Chapter 6: pn Junction Diode: I-V Characteristics

Where we are on the map. In Chapter 5 we studied the electrostatics of a pn junction: what the bands, fields and charges look like when the junction just sits there. Now we ask the question that matters for every circuit: if I push a voltage $V_{A}$ across this thing, how much current $I$ comes out? The answer, the $I$ - $V$ characteristic, is the “personality” of the diode. We build that personality in three acts: (1) an idealized diode where everything goes right, giving the famous Shockley equation; (2) a comparison with real diodes, where we find things going wrong (breakdown, extra recombination current, series resistance, high-level injection); and (3) two supplementary tools (charge control and the narrow-base diode) that will pay off when we reach transistors.

6.1 The Ideal Diode Equation

The ideal diode is a deliberately simplified diode. It is wonderful for insight, because it lets us see the machinery without the clutter. The goal of this section is to derive the ideal diode equation

I = I_{0} (e^{q V_{A} / k T} - 1)

but, as the author stresses, the journey (the solution strategy, the boundary conditions, the pictures of what carriers do) matters more than the destination. We first “derive” the result with no equations at all, using only the energy band diagram.

6.1.1 Qualitative Derivation

Setting the stage (Fig. 6.1a, equilibrium, $V_{A} = 0$ ). Draw the equilibrium band diagram of a pn junction. Sprinkle dots (electrons) and circles (holes) over it to show where the carriers live:

On the quasineutral n-side there are very many electrons near $E_{c}$ and very few holes. The electron “pyramid” of dots thins out roughly exponentially as you climb higher in the conduction band (a Boltzmann tail).
On the quasineutral p-side there are very many holes near $E_{v}$ and very few electrons. The hole population thins out roughly exponentially as you move down in the valence band.

Picture it: a mountain pass. Imagine two crowded valleys (n-side with electrons, p-side with holes) separated by a mountain ridge. People (carriers) jiggle around with thermal energy. Most electrons in the n-valley do not have enough energy to climb the ridge and get reflected back. But a few high-energy ones make it over. That trickle is diffusion.

Now consider the two kinds of carrier on each side:

Majority carriers (n-side electrons, p-side holes) see the potential hill. Only the energetic few surmount it and cross. This is diffusion current.
Minority carriers (p-side electrons, n-side holes) see no hill, because for them the potential energy slope runs downhill. If a minority electron wanders into the depletion region, the field sweeps it to the n-side immediately. This is a drift current.

At equilibrium these precisely cancel, for electrons and for holes separately:

I_{N}^{diff} = I_{N}^{drift}, I_{P}^{diff} = I_{P}^{drift} \Rightarrow I = 0.

Forward bias (Fig. 6.1b, $V_{A} > 0$ ). The most important change is that the potential hill is lowered. The minority carriers still wander in and fall down the hill at the same rate as before (the “waterfall” is unaffected). But now many more n-side electrons and p-side holes can surmount the lowered hill. They give an electron current $I_{N}$ and a hole current $I_{P}$ , both directed from p to n (conventional current), so the total $I = I_{N} + I_{P}$ flows in the correct direction for a forward-biased diode.

Why exponential? Two facts combine:

The hill is lowered by $q V_{A}$ , i.e. linearly in $V_{A}$ .
The number of carriers with energy greater than a given height varies exponentially with that height (Boltzmann factor $e^{- E / k T}$ ).

So the number that can surmount the hill grows as $e^{q V_{A} / k T}$ : forward current increases exponentially with $V_{A}$ (Fig. 6.1d).

Reverse bias (Fig. 6.1c, $V_{A} < 0$ ). The hill is raised. Even a modest reverse bias (more than a few $k T / q$ ) shuts majority-carrier diffusion down to a negligible level. The minority carriers wandering into the depletion region and being swept across are unaffected. So a small current flows from n to p, associated with minority carriers (so it is tiny), and because the minority-carrier drift current depends only on how many wander in per second and not on the height of the hill, it saturates.

Analogy: the waterfall. The amount of water going over a waterfall depends on how much water the river supplies, not on how tall the falls are. Raise the hill (more reverse bias) and the minority carriers fall farther, but the number that fall per second is unchanged.

Calling the saturation value $- I_{0}$ , the overall behavior must be

I = I_{0} (e^{V_{A} / V_{ref}} - 1) (6.1)

which equals the ideal diode equation if $V_{ref} = k T / q$ . We have found the form of the diode equation without writing a single transport equation. The same picture also explains rectification: forward bias lowers the hill, so large numbers of majority carriers are injected; reverse bias raises it, leaving only a residual minority-carrier current.

The “big picture” objection

The author recounts being asked: “Yes, but doesn’t injecting majority carriers (forward) or extracting minority carriers (reverse) cause a charge build-up in the device?” The short answer is that we are in steady state, so nothing builds up. The deeper answer is the “big picture” of Fig. 6.2, drawn for reverse bias (capacitor-like plates at the outer ends of the band diagram represent the ohmic contacts):

Minority electrons wander from the p-side into the depletion region and are swept to the n-side. Each one that leaves is replaced by thermal generation through an R-G center (an electron jumps $E_{v} \to E_{T} \to E_{c}$ on the p-side near the depletion region).
Similarly, each swept n-side hole is replaced by generation on the n-side.
The electrons that arrive on the n-side (and those generated there) create a slight local excess of majority electrons next to the depletion region. The tiny local electric field this excess creates pushes the adjacent electrons toward the contact, who push their neighbors, and so on. This displacement propagates with great speed through the whole in-line group of n-side electrons. The electron adjacent to the contact is pushed into the metal and out into the external circuit.
On the p-side the story is mirrored with holes: the excess at the depletion edge pushes in-line holes along, and holes at the contact recombine with electrons from the metal, completing the circle.

Analogy: a crowded room with two doors. Latecomers to a party force their way in through one door of an absolutely crammed room. A like number of people get pushed out the other door. Nobody runs across the room; each person only moves a little. The “signal” travels fast; the individual carriers move slowly.

A side observation from Fig. 6.2: both electrons and holes carry current through the depletion region, but far from the junction on the p-side the hole current dominates, and far on the n-side the electron current dominates. The total current is the same everywhere, while the electron and hole components vary with position.

6.1.2 Quantitative Solution Strategy

“Going into a football game, coaches always have a game plan.” The mathematics of the ideal diode derivation is short; the strategy is what you must understand, because you will reuse it for other problems and for modified theories.

General considerations: assumptions

The diode is operated under steady state conditions.
A nondegenerately doped step junction models the doping profile.
The diode is one-dimensional.
Low-level injection prevails in the quasineutral regions.
No processes other than drift, diffusion and thermal recombination-generation occur inside the diode. In particular $G_{L} = 0$ .

(These were all explicitly or implicitly used in establishing the pn junction electrostatics.)

Current relationships available

I = A J (A = cross-sectional area) (6.2)

J = J_{N} (x) + J_{P} (x) (6.3)

J_{N} = q μ_{n} n E + q D_{N} \frac{d n}{d x} (6.4a)

J_{P} = q μ_{p} p E - q D_{P} \frac{d p}{d x} (6.4b)

Eq. (6.3) says the total current density is constant throughout the diode, but its electron and hole parts change with position. If we had exact $E$ , $n$ , $p$ versus $x$ we would be done. We do not; we only have the approximate three-region electrostatic solution of Fig. 6.3:

Region	Extent	Field
Quasineutral p-region	$x \leq - x_{p}$	$E \approx 0$
Depletion region	$- x_{p} \leq x \leq x_{n}$	$E \neq = 0$
Quasineutral n-region	$x \geq x_{n}$	$E \approx 0$

In the quasineutral regions $E \approx 0$ and low-level injection holds, so the minority carrier diffusion equations can be used.

Quasineutral region considerations

Under steady state with $G_{L} = 0$ :

0 = D_{N} \frac{d ^{2} Δ n _{p}}{d x ^{2}} - \frac{Δ n _{p}}{τ _{n}} x \leq - x_{p} (6.5a)

0 = D_{P} \frac{d ^{2} Δ p _{n}}{d x ^{2}} - \frac{Δ p _{n}}{τ _{p}} x \geq x_{n} (6.5b)

Since $E = 0$ and $d n_{0} / d x = d p_{0} / d x = 0$ , Eqs. (6.4) reduce to (minority carrier currents are pure diffusion):

J_{N} = q D_{N} \frac{d Δ n _{p}}{d x} x \leq - x_{p} (6.6a)

J_{P} = - q D_{P} \frac{d Δ p _{n}}{d x} x \geq x_{n} (6.6b)

The catch. These give $J_{N} (x)$ only in the p-side quasineutral region and $J_{P} (x)$ only in the n-side quasineutral region: non-overlapping pieces. To evaluate $J = J_{N} + J_{P}$ we need at least one point where we know both, possibly by extrapolation.

Depletion region considerations

Here the full continuity equations must be used. With the assumptions above they become

0 = \frac{1}{q} \frac{d J _{N}}{d x} + \frac{\partial n}{\partial t}_{thermal R-G} (6.7a)

0 = - \frac{1}{q} \frac{d J _{P}}{d x} + \frac{\partial p}{\partial t}_{thermal R-G} (6.7b)

Now make the crucial additional assumption that thermal R-G is negligible throughout the depletion region: $\partial n / \partial t ∣_{R-G} = \partial p / \partial t ∣_{R-G} = 0$ . Then $d J_{N} / d x = 0$ and $d J_{P} / d x = 0$ . The carrier currents are constants across the depletion region, edges included:

J_{N} (- x_{p} \leq x \leq x_{n}) = J_{N} (- x_{p}) (6.8a)

J_{P} (- x_{p} \leq x \leq x_{n}) = J_{P} (x_{n}) (6.8b)

And $J_{N} (- x_{p})$ and $J_{P} (x_{n})$ can be obtained from the quasineutral solutions evaluated at the depletion edges. Hence

J = J_{N} (- x_{p}) + J_{P} (x_{n}) (6.9)

We now have a plan: solve for the minority carrier current densities in the quasineutral regions, evaluate them at the depletion edges, add, multiply by $A$ .

A word of honesty. “No thermal R-G in the depletion region” is really the defining property of the ideal diode. There is no a priori reason it should be true; it simply gives a simple answer. We will check it against experiment in Section 6.2 (and find it fails at small forward biases and all reverse biases in room-temperature Si).

Boundary conditions

Two boundary conditions are needed for each of Eqs. (6.5a) and (6.5b): the excess concentrations at the ohmic contacts and at the depletion region edges (Fig. 6.4a).

At the ohmic contacts. The ideal diode is taken to be a wide-base diode: the contacts are several minority-carrier diffusion lengths or more from the depletion edges. Any perturbation created at a depletion edge dies away before reaching the contact, so the contacts may be placed at $x = \pm \infty$ :

Δ n_{p} (x \to - \infty) = 0 (6.10a)

Δ p_{n} (x \to + \infty) = 0 (6.10b)

At the depletion region edges. Use the quasi-Fermi level formalism. Multiply the exponential forms of Eqs. (3.72a) and (3.72b), $n = n_{i} e^{(F_{N} - E_{i}) / k T}$ and $p = n_{i} e^{(E_{i} - F_{P}) / k T}$ :

n p = n_{i}^{2} e^{(F_{N} - F_{P}) / k T} (6.11)

This is valid everywhere in the diode under arbitrary operating conditions. We do not know $F_{N} (x)$ and $F_{P} (x)$ beforehand, but as envisioned in Fig. 6.4b it is reasonable that they vary monotonically from $F_{P} = E_{F}$ far on the p-side to $F_{N} = E_{F}$ far on the n-side. The Fermi level difference between the far ends is $q V_{A}$ , and monotonic variation makes $F_{N} - F_{P} \leq q V_{A}$ at all points. Assume the equal sign holds throughout the depletion region: the quasi-Fermi levels are constant across the depletion region, $F_{N} - F_{P} = q V_{A}$ . Then

n p = n_{i}^{2} e^{q V_{A} / k T}, - x_{p} \leq x \leq x_{n} (6.12)

Eq. (6.12) is called the law of the junction.

Evaluate at the p-edge ( $x = - x_{p}$ ), where $p \approx N_{A}$ (low-level injection: the majority concentration is unchanged):

n (- x_{p}) p (- x_{p}) = n (- x_{p}) N_{A} = n_{i}^{2} e^{q V_{A} / k T} (6.13)

n_{p} (- x_{p}) = \frac{n _{i}^{2}}{N _{A}} e^{q V_{A} / k T} (6.14)

Subtract the equilibrium value $n_{p 0} = n_{i}^{2} / N_{A}$ :

Δ n_{p} (- x_{p}) = \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1) (6.15)

At the n-edge ( $x = x_{n}$ ), where $n \approx N_{D}$ :

n (x_{n}) p (x_{n}) = N_{D} p (x_{n}) = n_{i}^{2} e^{q V_{A} / k T} (6.16)

p_{n} (x_{n}) = \frac{n _{i}^{2}}{N _{D}} e^{q V_{A} / k T} (6.17)

Δ p_{n} (x_{n}) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.18)

Meaning. Forward bias ( $V_{A} > 0$ ) raises the minority concentration at the depletion edge above equilibrium by an exponential factor. Reverse bias beyond a few $k T / q$ drives the edge minority concentration essentially to zero: $Δ n_{p} \to - n_{i}^{2} / N_{A} = - n_{p 0}$ . The depletion region acts as a “minority carrier sink.”

The assumption $F_{N} - F_{P} = q V_{A}$ across the depletion region is central and critical. It is equivalent to assuming $J_{N} \approx 0$ and $J_{P} \approx 0$ relative to their individual drift and diffusion components inside the depletion region (Problem 6.5 carries out this alternative, Shockley-style derivation). Like the constant quasi-Fermi assumption, and contrary to statements in some texts, there is no simple a priori justification of this assumption. Lengthy subsidiary analyses and comparison with experiment indicate it is typically justified.

”Game plan” summary

To get the current in an ideal diode versus $V_{A}$ :

Solve the minority carrier diffusion equations (6.5) with boundary conditions (6.10) and (6.15)/(6.18) to get $Δ n_{p}$ and $Δ p_{n}$ .
Compute the minority carrier current densities in the quasineutral regions using Eqs. (6.6).
Per Eq. (6.9), evaluate $J_{N} (x)$ and $J_{P} (x)$ at the depletion edges and add the two edge current densities. Finally, multiply by the cross-sectional area $A$ .

6.1.3 Derivation Proper

Time to execute the game plan. Start with the holes on the quasineutral n-side. To make the math pretty, shift the origin to the n-edge of the depletion region (Fig. 6.5a): $x^{'} = x - x_{n}$ , so the n-side quasineutral region is $x^{'} \geq 0$ .

Step 1: the differential equation and its boundary conditions.

0 = D_{P} \frac{d ^{2} Δ p _{n}}{d x ^{'2}} - \frac{Δ p _{n}}{τ _{p}} x^{'} \geq 0 (6.19)

Δ p_{n} (x^{'} \to \infty) = 0 (6.20a)

Δ p_{n} (x^{'} = 0) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.20b)

Step 2: general solution. Divide (6.19) by $D_{P}$ : $d^{2} Δ p_{n} / d x^{'2} = Δ p_{n} / L_{P}^{2}$ with

L_{P} = D_{P} τ_{p} (6.22)

This is special-case diffusion equation no. 1 of Table 3.2. Try $e^{λ x^{'}}$ : $λ^{2} = 1/ L_{P}^{2}$ , so $λ = \pm 1/ L_{P}$ :

Δ p_{n} (x^{'}) = A_{1} e^{- x^{'} / L_{P}} + A_{2} e^{x^{'} / L_{P}} x^{'} \geq 0 (6.21)

Step 3: apply boundary conditions.

(6.20a): $e^{x^{'} / L_{P}} \to \infty$ as $x^{'} \to \infty$ , so the only way to keep $Δ p_{n} \to 0$ is $A_{2} = 0$ .
(6.20b): at $x^{'} = 0$ , $A_{1} = Δ p_{n} (x^{'} = 0)$ .

Therefore

Δ p_{n} (x^{'}) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) e^{- x^{'} / L_{P}} x^{'} \geq 0 (6.23)

Step 4: hole current density from (6.6b), $J_{P} = - q D_{P} d Δ p_{n} / d x^{'}$ :

\frac{d Δ p _{n}}{d x ^{'}} J_{P} (x^{'}) = - \frac{1}{L _{P}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) e^{- x^{'} / L_{P}} = q \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) e^{- x^{'} / L_{P}} x^{'} \geq 0 (6.24)

Step 5: the p-side. With the coordinate $x^{''} = - x - x_{p}$ (distance measured into the p-region from the p-edge, Fig. 6.5b), the electron problem is the exact mirror image ( $D_{N}, τ_{n}, L_{N} = D_{N} τ_{n}$ , $N_{A}$ ):

Δ n_{p} (x^{''}) = \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1) e^{- x^{''} / L_{N}} x^{''} \geq 0 (6.25)

J_{N} (x^{''}) = q D_{N} \frac{d Δ n _{p}}{d x} \Rightarrow J_{N} (x^{''}) = q \frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1) e^{- x^{''} / L_{N}} x^{''} \geq 0 (6.26)

(The sign: $d x^{''} = - d x$ so $d / d x = - d / d x^{''}$ , and $d e^{- x^{''} / L_{N}} / d x^{''} = - e^{- x^{''} / L_{N}} / L_{N}$ ; the two minus signs combine to give a positive $J_{N}$ in our convention, in which both carrier currents are positive for forward bias.)

Step 6: evaluate at the depletion edges (set $x^{'} = 0$ , $x^{''} = 0$ ):

J_{N} (x = - x_{p}) = J_{N} (x^{''} = 0) = q \frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1) (6.27a)

J_{P} (x = x_{n}) = J_{P} (x^{'} = 0) = q \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.27b)

Step 7: add and multiply by $A$ (Eqs. 6.8, 6.9, 6.2):

I = A J = q A (\frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} + \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}}) (e^{q V_{A} / k T} - 1) (6.28)

I = I_{0} (e^{q V_{A} / k T} - 1) (6.29)

I_{0} = q A (\frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} + \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}}) (6.30)

Eq. (6.29) is the ideal diode equation, also called the Shockley equation. $I_{0}$ is the (reverse-bias) saturation current.

6.1.4 Examination of Results

Let us now work backward through the derivation and look at what we found, as a physician examines a patient: the final result first, then the internal organs (currents and carrier concentrations).

Ideal $I$ - $V$

(Fig. 6.6 summarizes it.)

Reverse bias with $∣ V_{A} ∣ ≫ k T / q$ (a few tenths of a volt at room temperature): the exponential vanishes and $I \to - I_{0}$ . The ideal theory says this saturation persists to reverse voltages of unlimited magnitude.
Forward bias with $V_{A} ≫ k T / q$ : the " $- 1$ " is negligible and $I \to I_{0} e^{q V_{A} / k T}$ .

Because of the exponential, the forward-bias characteristic is commonly drawn on a semilog scale (Fig. 6.6b):

ln I = ln I_{0} + \frac{q}{k T} V_{A} V_{A} ≫ \frac{k T}{q} (6.31)

A straight line with slope $q / k T$ and an extrapolated $V_{A} = 0$ intercept $ln I_{0}$ . That is how you read $I_{0}$ off a semilog plot.

Quick feel for the numbers. At 300 K, $k T / q = 0.0259$ V, so every $0.0259 ln 10 \approx 60$ mV of extra forward bias multiplies the current by 10. That is the famous “60 mV per decade.”

The saturation current: two big observations

(1) Material dependence. $I_{0}$ contains $n_{i}^{2}$ , which varies enormously with the semiconductor. At room temperature $n_{i} \approx 1 0^{10} cm^{- 3}$ in Si, but $n_{i} \approx 1 0^{13} cm^{- 3}$ in Ge. Therefore $n_{i}^{2}$ differs by a factor $\sim 1 0^{6}$ , and Ge diodes are expected to have a reverse saturation current roughly a million times larger than comparable Si diodes. (Think of $n_{i}^{2}$ as the “thermal leakiness” of the material: small band gap means easy thermal generation of carriers.)

(2) Asymmetric junctions. $I_{0}$ has two terms, each inversely proportional to the doping of its own side. In a $p^{+}$ - $n$ or $n^{+}$ - $p$ junction, the term for the heavily doped side is negligible:

I_{0} \approx q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} p^{+} - n diodes (6.32a)

I_{0} \approx q A \frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} n^{+} - p diodes (6.32b)

The lightly doped side controls everything. The same was true for the depletion width and other electrostatic variables. As a general rule: the heavily doped side of an asymmetrical junction can be ignored in determining its electrical characteristics. Whenever a problem says ” $p^{+}$ - $n$ ,” you may ignore the $p^{+}$ side.

It is fortuitous that the heavy side is negligible. In real diodes the $p^{+}$ or $n^{+}$ doping is degenerate; if its contribution mattered, the ideal theory would have to be modified to treat degenerate doping.

Exercise 6.1

Problem. Two ideal $p^{+}$ - $n$ step junction diodes at room temperature are identical except that $N_{D 1} = 1 0^{15} / cm^{3}$ and $N_{D 2} = 1 0^{16} / cm^{3}$ . Compare the $I$ - $V$ characteristics; sketch both on one set of axes.

Solution. For $p^{+}$ - $n$ diodes

I_{0} \approx q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} .

