 Supplemental Concepts

3.5.1 Diffusion Lengths

The situation of Sample Problem 2 (excess minority carriers appear on a plane, diffuse away, and fall off exponentially with decay length $L_P$) occurs so often that the characteristic length has a special name. Specifically,

L_P\equiv\sqrt{D_P\tau_p}\quad\text{(minority holes in an }n\text{-type material)}\tag{3.69a}

L_N\equiv\sqrt{D_N\tau_n}\quad\text{(minority electrons in a }p\text{-type material)}\tag{3.69b}

are the minority carrier diffusion lengths.

Physically, $L_P$ and $L_N$ represent the average distance minority carriers diffuse into a sea of majority carriers before being annihilated. This is consistent with Sample Problem 2, where the average position of the excess carriers is

\langle x\rangle=\frac{\displaystyle\int_0^\infty x\,\Delta p_n(x)\,dx}{\displaystyle\int_0^\infty \Delta p_n(x)\,dx}=\frac{L_P^2}{L_P}=L_P\tag{3.70}

Intuition (the piranha river): Minority carrier diffusion into a sea of majority carriers is like a small group of animals trying to cross a piranha-infested stretch of the Amazon. $L_P$ and $L_N$ are the average distance the animals get into the river before being eaten. A longer lifetime means slower piranhas; a larger $D$ means faster swimmers.

Size estimate. Take $T=300$ K, $N_D=10^{14}/{\text{cm}}^3$ Si, and ${\tau }_p=10^{-6}$ s. With ${\mu }_p\approx 458{\text{cm}}^2/\text{V-s}$ and the Einstein relation $D_P=(kT/q){\mu }_p$:

$\begin{array}{rl}{L}_P& =\sqrt{D_P{\tau }_p}=\sqrt{(kT/q){\mu }_p{\tau }_p}\\ & ={\left[(0.0259)(458)(10^{-6})\right]}^{1/2}\\ & =3.44\times 10^{-3}\text{cm}\end{array}$

That is about $34\mu$m. Although this is fairly representative, diffusion lengths can vary over several orders of magnitude because lifetimes vary widely.

3.5.2 Quasi-Fermi Levels

Quasi-Fermi levels are energy levels used to specify the carrier concentrations inside a semiconductor under nonequilibrium conditions.

Why we need them. In Sample Problem 1, equilibrium held before $t=0$ with $n_0=N_D=10^{14}/{\text{cm}}^3$ and $p_0=10^6/{\text{cm}}^3$. The band diagram in Fig. 3.28(a) shows this, and the Fermi level position alone tells us the concentrations:

n_0=n_ie^{(E_F-E_i)/kT}\tag{3.71a}

p_0=n_ie^{(E_i-E_F)/kT}\tag{3.71b}

Under equilibrium there is a one-to-one correspondence between $E_F$ and the carrier concentrations: knowing $E_F$ specifies $n_0$ and $p_0$, and vice versa.

Now consider the steady-state situation in Problem 1 for $t\gtrsim {\tau }_p$. Then $\Delta p=G_L{\tau }_p=10^{11}/{\text{cm}}^3$, $p=p_0+\Delta p\approx 10^{11}/{\text{cm}}^3$, and $n\approx n_0=10^{14}/{\text{cm}}^3$. Although $n$ is essentially unperturbed, $p$ has grown by five orders of magnitude, so the diagram of Fig. 3.28(a) no longer describes the system. The Fermi level is defined only for equilibrium and cannot be used to deduce carrier concentrations in a nonequilibrium system.

Definition. We restore the convenience by introducing two energies, $F_N$ (quasi-Fermi level for electrons) and $F_P$ (for holes), defined to relate to the nonequilibrium concentrations exactly the way $E_F$ relates to equilibrium ones. For a nondegenerate semiconductor:

n\equiv n_ie^{(F_N-E_i)/kT}\quad\text{or}\quad F_N=E_i+kT\ln\frac{n}{n_i}\tag{3.72a}

p\equiv n_ie^{(E_i-F_P)/kT}\quad\text{or}\quad F_P=E_i-kT\ln\frac{p}{n_i}\tag{3.72b}

$F_N$ and $F_P$ are conceptual constructs whose values are totally determined from prior knowledge of $n$ and $p$. They are defined so that when a perturbed system relaxes toward equilibrium, $F_N\to E_F$, $F_P\to E_F$, and Eqs. (3.72) $\to$ Eqs. (3.71).

Intuition: In equilibrium, one number ($E_F$) sets the “water level” for both electrons and holes. Out of equilibrium the two populations no longer share a single water level, so we give each its own: $F_N$ for electrons, $F_P$ for holes. The gap between $F_N$ and $F_P$ is a direct measure of how far from equilibrium we are (in equilibrium they coincide; $np=n_i^2{e}^{(F_N-F_P)/kT}$).