Use $L_{P} = D_{P} τ_{p}$ , so $D_{P} / L_{P} = D_{P} / τ_{p}$ . Also $D_{P} = (k T / q) μ_{p}$ . Then

\frac{I _{0, 1}}{I _{0, 2}} = \frac{D _{P 1} / τ _{p 1}}{D _{P 2} / τ _{p 2}} \cdot \frac{N _{D 2}}{N _{D 1}}

We have the same diode area, same $n_{i}$ (same semiconductor). The R-G center concentrations are taken to be identical, so $τ_{p 1} = τ_{p 2}$ . The mobilities at $1 0^{15}$ and $1 0^{16} cm^{- 3}$ differ only slightly in most materials (Fig. 3.5), so $μ_{p 1} \approx μ_{p 2}$ and $D_{P 1} \approx D_{P 2}$ . Hence

\frac{I _{0, 1}}{I _{0, 2}} \approx \frac{N _{D 2}}{N _{D 1}} = \frac{1 0 ^{16}}{1 0 ^{15}} = 10

Result. The diode-1 current is approximately ten times larger than the diode-2 current at all applied voltages (Fig. E6.1). On a linear plot the curves have the same shape (same exponent), with diode 1 scaled up by 10; on a semilog forward plot they are two parallel lines (same slope $q / k T$ ) displaced vertically by one decade; and the reverse saturation currents are $- I_{0, 1} = - 10 I_{0, 2}$ .

Exercise 6.2 (computer): Ideal $I$ - $V$ versus doping

Problem. An ideal Si $p^{+}$ - $n$ step junction diode at 300 K has $A = 1 0^{- 4} cm^{2}$ and $τ_{p} = 1 0^{- 6}$ s. Using the empirical-fit hole mobility from Exercise 3.1, write a MATLAB program that computes and plots the ideal $I$ - $V$ characteristic. The $n$ -side doping is an input variable. Use a linear plot with axis set to [-1, 0.2, -2*I0, 5*I0], where I0 is the reverse saturation current. Explore how the characteristic depends on doping.

The logic of the program.

Set constants: $k = 8.617 \times 1 0^{- 5}$ eV/K, $q = 1.6 \times 1 0^{- 19}$ C.
Set parameters: $A$ , $n_{i} = 1 0^{10}$ , $τ_{p}$ , $T = 300$ K, and ask for $N_{D}$ (a vector so several dopings can be plotted together by typing e.g. [1e16 2e16 5e16 1e17]).
Hole mobility: the empirical fit $μ_{p} = μ_{p, m i n} + μ_{p, 0} / [1 + (N_{D} / N_{r e f})^{α}]$ with $μ_{p, m i n} = 54.3$ , $μ_{p, 0} = 406.9$ , $N_{r e f} = 2.35 \times 1 0^{17}$ , $α = 0.88$ (units $cm^{2} / V \cdot s$ ). (The minority carrier mobility is assumed equal to the majority carrier mobility at the same doping.)
$D_{P} = (k T / q) μ_{p}$ (with $k$ in eV/K, $k T$ in eV equals $k T / q$ in volts), $L_{P} = D_{P} τ_{p}$ , and $I_{0} = q A (D_{P} / L_{P}) (n_{i}^{2} / N_{D})$ .
For each doping compute $I = I_{0} (e^{V_{A} / k T} - 1)$ over $V_{A} = - 1 \dots 0.2$ V using an outer product (a column of $I_{0}$ values times a row of exponentials) so that one matrix holds all curves.
Plot, set axes by the first (largest) $I_{0}$ , add axes lines and a legend by hand.

MATLAB program (cleaned from the noisy OCR):

% Variation of Ideal-Diode I-V with semiconductor doping.
% Si step junction, T = 300 K.
% In response to the "ND=" prompt type [ND1 ND2 ...] to input
% multiple doping values.
 
% Initialization and Universal Constants
clear
k = 8.617e-5;          % Boltzmann constant (eV/K)
q = 1.6e-19;           % electronic charge (C)
 
% Device, Material, and System Parameters
A = 1.0e-4;            % cm^2
ni = 1.0e10;           % cm^-3
taup = 1.0e-6;         % s
ND = input('Input the n-side doping concentration, ND= ');
T = 300;
 
% Hole Mobility Calculation
NAref = 2.35e17;
mupmin = 54.3;
mup0 = 406.9;
ap = 0.88;
mup = mupmin + mup0./(1 + (ND./NAref).^ap);
% The mobility calculation here assumes the hole minority carrier
% mobility is equal to the hole majority carrier mobility.
 
% I-V Calculation
VA = linspace(-1, 0.2);
DP = k.*T.*mup;
LP = sqrt(DP.*taup);
I0 = q.*A.*(DP./LP).*(ni^2./ND)       % no semicolon: prints I0 values
I = I0.'*(exp(VA./(k.*T)) - 1);       % each row = one doping
 
% Plotting Result
close
plot(VA, I); grid
ymin = -2*I0(1); ymax = 5*I0(1);
axis([-1, 0.2, ymin, ymax]);
xlabel('V_A (volts)'); ylabel('I (amps)');
 
% Adding axes, key
xx = [-1 0.2]; yx = [0 0];
xy = [0 0];    yy = [ymin, ymax];
hold on
plot(xx, yx, '-w', xy, yy, '-w');
j = length(ND);
for i = 1:j
    yput = (0.70 - 0.06*i)*ymax;
    yk(i,1) = yput; yk(i,2) = yput;
    text(-0.68, (0.69 - 0.06*i)*ymax, ['ND = ', num2str(ND(i)), '/cm^3']);
end
xk = [-0.8 -0.7];
plot(xk, yk);
text(-0.74, 0.75*ymax, 'Si, 300 K');
hold off

Results (Fig. E6.2). The program was run with $N_{D} = 1 0^{16}, 2 \times 1 0^{16}, 5 \times 1 0^{16}, 1 0^{17} cm^{- 3}$ . The reverse saturation currents are extremely small (of order $1 0^{- 16}$ A), and $I_{0}$ decreases as doping increases, following the same general $1/ N_{D}$ dependence noted in Exercise 6.1. Looking carefully at the numerical $I_{0}$ values printed in the Command window, there is a noticeable additional mobility dependence at these higher dopings, because $μ_{p}$ falls as $N_{D}$ rises (so $D_{P} / L_{P} = D_{P} / τ_{p}$ drops too): the $I_{0}$ ratio is a bit larger than the pure doping ratio.

Carrier currents

A sample plot of carrier and total current densities versus position in a forward-biased diode is Fig. 6.7. (The reverse-bias plot is essentially the same with all current densities negative.) The recipe for constructing Fig. 6.7:

From Eqs. (6.24) and (6.26): the minority carrier current densities in the quasineutral regions decay exponentially away from the depletion region edges (on the scale of $L_{P}$ or $L_{N}$ ).
Inside the depletion region, per Eqs. (6.8), $J_{N}$ and $J_{P}$ are drawn constant at their depletion-edge values. The total current density in the depletion region is the graphical sum $J_{N} + J_{P}$ .
The total current density is constant everywhere in the diode (Eq. 6.3), so the value found in the depletion region extends throughout the device.
Finally, the majority carrier current in each quasineutral region is obtained by graphically subtracting the minority current from the constant total: $J_{P} (x) = J - J_{N} (x)$ on the p-side, $J_{N} (x) = J - J_{P} (x)$ on the n-side.

Story. Picture a conveyor belt carrying a fixed total flow. Far on the p-side nearly all the flow is holes. As you approach the junction, holes recombine with injected electrons and “hand over” the flow to the electrons. Far on the n-side nearly all the flow is electrons. The hand-over is the recombination zone, a few diffusion lengths wide.

This is consistent with the qualitative observation: both carriers contribute through the depletion region; holes dominate far on the p-side, electrons far on the n-side.

Carrier concentrations

Eqs. (6.23) and (6.25) tell how the minority concentrations deviate from equilibrium in the quasineutral regions:

Forward bias increases minority concentrations above equilibrium (a pile-up, $Δ n_{p} > 0, Δ p_{n} > 0$ ).
Reverse bias decreases them below equilibrium (a deficit).
In both cases the perturbation decays exponentially with distance from the depletion edge. After several diffusion lengths it vanishes: $n_{p} \to n_{i}^{2} / N_{A}$ as $x \to - \infty$ , $p_{n} \to n_{i}^{2} / N_{D}$ as $x \to + \infty$ , independent of bias.
Because low-level injection prevails, the majority carrier concentrations stay approximately at their equilibrium values, $p_{p} \approx N_{A}$ and $n_{n} \approx N_{D}$ , regardless of bias.

Fig. 6.8 shows these (forward bias on a semilog scale, where exponentials appear as straight lines; reverse bias on a linear scale; $N_{A} > N_{D}$ assumed).

Forward bias, physical story. Majority carriers are injected in large numbers over the lowered hill. On the other side they become minority carriers and are progressively eliminated by recombination as they try to diffuse deeper (this is the same math as Sample Problem No. 2 of Chapter 3). The net result is a pile-up of excess minority carriers adjacent to the depletion region. This pile-up, a consequence of forward biasing, will be important in later chapters (stored charge, diffusion capacitance).

Reverse bias, physical story. The depletion region acts as a sink for minority carriers, draining adjacent regions. A reverse bias of only a few $k T / q$ drives the edge concentrations to essentially zero, and larger reverse biases barely change the distributions, which is consistent with the saturation of current (the current is proportional to the slope of the distribution at the depletion edge). Overall, reverse bias creates a modest deficit of minority carriers near the depletion region.

Exercise 6.3

Problem. Fig. E6.3 is a dimensioned plot of the steady state carrier concentrations (log scale) inside a pn junction diode at room temperature. Reading the figure: $n_{n} (far) = N_{D} = 1 0^{15} cm^{- 3}$ on the n-side; at the depletion edge $p_{n} (x_{n}) = 1 0^{10} cm^{- 3}$ ; far away $p_{n} \to 1 0^{5} cm^{- 3}$ (so $n_{i} = 1 0^{10}$ ); $p_{n}$ falls to $1 0^{8}$ at the point $x^{'}$ marked in the figure. (a) Is the diode forward or reverse biased? Explain. (b) Do low-level injection conditions prevail in the quasineutral regions? Explain. (c) Determine the applied voltage $V_{A}$ . (d) Determine the hole diffusion length $L_{P}$ .

Solution.

(a) Forward biased. There is a pile-up (excess) of minority carriers at the depletion region edges ( $Δ n_{p} > 0$ , $Δ p_{n} > 0$ ): the minority concentration near the edge exceeds its far-away equilibrium value. A deficit would indicate reverse bias.

(b) Yes. $Δ p_{n} ≪ n_{n}$ and $Δ n_{p} ≪ p_{p}$ everywhere in the quasineutral regions: the highest minority concentration ( $1 0^{10}$ at the edge) is five orders of magnitude below the majority concentration ( $1 0^{15}$ ).

n p = n_{i}^{2} e^{q V_{A} / k T} \Rightarrow V_{A} = \frac{k T}{q} ln \frac{n p}{n _{i}^{2}}, - x_{p} \leq x \leq x_{n}

Evaluate at the n-edge of the depletion region and note $n_{i}^{2} = n_{n} (\infty) p_{n} (\infty)$ (far from the junction, equilibrium):

V_{A} = \frac{k T}{q} ln \frac{n _{n} ( x _{n} ) p _{n} ( x _{n} )}{n _{n} ( \infty ) p _{n} ( \infty )} = (0.0259) ln \frac{1 0 ^{15} \cdot 1 0 ^{10}}{1 0 ^{15} \cdot 1 0 ^{5}} = (0.0259) ln (1 0^{5}) = (0.0259) (11.51) \approx 0.3 V

(d) Rewrite Eq. (6.23) in terms of $p_{n}$ . Near the depletion edge $Δ p_{n} \approx p_{n}$ (the pile-up far exceeds the equilibrium value there):

p_{n} (x^{'}) ln \frac{p _{n} ( x ^{'} )}{p _{n} ( 0 )} L_{P} = p_{n} (0) e^{- x^{'} / L_{P}} = - \frac{x ^{'}}{L _{P}} = \frac{x ^{'}}{ln [ p _{n} ( 0 ) / p _{n} ( x ^{'} ) ]}

From the figure the concentration drops from $1 0^{10}$ to $1 0^{8}$ over $x^{'} = 1.6 \times 1 0^{- 2} cm$ (measured from the depletion edge):

L_{P} = \frac{1.6 \times 1 0 ^{- 2}}{ln ( 1 0 ^{10} /1 0 ^{8} )} = \frac{1.6 \times 1 0 ^{- 2}}{ln 100} = \frac{1.6 \times 1 0 ^{- 2}}{4.605} = 3.47 \times 1 0^{- 3} cm

(Reading assumption: the OCR of the figure axis labels is partly garbled; the numbers above reproduce the book’s quoted answers $V_{A} \approx 0.3$ V and $L_{P} = 3.47 \times 1 0^{- 3}$ cm.)

Exercise 6.4

Problem. When reverse biased by more than a few $k T / q$ , the diode current equals $q$ times the number of minority carriers per second that wander into the depletion region and are swept across. Under steady state this must equal the number generated per second in the quasineutral regions. So the saturation current may alternatively be viewed as arising from minority carrier generation in the quasineutral regions, $I_{0} = q \times$ (carriers generated per second). Suppose the actual reverse-bias minority carrier distributions are approximated by squared-off distributions (Fig. E6.4): all minority carriers are depleted a minority carrier diffusion length to either side of the depletion region. Remembering low-level injection holds in the quasineutral regions, derive an expression for the reverse-bias saturation current.

Solution. In a low-level quasineutral region, the thermal net R-G rate is $- Δ n / τ_{n}$ (and similarly for holes). Under reverse bias the edge concentrations are driven to zero, so $Δ n_{p} = - n_{p 0} = - n_{i}^{2} / N_{A}$ and $Δ p_{n} = - n_{p 0} \to - n_{i}^{2} / N_{D}$ in the depleted zones. Thus

\frac{\partial n}{\partial t}_{R-G} = - \frac{Δ n _{p}}{τ _{n}} = + \frac{n _{i}^{2}}{N _{A} τ _{n}} (- L_{N} - x_{p} \leq x \leq - x_{p})

\frac{\partial p}{\partial t}_{R-G} = - \frac{Δ p _{n}}{τ _{p}} = + \frac{n _{i}^{2}}{N _{D} τ _{p}} (x_{n} \leq x \leq x_{n} + L_{P})

(positive means net generation). Generation takes place in volumes $A L_{N}$ and $A L_{P}$ on the p- and n-sides. Then

I_{0} = q (A L_{N}) \frac{n _{i}^{2}}{N _{A} τ _{n}} + q (A L_{P}) \frac{n _{i}^{2}}{N _{D} τ _{p}} = q A (\frac{L _{N}}{τ _{n}} \frac{n _{i}^{2}}{N _{A}} + \frac{L _{P}}{τ _{p}} \frac{n _{i}^{2}}{N _{D}})

Because $L_{N} / τ_{n} = D_{N} / L_{N}$ (from $L_{N}^{2} = D_{N} τ_{n}$ ) and $L_{P} / τ_{p} = D_{P} / L_{P}$ ,

I_{0} = q A (\frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} + \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}})

One obtains the usual $I_{0}$ expression. The squaring-off trick (replace an exponential by a rectangle of height equal to the peak and width equal to one decay length, enclosing the same area) is a very useful analytical tool, and we will use it again in the charge-control approach of Section 6.3.

6.2 Deviations from the Ideal

The ideal diode derivation used a fairly long list of assumptions, some made with no a priori justification. It would be surprising if a careful experiment did not find discrepancies. Here we first compare experiment with theory to spot the major deviations, and then treat each one: its cause and the modification needed.

Do not be dismayed by this “band-aid” style of progress. Formulate a simple theory, compare with experiment, revise, compare again is how engineers and scientists tackle every complicated real problem.

6.2.1 Ideal Theory Versus Experiment

Fig. 6.9 shows the measured $I$ - $V$ of a commercial Si diode at room temperature on a $0.1 μ A$ /division linear scale (note the change of voltage scale between forward and reverse). Largely the characteristic looks as predicted: forward current rises rapidly and reverse current is vanishingly small over most of the range. But one obvious feature is not in the ideal theory: a large reverse current flows when the reverse voltage exceeds a certain value. This is breakdown (Subsection 6.2.2). It occurs at voltages that range over several orders of magnitude from diode to diode, but is common to all pn diodes.

To see further deviations, look at the data on better scales (Fig. 6.10, measured with an HP 4145B parameter analyzer):

Forward bias (semilog, Fig. 6.10a).

Between roughly 0.35 V and 0.7 V the slope is the expected $q / k T$ (ideal behavior).
Above 0.7 V the slope progressively decreases: the characteristic “slopes over”. This is related to the high current flowing as $V_{A} \to V_{bi}$ (Subsection 6.2.4).
Below 0.35 V the current is quite small, far larger than the ideal expectation, and the slope is curiously close to $q /2 k T$ . A current component has been overlooked.

Reverse bias (Fig. 6.10b, $50$ pA/division).

The estimated $I_{0} \sim 1 0^{- 14}$ A (from the ideal forward region) is some three orders of magnitude smaller than the reverse current at $V_{A} = - 5$ V!
The reverse current does not saturate; it continually increases with reverse bias.
(The wiggles are noise in the device and measuring system.)

As Subsection 6.2.3 shows, the small-forward-bias and the reverse-bias discrepancies are both due to thermal carrier recombination-generation in the depletion region, which the ideal diode derivation assumed negligible.

Features vary from device to device, and other nonidealities can occur, but this is fairly representative of commercial Si diodes at room temperature. GaAs diodes are similar. Ge diodes at room temperature and Si diodes at elevated temperatures are often closer to ideal (saturating reverse characteristics): an explanation follows in the R-G discussion.

Region of $I$ - $V$	Observed	Cause (where treated)
Large reverse bias	Current $\to \infty$	Breakdown: avalanche / Zener (6.2.2)
All reverse bias, and small forward bias	Excess current, no saturation, slope $q /2 k T$	Depletion-region R-G (6.2.3)
Moderate forward bias (0.35-0.7 V in the example)	Ideal slope $q / k T$	Diffusion current
$V_{A} \to V_{bi}$	Slope-over (and possibly $q /2 k T$ again)	Series resistance, high-level injection (6.2.4)

6.2.2 Reverse-Bias Breakdown

Though called “breakdown,” the large reverse current is a completely reversible process: it does not damage the diode (provided the current is limited to avoid overheating). The absolute value of the reverse voltage at which current “goes off to infinity” is the breakdown voltage, $V_{B R}$ . In practice $V_{B R}$ measurements quote the voltage at which the current reaches a preselected value such as $1 μ$ A or $1$ mA.

Fig. 6.11 plots the expected $V_{B R}$ of planar (one-dimensional) $p^{+}$ - $n$ and $n^{+}$ - $p$ step junctions versus the doping on the lightly doped side, for Ge, Si, and GaAs. Observations:

At a given doping, $V_{B R}$ increases with band gap ( $E_{G}$ : Ge < Si < GaAs).
More importantly, the light-side doping can vary $V_{B R}$ from a few volts to over a thousand volts.
Above the dashed line in Fig. 6.11 (avalanche dominated) the dependence is roughly

V_{B R} \propto \frac{1}{N _{B}} (roughly; more accurately N_{B}^{- 3/4}) (6.33)

where $N_{B}$ is the doping on the lightly doped side. (The book writes the approximate dependence in terms of a power law; Problem 6.14 uses $V_{B R} \approx 60 (N_{B} /1 0^{16})^{- 3/4}$ volts for Si.)

Theoretically breakdown comes from failure of the “no other processes” assumption. Two “other processes” can cause it: avalanching and the Zener process. Avalanching typically dominates; the Zener process is important only when both sides of the junction are heavily doped.

Avalanching

Start with a small reverse bias far below breakdown. The reverse current is minority carriers randomly entering the depletion region and being accelerated by the field. The acceleration is not continuous: it is interrupted by energy-losing collisions with the lattice (Fig. 6.12a). The mean free path between collisions is $\sim 1 0^{- 6}$ cm and a median depletion width is $\sim 1 0^{- 4}$ cm, so a carrier suffers tens to thousands of collisions while crossing. The energy lost per collision is small at small bias, simply heating the lattice a little (lattice vibrations, readily dissipated).

With increasing reverse bias the energy transferred per collision grows. Near $V_{B R}$ it becomes enough to ionize a semiconductor atom: the collision frees a valence electron, creating an electron-hole pair. This is impact ionization (Fig. 3.15f). The new carriers are accelerated by the same field and make further collisions, creating still more carriers (Fig. 6.12b): a snowballing multiplication much like a mountainside snow avalanche. At $V_{B R}$ the carrier creation and reverse current effectively go to infinity.

Analogy: a chain letter / rolling snowball. Each ionizing collision turns one carrier into two, each of which can then ionize again. If on average each carrier causes more than one new ionization while crossing, the population explodes.

Two clarifying comments:

Fig. 6.12b cannot realistically be drawn to scale. If $V_{B R} = 100$ V, the separation of the p-side and n-side Fermi levels would be roughly 100 times the $E_{c} - E_{v}$ gap distance.
Avalanche does not switch on sharply at $∣ V_{A} ∣ = V_{B R}$ . Carefully examined (Fig. 6.10b) the approach to breakdown is sloping: there is considerable carrier multiplication a few volts before $V_{B R}$ and even some far below. The reason is that the distance between collisions is a random variable distributed about its mean, so carriers occasionally gain ionizing energy at voltages far below $V_{B R}$ ; a few volts below breakdown the number of such carriers becomes quite large.

The current increase due to carrier multiplication is modeled by a multiplication factor $M$ . If $I_{0}^{'}$ is the current without multiplication,

M = \frac{I}{I _{0}^{'}} (6.34)

An empirical fit to data:

M = \frac{1}{1 - ( ∣ V _{A} ∣/ V _{B R} ) ^{m}} (6.35)

with $m$ between 3 and 6 depending on the semiconductor. The multiplication factor can be used to correct the ideal diode equation to include avalanching: multiply the reverse current by $M$ .

Why does $V_{B R}$ depend on doping this way? Breakdown occurs when carriers gain an ionizing energy over a lattice-scattering mean free path. That requirement is independent of junction doping, but a specific energy gain over a given distance corresponds to a specific electric field. So breakdown occurs when the maximum depletion field reaches a critical value $E_{C R}$ , essentially independent of doping.

For a step junction, from Eqs. (5.35) and (5.37), the field at $x = 0$ (where it is largest) is

∣ E (0) ∣ = \frac{q N _{D} x _{n}}{K _{S} ε _{0}} = [\frac{2 q}{K _{S} ε _{0}} \frac{N _{A} N _{D}}{N _{A} + N _{D}} (V_{bi} - V_{A})]^{1/2} (6.36)

Square it, and set $∣ E (0) ∣ \to E_{C R}$ when $V_{bi} - V_{A} \to V_{bi} + V_{B R} \approx V_{B R}$ (since $V_{B R} ≫ V_{bi}$ ):

E_{C R}^{2} = \frac{2 q}{K _{S} ε _{0}} \frac{N _{A} N _{D}}{N _{A} + N _{D}} V_{B R} (6.37)

Since $E_{C R}$ is independent of doping, the right-hand side must be too, which requires

V_{B R} \propto \frac{N _{A} + N _{D}}{N _{A} N _{D}} (6.38)

or, for an asymmetric junction with light-side doping $N_{B}$ ,

V_{B R} \propto \frac{1}{N _{B}} (6.39)

This is not exactly what is observed (exponent nearer $- 3/4$ ), but is acceptably close for such a simple argument.

Other dependences.

Band gap. Required ionization energy increases with $E_{G}$ while the mean free path varies only slightly across Ge, Si, GaAs. So $V_{B R}$ increases Ge to Si to GaAs, as in Fig. 6.11.
Temperature. Avalanche $V_{B R}$ increases with $T$ : lattice scattering grows with temperature (Subsection 3.1.3), shortening the mean free path, so a larger critical field (and voltage) is needed.
Geometry. The discussion is for planar step junctions. Linearly graded diodes have their own curves. Diffused or implanted junctions have curved lateral edges where the field is greater for a given voltage, so breakdown occurs there first, lowering $V_{B R}$ . This “curvature effect” gets worse as junction depth shrinks; special edge modifications (guard rings and the like) minimize it.

Exercise 6.5 (computer): the multiplication factor

Problem. Compute and plot $M$ versus $∣ V_{A} ∣/ V_{B R}$ for $m = 3$ and $m = 6$ . Orient axes to get a reverse-current-like plot ( $M$ increasing downward, $∣ V_{A} ∣/ V_{B R}$ increasing to the left). Discuss.

Logic. Let $x = ∣ V_{A} ∣/ V_{B R}$ run from 0 to 0.99 (never exactly 1, where $M$ is infinite). Compute $M = 1/ (1 - x^{m})$ . To get the reverse-current-looking orientation in MATLAB, the author plotted $- M$ against $- x$ .

% Exercise 6.5: Multiplication factor
% Initialization
close
clear
% M calculation
x = linspace(0, .99);    % x = |VA|/VBR
M3 = 1./(1 - x.^3);      % M when m = 3
M6 = 1./(1 - x.^6);      % M when m = 6
% Plotting result
plot(-x, -M3, -x, -M6); grid
axis([-1 0 -10 0])
xlabel('-|VA|/VBR')
ylabel('-Multiplication factor')
text(-0.8, -2.5, 'm=3')
text(-0.95, -1.5, 'm=6')

Discussion of Fig. E6.5. The plot closely resembles the left-hand side of Fig. 6.10(b). A perhaps surprising result: the current doubles ( $M = 2$ ) well before $V_{B R}$ :

M = 2 \Rightarrow x^{m} = \frac{1}{2} \Rightarrow x = 0. 5^{1/ m} = {0.794 0.891 m = 3 m = 6

so at about 80% (for $m = 3$ ) and 90% (for $m = 6$ ) of the way to breakdown, the current has already doubled. There is already a $\sim 10%$ enhancement ( $M = 1.1$ ) at

x = (1 - \frac{1}{1.1})^{1/ m} = (0.0909)^{1/ m} = {0.45 0.67 m = 3 m = 6

i.e., $0.45 V_{B R}$ for $m = 3$ and about $0.65 V_{B R}$ for $m = 6$ . The larger $m$ , the sharper the approach to breakdown.

Zener Process

The Zener process is tunneling in a reverse-biased diode. Tunneling is the first genuinely quantum-mechanical phenomenon we meet: it has no classical analog.

Thought experiment: a ball and a hill. Classically a ball with less energy than the hill’s top can never get to the other side. A quantum particle has a nonzero probability to appear on the other side, “tunneling through” the barrier, without gaining energy (its energy is constant during tunneling) and without damaging itself or the barrier.

Two requirements for significant tunneling (Fig. 6.13):

There must be filled states on one side of the barrier and empty states on the other side at the same energy. Tunneling cannot take place into a region void of allowed states.
The barrier width $d$ must be very thin: tunneling is significant only if $d ≲ 100 \overset{˚}{A} = 1 0^{- 6}$ cm.

In the reverse-biased diode (Fig. 6.14), the tunneling particles are valence band electrons on the p-side, going through the barrier to empty conduction band states on the n-side at the same energy. The larger the reverse bias, the more filled valence states on the p-side face empty conduction states on the n-side, so the tunneling current increases with reverse bias.

The barrier thickness is roughly the depletion width, which must be $≲ 1 0^{- 6}$ cm. From Fig. E5.3, Si depletion widths $< 1 0^{- 6}$ cm need doping above about $1 0^{18} cm^{- 3}$ on the “lightly” doped side as well. So the Zener process matters only for diodes heavily doped on both sides, which have correspondingly small breakdown voltages. The Zener process contributes significantly when $V_{B R} < 6 E_{G} / q$ ( $\approx 6.7$ V in Si at 300 K) and dominates when $V_{B R} < 4 E_{G} / q$ ( $\approx 4.5$ V in Si at 300 K).

How to tell avalanche from Zener experimentally:

	Avalanche	Zener
$V_{B R}$ vs. $T$	increases with $T$	decreases with $T$
Shape of breakdown	fairly sharp	very “soft”, slow approach even on a coarse scale

Historically the Zener process was the first proposed to explain reverse breakdown, so all diodes specifically fabricated to use breakdown became known as Zener diodes. The name persists even for those with $V_{B R} > 6 E_{G} / q$ , which are functionally avalanche diodes.

6.2.3 The R-G Current

In comparing theory and experiment, a current far exceeding the ideal prediction was found at small forward biases and at all reverse biases in room-temperature Si diodes. It arises from thermal carrier recombination-generation in the depletion region, assumed negligible in the ideal derivation. (Note: “depletion region” does not mean the carrier concentrations there are zero, or automatically less than equilibrium. They are merely small compared to the background doping.)

Reverse bias (Fig. 6.15a). Until now we thought of the reverse current as minority carriers wandering in from the two sides. But when reverse biased, the carrier concentrations in the depletion region fall below their equilibrium values, so $n p < n_{i}^{2}$ , and the system tries to restore equilibrium by thermal generation of electron-hole pairs throughout the depletion region. The strong field sweeps the generated carriers out into the quasineutral regions, adding to the reverse current.

Forward bias (Fig. 6.15b). Forward bias raises the carrier concentrations in the depletion region above equilibrium ( $n p > n_{i}^{2}$ ), giving recombination there. Viewed on the band diagram, the added forward current is the carriers that cannot make it over the potential hill being partially eliminated by recombination at R-G centers in the depletion region. (They flowed in, but never got out the other side; the contacts supply replacements, so the external current includes them.)

Quantitative expression

For every electron-hole pair created or destroyed per second in the depletion region, one electron per second flows out of or into the diode contacts. So sum the electrons (or holes) created/destroyed per second throughout the depletion region and multiply by $q$ to get the magnitude of the added current. With the right sign convention (generation gives negative current, recombination positive):

I_{R-G} = - q A \int_{- x_{p}}^{x_{n}} \frac{\partial n}{\partial t}_{thermal R-G} d x (6.40)

In the depletion region the familiar low-level shortcut $\partial n / \partial t = - Δ n / τ_{n}$ does not apply. We must use the general-case result, Eq. (3.35):

\frac{\partial n}{\partial t}_{thermal R-G} = - \frac{n p - n _{i}^{2}}{τ _{p} ( n + n _{1} ) + τ _{n} ( p + p _{1} )} (6.41)

so that

I_{R-G} = q A \int_{- x_{p}}^{x_{n}} \frac{n p - n _{i}^{2}}{τ _{p} ( n + n _{1} ) + τ _{n} ( p + p _{1} )} d x (6.42)

Reverse bias (beyond a few $k T / q$ ). The carrier concentrations are quite small through most of the depletion region ( $n \to 0$ , $p \to 0$ ), so $n p \to 0$ , and the denominator reduces to $τ_{p} n_{1} + τ_{n} p_{1}$ . With $n_{1} = n_{i} e^{(E_{T} - E_{i}) / k T}$ and $p_{1} = n_{i} e^{(E_{i} - E_{T}) / k T}$ , the integrand is constant and the integral is trivial:

I_{R-G} \approx - q A \int_{0}^{W} \frac{n _{i}^{2}}{τ _{p} n _{1} + τ _{n} p _{1}} d x = - \frac{q A n _{i}^{2} W}{τ _{p} n _{1} + τ _{n} p _{1}} = - \frac{q A n _{i} W}{2 τ _{0}} reverse biases > a few k T / q (6.43)

where

τ_{0} = \frac{τ _{p} n _{1} + τ _{n} p _{1}}{2 n _{i}} = \frac{1}{2} [τ_{p} e^{(E_{T} - E_{i}) / k T} + τ_{n} e^{- (E_{T} - E_{i}) / k T}] (6.44)

(For $τ_{n} = τ_{p} = τ_{0}$ and $E_{T} \approx E_{i}$ , this reduces to $τ_{0}$ equal to the common lifetime; $τ_{0}$ is a generation lifetime.) Notice three things: $I_{R-G} \propto n_{i}$ (not $n_{i}^{2}$ ), $I_{R-G} \propto W$ , and since $W \propto (V_{bi} - V_{A})^{1/2}$ , the reverse R-G current increases (as a square root) with reverse bias; it never saturates.

Forward bias. The concentrations can’t be neglected and even an approximate evaluation of (6.42) is involved. We note only that $I_{R-G}$ is expected to vary as $exp (q V_{A} / η k T)$ with $1 < η \leq 2$ at forward biases beyond a few $k T / q$ . Typically $η$ is close to 2 (the maximum recombination rate, where $n \approx p$ , occurs where the quasi-Fermi-level split is half the total: $n p = n_{i}^{2} e^{q V_{A} / k T}$ with $n \approx p$ gives $n = n_{i} e^{q V_{A} /2 k T}$ ). The combined forward and reverse dependence is approximately described by

I_{R-G} \approx \frac{q A n _{i} W}{2 τ _{0}} \frac{e ^{q V_{A} / k T} - 1}{1 + \frac{V _{bi} - V _{A}}{2 k T / q} e ^{q V_{A} /2 k T}} (6.45)

(Sanity checks. Reverse: the exponential terms drop, giving $- q A n_{i} W /2 τ_{0}$ , matching 6.43. Forward, $V_{A} ≫ k T / q$ : $I_{R-G} \to \frac{q A n _{i} W}{τ _{0}} \frac{k T / q}{V _{bi} - V _{A}} e^{q V_{A} /2 k T}$ , the " $q /2 k T$ " behavior observed. This is the form used in Exercise 6.6 and its program.)

Total current and which one wins

With a second component, the ideal-diode-equation current is now called the diffusion current, $I_{DIFF}$ :

I_{DIFF} = q A (\frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} + \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}}) (e^{q V_{A} / k T} - 1) (6.46)

I = I_{DIFF} + I_{R-G} (6.47)

Now the experimental observations are explained:

In Si at room temperature, $q A n_{i} W /2 τ_{0} ≫ I_{0}$ , so the R-G current dominates at reverse biases and small forward biases. In reverse bias it never saturates (since $W$ grows). The forward R-G current varies as $exp (q V_{A} /2 k T)$ for $V_{A} >$ a few $k T / q$ , consistent with the observed slope of $q /2 k T$ .
With increasing forward bias, $I_{DIFF}$ , which rises faster ( $e^{q V_{A} / k T}$ vs. $e^{q V_{A} /2 k T}$ ), eventually overtakes $I_{R-G}$ , giving the $q / k T$ region in the semilog plot.

Why Ge and hot Si are closer to ideal. Compare the sizes of the two prefactors:

$I_{0} \propto n_{i}^{2}$ , while $I_{R-G} \propto n_{i}$ . The larger $n_{i}$ of Ge makes $I_{0} > q A n_{i} W /2 τ_{0}$ in Ge diodes at room temperature, even though $q A n_{i} W /2 τ_{0} > I_{0}$ in Si and GaAs.
Since $I_{0} \propto n_{i}^{2}$ but reverse $I_{R-G} \propto n_{i}$ , the reverse-bias diffusion component rises faster with temperature. At a sufficiently elevated temperature the diffusion component dominates (see Problem 6.16 for “how elevated”).

Thus Ge diodes at room temperature, and Si diodes at elevated temperature, show saturating reverse characteristics closer to the ideal.

Exercise 6.6 (computer): $I_{01}$ and $I_{02}$ for Si and Ge

Problem. If the factor $e^{q V_{A}}$ … more precisely, if the $e^{q V_{A} /2 k T}$ term in the denominator of the R-G expression is large (forward bias), the forward-bias current for a $p^{+}$ - $n$ diode beyond a few $k T / q$ may be approximately modeled by

I = I_{01} e^{q V_{A} / k T} + I_{02} e^{q V_{A} /2 k T}

(the first term is $I_{DIFF}$ , the second $I_{R-G}$ ), with $I_{01}$ and $I_{02}$ voltage-independent constants.

(a) Compute $I_{01}$ and $I_{02}$ for a Si $p^{+}$ - $n$ step junction diode at 300 K with $A = 1 0^{- 4} cm^{2}$ , $N_{D} = 1 0^{16} cm^{- 3}$ and $τ_{n} = τ_{p} = τ_{0} = 1 0^{- 6}$ s. In the $I_{02}$ computation, assume the $e^{q V_{A} /2 k T}$ term in the denominator of the R-G expression is much greater than unity, and set $V_{A} = V_{bi} /4$ in evaluating voltage-dependent terms. Compute $V_{bi} = (E_{G} /2 q) + (k T / q) ln (N_{D} / n_{i})$ . (b) Repeat for a Ge $p^{+}$ - $n$ diode at 300 K with the same parameters except $μ_{p} = 1500 cm^{2} / V \cdot s$ , $n_{i} = 2.5 \times 1 0^{13} cm^{- 3}$ , $K_{S} = 16$ . (c) Using the simplified relationship, construct a semilog plot of $I$ vs $V_{A}$ simultaneously displaying the forward-bias characteristics of the Si and Ge diodes. Limit axes to $0 \leq V_{A} \leq 1$ V and $1 0^{- 12} \leq I \leq 1 0^{- 3}$ A. Begin the calculation at $V_{A} = 0.1$ V. Discuss.

Formulas used.

I_{01} = q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}}, I_{02} = \frac{q A n _{i} W}{τ _{0}} \frac{k T / q}{V _{bi} - V _{A}}_{V_{A} = V_{bi} /4}, W = \frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})

(taking $W$ at the same $V_{A} = V_{bi} /4$ ; with $V_{bi} - V_{A} = 3 V_{bi} /4$ .)

Step-by-step for Si (all voltages in volts, $k T / q = 0.02585$ V):

μ_{p} (1 0^{16}) L_{P} I_{01} \approx 437 cm^{2} / V s (from the fit), = D_{P} τ_{p} = (11.3) (1 0^{- 6}) = 3.36 \times 1 0^{- 3} cm = (1.6 \times 1 0^{- 19}) (1 0^{- 4}) (3.36 \times 1 0^{3}) \frac{1 0 ^{20}}{1 0 ^{16}} D_{P} \frac{D _{P}}{L _{P}} = (0.02585) (437) = 11.3 cm^{2} / s = 3.36 \times 1 0^{3} cm/s = 5.38 \times 1 0^{- 16} A

V_{bi} V_{A} W I_{02} = 0.56 + 0.02585 ln \frac{1 0 ^{16}}{1 0 ^{10}} = 0.56 + 0.357 = 0.92 V = V_{bi} /4 = 0.23 V, V_{bi} - V_{A} = 0.69 V = \frac{2 ( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} )} (0.69) \approx 3.0 \times 1 0^{- 5} cm = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{- 4} ) ( 1 0 ^{10} ) ( 3.0 \times 1 0 ^{- 5} )}{1 0 ^{- 6}} \cdot \frac{0.02585}{0.69} \approx 2 \times 1 0^{- 13} A

in agreement with the book’s value (table below).

Results table (book’s values).

	Silicon	Germanium
$I_{01}$	$5.38 \times 1 0^{- 16}$ A	$6.23 \times 1 0^{- 9}$ A
$I_{02}$	$2.08 \times 1 0^{- 13}$ A	$8.34 \times 1 0^{- 10}$ A

As expected, $I_{02} > I_{01}$ for Si, while $I_{01} > I_{02}$ for Ge.

MATLAB program (cleaned):

% Comparison of forward bias I-VA for Si and Ge diodes at 300 K.
% This program uses a simplified formula for the current:
%   I = I01*exp(qVA/(kT)) + I02*exp(qVA/(2kT)).
% Also, Vbi = EG/2q + (kT/q)*ln(ND/ni).
 
% Initialization
close
clear
% 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 (C)
KS = [11.8 16];        % Dielectric constant [Si Ge]
ni = [1.0e10 2.5e13];  % intrinsic carrier conc. at 300 K [Si Ge]
mup = [437 1500];      % hole mobility [Si Ge]
EG = [1.12 0.66];      % band gap [Si Ge]
% Given Constants
A = 1.0e-4;            % cm^2
ND = 1.0e16;           % cm^-3
taun = 1.0e-6;         % seconds
taup = 1.0e-6;         % seconds
 
% I01
DP = k*T.*mup;
LP = sqrt(DP.*taup);
I01 = q*A.*(DP./LP.*ni.^2./ND);
 
% I02
Vbi = EG/2 + k*T.*log(ND./ni);
W = sqrt(2.*KS*e0/(q*ND).*Vbi);          % evaluated using Vbi - 0.25Vbi
I02 = q*A.*ni/sqrt(taun*taup).*W.*(k*T)./(3.*Vbi./4);
% (W factor uses 3*Vbi/4 per the stated assumption; see text)
 
% Currents for both Silicon (ISi) and Germanium (IGe)
VA = linspace(0.1, 1);
ISi = I01(1).*exp(VA./(k*T)) + I02(1).*exp(VA./(2*k*T));
IGe = I01(2).*exp(VA./(k*T)) + I02(2).*exp(VA./(2*k*T));
 
% Plot
semilogy(VA, ISi, VA, IGe, '--'); grid
axis([0 1 1.0e-12 1.0e-3]);
xlabel('VA (volts)');
ylabel('I (A)');
text(.7, 1.4e-9, 'T=300K');
text(.7, 4.0e-10, 'ND=1.0e16/cm^3');
text(.25, 1.4e-5, 'Ge');
text(.48, 1.4e-8, 'Si');

(In the book’s script $W$ is computed from $V_{bi}$ via the W = sqrt(2.*KS*e0/(q*ND).*Vbi) line; to follow the problem statement exactly, one would use $3 V_{bi} /4$ inside the square root. The difference changes $I_{02}$ by only $4/3$ .)

Discussion (Fig. E6.6). Because of the small band gap and large $n_{i}$ of Ge, large forward currents flow in the Ge diode at relatively small voltages, making it virtually impossible to observe the R-G component in Ge: $I_{DIFF}$ takes over almost immediately. In the Si diode, both the R-G ( $q /2 k T$ ) region at small bias and the diffusion ( $q / k T$ ) region at larger bias are clearly visible on the semilog plot, with a crossover around 0.4-0.5 V. The Ge curve sits several orders of magnitude to the left (lower voltage for the same current): for a given current, the Ge diode needs a much smaller forward voltage, which is why Ge diodes were once prized as low-drop rectifiers.