Application. Consider the $t\gtrsim {\tau }_p$ state of Sample Problem 1. Since $n\approx n_0$, $F_N\approx E_F$. Substituting $p=10^{11}/{\text{cm}}^3$, $n_i=10^{10}/{\text{cm}}^3$, $kT=0.0259$ eV into (3.72b):

$\begin{array}{rl}{F}_P& =E_i-kT\ln \frac{p}{n_i}\\ & =E_i-(0.0259)\ln (10)\text{eV}\\ & \approx E_i-0.06\text{eV}\end{array}$

Erasing $E_F$ from the band diagram and drawing lines at $F_N$ and $F_P$ gives Fig. 3.28(b). This diagram tells any observer at a glance that the system is out of equilibrium (two separate levels), and, compared with part (a), that low-level injection is occurring (only $F_P$ moved appreciably; $F_N$ stays put), creating a minority hole concentration in excess of $p_0$.

Compact current relations. Quasi-Fermi levels also recast the current equations elegantly. The standard total hole current is

\vec J_P=q\mu_pp\vec{\mathscr E}-qD_P\nabla p\qquad\text{(same as 3.18a)}\tag{3.73}

Differentiating (3.72b):

\nabla p=\frac{n_i}{kT}e^{(E_i-F_P)/kT}\left(\nabla E_i-\nabla F_P\right)=\frac{p}{kT}\left(\nabla E_i-\nabla F_P\right)\tag{3.74}

The identity $\vec{\mathcal{E}}=\nabla E_i/q$ (3-D version of Eq. 3.15) turns (3.74) into $\nabla p=\frac{p}{kT}\left(q\vec{\mathcal{E}}-\nabla F_P\right)$. Eliminating $\nabla p$ from (3.73):

\begin{aligned}\vec J_P&=q\mu_pp\vec{\mathscr E}-qD_P\frac{p}{kT}\left(q\vec{\mathscr E}-\nabla F_P\right)\&=qp\vec{\mathscr E}\left(\mu_p-\frac{qD_P}{kT}\right)+\frac{qD_P}{kT}\,p\,\nabla F_P\end{aligned}\tag{3.75}

From the Einstein relationship $q{D}_P/kT={\mu }_p$, so the field terms cancel exactly and

\boxed{\vec J_P=\mu_p\,p\,\nabla F_P}\tag{3.76a}

and similarly

\boxed{\vec J_N=\mu_n\,n\,\nabla F_N}\tag{3.76b}

Since ${\vec{J}}_P\propto \nabla F_P$ and ${\vec{J}}_N\propto \nabla F_N$, we arrive at a general interpretation of band diagrams with quasi-Fermi levels: a quasi-Fermi level that varies with position ($d{F}_P/dx\ne 0$ or $d{F}_N/dx\ne 0$) shows at a glance that current is flowing in the semiconductor. (In equilibrium $E_F$ is flat and no net current flows, the same statement with $F_N=F_P=E_F$.)

Intuition: $F$ plays the role of a chemical-plus-electrical “pressure” for carriers, like the height of water in connected reservoirs. Current flows only where the level has a slope, and how much flows depends on how many carriers ($n$ or $p$) and how mobile they are.

Exercise 3.5

Problem. In Sample Problem 2, nonpenetrating illumination of a semiconductor bar caused a steady-state excess hole concentration $\Delta p_n(x)=\Delta p_{n0}\exp (-x/L_P)$. Given low-level injection and $p=p_0+\Delta p$, we can state for the illuminated sample

$n\approx n_0,p=p_0+\Delta p_{n0}{e}^{-x/L_P}$

(a) Using Eqs. (3.72), establish relationships for $F_N$ and $F_P$ in the illuminated bar. (b) Simplify the $F_P$ expression in the region where $\Delta p_n(x)\gg p_0$. (c) Using parts (a) and (b), sketch the band diagrams for the bar of Sample Problem 2 under equilibrium and illuminated steady-state conditions (assume $\mathcal{E}=0$ in the illuminated bar). (d) Is there a hole current in the illuminated bar under steady state? Explain. (e) Is there an electron current in the illuminated bar under steady state? Explain.

Solution.

(a) Since $n\approx n_0$, Eq. (3.72a) gives $F_N\approx E_F$ (the equilibrium value, independent of $x$). Substituting the $p$-expression into (3.72b):

$\begin{array}{rl}{F}_P& =E_i-kT\ln \frac{p}{n_i}\\ & =E_i-kT\ln \left[\frac{p_0}{n_i}+\frac{\Delta p_{n0}}{n_i}{e}^{-x/L_P}\right]\end{array}$

(b) If $\Delta p_n(x)\gg p_0$, then $(\Delta p_{n0}/n_i)e^{-x/L_P}\gg p_0/n_i$, so

$\begin{array}{rl}{F}_P& \approx E_i-kT\ln \left[\frac{\Delta p_{n0}}{n_i}{e}^{-x/L_P}\right]\\ & =E_i-kT\ln \frac{\Delta p_{n0}}{n_i}+kT\frac{x}{L_P}\end{array}$

That is, $F_P$ is a linear function of $x$ with slope $kT/L_P$.