Exercise 6.7: generalized forward-bias model

A generalization of the forward-bias model of Exercise 6.6 is common in analyzing experimental data:

I = I_{01} e^{q V_{A} / n_{1} k T} + I_{02} e^{q V_{A} / n_{2} k T}

(first term $I_{DIFF}$ , second $I_{R-G}$ ) with $n_{1}$ , $n_{2}$ arbitrary constants determined by matching the near-ideal and R-G regions.

Problem. Determine $I_{01}$ , $I_{02}$ , $n_{1}$ , $n_{2}$ giving an optimal match to the forward data of Fig. 6.10(a).

Solution. Draw straight lines through the two linear regions of the semilog plot (Fig. E6.7). Each line is, for $j = 1, 2$ ,

ln I = ln I_{0 j} + \frac{q}{n _{j} k T} V_{A}

The $V_{A} = 0$ intercept of line $j$ gives $I_{0 j}$ .
The slope gives $n_{j}$ : for two points $(V_{A 1}, I_{1})$ and $(V_{A 2}, I_{2})$ on line $j$ ,

n_{j} = \frac{V _{A 2} - V _{A 1}}{( k T / q ) ln ( I _{2} / I _{1} )}

From the graph:

I_{01} \approx 1 0^{- 14} A, n_{1} \approx 1.1; I_{02} \approx 1 0^{- 12} A, n_{2} \approx 1.5

The $I_{DIFF}$ component is almost uncharacteristically ideal. The observed $n_{2}$ is on the low side but within the normally observed range ( $1 < n_{2} \leq 2$ ), and $I_{02} > I_{01}$ , as expected.

6.2.4 $V_{A} \to V_{bi}$ High-Current Phenomena

A significant current starts to flow when the forward bias approaches $V_{bi}$ . A large current creates conditions inconsistent with several ideal-diode assumptions. First, “all the applied voltage drops across the depletion region” becomes doubtful: with $I = 0.1$ A, even 1 ohm of resistance outside the depletion region drops a significant 0.1 V. A large current may also cause high-level injection. We take these in turn.

Series resistance

The quasineutral regions have an inherent resistance set by doping and dimensions, plus a residual contact resistance. Together they form a resistance $R_{S}$ in series with the current flowing across the junction (Fig. 6.16a). At low current $I R_{S} ≪ V_{A}$ , so the voltage across the depletion region, the “junction voltage” $V_{J}$ , equals $V_{A}$ (as assumed so far). But when $I R_{S}$ becomes comparable with $V_{A}$ , the voltage across the depletion region falls:

V_{J} = V_{A} - I R_{S} (6.48)

Part of the applied voltage is wasted: a larger $V_{A}$ is needed for the same current, and the characteristic slopes over (Fig. 6.10a). The fix is to replace $V_{A}$ by $V_{J} = V_{A} - I R_{S}$ in earlier relations. As diffusion current dominates at these currents,

I = I_{0} e^{q V_{J} / k T} = I_{0} e^{q (V_{A} - I R_{S}) / k T} V_{A} \to V_{bi} (6.49)

This is transcendental in $I$ : you cannot solve for $I$ as a function of $V_{A}$ in closed form. But $I$ vs. $V_{A}$ is easy to construct parametrically: choose $V_{J}$ , compute $I = I_{0} e^{q V_{J} / k T}$ , then compute $V_{A} = V_{J} + I R_{S}$ (Eq. 6.48). (Requires knowing $R_{S}$ .)

Measuring $R_{S}$ (Fig. 6.16b, c). On a forward-bias semilog plot like Fig. 6.10a, extend the ideal ( $q / k T$ ) portion into the slope-over region and measure the voltage displacement $Δ V$ between the two curves at each $I$ . Since $Δ V = V_{A} - V_{J} = I R_{S}$ , a plot of $Δ V$ versus $I$ is a straight line whose slope is $R_{S}$ .

Exercise 6.8

Problem. Crudely estimate $R_{S}$ of the diode of Fig. 6.10(a).

Solution. A decent $Δ V$ vs. $I$ plot would need a blow-up of the slope-over region; instead use $Δ V$ at a given $I$ . The curve terminates at $I \approx 80$ mA with $V_{A} = 0.85$ V. In Fig. E6.7, extending the $q / k T$ line passes through 80 mA at about 0.78 V. Thus

Δ V \approx 0.85 - 0.78 = 0.07 V, R_{S} = \frac{Δ V}{I} \approx \frac{0.07}{0.08} \approx 0.9 \approx 1 Ω

High-level injection

The low-level injection assumption fails when the minority carrier concentration at the depletion edge on the lightly doped side approaches the doping concentration. In Si at room temperature this typically happens a few tenths of a volt below $V_{bi}$ . Beyond that, the injection is high level: both minority and majority carrier concentrations adjacent to the depletion region are perturbed (Fig. 6.17a). The majority concentration must rise to maintain approximate charge neutrality ( $Δ n \approx Δ p$ ).

An analysis of high-level injection predicts a current varying roughly as $exp (q V_{A} /2 k T)$ : a high-current $q /2 k T$ region on the semilog plot (Fig. 6.17b). Intuitively, with $n \approx p$ at the edge, the law of the junction $n p = n_{i}^{2} e^{q V_{A} / k T}$ gives $p \propto e^{q V_{A} /2 k T}$ , so the injected carrier density, hence the current, goes as $e^{q V_{A} /2 k T}$ .

This $q /2 k T$ high-current region is seldom observed because it is obscured by the series resistance slope-over. Still, the enhanced carrier concentration of high-level injection can reduce the observed series resistance: the drop in resistivity from heavy carrier injection is called conductivity modulation. (Think of a narrow road suddenly receiving an extra lane of cars: the “resistance” to flow drops because there are more carriers to carry current.)

Exercise 6.9: diagnose the nonidealities

Problem. The measured $I$ - $V$ of a room-temperature Si diode is crudely sketched in Fig. E6.9 (log scale for forward, linear scale for reverse). Nonidealities are labeled A to E. Possible sources are listed (1 photogeneration; 2 thermal recombination in the depletion region; 3 avalanching and/or Zener process; 4 low-level injection; 5 depletion approximation; 6 thermal generation in the depletion region; 7 band bending; 8 series resistance; 9 $V_{A} > V_{bi}$ ; 10 high-level injection). Identify the cause of each nonideal feature.

Solution. (The OCR of the list is partly scrambled; the book’s answer is A-3, B-6, C-2, D-10, E-8.)

Label	Feature on sketch	Cause
A	Abrupt current increase at large reverse bias	3. Avalanching and/or Zener process (breakdown)
B	Reverse current that creeps up, does not saturate	6. Thermal generation in the depletion region
C	Excess forward current at small $V_{A}$ , slope $q /2 k T$	2. Thermal recombination in the depletion region
D	High-current region of slope $\approx q /2 k T$ above the ideal region	10. High-level injection
E	Slope-over at the highest currents	8. Series resistance

6.3 Special Considerations

The two topics here are supplementary to the basic d.c. response. They could be delayed until needed, but that would hide their importance and weaken the linkage of ideas. The charge control approach is a “big picture” method that often gives a reasonably accurate answer with minimal mathematics. The narrow-base diode is both a deviation from the ideal and a conceptual bridge to the bipolar junction transistor.

6.3.1 Charge Control Approach

In the charge control approach the basic carrier variable is the total charge of the minority carrier excess (or deficit) within an entire quasineutral region. Consider a forward-biased $p^{+}$ - $n$ diode. Let $Δ p_{n} (x, t)$ be the excess hole concentration on the n-side at time $t$ and position $x_{n} \leq x < \infty$ . The total excess hole charge is

Q_{P} = q A \int_{x_{n}}^{\infty} Δ p_{n} (x, t) d x (6.50)

(For reverse bias $Q_{P} < 0$ is a deficit.) $Q_{P}$ is independent of $x$ but may depend on time.

Goal: a relation governing the combined behavior of this charge. Start from the minority carrier diffusion equation (3.54b) with $G_{L} = 0$ :

\frac{\partial Δ p _{n}}{\partial t} = D_{P} \frac{\partial ^{2} Δ p _{n}}{\partial x ^{2}} - \frac{Δ p _{n}}{τ _{p}} (6.51)

In a $E \approx 0$ region the hole current density is

J_{P} = - q D_{P} \frac{\partial Δ p _{n}}{\partial x} (6.52)

so $q D_{P} \partial^{2} Δ p_{n} / \partial x^{2} = - \partial J_{P} / \partial x$ , and multiplying (6.51) by $q$ :

\frac{\partial ( q Δ p _{n} )}{\partial t} = - \frac{\partial J _{P}}{\partial x} - \frac{q Δ p _{n}}{τ _{p}} (6.53)

Integrate every term over the volume ( $A \int d x$ ) of the n-side quasineutral region, $x_{n}$ to $\infty$ :

\frac{d}{d t} [q A \int_{x_{n}}^{\infty} Δ p_{n} d x] = - A \int_{x_{n}}^{\infty} \frac{\partial J _{P}}{\partial x} d x - \frac{1}{τ _{p}} [q A \int_{x_{n}}^{\infty} Δ p_{n} d x] (6.54)

The bracketed quantities are $Q_{P}$ . Evaluate the current integral:

- A \int_{J_{P} (x_{n})}^{J_{P} (\infty)} d J_{P} = - A J_{P} (\infty) + A J_{P} (x_{n}) = A J_{P} (x_{n}) \equiv A J_{DIFF} = i_{DIFF} (6.55)

because the hole current goes to zero far into the n-side, $J_{P} (\infty) = 0$ . (Lowercase $i$ because the current may in general vary with time.) Therefore

\frac{d Q _{P}}{d t} = i_{DIFF} - \frac{Q _{P}}{τ _{p}} (6.56)

Interpretation. This is effectively a continuity equation for the excess hole charge. There are only two ways to change the excess charge in a region: holes can flow in or out ( $i_{DIFF}$ ), or the excess charge can be changed by recombination-generation within the region ( $- Q_{P} / τ_{p}$ ).

Analogy: a bathtub. $Q_{P}$ is the water level; $i_{DIFF}$ is the tap; $Q_{P} / τ_{p}$ is the leak in the bottom. In steady state the tap flow exactly equals the leak.

Steady-state use. For a $p^{+}$ - $n$ diode in steady state, $d Q_{P} / d t = 0$ , $i_{DIFF} = I_{DIFF}$ :

I_{DIFF} = \frac{Q _{P}}{τ _{p}} (6.57)

Now suppose we could not solve for $Δ p_{n} (x)$ . We know $Δ p_{n} (0)$ from (6.18), and suspect the solution is exponential with decay length $L_{P}$ . Use the squared-off approximation of Exercise 6.4: equate $Q_{P}$ to the charge of a rectangle of width $L_{P}$ and height $Δ p_{n} (x^{'} = 0)$ :

Q_{P} \approx q (A L_{P}) Δ p_{n} (x^{'} = 0) = q A L_{P} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.58)

I_{DIFF} = \frac{Q _{P}}{τ _{p}} = q A \frac{L _{P}}{τ _{p}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.59)

Since $L_{P} / τ_{p} = D_{P} / L_{P}$ , (6.59) is the usual $p^{+}$ - $n$ result. (Here the approximation happens to be exact, because the exact distribution is a pure exponential, whose area is exactly peak times $L_{P}$ .) The real power of charge control appears in transient problems, where solving the PDE is hard but (6.56) is easy.

6.3.2 Narrow-Base Diode

Current derivation

The ideal diode assumed contacts several diffusion lengths from the depletion edges ( $Δ n_{p} (- \infty) = 0$ , $Δ p_{n} (\infty) = 0$ ). For a diode whose lightly doped side is a substrate/wafer thickness, that is typically valid. It is questionable if the diode is made in a thin epitaxial layer. A diode whose quasineutral width on the lightly doped side is comparable to or less than a diffusion length is a narrow-base diode.

Take a $p^{+}$ - $n$ step junction identical to the ideal diode except that the n-side quasineutral width may be $≲ L_{P}$ . Define (Fig. 6.18): $x_{c}$ as the distance from the metallurgical junction to the n-side contact, and $x_{c}^{'} = x_{c} - x_{n}$ is the distance from the n-edge of the depletion region to the contact. The contact is ohmic, so $Δ p_{n} = 0$ at $x^{'} = x_{c}^{'}$ (an ohmic contact has a very high R-G rate, pinning the minority concentration to equilibrium).

Paralleling the ideal derivation, solve

0 = D_{P} \frac{d ^{2} Δ p _{n}}{d x ^{'2}} - \frac{Δ p _{n}}{τ _{p}} 0 \leq x^{'} \leq x_{c}^{'} (6.60)

with

Δ p_{n} (x^{'} = 0) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) (6.61a)

Δ p_{n} (x^{'} = x_{c}^{'}) = 0 (6.61b)

General solution:

Δ p_{n} (x^{'}) = A_{1} e^{- x^{'} / L_{P}} + A_{2} e^{x^{'} / L_{P}} 0 \leq x^{'} \leq x_{c}^{'} (6.62)

This time $A_{2}$ need not vanish. Apply the boundary conditions:

Δ p_{n} (0) = A_{1} + A_{2} (6.63a)

0 = A_{1} e^{- x_{c}^{'} / L_{P}} + A_{2} e^{x_{c}^{'} / L_{P}} (6.63b)

From (6.63b): $A_{1} = - A_{2} e^{2 x_{c}^{'} / L_{P}}$ . Substitute in (6.63a): $Δ p_{n} (0) = A_{2} (1 - e^{2 x_{c}^{'} / L_{P}})$ , so

A_{2} = \frac{Δ p _{n} ( 0 )}{1 - e ^{2 x_{c}^{'} / L_{P}}}, A_{1} = - \frac{Δ p _{n} ( 0 ) e ^{2 x_{c}^{'} / L_{P}}}{1 - e ^{2 x_{c}^{'} / L_{P}}}

Putting back into (6.62):

Δ p_{n} (x^{'}) = Δ p_{n} (0) \frac{e ^{(x_{c}^{'} - x^{'}) / L_{P}} - e ^{- (x_{c}^{'} - x^{'}) / L_{P}}}{e ^{x_{c}^{'} / L_{P}} - e ^{- x_{c}^{'} / L_{P}}} 0 \leq x^{'} \leq x_{c}^{'} (6.64)

(multiply numerator and denominator by $e^{- x_{c}^{'} / L_{P}}$ to confirm). In compact form, with $sinh ξ \equiv (e^{ξ} - e^{- ξ}) /2$ :

Δ p_{n} (x^{'}) = Δ p_{n} (0) \frac{sinh [ ( x _{c}^{'} - x ^{'} ) / L _{P} ]}{sinh [ x _{c}^{'} / L _{P} ]} (6.65, 6.66)

Current:

I_{DIFF} = A J_{P} (x^{'} = 0) = - q A D_{P} \frac{d Δ p _{n}}{d x ^{'}}_{x^{'} = 0} (6.67)

Differentiate (6.65): $\frac{d}{d x ^{'}} sinh [\frac{x _{c}^{'} - x ^{'}}{L _{P}}] = - \frac{1}{L _{P}} cosh [\frac{x _{c}^{'} - x ^{'}}{L _{P}}]$ ; at $x^{'} = 0$ this is $- cosh (x_{c}^{'} / L_{P}) / L_{P}$ . Hence

I_{DIFF} = I_{0}^{'} (e^{q V_{A} / k T} - 1) (6.68)

I_{0}^{'} = q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} coth (\frac{x _{c}^{'}}{L _{P}}) = q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} \frac{cosh ( x _{c}^{'} / L _{P} )}{sinh ( x _{c}^{'} / L _{P} )} (6.69)

with $cosh ξ \equiv (e^{ξ} + e^{- ξ}) /2$ (6.70).

Limiting cases and punch-through

Useful limits of the hyperbolic functions:

sinh ξ \to ξ (ξ \to 0) (6.71 a), sinh ξ \to \frac{1}{2} e^{ξ} (ξ \to \infty) (6.71 b)

cosh ξ \to 1 + \frac{ξ ^{2}}{2} \to 1 (ξ \to 0) (6.72 a), cosh ξ \to \frac{1}{2} e^{ξ} (ξ \to \infty) (6.72 b)

Wide-base limit ( $x_{c}^{'} \to \infty$ , $x_{c}^{'} / L_{P} ≫ 1$ ). The sinh ratio in (6.65) becomes $e^{- x^{'} / L_{P}}$ (since $sinh [(x_{c}^{'} - x^{'}) / L_{P}] / sinh [x_{c}^{'} / L_{P}] \to e^{- x^{'} / L_{P}}$ ), recovering Eq. (6.23). And $cosh / sinh \to 1$ , so (6.69) reduces to the standard ideal $p^{+}$ - $n$ result. The narrow-base analysis is thus a generalization of the ideal diode to a quasineutral region of arbitrary thickness. Mathematically reassuring but nothing new.

Narrow-base limit ( $x_{c}^{'} ≪ L_{P}$ ). Much more interesting.

Using $sinh ξ \approx ξ$ for both numerator and denominator:

Δ p_{n} (x^{'}) = Δ p_{n} (0) (1 - \frac{x ^{'}}{x _{c}^{'}}) (6.73)

a straight line between the two end values set by the boundary conditions. This linearity comes from negligible thermal R-G in a region much shorter than a diffusion length. In fact it follows immediately by dropping the $- Δ p_{n} / τ_{p}$ term of the diffusion equation ( $d^{2} Δ p_{n} / d x^{'2} = 0$ ). That justifies neglecting the thermal R-G term whenever the quasineutral width is small compared with a diffusion length.

The current: $coth (x_{c}^{'} / L_{P}) \to L_{P} / x_{c}^{'}$ , so

I_{DIFF} \to q A \frac{n _{i}^{2}}{N _{D}} \frac{D _{P}}{x _{c}^{'}} (e^{q V_{A} / k T} - 1)

(no $L_{P}$ : all the injected holes reach the contact, with none lost to recombination). The width $x_{c}^{'} = x_{c} - x_{n}$ shrinks with increasing reverse bias because the depletion width grows. Hence $I_{DIFF} (V_{A} < 0) \propto 1/ x_{c}^{'}$ does not saturate but increases systematically with reverse bias.

If $x_{c}$ is small enough, $x_{c}^{'} \to 0$ is possible: the entire quasineutral region becomes depleted. This is called punch-through. From (6.68)/(6.69), $I_{DIFF} (V_{A} < 0) \to - \infty$ as $x_{c}^{'} \to 0$ . In reality the depletion approximation (sharp boundaries) breaks down in that limit; experimentally, and in more precise theories, the diffusion current stays finite at punch-through unless the field is high enough to cause avalanche breakdown.

Exercise 6.10

Problem. A planar Si $p^{+}$ - $n$ step junction at room temperature has $N_{D} = 1 0^{16} cm^{- 3}$ and an n-side thickness $x_{c} = 2 μ m$ . Using the depletion approximation, find the punch-through voltage: the voltage required to completely deplete the n-side.

Solution. At punch-through $x_{n} = x_{c}$ . From Eq. (5.37), $x_{n} = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}$ (the $p^{+}$ side contributes negligibly). Then

x_{c}^{2} V_{A} = \frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A}) = V_{bi} - \frac{q N _{D} x _{c}^{2}}{2 K _{S} ε _{0}} = 0.92 - \frac{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} ) ( 2 \times 1 0 ^{- 4} ) ^{2}}{2 ( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )} = 0.92 - \frac{6.4 \times 1 0 ^{- 11}}{2.09 \times 1 0 ^{- 12}} = 0.92 - 30.6 = - 29.7 V

where $V_{bi} = 0.92$ V was read from Fig. 5.11 (equivalently $V_{bi} = E_{G} /2 q + (k T / q) ln (N_{D} / n_{i}) = 0.56 + 0.357$ ). So the punch-through voltage is 29.7 V reverse. From Fig. 6.11, avalanche breakdown in a comparable wide-base diode occurs at $V_{B R} \approx 55$ V. Thus punch-through takes place prior to breakdown in the given narrow-base diode.

6.4 Summary and Concluding Comments

The chapter modeled the steady state response of the pn junction diode.

Qualitative picture. Using the energy band diagram:

Reverse-bias current in an ideal diode: minority carriers wander into the depletion region and are accelerated to the other side.
Forward bias lowers the hill between the two sides and enhances majority carrier injection into the opposite-side quasineutral region.
Nothing builds up in steady state: R-G stabilizes the minority concentrations while very rapid rearrangement and interchange with the contacts maintains the majority concentrations.

The ideal diode theory. We formulated, derived and examined it. Although the predicted $I$ - $V$ has deficiencies against experiment, it gives valuable insight and an analytical base for more exact treatments, and it is widely used for first-order predictions. The ideal diode equation is only a small part of the ideal diode theory, which also includes the solution strategy and approximations, the mathematical steps, and predictions about carrier concentrations, carrier currents, material dependences, doping dependences, etc.

Deviations from the ideal (Si, room temperature):

Deviation	Cause	Key information
Reverse breakdown	Avalanche or Zener process	avalanche $V_{B R}$ varies roughly inversely with light-side doping; carrier multiplication gives enhanced current far below $V_{B R}$
Excess current at small forward and all reverse biases	Thermal R-G in the depletion region	forward $\sim exp (q V_{A} /2 k T)$ ( $η \to 2$ ); reverse $\propto W$ ; $I_{R-G} \propto n_{i}$ while $I_{DIFF} \propto n_{i}^{2}$
Slope-over at high forward current	Voltage drop across series resistance	$V_{J} = V_{A} - I R_{S}$
Possible $q /2 k T$ at high current	High-level injection	usually obscured by $R_{S}$ ; conductivity modulation reduces observed $R_{S}$

(The development concentrated on the step junction; other profiles may need analytical modifications.)

Special considerations. Charge control (a different analytical approach) and the narrow-base diode (the ideal analysis generalized to arbitrary base width). For widths small compared to a diffusion length the perturbed carrier distribution becomes linear; for extremely small widths there is the possibility of punch-through at a reverse bias below $V_{B R}$ .

Equations worth memorizing: the ideal diode solution (6.29/6.30), the law of the junction (6.12), the depletion-edge boundary conditions (6.15/6.18), and the reverse-bias R-G current (6.43). Retain the book for the rest.

Problems (Chapter 6)

Problem information table (from the book): suggested point weightings and difficulty levels.

Problem	Complete after	Difficulty	Points	Short description
6.1	6.1.4	1	10 (1 each part)	Ideal diode quiz
6.2	6.2.4	1	10	Compare actual and ideal $I$ - $V$
6.3	6.1.1	1	9 (3 each part)	Band diagram sketches
6.4	6.1.1	2	8	Forward-bias big picture
6.5	6.1.2	2-3	10	Derive Eq. (6.15) using $J_{N} = 0$
6.6	6.1.3	2	10 (5 each part)	Typical $I$ values using diary
6.7	6.1.3	2	16 (a-10, b-5, c-1)	Ideal $I$ - $V$ variation with $T$
6.8	6.1.3	3	16 (a-3, b-10, c-3)	Photodiode $I$ - $V$ derivation
6.9	6.1.3	1	5	Photodiode $I$ - $V$ plot
6.10	6.1.4	2	10 (a-c: 2 each, d-4)	Deduce info from concentration plot
6.11	6.1.4	3	12	Deduce, plot carrier concentrations
6.12	6.1.4	3	12 (a-9, b-3)	pn diode temperature sensor
6.13	6.2.2	2	6 (2 each part)	$V_{B R}$ -related calculations
6.14	6.2.2	2	10 (a-8, b-2)	Breakdown $I$ - $V$ ( $M I_{0}$ )
6.15	6.2.3	3	12	Breakdown $I$ - $V$ ( $M I_{R-G}$ )
6.16	6.2.3	3	8	Determine elevated $T$ point
6.17	6.2.3	3	10	Modified $I_{R-G}$
6.18	6.2.4	3	9 (a-6, b-3)	$R_{S}$ -related computations
6.19	6.2.4	3	20	Deduce $I_{0 j}$ , $n_{j}$ from $I$ - $V$ (10 line fit, 8 least-squares fit, 2 discuss)
6.20	6.2.4	2-3	15	Deduce $R_{S}$ from $I$ - $V$
6.21	6.2.4	4	40	Multiple nonideality $I$ - $V$
6.22	6.3.2	3	10	Pseudo-narrow-base diode
6.23	6.3.2	3	15	Narrow/wide diode combination
6.24	6.3.2	4	34	Multiple considerations

A note on data quality. The OCR of several problem statements lost digits or whole phrases (for example the doping in 6.6, 6.13, 6.16 and the figure values in 6.10 and 6.24). Wherever that happened, I state my assumption explicitly, give the solution symbolically first, and then evaluate with the stated assumption, so you can re-plug the true numbers from your textbook.

Standard numbers used throughout (Si, 300 K): $k T / q = 0.0259$ V, $n_{i} = 1 0^{10} cm^{- 3}$ , $K_{S} = 11.8$ , $ε_{0} = 8.85 \times 1 0^{- 14}$ F/cm, $q = 1.6 \times 1 0^{- 19}$ C, $E_{G} = 1.12$ eV. Mobility fits (Exercise 3.1 form): $μ_{p} = 54.3 + 406.9/ [1 + (N /2.35 \times 1 0^{17})^{0.88}]$ ; $μ_{n} = 92 + 1268/ [1 + (N /1.3 \times 1 0^{17})^{0.91}]$ (cm $^{2}$ /V s).

Problem 6.1 Ideal Diode Quiz

Statement. Answer concisely, assuming an ideal diode. (a) The forward-bias current is associated with what type of carrier activity? (b) The reverse-bias current? (c) Why is the reverse-bias current expected to be small and to saturate at small reverse voltage? (d) Under reverse bias, what processes occur in the quasineutral regions adjacent to the depletion region edges: drift and diffusion, diffusion and recombination, generation and diffusion, or generation and drift? (e) Why can’t the minority carrier diffusion equations be used to find the minority carrier concentrations and currents in the depletion region? (f) What a priori justification is there for assuming R-G is negligible throughout the depletion region? (g) What exactly is a “wide-base” diode? (h) What is the “law of the junction”? (i) Given a semilog plot of the forward-bias $I$ - $V$ , how does one determine $I_{0}$ ? (j) True or false: the reverse saturation current may be viewed as arising from minority carrier generation in the quasineutral regions.

Solution.

(a) Injection and diffusion of majority carriers over the lowered potential hill; once across they are minority carriers that diffuse and recombine in the opposite quasineutral region (diffusion current).

(b) Minority carriers wandering into the depletion region from the quasineutral regions and being swept across by the field (minority carrier drift/collection).

(c) It is carried by minority carriers, which are few ( $n_{p 0} = n_{i}^{2} / N_{A}$ , $p_{n 0} = n_{i}^{2} / N_{D}$ ), hence small. It saturates because, once the reverse bias exceeds a few $k T / q$ , majority carrier diffusion is cut off and the current is set by the rate at which minority carriers arrive at the depletion edge, which does not depend on the barrier height (waterfall analogy).

(d) Generation and diffusion. The depletion edges act as a sink that depletes minority carriers, so thermal generation in the quasineutral regions replenishes them, and the generated minority carriers diffuse toward the depletion edge (net generation, since $Δ n < 0$ ).

(e) The diffusion equations were derived assuming $E \approx 0$ and low-level injection. Neither holds in the depletion region, where $E \neq = 0$ is large (drift is important) and the carrier concentrations vary rapidly; there the full continuity equations are needed.

(f) None. It is simply an assumption that leads to a simple solution (it makes $J_{N}$ and $J_{P}$ constant across the depletion region). Its validity for real diodes is tested by comparison with experiment (it fails at small forward and all reverse biases in room-temperature Si).

(g) A diode whose contacts are several minority carrier diffusion lengths or more from the edges of the depletion region, so that perturbations die out before reaching the contacts (boundary conditions $Δ n_{p} (- \infty) = 0$ , $Δ p_{n} (+ \infty) = 0$ ).

(h) $n p = n_{i}^{2} e^{q V_{A} / k T}$ for $- x_{p} \leq x \leq x_{n}$ : the product of the carrier concentrations inside the depletion region (it follows from assuming the quasi-Fermi levels are constant across the depletion region, $F_{N} - F_{P} = q V_{A}$ ). It gives the depletion-edge boundary conditions.

(i) Extend the linear (slope $q / k T$ ) region of the semilog plot back to $V_{A} = 0$ ; the intercept on the current axis is $I_{0}$ .

(j) True. Under reverse bias the minority carriers swept out of the quasineutral regions are replaced by generation there, so $I_{0} = q \times$ (generation rate in $\sim L_{N}$ and $L_{P}$ thick layers), as in Exercise 6.4.

Problem 6.2 Actual versus ideal characteristic

Statement. Compare (with sketches) the $I$ - $V$ of an actual room-temperature Si pn diode with the ideal characteristic. Note and clarify all deviations; briefly identify the cause of each (a sentence or two each).

Solution. Sketch description: on a linear plot, the ideal curve is flat at $- I_{0}$ for reverse bias, and rises exponentially for forward bias. The actual curve is the same shape at forward bias but with a gentler rise at the very high-current end, and in the reverse direction creeps down and then plunges. On a semilog forward plot, the ideal curve is a single straight line of slope $q / k T$ ; the actual curve has three segments: slope $\approx q /2 k T$ at low $V_{A}$ (above the extrapolated ideal line), the ideal $q / k T$ segment, then a flattening at high $V_{A}$ .

Region	Actual vs. ideal	Cause
Large reverse bias	Current $\to \infty$ at $V_{B R}$ (ideal: saturates forever); a sloping approach with carrier multiplication $M=1/[1-(	V_A
Moderate reverse bias	Current larger than $I_{0}$ by orders of magnitude and increases with $	V_A
Small forward bias	Current above ideal, slope $q /2 k T$	Thermal recombination in the depletion region, $I \propto e^{q V_{A} /2 k T}$
Moderate forward bias	Ideal $q / k T$ slope: matches theory	Diffusion current
$V_{A} \to V_{bi}$	Slope-over: current below the ideal extrapolation	Series resistance ( $V_{J} = V_{A} - I R_{S}$ ); possibly an obscured $q /2 k T$ region from high-level injection

Problem 6.3 Band diagrams

Statement. Sketch the energy band diagram for an ideal $p^{+}$ - $n$ step junction diode showing carrier activity in and near the depletion region when (a) $V_{A} = 0$ , (b) $V_{A} > 0$ , (c) $V_{A} < 0$ .

Solution. (Descriptions of the sketches; follow Fig. 6.1.)

(a) $V_{A} = 0$ . Flat $E_{F}$ through the entire diode. Bands bend by $q V_{bi}$ across the depletion region; since $p^{+}$ - $n$ , $E_{F}$ sits very close to $E_{v}$ on the p-side and well above midgap on the n-side. Draw:

n-side electrons (many) near $E_{c}$ ; a few of the energetic ones climb the hill and cross: arrow right-to-left (n to p) labeled “diffusion of electrons.”
p-side holes (many, extremely many since $p^{+}$ ) near $E_{v}$ ; the energetic few cross to the n-side: “diffusion of holes.”
p-side minority electrons (very few) rolling “downhill” to the n-side: “drift”; n-side minority holes floating “uphill” (holes float up on the valence band) to the p-side: “drift.”
Equal and opposite arrows: net current zero.

(b) $V_{A} > 0$ . $E_{F n}$ above $E_{F p}$ by $q V_{A}$ (the n-side shifted up relative to the p-side). The barrier is $q (V_{bi} - V_{A})$ , lower. Show a larger number of crossing arrows of majority carriers (electrons n to p, holes p to n), with the hole arrow much thicker than the electron arrow (since $p^{+}$ - $n$ : $I_{P} ≫ I_{N}$ ). The minority drift arrows are unchanged. Show the injected carriers as an excess pile-up on the far sides (shaded), and the quasi-Fermi levels $F_{N}$ , $F_{P}$ flat across the depletion region.

(c) $V_{A} < 0$ . $E_{F p}$ above $E_{F n}$ by $q ∣ V_{A} ∣$ ; barrier $q (V_{bi} + ∣ V_{A} ∣)$ , taller. Majority diffusion arrows reduced to negligible (draw tiny or crossed out). Minority carrier drift arrows (p-side electrons going to the n-side; n-side holes going to the p-side) still present and unchanged. Add R-G centers $E_{T}$ in the quasineutral regions near the depletion edges showing generation (valence electron up to $E_{T}$ and to $E_{c}$ ) that replaces the lost minority carriers. Total current is the small saturation current $- I_{0}$ from n to p.

Problem 6.4 The big picture, forward bias

Statement. Construct a composite energy-band/circuit diagram analogous to Fig. 6.2 that gives an overall view of carrier activity in a forward-biased pn diode.

Solution. The diagram is the band diagram of Fig. 6.1(b) with ohmic contacts at both ends (capacitor-like plates), connected through a battery of voltage $V_{A}$ (positive terminal at the p-contact). Add the following arrows, left (p-contact) to right (n-contact), as the mirror image of Fig. 6.2:

p-contact. Holes enter the p-quasineutral region from the contact, pushed in by the battery: the holes in-line shift slightly toward the junction (displacement propagates rapidly, like latecomers pushing into a packed room). Deep in the p-side the current is almost all hole drift, driven by a tiny field.
p-side quasineutral region. As holes approach the junction, some of them recombine with electrons that were injected from the n-side (via R-G centers $E_{T}$ : an electron falls to $E_{T}$ and then to $E_{v}$ , filling a hole). The hole current decays, the electron current (injected electrons diffusing leftward) grows toward the junction.
Depletion region. Majority carriers surmount the lowered hill: holes p to n, electrons n to p. Both carriers pass through the depletion region (draw both arrows; no R-G assumed in an ideal diode, so each carrier current is constant across it).
n-side quasineutral region. The injected holes diffuse into the n-region as minority carriers and recombine with electrons (via $E_{T}$ ). Each recombination consumes an electron, which is resupplied from the n-contact: electrons enter from the n-contact (pushed by the battery) and move toward the junction by drift, becoming the dominant current carriers far from the junction.
Circuit. At the p-contact each hole that arrives is “created” by an electron leaving the semiconductor into the metal; at the n-contact, electrons are supplied from the metal. In the external circuit, electrons flow from the battery’s negative terminal to the n-contact and from the p-contact to the positive terminal, i.e. conventional current flows in the forward direction through the diode.
Steady state. No charge accumulates: every electron-hole pair recombined in the quasineutral regions is replaced by carriers arriving from the contacts; this is the forward-bias counterpart of generation-replacement in Fig. 6.2.

Summary label on the diagram: $I =$ (hole current dominating at the p-end) $=$ (electron current dominating at the n-end), constant overall.

Problem 6.5 Boundary condition via the “Videri” ( $J_{N} = 0$ ) approach

Statement. Referring to the derivation in Subsection 5.1.4 (equilibrium, $J_{N} = 0$ ), show that Eq. (6.15) can be derived by continuing to assume $J_{N} = 0$ when $V_{A} \neq = 0$ . Use the Eq. (5.10) result to eliminate $V_{bi}$ in your expression for $n (- x_{p})$ .

Solution. Assume (as in 5.1.4) that the electron current is negligible compared with its drift and diffusion components inside the depletion region, i.e. $J_{N} \approx 0$ there, even for $V_{A} \neq = 0$ . Start with

J_{N} = q μ_{n} n E + q D_{N} \frac{d n}{d x} = 0

Use $E = - d ψ / d x$ ( $ψ$ electrostatic potential) and the Einstein relation $D_{N} = (k T / q) μ_{n}$ :

0 \frac{d n}{n} = - q μ_{n} n \frac{d ψ}{d x} + q μ_{n} \frac{k T}{q} \frac{d n}{d x} = \frac{q}{k T} d ψ

Integrate across the depletion region from $x = - x_{p}$ to $x = x_{n}$ :

ln \frac{n ( x _{n} )}{n ( - x _{p} )} = \frac{q}{k T} [ψ (x_{n}) - ψ (- x_{p})]

The potential difference across the depletion region is the built-in potential reduced by the applied voltage: $ψ (x_{n}) - ψ (- x_{p}) = V_{bi} - V_{A}$ . Therefore

n (- x_{p}) = n (x_{n}) e^{- q (V_{bi} - V_{A}) / k T}

Low-level injection on the n-side gives $n (x_{n}) \approx N_{D}$ :

n (- x_{p}) = N_{D} e^{- q V_{bi} / k T} e^{q V_{A} / k T}

Eq. (5.10): $V_{bi} = \frac{k T}{q} ln \frac{N _{A} N _{D}}{n _{i}^{2}}$ , so $e^{- q V_{bi} / k T} = \frac{n _{i}^{2}}{N _{A} N _{D}}$ . Hence

n (- x_{p}) = N_{D} \frac{n _{i}^{2}}{N _{A} N _{D}} e^{q V_{A} / k T} = \frac{n _{i}^{2}}{N _{A}} e^{q V_{A} / k T}

which is Eq. (6.14); subtracting $n_{p 0} = n_{i}^{2} / N_{A}$ gives

Δ n_{p} (- x_{p}) = \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1) ■

Problem 6.6 Typical ideal currents (computer)

Statement. An $n^{+}$ - $p$ Si step junction ( $N_{A} = 1 0^{15} cm^{- 3}$ [assumed; OCR shows “10?” with the exponent lost], $τ_{n} = 1 0^{- 6}$ s, $A = 1 0^{- 3} cm^{2}$ ) is ideal. Compute typical currents with MATLAB (record using diary). (a) Ideal diode current at $T = 300$ K for (i) $V_{A} = - 50$ V, (ii) $- 0.1$ V, (iii) $0.1$ V, (iv) $0.5$ V. (b) Assuming $τ_{n}$ is temperature independent, repeat at $T = 500$ K.

Solution. For an $n^{+}$ - $p$ diode only the p-side matters:

I_{0} = q A \frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}}, I = I_{0} (e^{q V_{A} / k T} - 1), L_{N} = D_{N} τ_{n}, D_{N} = \frac{k T}{q} μ_{n}

Temperature dependence of $n_{i}$ (used for 500 K): $n_{i} (T) = n_{i} (300) (\frac{T}{300})^{3/2} exp [\frac{E _{G}}{2 k} (\frac{1}{300} - \frac{1}{T})]$ with $E_{G} = 1.12$ eV, which gives $n_{i} (500 K) \approx 1.25 \times 1 0^{14} cm^{- 3}$ . The mobility is taken as that of 300 K (the problem does not give a $T$ dependence; with a lattice-scattering $T^{- 3/2}$ it would shift $I_{0}$ by only a small factor compared with the $n_{i}^{2}$ change of $1 0^{8}$ ).

(a) $T = 300$ K. $μ_{n} (1 0^{15}) \approx 1345 cm^{2} / V s$ :

D_{N} L_{N} I_{0} = (0.02585) (1345) = 34.8 cm^{2} / s = (34.8) (1 0^{- 6}) = 5.9 \times 1 0^{- 3} cm = (1.6 \times 1 0^{- 19}) (1 0^{- 3}) \frac{34.8}{5.9 \times 1 0 ^{- 3}} \frac{1 0 ^{20}}{1 0 ^{15}} = 9.4 \times 1 0^{- 14} A

$V_{A}$	$e^{q V_{A} / k T} - 1$	$I$
$- 50$ V	$\approx - 1$	$- 9.4 \times 1 0^{- 14}$ A
$- 0.1$ V	$- 0.979$	$- 9.2 \times 1 0^{- 14}$ A
$+ 0.1$ V	$47.0$	$4.4 \times 1 0^{- 12}$ A
$+ 0.5$ V	$2.5 \times 1 0^{8}$	$2.4 \times 1 0^{- 5}$ A

(b) $T = 500$ K. $k T / q = 0.0431$ V; $D_{N} = (0.0431) (1345) = 58$ cm $^{2}$ /s (same mobility), $L_{N} = 7.6 \times 1 0^{- 3}$ cm:

I_{0} = (1.6 \times 1 0^{- 19}) (1 0^{- 3}) \frac{58}{7.6 \times 1 0 ^{- 3}} \frac{( 1.25 \times 1 0 ^{14} ) ^{2}}{1 0 ^{15}} \approx 1.9 \times 1 0^{- 5} A

$V_{A}$	$e^{q V_{A} / k T} - 1$	$I$
$- 50$ V	$- 1$	$- 1.9 \times 1 0^{- 5}$ A
$- 0.1$ V	$- 0.90$	$- 1.7 \times 1 0^{- 5}$ A
$+ 0.1$ V	$9.2$	$1.7 \times 1 0^{- 4}$ A
$+ 0.5$ V	$1.1 \times 1 0^{5}$	about $2$ A

(The $0.5$ V, 500 K result is of course unrealistically large: a real diode would be limited by series resistance, as discussed in 6.2.4.) The key lesson: a rise of 200 K increases $I_{0}$ by about eight orders of magnitude, because of the $n_{i}^{2} \propto T^{3} e^{- E_{G} / k T}$ dependence.

MATLAB record (diary) program:

diary ideal6_6.txt
clear; k = 8.617e-5; q = 1.6e-19;
A = 1e-3; NA = 1e15; taun = 1e-6; EG = 1.12;
mun = 92 + 1268/(1 + (NA/1.3e17)^0.91);       % minority electron mobility
for T = [300 500]
    ni = 1e10*(T/300)^1.5*exp(EG/(2*k)*(1/300 - 1/T));
    DN = k*T*mun;  LN = sqrt(DN*taun);
    I0 = q*A*(DN/LN)*ni^2/NA;
    VA = [-50 -0.1 0.1 0.5];
    I  = I0*(exp(VA/(k*T)) - 1);
    fprintf('T = %g K, I0 = %.3e A\n', T, I0);
    disp([VA' I'])
end
diary off

Problem 6.7 Ideal $I$ - $V$ versus temperature (computer)

Statement. Ideal Si $p^{+}$ - $n$ diode, $A = 1 0^{- 4} cm^{2}$ , $N_{D} = 1 0^{16} cm^{- 3}$ , room-temperature $τ_{p} = 1 0^{- 6}$ s. (a) Using the empirical $μ_{p}$ fit (Exercise 3.1), the $n_{i} (T)$ fit (Exercise 2.4b), and assuming $τ_{p}$ is temperature independent, write a MATLAB program computing and plotting the ideal $I$ - $V$ with $T$ as an input; print a plot with $T = 295$ , $300$ , $305$ K, limiting axes to $- 1 \leq V_{A} \leq 0.2$ V and $- 20 I_{0} \leq I \leq 50 I_{0}$ (use the largest $I_{0}$ to set limits). (b) Modify to get a semilog plot of $I_{0}$ versus $T$ for $300 \leq T \leq 400$ K. (c) Comment.

Solution.