(c) Numbers: $\Delta p_{n0}=10^{10}/{\text{cm}}^3$, $n_i=10^{10}/{\text{cm}}^3$, $n_0=N_D=10^{14}/{\text{cm}}^3$, $p_0=n_i^2/N_D=10^6/{\text{cm}}^3$.

• (i) Near $x=0$, $\Delta p_n\gg p_0$ and $F_P$ is a linear function of $x$.
• (ii) At $x=0$, $\Delta p_{n0}=n_i$, so the part (b) result gives $F_P=E_i$.
• (iii) For large $x$, $F_P$ eventually approaches $F_N=E_F$ (where $\Delta p_n\lesssim p_0$; this happens when $10^{10}{e}^{-x/L_P}\approx 10^6$, i.e. $x\approx L_P\ln 10^4\approx 9.2L_P$).
• (iv) $F_N-E_i\approx E_F-E_i=kT\ln (N_D/n_i)=0.0259\ln (10^4)\approx 0.24$ eV.

(The book’s printed number, 0.30 eV, corresponds to a slightly different $N_D/n_i$ ratio; with the stated values of the problem the result is 0.24 eV. It is only a sketch quantity.)

Utilizing this information: in the equilibrium diagram, a flat $E_F$ lies $0.24$ eV above the flat $E_i$ (a band diagram with $E_C$, $E_i$, $E_F$, $E_V$ all horizontal). In the illuminated steady-state diagram, $E_C$, $E_i$, $E_V$ remain flat (no field), $F_N$ stays flat at $0.24$ eV above $E_i$, while $F_P$ starts at $E_i$ at $x=0$ and rises linearly with slope $kT/L_P$ until it merges with $F_N$ far into the bar.

(d) Assuming $p\ne 0$, Eq. (3.76a) says there is a hole current wherever $d{F}_P/dx\ne 0$. There is obviously a hole current in the illuminated bar near $x=0$ (numerically $J_P={\mu }_{p}pkT/L_P$ in magnitude, equivalent to the diffusion current $q{D}_P\Delta p_{n0}/L_P$ at $x=0$). Holes diffuse into the bar and recombine there.

(e) Appearances can deceive: $J_N\ne 0$ near $x=0$. One might conclude from part (c) that $d{F}_N/dx=0$ and hence $J_N=0$. But under steady state the total current $J_N+J_P$ must be the same at all points in the bar, and since no current leaves the bar at $x=0$, the total must vanish there, so $J_N(x)=-J_P(x)$ and we know $J_P\ne 0$ near $x=0$. The apparent discrepancy comes from the fact that $J_N$ is proportional to both $n$ and $\nabla F_N$. Since the majority electron concentration is much larger than the minority hole concentration, $d{F}_N/dx$ must be correspondingly much smaller than $d{F}_P/dx$: the slope of $F_N$ is simply too small to be detected by eye on the band diagram.

3.6 Summary and Concluding Comments

Most of the chapter was devoted to the three primary types of carrier action in semiconductors: drift, diffusion, and recombination-generation. In each case the action was defined and visualized. Drift is charged-particle motion in response to an applied electric field; diffusion is the net movement of particles from regions of high concentration to regions of low concentration due to the random thermal motion of the carriers; recombination and generation are respectively the annihilation and creation of carriers. The quantitative effect of each action was then analyzed: drift and diffusion give rise to particle currents (Eqs. 3.4 and 3.17–3.19); recombination-generation changes the carrier concentrations as a function of time (Eqs. 3.34).

Associated with each carrier action is a “constant of the motion,” a material-based parameter that specifies its magnitude: the carrier mobilities (drift), the diffusion coefficients (diffusion), and the minority carrier lifetimes (R-G). In the major semiconductors the mobilities are well characterized as functions of doping and temperature (data in Figs. 3.5 and 3.7; an empirical fit for Si in Exercise 3.1). Diffusion coefficients in nondegenerate semiconductors follow from the mobilities via the Einstein relationship (Eqs. 3.25). Lifetimes are normally measured for a given sample.

Although introduced separately, these actions occur simultaneously. Mathematically combining them leads to the continuity equations (Eqs. 3.46), which simplify to the minority carrier diffusion equations (Eqs. 3.54) under conditions common in practice. Further simplifications are in Table 3.1 and widely used solutions in Table 3.2. The continuity or minority-carrier diffusion equations, combined with the other relationships of the chapter, let one model the state of a semiconductor subject to an external perturbation. The more important working relationships are organized in Table 3.3.