Physics. $I_{0} (T) = q A D_{P} / τ_{p} n_{i}^{2} (T) / N_{D}$ with $D_{P} = (k T / q) μ_{p}$ . Since $n_{i}^{2} \propto T^{3} e^{- E_{G} / k T}$ dominates:

\frac{1}{I _{0}} \frac{d I _{0}}{d T} \approx \frac{3}{T} + \frac{E _{G}}{k T ^{2}} = \frac{3 + E _{G} / k T}{T} = \frac{3 + 43.3}{300} \approx 0.154 K^{- 1}

i.e. about 15% per kelvin. (For the $n_{i} (T)$ fit of Exercise 2.4(b) I use the physically equivalent form $n_{i} = 1 0^{10} (T /300)^{3/2} exp [(E_{G} /2 k) (1/300 - 1/ T)]$ with $E_{G} = 1.12$ eV, since the book’s fit formula is not reproduced in this chapter.)

(a) Program.

% Problem 6.7(a): ideal I-V versus temperature, Si p+-n
clear; close
k = 8.617e-5; q = 1.6e-19; A = 1e-4; ND = 1e16; taup = 1e-6; EG = 1.12;
T = input('Enter temperatures in K, e.g. [295 300 305]: ');
mup = 54.3 + 406.9/(1 + (ND/2.35e17)^0.88);   % mobility at ND (T-independent here)
VA = linspace(-1, 0.2, 400);
I = zeros(length(T), length(VA));  I0 = zeros(size(T));
for i = 1:length(T)
    ni = 1e10*(T(i)/300)^1.5*exp(EG/(2*k)*(1/300 - 1/T(i)));
    DP = k*T(i)*mup;  LP = sqrt(DP*taup);
    I0(i) = q*A*(DP/LP)*ni^2/ND;
    I(i,:) = I0(i)*(exp(VA/(k*T(i))) - 1);
end
plot(VA, I'); grid
axis([-1 0.2 -20*max(I0) 50*max(I0)])      % limits set from the largest I0
xlabel('V_A (volts)'); ylabel('I (amps)')
legend(num2str(T'))

At 300 K, $μ_{p} = 437$ , $D_{P} = 11.3$ , $L_{P} = 3.4 \times 1 0^{- 3}$ cm and $I_{0} = 5.4 \times 1 0^{- 16}$ A (the same value as in Exercise 6.6). With the $I_{0}$ scaling of about 15% per kelvin, the 295 K and 305 K curves differ from the 300 K one by $I_{0} (305) / I_{0} (300) \approx 2.1$ and $I_{0} (295) / I_{0} (300) \approx 0.47$ .

(b) Semilog $I_{0} (T)$ .

T = 300:2:400;
ni = 1e10*(T/300).^1.5.*exp(EG/(2*k)*(1./300 - 1./T));
DP = k*T*mup; LP = sqrt(DP*taup);
I0 = q*A*(DP./LP).*ni.^2/ND;
semilogy(T, I0); grid; xlabel('T (K)'); ylabel('I_0 (A)')

(c) Comments. (i) Under reverse bias, $I \approx - I_{0}$ moves strongly: the 295/300/305 K curves are widely separated in the reverse direction, and for the forward direction they shift to lower $V_{A}$ at higher $T$ for the same current (the exponential $e^{q V_{A} / k T}$ is reduced slightly by the larger $T$ , but this is overwhelmed by the huge growth of $I_{0}$ ). (ii) $I_{0}$ vs. $T$ is nearly a straight line on the semilog plot (an Arrhenius-like law dominated by $e^{- E_{G} / k T}$ ), increasing by roughly four orders of magnitude between 300 K and 400 K. (iii) Practical consequences: diode characteristics are very temperature sensitive; this is the basis of the diode thermometer (Problem 6.12) but also a hazard for circuit design.

Problem 6.8 Photodiode / solar cell

Statement. A photodiode is a pn diode packaged to admit light. Consider an ideal $p^{+}$ - $n$ step diode in which light is uniformly absorbed throughout the device with generation rate $G_{L}$ (EHP/cm $^{3}$ s). Assume low-level injection. (a) What is the excess minority carrier concentration on the n-side far from the junction ( $x \to \infty$ )? [Note: $Δ p_{n} (x \to \infty) \neq = 0$ .] (b) The usual boundary conditions (6.15), (6.18) still hold at the depletion edges. Using the revised far-field condition of (a) and (6.18), derive the $I$ - $V$ of the $p^{+}$ - $n$ diode under illumination. As before ignore all R-G (including photogeneration) in the depletion region. (c) Sketch the general form of the photodiode $I$ - $V$ for $G_{L} = 0$ , $G_{L 0}$ , $2 G_{L 0}$ , $4 G_{L 0}$ . Assume the light intensity is high enough to perturb the characteristics significantly when $G_{L} = G_{L 0}$ (Problem 6.9 may be substituted).

Solution.

(a) Far from the junction nothing depends on $x$ , so $d^{2} Δ p_{n} / d x^{2} = 0$ and the diffusion equation with generation, $0 = D_{P} \frac{d ^{2} Δ p _{n}}{d x ^{2}} - \frac{Δ p _{n}}{τ _{p}} + G_{L}$ , gives

0 = - \frac{Δ p _{n} ( \infty )}{τ _{p}} + G_{L} \Rightarrow Δ p_{n} (x \to \infty) = G_{L} τ_{p}

(generation balanced by recombination).

(b) On the n-side ( $x^{'} \geq 0$ ) solve

0 = D_{P} \frac{d ^{2} Δ p _{n}}{d x ^{'2}} - \frac{Δ p _{n}}{τ _{p}} + G_{L}

Let $u = Δ p_{n} - G_{L} τ_{p}$ . Then $d^{2} u / d x^{'2} = u / L_{P}^{2}$ (the constant cancels), with boundary conditions $u (\infty) = 0$ and

u (0) = Δ p_{n} (0) - G_{L} τ_{p} = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) - G_{L} τ_{p}

By the same argument as before ( $A_{2} = 0$ ):

Δ p_{n} (x^{'}) = G_{L} τ_{p} + [\frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) - G_{L} τ_{p}] e^{- x^{'} / L_{P}}

Hole current density at the depletion edge, $J_{P} (0) = - q D_{P} d Δ p_{n} / d x^{'} ∣_{0}$ :

J_{P} (0) = \frac{q D _{P}}{L _{P}} [\frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) - G_{L} τ_{p}]

Since $p^{+}$ - $n$ means the electron term from the $p^{+}$ side is neglected, $I = A J_{P} (0)$ :

I = q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1) - q A \frac{D _{P} τ _{p}}{L _{P}} G_{L}

Use $D_{P} τ_{p} / L_{P} = L_{P}^{2} / L_{P} = L_{P}$ :

I = I_{0} (e^{q V_{A} / k T} - 1) + I_{L}, I_{L} = - q A L_{P} G_{L}

with $I_{0} = q A (D_{P} / L_{P}) (n_{i}^{2} / N_{D})$ . The light adds a constant, negative (reverse-direction) photocurrent, equal to $q$ times the generation rate in a volume $A L_{P}$ : only carriers generated within about a diffusion length of the junction get collected.

(c) The photodiode characteristic is the dark diode curve shifted downward by $∣ I_{L} ∣ \propto G_{L}$ . Sketch: four copies of the ordinary exponential $I$ - $V$ curve, each displaced down by $0$ , $∣ I_{L 0} ∣$ , $2∣ I_{L 0} ∣$ , $4∣ I_{L 0} ∣$ . Each passes through the fourth quadrant (voltage positive, current negative): this is the power-generating (solar-cell) quadrant. The curves cross the $V_{A}$ axis at the open-circuit voltage $V_{O C} = (k T / q) ln (1 + ∣ I_{L} ∣/ I_{0})$ , which increases logarithmically with $G_{L}$ (so $V_{O C}$ for $4 G_{L 0}$ is only about $(k T / q) ln 4 = 36$ mV larger than for $2 G_{L 0}$ plus the base difference), and cross the current axis at the short-circuit current $I_{S C} = I_{L}$ , which is directly proportional to $G_{L}$ . In reverse bias ( $V_{A} < 0$ , third quadrant), the current is $- I_{0} + I_{L} \approx I_{L}$ : the photodiode operated this way gives a current proportional to light intensity.

Problem 6.9 Solar cell characteristic (computer)

Statement. The $I$ - $V$ of a solar cell is approximately $I = I_{0} (e^{q V_{A} / k T} - 1) + I_{L}$ , with $I_{L}$ always negative and a voltage-independent constant for a given illumination. Plot the general nature of the characteristics: at $T = 300$ K, simultaneously plot $I$ vs. $V_{A}$ for $I_{L} / I_{0} = 0, - 1, - 2, - 4$ [assumed: the OCR gives " $I_{L 0} = 0, - 1, - 2, - 4$ " with the unit lost; I plot $I / I_{0}$ so the choice of $I_{0}$ does not matter, and take $I_{L}$ in units of $I_{0}$ with an optional scale factor s to emphasize the effect]. Limit $V_{A}$ to $- 0.5 \leq V_{A} \leq 0.1$ V.

Solution.

% Problem 6.9: solar-cell I-V family
clear; close
k = 8.617e-5; T = 300;
s  = 1e3;                       % scale: IL = s*[0 -1 -2 -4]*I0 (use s=1 for the literal reading)
IL = s*[0 -1 -2 -4];            % in units of I0
VA = linspace(-0.5, 0.1, 300);
hold on
for j = 1:4
    plot(VA, exp(VA/(k*T)) - 1 + IL(j));
end
grid on; hold off
xlabel('V_A (volts)'); ylabel('I / I_0')
axis([-0.5 0.1 -5*s 5])

Result: four identical exponential curves shifted down by $∣ I_{L} ∣$ . Reverse bias gives $I \approx I_{L} - I_{0}$ (flat, light-proportional); the curves cross the voltage axis at $V_{O C} = (k T / q) ln (1 - I_{L} / I_{0})$ (for $s = 1 0^{3}$ : $V_{O C} = 0.0259 ln (1001) \approx 0.18$ V, 0.20 V, 0.22 V, … the larger the light, the larger $V_{O C}$ , but only logarithmically). The 4th quadrant region between the curve and the axes is where power $∣ I V ∣$ is delivered; the maximum power point lies on the knee just below $V_{O C}$ .

Problem 6.10 Reading information from a concentration plot

Statement. Fig. P6.10 is a dimensioned plot of the steady-state carrier concentrations (log scale) in a pn step diode at room temperature. (a) Forward or reverse biased? Explain. (b) Does low-level injection prevail in the quasineutral regions? Explain. (c) What are the p-side and n-side dopings? (d) Determine $V_{A}$ .

Reading the figure. The OCR of the figure labels is scrambled (only fragments “10”, “10^4”, “10^6”, “10^3” survive). I reconstruct the most consistent reading, which parallels Exercise 6.3: $n_{i} = 1 0^{10} cm^{- 3}$ ; far-field minority levels $p_{n 0} = 1 0^{4} cm^{- 3}$ (n-side) and $n_{p 0} = 1 0^{3} cm^{- 3}$ (p-side); edge minority levels $p_{n} (x_{n}) = 1 0^{6} cm^{- 3}$ and $n_{p} (- x_{p}) = 1 0^{5} cm^{- 3}$ ; majority levels $n_{n} = N_{D}$ , $p_{p} = N_{A}$ far above. If your figure has different numbers, use the same formulas.

Solution.

(a) Forward biased. The minority concentrations at the depletion edges ( $p_{n} (x_{n}) = 1 0^{6}$ , $n_{p} (- x_{p}) = 1 0^{5}$ ) are above their equilibrium far-field values ( $1 0^{4}$ , $1 0^{3}$ ): there is a pile-up of minority carriers. A reverse bias would give a deficit.

(b) Yes. Low-level injection requires the excess minority concentration to be much smaller than the majority concentration: $Δ p_{n} \leq 1 0^{6} ≪ n_{n} \sim 1 0^{16}$ and $Δ n_{p} \leq 1 0^{5} ≪ p_{p} \sim 1 0^{17}$ .

(c) Far from the junction the minority carriers are at equilibrium: $p_{n 0} = n_{i}^{2} / N_{D}$ and $n_{p 0} = n_{i}^{2} / N_{A}$ :

N_{D} = \frac{n _{i}^{2}}{p _{n 0}} = \frac{1 0 ^{20}}{1 0 ^{4}} = 1 0^{16} cm^{- 3}, N_{A} = \frac{n _{i}^{2}}{n _{p 0}} = \frac{1 0 ^{20}}{1 0 ^{3}} = 1 0^{17} cm^{- 3}

(consistent with the majority levels $n_{n} = 1 0^{16}$ , $p_{p} = 1 0^{17}$ on the plot).

(d) Law of the junction at the n-edge (or equivalently the depletion edge boundary condition):

p_{n} (x_{n}) = p_{n 0} e^{q V_{A} / k T} \Rightarrow V_{A} = \frac{k T}{q} ln \frac{p _{n} ( x _{n} )}{p _{n 0}} = (0.0259) ln \frac{1 0 ^{6}}{1 0 ^{4}} = (0.0259) ln 100 = (0.0259) (4.605) = 0.119 V

Check at the p-edge: $(0.0259) ln (1 0^{5} /1 0^{3}) = 0.119$ V. $✓$

Problem 6.11 Sketching carrier concentrations

Statement. $V_{A} = 23.03 (k T / q)$ is applied to a step junction diode with $N_{A} = 1 0^{17} cm^{- 3}$ , $N_{D} = 1 0^{16} cm^{- 3}$ , $n_{i} = 1 0^{10} cm^{- 3}$ . Make a dimensioned $lo g (p and n)$ versus $x$ sketch of both majority and minority concentrations in the quasineutral regions. Identify points 10 and 20 diffusion lengths from the depletion region edges on your sketch.

Solution.

e^{q V_{A} / k T} = e^{23.03} = 1 0^{10} (since ln 10 = 2.303, 23.03/2.303 = 10)

Equilibrium minority levels: $n_{p 0} = n_{i}^{2} / N_{A} = 1 0^{3}$ , $p_{n 0} = n_{i}^{2} / N_{D} = 1 0^{4} cm^{- 3}$ .

Edge values (law of the junction):

n_{p} (- x_{p}) = n_{p 0} e^{q V_{A} / k T} = 1 0^{3} \cdot 1 0^{10} = 1 0^{13} cm^{- 3}, p_{n} (x_{n}) = 1 0^{4} \cdot 1 0^{10} = 1 0^{14} cm^{- 3}

Low-level injection check: $1 0^{14} ≪ N_{D} = 1 0^{16}$ and $1 0^{13} ≪ N_{A} = 1 0^{17}$ . So the majority concentrations stay flat: $p_{p} = 1 0^{17}$ , $n_{n} = 1 0^{16}$ .

Decay: $Δ p_{n} = Δ p_{n} (0) e^{- x^{'} / L_{P}}$ , $Δ n_{p} = Δ n_{p} (0) e^{- x^{''} / L_{N}}$ with $Δ (0) \approx 1 0^{14}, 1 0^{13}$ (the " $- 1$ " is negligible).

Distance from depletion edge	$e^{- d / L}$	$p_{n}$ (n-side)	$n_{p}$ (p-side)
0	1	$1 0^{14}$	$1 0^{13}$
$10 L$	$4.54 \times 1 0^{- 5}$	$4.5 \times 1 0^{9} + 1 0^{4} \approx 4.5 \times 1 0^{9}$	$4.5 \times 1 0^{8}$
$20 L$	$2.06 \times 1 0^{- 9}$	$2.1 \times 1 0^{5} + 1 0^{4} \approx 2.2 \times 1 0^{5}$	$2.1 \times 1 0^{4} + 1 0^{3} \approx 2.2 \times 1 0^{4}$
$\infty$	0	$1 0^{4}$	$1 0^{3}$

Sketch: vertical axis log from $1 0^{3}$ to $1 0^{17}$ ; horizontal axis $x$ with the depletion region in the middle. Draw $p_{p} = 1 0^{17}$ (flat, p-side) and $n_{n} = 1 0^{16}$ (flat, n-side) at the top. The minority curves are straight lines on the log plot descending from $1 0^{13}$ ( $n_{p}$ , at $- x_{p}$ , going left) and $1 0^{14}$ ( $p_{n}$ , at $x_{n}$ , going right), with a slope of $- 1/ (2.303 L)$ decades per unit length, i.e. one decade per $2.303 L$ . They reach $\approx 4.5 \times 1 0^{8}$ and $4.5 \times 1 0^{9}$ at $10 L$ and then flatten out near $1 0^{3}$ and $1 0^{4}$ (the equilibrium values) after about $20 L$ ( $1 0^{13} \to 1 0^{4}$ takes $ln 1 0^{9} = 20.7 L$ for $n_{p}$ ; $1 0^{14} \to 1 0^{4}$ takes $23.0 L$ for $p_{n}$ ).

Problem 6.12 Diode temperature sensor

Statement. The diode is forward biased with a constant current source and $V_{A}$ is monitored as a function of $T$ (Fig. P6.12). Assume $I = I_{0} exp (q V_{A} / k T)$ throughout, with intended range $0 \leq T (^{\circ} C) \leq 100$ . (a) With $I$ as an input variable, modify the 6.7(b) program to compute and plot $V_{A}$ vs. $T$ ; print a plot for $I = 1 0^{- 4}$ A. (b) Considering the sensitivity ( $d V_{A} / d T$ in mV/ $^{\circ}$ C), would it be preferable to monitor $T$ using $I = 1 0^{- 4}$ A or $I = 1 0^{- ?}$ A? [OCR lost the exponent of the second current; I compare against a smaller current $I = 1 0^{- 6}$ A, and the general conclusion holds for any pair.] Support your answer.

Solution.

(a) Solve for $V_{A}$ :

V_{A} = \frac{k T}{q} ln \frac{I}{I _{0} ( T )}

% Problem 6.12(a): V_A vs T at constant current
clear; close
k = 8.617e-5; q = 1.6e-19; A = 1e-4; ND = 1e16; taup = 1e-6; EG = 1.12;
I = input('Enter the constant current I (A): ');
Tc = 0:2:100;  T = Tc + 273.15;
mup = 54.3 + 406.9/(1 + (ND/2.35e17)^0.88);
ni = 1e10*(T/300).^1.5.*exp(EG/(2*k)*(1/300 - 1./T));
DP = k*T*mup; LP = sqrt(DP*taup);
I0 = q*A*(DP./LP).*ni.^2/ND;
VA = k*T.*log(I./I0);
plot(Tc, VA); grid; xlabel('T (^oC)'); ylabel('V_A (V)')

For $I = 1 0^{- 4}$ A: at 300 K, $I_{0} = 5.4 \times 1 0^{- 16}$ A and $V_{A} = 0.0259 ln (1 0^{- 4} /5.4 \times 1 0^{- 16}) = 0.0259 ln (1.86 \times 1 0^{11}) = 0.672$ V. The plot is an almost straight, decreasing line from about 0.72 V at 0 $^{\circ}$ C to about 0.53 V at 100 $^{\circ}$ C (about $- 1.9$ mV/ $^{\circ}$ C).

(b) Differentiate $V_{A} = (k T / q) ln (I / I_{0})$ with $I$ constant:

\frac{d V _{A}}{d T} = \frac{V _{A}}{T} - \frac{k T}{q} \frac{1}{I _{0}} \frac{d I _{0}}{d T}

and, since $I_{0} \propto n_{i}^{2} \propto T^{3} e^{- E_{G} / k T}$ (neglecting the weaker mobility and lifetime dependences),

\frac{1}{I _{0}} \frac{d I _{0}}{d T} = \frac{3}{T} + \frac{E _{G}}{k T ^{2}} \Rightarrow \frac{d V _{A}}{d T} = \frac{V _{A}}{T} - \frac{3 k}{q} - \frac{E _{G}}{q T} = - \frac{E _{G} / q - V _{A}}{T} - \frac{3 k}{q}

At 300 K:

$I = 1 0^{- 4}$ A: $V_{A} = 0.672$ V, $\frac{d V _{A}}{d T} = - \frac{1.12 - 0.672}{300} - 0.26 \times 1 0^{- 3} = - 1.49 \times 1 0^{- 3} - 0.26 \times 1 0^{- 3} \approx - 1.75$ mV/ $^{\circ}$ C.
$I = 1 0^{- 6}$ A: $V_{A} = 0.0259 ln (1.86 \times 1 0^{9}) = 0.553$ V, $\frac{d V _{A}}{d T} = - \frac{1.12 - 0.553}{300} - 0.26 \times 1 0^{- 3} \approx - 2.15$ mV/ $^{\circ}$ C.

Conclusion. A smaller current gives a larger-magnitude sensitivity (more mV per $^{\circ}$ C), because $V_{A}$ is smaller and so $E_{G} / q - V_{A}$ is larger. If maximum sensitivity were the only issue, choose the smaller current. (Practical caveats: at very small currents the R-G current ( $q /2 k T$ slope) and noise spoil the “ideal” assumption; at large currents series resistance matters. Self-heating is also smaller at small currents.) The extrapolation $V_{A} \to E_{G} / q$ at $T \to 0$ also shows why all curves of $V_{A} (T)$ extrapolate to roughly 1.2 V.

Problem 6.13 Breakdown calculations

Statement. Given a planar $p^{+}$ - $n$ Si step junction with $N_{D} = 1 0^{16} cm^{- 3}$ [assumed; the OCR shows “1/cm” with the exponent lost; this is the same diode as Exercise 6.10] and $T = 300$ K, determine (a) the approximate $V_{B R}$ ; (b) the depletion width at breakdown; (c) the maximum electric field in the depletion region at breakdown.

Solution.

(a) From Fig. 6.11 (Si curve at $N_{D} = 1 0^{16}$ ): $V_{B R} \approx 55$ V. (The empirical fit of Problem 6.14, $60 (N_{B} /1 0^{16})^{- 3/4}$ , gives 60 V, consistent.) Take $V_{B R} = 55$ V.

(b) Breakdown occurs at $V_{A} = - V_{B R}$ , so $V_{bi} - V_{A} = V_{bi} + V_{B R} = 0.92 + 55 = 55.9$ V. For a $p^{+}$ - $n$ junction $W \approx x_{n}$ :

W = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} + V_{B R})]^{1/2} = [\frac{2 ( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} )} (55.9)]^{1/2} = [(1.305 \times 1 0^{- 9}) (55.9)]^{1/2} = [7.30 \times 1 0^{- 8}]^{1/2} = 2.70 \times 1 0^{- 4} cm \approx 2.7 μ m

(c) The maximum field is at the metallurgical junction:

∣ E ∣_{m a x} = \frac{q N _{D} W}{K _{S} ε _{0}} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} ) ( 2.70 \times 1 0 ^{- 4} )}{( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )} = \frac{4.32 \times 1 0 ^{- 7}}{1.044 \times 1 0 ^{- 12}} \approx 4.1 \times 1 0^{5} V/cm

(Equivalently $∣ E ∣_{m a x} = 2 (V_{bi} + V_{B R}) / W = 2 (55.9) /2.70 \times 1 0^{- 4} = 4.1 \times 1 0^{5}$ V/cm.) This is the critical field $E_{C R}$ for Si avalanche, a few $\times 1 0^{5}$ V/cm. $✓$

Problem 6.14 Avalanche multiplication in an otherwise ideal diode (computer)

Statement. Model carrier multiplication and avalanche breakdown in an otherwise ideal diode. Eq. (6.35), $M = 1/ [1 - (∣ V_{A} ∣/ V_{B R})^{m}]$ , corrects the ideal diode equation: $I = M I_{0}$ in reverse bias. The avalanche $V_{B R}$ varies roughly as $N_{B}^{- 3/4}$ ; a reasonable fit to Fig. 6.11 for Si is $V_{B R} \approx 60 (N_{B} /1 0^{16})^{- 3/4}$ (with $N_{B}$ in cm $^{- 3}$ , $V_{B R}$ in volts). Take a near-ideal $p^{+}$ - $n$ Si step junction at 300 K. (a) Construct a MATLAB program computing and plotting the modified reverse-bias $I / I_{0}$ vs. $V_{A}$ from $- V_{B R}$ to 0 (note the $I_{0}$ normalization). Use $m = 6$ and set the axis function parameters to $[1.1 V_{B R}, 0, - 5, 0]$ [that is, plotting $- ∣ V_{A} ∣$ on the abscissa and $- I / I_{0}$ on the ordinate; the orientation resembles Fig. 6.10(b)]. Print a sample for $N_{D} = 2 \times 1 0^{16} cm^{- 3}$ . Is the plot consistent with the curving approach to breakdown in Fig. 6.10(b) and Exercise 6.5? (b) Explore how the characteristic varies with $m$ and doping.

Solution.

Logic. (1) Read $N_{D}$ and $m$ . (2) Compute $V_{B R} = 60 (N_{D} /1 0^{16})^{- 0.75}$ (for $N_{D} = 2 \times 1 0^{16}$ : $V_{B R} = 60 \cdot 2^{- 0.75} = 35.7$ V). (3) Make a reverse-bias vector $V_{A}$ from $- 0.999 V_{B R}$ to 0. (4) Compute $M = 1/ (1 - (∣ V_{A} ∣/ V_{B R})^{m})$ and the reverse current ratio $I / I_{0} = - M$ (the ideal reverse current $- I_{0}$ times $M$ ; $e^{q V_{A} / k T} ≪ 1$ for all but the smallest $∣ V_{A} ∣$ ; to be complete use $I / I_{0} = M (e^{q V_{A} / k T} - 1)$ ). (5) Plot $I / I_{0}$ against $V_{A}$ and set axes.

% Problem 6.14: Reverse I/I0 including avalanche multiplication
clear; close
k = 8.617e-5; T = 300;
ND = input('ND (cm^-3) = ');      % e.g. 2e16
m  = input('m = ');               % e.g. 6
VBR = 60*(ND/1e16)^(-0.75);
VA = linspace(-0.999*VBR, 0, 500);
M  = 1./(1 - (abs(VA)/VBR).^m);
I_I0 = M.*(exp(VA/(k*T)) - 1);    % ~ -M for |VA| > a few kT/q
plot(VA, I_I0); grid
axis([-1.1*VBR 0 -5 0])
xlabel('V_A (volts)'); ylabel('I / I_0')
title(['N_D = ', num2str(ND), ' cm^{-3},  m = ', num2str(m), ',  V_{BR} = ', num2str(VBR,3), ' V'])

(a) Results. For $N_{D} = 2 \times 1 0^{16}$ , $m = 6$ , $V_{B R} = 35.7$ V. $I / I_{0} = - 1$ (ideal saturation) for small $∣ V_{A} ∣$ ; $I / I_{0} = - 2$ at $∣ V_{A} ∣ = 0.89 V_{B R} = 31.8$ V; $- 5$ at $0.97 V_{B R}$ ; diverging at $V_{B R}$ . The curve is flat over most of the range and then curves sharply downward. Yes, it is consistent with the sloping approach to breakdown in Fig. 6.10(b) and the discussion of Exercise 6.5: current is enhanced by carrier multiplication even a few volts before $V_{B R}$ . (Note the plot omits the R-G current, which in the real diode adds the gradual creep at small $∣ V_{A} ∣$ .)

(b) Exploration. Rerun with other $m$ and $N_{D}$ : (i) Larger $m$ gives a sharper knee: $M = 2$ occurs at $∣ V_{A} ∣/ V_{B R} = 0. 5^{1/ m}$ (0.79 for $m = 3$ , 0.89 for $m = 6$ ) and the $M \approx 1$ plateau extends farther. (ii) Higher $N_{D}$ lowers $V_{B R}$ as $N_{D}^{- 3/4}$ , scaling the voltage axis (the normalized plot as a function of $∣ V_{A} ∣/ V_{B R}$ is unchanged, since $M$ depends only on $∣ V_{A} ∣/ V_{B R}$ ). Doubling $N_{D}$ reduces $V_{B R}$ by $2^{- 0.75} = 0.59$ .

Problem 6.15 Breakdown with multiplied R-G current (computer)

Statement. Repeat 6.14 replacing $I_{0}$ by the reverse-bias R-G current $I_{R-G}$ . Normalize the current axis to the magnitude of the current at $V_{A} = - V_{B R} /2$ . Run with $N_{D} = 2 \times 1 0^{16} cm^{- 3}$ . Compare with Fig. 6.10(b).

Solution. With multiplication the reverse current is $I = M I_{R-G}$ , and from Eq. (6.43) $∣ I_{R-G} ∣ \propto W \propto (V_{bi} - V_{A})^{1/2} = (V_{bi} + ∣ V_{A} ∣)^{1/2}$ . The proportionality constant cancels in the normalization:

\frac{I}{∣ I ( - V _{B R} /2 ) ∣} = - \frac{M ( V _{A} ) V _{bi} + ∣ V _{A} ∣}{M ( V _{B R} /2 ) V _{bi} + V _{B R} /2}, V_{bi} = \frac{E _{G}}{2 q} + \frac{k T}{q} ln \frac{N _{D}}{n _{i}}

With $N_{D} = 2 \times 1 0^{16}$ : $V_{bi} = 0.56 + 0.0259 ln (2 \times 1 0^{6}) = 0.936$ V, $V_{B R} = 35.7$ V, so $V_{B R} /2 = 17.85$ V, $M (17.85) = 1/ (1 - 0. 5^{6}) = 1.016$ .

% Problem 6.15: reverse current = M * I_RG, normalized at VA = -VBR/2
clear; close
k = 8.617e-5; T = 300; EG = 1.12; ni = 1e10;
ND = 2e16; m = 6;
VBR = 60*(ND/1e16)^(-0.75);
Vbi = EG/2 + k*T*log(ND/ni);
VA = linspace(-0.999*VBR, 0, 500);
M  = 1./(1 - (abs(VA)/VBR).^m);
IRG = sqrt(Vbi - VA);                         % proportional to W
I   = -M.*IRG;
Vref = -VBR/2;
Iref = abs((1/(1-(abs(Vref)/VBR)^m))*sqrt(Vbi - Vref));
plot(VA, I/Iref); grid
axis([-1.1*VBR 0 -5 0])
xlabel('V_A (volts)'); ylabel('I / |I(-V_{BR}/2)|')

Comparison with Fig. 6.10(b). The computed curve now has the slowly growing, non-saturating reverse current at small $∣ V_{A} ∣$ (the $V_{bi} + ∣ V_{A} ∣$ factor takes it from about 0.23 of the reference at 0 V to 1 at $V_{B R} /2$ ) followed by the sharp avalanche upturn near $V_{B R}$ . This reproduces both the non-saturating creep and the sloping breakdown seen in the measured Fig. 6.10(b), and is a better match than the $M I_{0}$ (6.14) curve, which is flat at small bias.

Problem 6.16 When does the diffusion current dominate?

Statement. The diffusion component is expected to dominate in Si diodes at sufficiently elevated temperatures. How elevated? Consider a Si pn step junction with $N_{D} = 1 0^{15} cm^{- 3}$ [assumed; digits lost], $τ_{p} = τ_{0} = 1 0^{- 6}$ s and $L_{P} \approx 1 0^{- 2}$ cm [as printed], for $300 \leq T \leq 500$ K. Determine the temperature where $I_{DIFF} = I_{R-G}$ at a reverse bias of $V_{bi} - V_{A} = V_{B R} /2$ .

Solution. Reverse-bias magnitudes (a $p^{+}$ - $n$ diode, $τ_{n} = τ_{p} = τ_{0}$ ):

∣ I_{DIFF} ∣ = q A \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}}, ∣ I_{R-G} ∣ = \frac{q A n _{i} W}{2 τ _{0}}

Set them equal and cancel $q A$ and one $n_{i}$ :

\frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}} = \frac{n _{i} W}{2 τ _{0}} \Rightarrow n_{i} (T^{*}) = \frac{W L _{P} N _{D}}{2 τ _{0} D _{P}}

Step 1: depletion width. $V_{B R} \approx 60 (N_{D} /1 0^{16})^{- 3/4} = 60 (0.1)^{- 0.75} = 337$ V, so $V_{bi} - V_{A} = V_{B R} /2 = 168$ V:

W = [\frac{2 ( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{15} )} (168)]^{1/2} = [2.19 \times 1 0^{- 6}]^{1/2} = 1.48 \times 1 0^{- 3} cm

(W does not change with temperature to first order.)

Step 2: required $n_{i}$ . With $L_{P} = 1 0^{- 2}$ cm, $N_{D} = 1 0^{15}$ , $τ_{0} = 1 0^{- 6}$ s and $D_{P} \approx 20 cm^{2} / s$ (its value near the crossover temperature):

n_{i}^{*} = \frac{W L _{P} N _{D}}{2 τ _{0} D _{P}} = \frac{( 1.48 \times 1 0 ^{- 3} ) ( 1 0 ^{- 2} ) ( 1 0 ^{15} )}{2 ( 1 0 ^{- 6} ) ( 20 )} = \frac{1.48 \times 1 0 ^{10}}{4 \times 1 0 ^{- 5}} \approx 3.7 \times 1 0^{14} cm^{- 3}

Step 3: temperature. Use $n_{i} (T) = 1 0^{10} (T /300)^{3/2} exp [\frac{E _{G}}{2 k} (\frac{1}{300} - \frac{1}{T})]$ :

$T$ (K)	$n_{i}$ ( $cm^{- 3}$ )	$D_{P}$	$I_{DIFF} / I_{R-G}$
300	$1.0 \times 1 0^{10}$	11.8	$1.6 \times 1 0^{- 5}$
400	$3.5 \times 1 0^{12}$	15.8	$7 \times 1 0^{- 3}$
460	$3.6 \times 1 0^{13}$	18.2	0.09
500	$1.2 \times 1 0^{14}$	19.7	0.33
540	$3.7 \times 1 0^{14}$	21.3	1.05

The ratio equals unity at $T^{*} \approx 540$ K with $L_{P}$ taken as the printed $1 0^{- 2}$ cm. (If instead one uses the computed $L_{P} = D_{P} τ_{p} \approx 3.5 \times 1 0^{- 3}$ cm, $I_{DIFF}$ is $\sim 2.9$ times larger and the crossover moves to about $T^{*} \approx 500$ K, inside the quoted range.) Answer: roughly 500-550 K (about 230-270 $^{\circ}$ C). Below that, the reverse current of a room-temperature-like Si diode is R-G dominated; the diffusion current wins only at very high temperature because $I_{DIFF} \propto n_{i}^{2}$ grows faster than $I_{R-G} \propto n_{i}$ .

Problem 6.17 R-G current with a special region

Statement. Inside a Si diode at 300 K a region of width $d$ contains three times as many R-G centers as the adjoining regions. This region lies totally inside the depletion region when the diode is zero biased (Fig. P6.17). Derive an expression for the R-G current to be expected from the diode when reverse biased by more than a few $k T / q$ . Record all derivational steps.

Solution.

Step 1: reverse-bias R-G integral. From (6.42) with $n \to 0$ , $p \to 0$ (negligible, for $∣ V_{A} ∣ ≫ k T / q$ ) and $n p \to 0$ :

I_{R-G} = - q A \int_{0}^{W} \frac{n _{i}^{2}}{τ _{p} n _{1} + τ _{n} p _{1}} d x = - q A \int_{0}^{W} \frac{n _{i}}{2 τ _{0} ( x )} d x

where $τ_{0} (x) = (τ_{p} n_{1} + τ_{n} p_{1}) / (2 n_{i})$ (6.44) is the local generation lifetime.

Step 2: how $τ_{0}$ depends on the R-G center concentration. Lifetimes are inversely proportional to the trap concentration: $τ_{n} = 1/ (c_{n} N_{T})$ , $τ_{p} = 1/ (c_{p} N_{T})$ (capture coefficients $c$ are fixed properties of the center). All centers are assumed to be the same type (same $E_{T}$ ), so $n_{1}$ , $p_{1}$ are unchanged. Hence $τ_{0} \propto 1/ N_{T}$ . With three times as many centers in the special region:

τ_{0}^{special} = \frac{τ _{0}}{3}

Step 3: split the integral into the normal part ( $W - d$ ) and the special part ( $d$ ):

∣ I_{R-G} ∣ = q A n_{i} [\frac{W - d}{2 τ _{0}} + \frac{d}{2 ( τ _{0} /3 )}] = \frac{q A n _{i}}{2 τ _{0}} [(W - d) + 3 d]

I_{R-G} = - \frac{q A n _{i}}{2 τ _{0}} (W + 2 d)

with $W = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}$ (for the $p^{+}$ - $n$ junction of Fig. P6.17) and $d$ the width of the special region, which is inside the depletion region. The special region adds the equivalent of an extra depletion width $2 d$ : it behaves as if the depletion region were $2 d$ wider. For $d \to 0$ the result reduces to (6.43), and for $d = W$ to $3 \times$ the ideal.

Problem 6.18 Series resistance computations

Statement. [The first lines of the statement are cut off by the OCR. What survives: ”… negligible. $T = 300$ K.” and (a) determine the forward-bias current at which the applied voltage differs from the ideal by 10%; (b) repeat part (a) with $R_{S} = 20 Ω$ .] Assumptions for a complete solution: an ideal $p^{+}$ - $n$ Si diode with $I_{0} = 1 0^{- 14}$ A [assumed], series resistance $R_{S} = 5 Ω$ in part (a), and R-G and high-level injection negligible, so $I = I_{0} e^{q V_{J} / k T}$ with $V_{J} = V_{A} - I R_{S}$ . “Applied voltage differs from the ideal by 10%” means $V_{A}$ exceeds the ideal-diode voltage $V_{J}$ needed for the same current by 10% of $V_{A}$ , i.e. $I R_{S} = 0.1 V_{A}$ .

Solution (symbolic).

I R_{S} = 0.1 V_{A} = 0.1 (V_{J} + I R_{S}) \Rightarrow 0.9 I R_{S} = 0.1 V_{J} \Rightarrow I R_{S} = \frac{V _{J}}{9}

and $V_{J} = (k T / q) ln (I / I_{0})$ , so

I = \frac{k T / q}{9 R _{S}} ln \frac{I}{I _{0}} (transcendental; solve by iteration)

(a) $R_{S} = 5 Ω$ : $I = \frac{0.0259}{45} ln \frac{I}{1 0 ^{- 14}} = 5.76 \times 1 0^{- 4} ln \frac{I}{1 0 ^{- 14}}$ . Iterate: start $I = 0.02$ A: RHS $= 5.76 \times 1 0^{- 4} ln (2 \times 1 0^{12}) = 5.76 \times 1 0^{- 4} (28.32) = 1.63 \times 1 0^{- 2}$ ; $I = 0.0163$ : $ln (1.63 \times 1 0^{12}) = 28.12$ , RHS $= 1.62 \times 1 0^{- 2}$ . Converged:

I \approx 16 mA (V_{J} = 0.728 V, I R_{S} = 0.081 V, V_{A} = 0.809 V; I R_{S} / V_{A} = 10% ✓)

(b) $R_{S} = 20 Ω$ : $I = \frac{0.0259}{180} ln \frac{I}{1 0 ^{- 14}} = 1.44 \times 1 0^{- 4} ln \frac{I}{1 0 ^{- 14}}$ . Start $I = 4$ mA: $ln (4 \times 1 0^{11}) = 26.72$ , RHS $= 3.85$ mA; $I = 3.85$ mA: $ln = 26.67$ , RHS $= 3.84$ mA. Converged:

I \approx 3.8 mA (V_{J} = 0.69 V, I R_{S} = 0.077 V, V_{A} = 0.77 V)

A fourfold larger $R_{S}$ makes the 10% error appear at roughly a 4 $\times$ lower current ( $I \approx V_{J} /9 R_{S}$ , with a weakly varying $V_{J}$ ).

Problem 6.19 Extracting $I_{0 j}$ and $n_{j}$ from data

Statement. Table P6.19 lists forward-bias data for a 1N757 Si diode at room temperature (plotted in Fig. P6.19). Following Exercise 6.7, determine $I_{01}$ , $I_{02}$ , $n_{1}$ , $n_{2}$ . Because line drawing is subjective, also apply a least-squares fit (MATLAB) to appropriate segments of the data. Discuss.

Solution.

Method. The model is $I = I_{01} e^{q V_{A} / n_{1} k T} + I_{02} e^{q V_{A} / n_{2} k T}$ . On the semilog plot each exponential is a straight line: $ln I = ln I_{0 j} + q V_{A} / (n_{j} k T)$ .

Choosing the segments. The local ideality factor $n (V) = [(k T / q) d ln I / d V]^{- 1}$ computed from adjacent data points is about 2.2-2.5 for $0.15 ≲ V_{A} ≲ 0.5$ V (R-G region), then drops to a minimum of $\approx 1.16$ near $0.64$ - $0.70$ V (the diffusion region), and then rises again (slope-over from $R_{S}$ ) beyond about 0.72 V. So the two fitting segments are:

R-G region: $0.10 \leq V_{A} \leq 0.40$ V.
Diffusion region: $0.64 \leq V_{A} \leq 0.70$ V.

Line-drawing (two-point) estimates:

Diffusion line through $(0.64 V, 1.359 \times 1 0^{- 4} A)$ and $(0.70 V, 9.444 \times 1 0^{- 4} A)$ :

n_{1} = \frac{0.70 - 0.64}{0.0259 ln ( 9.444 \times 1 0 ^{- 4} /1.359 \times 1 0 ^{- 4} )} = \frac{0.06}{0.0259 ( 1.939 )} = 1.20

ln I_{01} = ln (1.359 \times 1 0^{- 4}) - \frac{0.64}{1.20 ( 0.0259 )} = - 8.90 - 20.6 = - 29.5 \Rightarrow I_{01} \approx 1.4 \times 1 0^{- 13} A

R-G line through $(0.10, 5.417 \times 1 0^{- 9})$ and $(0.30, 1.666 \times 1 0^{- 7})$ :

n_{2} = \frac{0.30 - 0.10}{0.0259 ln ( 1.666 \times 1 0 ^{- 7} /5.417 \times 1 0 ^{- 9} )} = \frac{0.20}{0.0259 ( 3.426 )} = 2.25

ln I_{02} = ln (5.417 \times 1 0^{- 9}) - \frac{0.10}{2.25 ( 0.0259 )} = - 19.03 - 1.72 = - 20.75 \Rightarrow I_{02} \approx 1 \times 1 0^{- 9} A

Least-squares fits (MATLAB polyfit(V, log(I), 1); slope $s$ gives $n = q / (s k T) \to 1/ (0.0259 s)$ ; intercept $b$ gives $I_{0} = e^{b}$ ):

V = [0.02:0.02:0.90];                 % column 1 and 3 of Table P6.19, merged
I = [6.070e-10 1.203e-9 2.165e-9 3.508e-9 5.417e-9 8.210e-9 1.183e-8 1.730e-8 ...
     2.449e-8 3.416e-8 4.764e-8 6.501e-8 8.866e-8 1.209e-7 1.666e-7 2.305e-7 ...
     3.201e-7 4.462e-7 6.285e-7 8.845e-7 1.249e-6 1.776e-6 2.527e-6 3.615e-6 ...
     5.200e-6 7.576e-6 1.122e-5 1.711e-5 2.694e-5 4.426e-5 7.587e-5 1.359e-4 ...
     2.531e-4 4.852e-4 9.444e-4 1.841e-3 3.518e-3 6.433e-3 1.103e-2 1.752e-2 ...
     2.585e-2 3.579e-2 4.706e-2 5.941e-2 7.264e-2];
kT = 0.0259;
j1 = V >= 0.64 & V <= 0.70;           % diffusion segment
p1 = polyfit(V(j1), log(I(j1)), 1);
n1 = 1/(kT*p1(1));  I01 = exp(p1(2));
j2 = V >= 0.10 & V <= 0.40;           % R-G segment
p2 = polyfit(V(j2), log(I(j2)), 1);
n2 = 1/(kT*p2(1));  I02 = exp(p2(2));
semilogy(V, I, 'o', V, I01*exp(V/(n1*kT)), '-', V, I02*exp(V/(n2*kT)), '--')
xlabel('V_A (V)'); ylabel('I (A)'); grid

Method	$I_{01}$ (A)	$n_{1}$	$I_{02}$ (A)	$n_{2}$
Line through two points	$1.4 \times 1 0^{- 13}$	1.20	$\approx 1 \times 1 0^{- 9}$	2.25
Least squares (my run of the data)	$1.4 \times 1 0^{- 13}$	1.20	$1.2 \times 1 0^{- 9}$	2.3

Discussion. (i) The two methods agree within experimental scatter. The diffusion component is quite near-ideal ( $n_{1} = 1.2$ ; “ideal” would be 1), consistent with a good commercial Si diode. (ii) The R-G ideality factor $n_{2} \approx 2.3$ is a bit above the theoretical limit $n_{2} \leq 2$ and the $I_{02}$ is large compared with the first-principles estimate of Exercise 6.6 (about $2 \times 1 0^{- 13}$ A for a diode with a larger area); this indicates extra leakage (surface or perimeter leakage, shunt paths) in addition to the simple R-G current. (iii) The fits are sensitive to the segment boundaries, since the regions are not cleanly separated: between roughly 0.4 and 0.64 V both components are comparable, and above 0.72 V series resistance corrupts the fit. Least squares removes subjectivity but not model error.

Problem 6.20 Series resistance from the 1N757 data

Statement. Following the procedure of Fig. 6.16, determine $R_{S}$ of the 1N757 diode characterized by Table P6.19.

Solution.

Step 1. Take the ideal (diffusion) line fitted in 6.19: $ln I = ln I_{01} + V_{J} / (n_{1} k T)$ , with $I_{01} = 1.4 \times 1 0^{- 13}$ A and $n_{1} = 1.2$ (it describes the junction voltage $V_{J}$ at each current, assuming the extrapolation of the diffusion region holds at higher currents).

Step 2. For each high-current data point compute the junction voltage that the line predicts for that current,

V_{J} (I) = n_{1} \frac{k T}{q} ln \frac{I}{I _{01}}

and then $Δ V = V_{A} - V_{J} (I)$ ( $= I R_{S}$ ).

$V_{A}$ (V)	$I$ (A)	$V_{J}$ from line (V)	$Δ V$ (V)	$Δ V / I$ ( $Ω$ )
0.76	$6.43 \times 1 0^{- 3}$	0.760	0.0003	0.05
0.78	$1.103 \times 1 0^{- 2}$	0.776	0.0037	0.33
0.80	$1.752 \times 1 0^{- 2}$	0.791	0.0094	0.53
0.82	$2.585 \times 1 0^{- 2}$	0.803	0.0173	0.67
0.84	$3.579 \times 1 0^{- 2}$	0.813	0.0273	0.76
0.86	$4.706 \times 1 0^{- 2}$	0.821	0.0388	0.82
0.88	$5.941 \times 1 0^{- 2}$	0.828	0.0516	0.87
0.90	$7.264 \times 1 0^{- 2}$	0.835	0.0654	0.90

Step 3. Plot $Δ V$ versus $I$ ; the slope of the (essentially straight) high-current portion is $R_{S}$ . A least-squares line through the points with $I \geq 1.7 \times 1 0^{- 2}$ A gives a slope of

R_{S} = \frac{d ( Δ V )}{d I} \approx 0.95 Ω \approx 1 Ω

(The $Δ V / I$ column converges toward the same value as $I$ grows, which independently confirms $R_{S} \approx 0.9$ - $1 Ω$ ; the smaller ratios at lower current reflect the uncertainty in the extrapolated line.) This agrees with the rough estimate of $\approx 1 Ω$ obtained in Exercise 6.8 for a similar diode. $R_{S} \approx 1 Ω$ .

Problem 6.21 A generalized diode model (computer)

Statement. The model of Fig. P6.21 is a generalized representation of real pn diodes. The diode symbol is the standard model (diffusion and R-G components of the junction current); $R_{S}$ is the series resistance (quasineutral regions and contacts at high current); $R_{S H}$ is a shunt resistor accounting for a possible leakage current outside the pn junction. The current through the junction proper and its voltage are redefined as $I_{J}$ and $V_{J}$ .

(a) If the junction current is $I_{J} = I_{01} (e^{q V_{J} / n_{1} k T} - 1) + I_{02} (e^{q V_{J} / n_{2} k T} - 1)$ with constants $I_{01}$ , $I_{02}$ and typically $n_{1} \leq 1$ …, more precisely $n_{1} ≲ n_{2}$ [OCR: ” $n_{1} \leq 1$ and $n_{2}$ ”; read as $n_{1} \approx 1$ , $n_{2} \approx 2$ ], confirm that the total current is $I = I_{01} (e^{q V_{J} / n_{1} k T} - 1) + I_{02} (e^{q V_{J} / n_{2} k T} - 1) + \frac{V _{J}}{R _{S H}} with V_{A} = V_{J} + I R_{S}$ (b) Construct a program to investigate the effect of $R_{S}$ , $R_{S H}$ and the sizes of the current prefactors on the observed forward-bias $I$ - $V_{A}$ . Use evenly stepped $V_{J}$ values. Limit the plotted $lo g I$ vs. $V_{A}$ to $0 \leq V_{A} \leq 1$ V and $1 0^{- 10} \leq I \leq 1 0^{- 2}$ A. Use $I_{01} = 1 0^{- 13}$ A, $I_{02} = 1 0^{- 9}$ A, $n_{1} = 1$ , $n_{2} = 2$ , $R_{S} = 1 Ω$ , $R_{S H} = 1 0^{12} Ω$ [OCR shows $1 0^{2}$ -something; I use the large value so the shunt is initially irrelevant; see the sweeps in (c)], $T = 300$ K. (c) With a manually controlled switch, modify the program to display four characteristics for four values of one of $R_{S H}$ , $R_{S}$ , $I_{02}$ , $I_{01}$ : (i) $R_{S H} = 1 0^{12}, 1 0^{6}, 1 0^{5}, 1 0^{3} Ω$ [as reconstructed] ( $I_{01} = 1 0^{- 13}$ , $I_{02} = 1 0^{- 9}$ , $R_{S} = 1$ ); (ii) $R_{S} = 1, 10, 1 0^{2}, 1 0^{3} Ω$ ( $R_{S H} = 1 0^{12}$ ); (iii) $I_{02} = 1 0^{- 9}, 1 0^{- 10}, 1 0^{- 11}, 1 0^{- 12}$ A; (iv) $I_{01} = 1 0^{- 13}, 1 0^{- 12}, 1 0^{- 11}, 1 0^{- 10}$ A. (d) Following Exercise 6.6, construct a subsidiary program computing $I_{01}$ and $I_{02}$ from first principles for an $n^{+}$ - $p$ step junction; join it to the main program. (e) Make a linear-plot version ( $0 \leq I \leq 1 0^{- 2}$ A); compare with (c). (f) Comment.

Solution.

(a) Proof. In Fig. P6.21 the junction proper (current $I_{J}$ ) is in parallel with $R_{S H}$ , and this parallel combination is in series with $R_{S}$ .

Parallel combination: both elements see the same voltage $V_{J}$ . Ohm’s law for the shunt: $I_{S H} = V_{J} / R_{S H}$ .
Kirchhoff’s current law at the top node of the parallel pair: the total current entering from the external circuit splits into $I_{J}$ and $I_{S H}$ : $I = I_{J} + I_{S H} = I_{01} (e^{q V_{J} / n_{1} k T} - 1) + I_{02} (e^{q V_{J} / n_{2} k T} - 1) + \frac{V _{J}}{R _{S H}}$
The same total $I$ flows through $R_{S}$ , so Kirchhoff’s voltage law around the outer loop gives $V_{A} = V_{J} + I R_{S}$ . $■$

(b) Program logic. Because $V_{A}$ depends on $I$ , which depends on $V_{J}$ , the equations are transcendental in $V_{A}$ . The trick (Eqs. 6.48-6.49 discussion) is to step $V_{J}$ evenly, compute $I (V_{J})$ from (a), then compute $V_{A} = V_{J} + I R_{S}$ , and plot $lo g I$ against that $V_{A}$ .

% Problem 6.21 (b)-(e): generalized diode with Rs and Rsh
clear; close
k = 8.617e-5; T = 300; kT = k*T;
sw = input('Sweep: 0=none, 1=Rsh, 2=Rs, 3=I02, 4=I01 : ');
I01 = 1e-13; I02 = 1e-9; n1 = 1; n2 = 2; Rs = 1; Rsh = 1e12;
% (d) Optional: first-principles I01, I02 for an n+-p step junction (see below)
Vj = linspace(0.0, 1.0, 2000);          % evenly stepped junction voltage
switch sw
    case 0, P = Rsh;                    nm = 'Rsh';
    case 1, P = [1e12 1e6 1e5 1e3];     nm = 'Rsh';
    case 2, P = [1 10 100 1000];        nm = 'Rs';
    case 3, P = [1e-9 1e-10 1e-11 1e-12]; nm = 'I02';
    case 4, P = [1e-13 1e-12 1e-11 1e-10]; nm = 'I01';
end
figure; hold on
for j = 1:length(P)
    switch nm
        case 'Rsh', Rsh = P(j);
        case 'Rs',  Rs  = P(j);
        case 'I02', I02 = P(j);
        case 'I01', I01 = P(j);
    end
    I  = I01*(exp(Vj/(n1*kT)) - 1) + I02*(exp(Vj/(n2*kT)) - 1) + Vj/Rsh;
    VA = Vj + I*Rs;
    semilogy(VA, I)                    % for the linear plot of (e): plot(VA, I)
end
axis([0 1 1e-10 1e-2]); grid on
xlabel('V_A (V)'); ylabel('I (A)')

(For the linear plot of part (e), replace semilogy by plot and set axis([0 1 0 1e-2]).)

(d) First-principles prefactors for an $n^{+}$ - $p$ step junction (assumed: $A = 1 0^{- 4} cm^{2}$ , $N_{A} = 1 0^{16} cm^{- 3}$ [parameters truncated in the OCR], $τ_{n} = τ_{p} = 1 0^{- 6}$ s, Si at 300 K, $V_{A} = V_{bi} /4$ in the voltage-dependent R-G terms as in Exercise 6.6):

% Subsidiary program: I01 and I02 for an n+-p step junction (Si, 300 K)
q = 1.6e-19; eps0 = 8.85e-14; KS = 11.8; ni = 1e10; EG = 1.12;
A = 1e-4; NA = 1e16; tau0 = 1e-6;
mun = 92 + 1268/(1 + (NA/1.3e17)^0.91);
DN = kT*mun; LN = sqrt(DN*tau0);
I01 = q*A*(DN/LN)*ni^2/NA;                          % diffusion prefactor
Vbi = EG/2 + kT*log(NA/ni);
W   = sqrt(2*KS*eps0/(q*NA)*(0.75*Vbi));            % at VA = Vbi/4
I02 = q*A*ni*W/tau0*kT/(0.75*Vbi);                  % R-G prefactor

For these values $I_{01} \approx 9 \times 1 0^{- 16}$ A and $I_{02} \approx 2 \times 1 0^{- 13}$ A: the same order as the Si diode of Exercise 6.6 (the diffusion prefactor in practice is also set by the lighter-doped side’s $D / L$ ).

(c), (e), (f) Expected results and comments.

Base case ( $I_{01} = 1 0^{- 13}$ , $I_{02} = 1 0^{- 9}$ , $n_{1} = 1$ , $n_{2} = 2$ , $R_{S} = 1 Ω$ , $R_{S H} = 1 0^{12} Ω$ ). On the semilog plot: a $q /2 k T$ R-G line at small $V_{A}$ (up to about 0.4 V), a $q / k T$ diffusion region around 0.5-0.75 V, and a slope-over above about 0.75 V when $I R_{S} \to$ a significant fraction of $V_{A}$ . Crossover where $I_{01} e^{q V / k T} = I_{02} e^{q V /2 k T}$ : $e^{q V /2 k T} = I_{02} / I_{01} = 1 0^{4}$ , $V = 2 (0.0259) ln 1 0^{4} = 0.48$ V.

(i) Varying $R_{S H}$ ( $1 0^{12}, 1 0^{6}, 1 0^{5}, 1 0^{3} Ω$ ): the shunt adds a linear (ohmic) current $V_{J} / R_{S H}$ that dominates at small voltages: for $R_{S H} = 1 0^{3} Ω$ , $I \approx V /1 0^{3}$ is 1 mA at 1 V, and on the semilog plot it appears as a line of slope 1 on the log-log scale (a curve bulging above the exponential at small voltage). Smaller $R_{S H}$ masks the R-G region and the reverse-like leakage: the first section of the plot becomes a straight ohmic rise. $R_{S H} = 1 0^{12}$ has no visible effect.

(ii) Varying $R_{S}$ ( $1, 10, 1 0^{2}, 1 0^{3} Ω$ ): the slope-over appears at lower and lower current/voltage. For $R_{S} = 1 0^{3} Ω$ the characteristic is essentially linear above about 0.5-0.6 V, with $I \approx (V_{A} - V_{o n}) / R_{S}$ .

(iii) Varying $I_{02}$ ( $1 0^{- 9} \to 1 0^{- 12}$ A): the R-G region shrinks and moves lower and finally disappears (at $I_{02} = 1 0^{- 12}$ the crossover with the diffusion line is at $0.0518 ln (10) = 0.12$ V); the characteristic becomes closer to the ideal.

(iv) Varying $I_{01}$ ( $1 0^{- 13} \to 1 0^{- 10}$ A): the diffusion line shifts left (a given current at lower $V_{A}$ ), and the R-G region is overtaken sooner: for $I_{01} = 1 0^{- 10}$ the crossover voltage is $0.0518 ln (10) = 0.12$ V, so essentially no R-G region is visible.

(e) Linear plot. On a linear scale, all of the sub-threshold structure (R-G region, shunt leakage, and so on) is invisible, because the currents are $≪ 1 0^{- 2}$ A over most of the voltage range; one sees only a “knee” at about $0.6$ - $0.8$ V followed by a straight line of slope $1/ R_{S}$ at high current. The effect of $R_{S}$ is the main thing visible on a linear plot, while $R_{S H}$ , $I_{01}$ , $I_{02}$ changes only shift the knee slightly.

(f) Overall comment. The semilog plot is a powerful diagnostic instrument: different nonidealities dominate in different voltage windows (shunt leakage at lowest bias, R-G at small bias, diffusion at moderate bias, series resistance at the highest bias), each leaving a distinct slope signature. A linear plot hides all but the series resistance. The model parameters can be extracted window by window, as done in Problem 6.19.

Problem 6.22 A pseudo-narrow-base diode

Statement. Consider the special Si $p^{+}$ - $n$ step junction diode of Fig. P6.22: $τ_{p} = \infty$ for $0 \leq x \leq x_{b}$ and $τ_{p} = 0$ for $x_{b} \leq x \leq x_{c}$ (the n-side of the diode has a segment with no recombination followed by a segment with infinitely fast recombination; $x = 0$ is the metallurgical junction). Excluding biases that cause high-level injection or breakdown, develop an expression for the room-temperature $I$ - $V$ characteristic. Assume the depletion width $W$ never exceeds $x_{b}$ for all biases of interest.

Solution. The p $^{+}$ side is ignored, and $x_{n} = W$ ( $p^{+}$ - $n$ ).

Setting up. In the region $x_{n} \leq x \leq x_{b}$ there is no recombination ( $τ_{p} = \infty$ ): the diffusion equation is $D_{P} d^{2} Δ p_{n} / d x^{2} = 0$ with straight-line solutions. At $x = x_{b}$ the lifetime is zero, which is equivalent to an infinite recombination rate: the excess holes are annihilated immediately, so $Δ p_{n} (x_{b}) = 0$ . (Equivalently, think of $x_{b}$ as an ideal ohmic contact.) With $x^{'} = x - x_{n}$ and the effective width $x_{b} - x_{n} = x_{b} - W$ :

\frac{d ^{2} Δ p _{n}}{d x ^{'2}} = 0, Δ p_{n} (0) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1), Δ p_{n} (x_{b} - x_{n}) = 0

Δ p_{n} (x^{'}) = Δ p_{n} (0) (1 - \frac{x ^{'}}{x _{b} - x _{n}})

Current:

I = A J_{P} (0) = - q A D_{P} \frac{d Δ p _{n}}{d x ^{'}}_{0} = q A \frac{D _{P}}{x _{b} - x _{n}} \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1)

with

I = \frac{q A D _{P} n _{i}^{2}}{N _{D} ( x _{b} - x _{n} )} (e^{q V_{A} / k T} - 1), x_{n} = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}

This is the narrow-base limit of Eq. (6.68)/(6.69) with $x_{c}^{'} \to x_{b} - x_{n}$ and $coth (ξ) \to 1/ ξ$ (here exactly, because the lifetime is infinite so no recombination loss occurs). Under reverse bias $x_{n}$ grows, $x_{b} - x_{n}$ shrinks, and the current increases in magnitude with $∣ V_{A} ∣$ instead of saturating (and diverges at punch-through, $x_{n} \to x_{b}$ , outside the stated validity range because $W < x_{b}$ ). Under forward bias, $x_{n}$ shrinks and the factor $1/ (x_{b} - x_{n})$ varies weakly, so the usual exponential dependence is retained. Room temperature enters via $D_{P} = (k T / q) μ_{p}$ , $n_{i} = 1 0^{10} cm^{- 3}$ and $k T / q = 0.0259$ V.

Problem 6.23 Narrow base with a finite-lifetime lossy layer

Statement. Reconsider the diode of Fig. P6.22. Instead of $τ_{p} = 0$ in the $x_{b} \leq x \leq x_{c}$ region, take $τ_{p}$ nonzero but still sufficiently small that $L_{P} ≪ x_{c} - x_{b}$ . Assuming $W$ never exceeds $x_{b}$ for all biases of interest, and excluding high-level injection and breakdown, derive an expression for the room-temperature $I$ - $V$ characteristic. (As in 6.22, $τ_{p} = \infty$ for $0 \leq x \leq x_{b}$ .)

Solution. Now the n-side is two layers:

Region 1 ( $x_{n} \leq x \leq x_{b}$ , width $w = x_{b} - x_{n}$ ): $τ_{p} = \infty$ , no recombination: $Δ p_{n}^{''} = 0$ , so $Δ p_{n}$ is linear and the hole current density is constant.
Region 2 ( $x \geq x_{b}$ with $L_{P} ≪ x_{c} - x_{b}$ ): the layer is effectively semi-infinite: $Δ p_{n} = Δ p_{n} (x_{b}) e^{- (x - x_{b}) / L_{P}}$ with $L_{P}$ the (short) diffusion length in this layer, because the contact lies many diffusion lengths away. Its current at its left edge is

J_{P} (x_{b}) = q D_{P} \frac{Δ p _{n} ( x _{b} )}{L _{P}}

(From $J_{P} = - q D_{P} d Δ p_{n} / d x$ with the exponential.) We assume both regions have the same $D_{P}$ .

Matching at $x_{b}$ : continuity of both $Δ p_{n}$ and $J_{P}$ . Region 1 is a straight line from $Δ p_{n} (0)$ at $x_{n}$ to $Δ p_{n} (x_{b}) \equiv Δ_{b}$ at $x_{b}$ :

J_{P} = q D_{P} \frac{Δ p _{n} ( 0 ) - Δ _{b}}{w}

Equate to region 2’s current:

\frac{Δ p _{n} ( 0 ) - Δ _{b}}{w} Δ p_{n} (0) Δ_{b} = \frac{Δ _{b}}{L _{P}} = Δ_{b} (1 + \frac{w}{L _{P}}) = \frac{Δ p _{n} ( 0 )}{1 + w / L _{P}}

J_{P} = q D_{P} \frac{Δ _{b}}{L _{P}} = \frac{q D _{P} Δ p _{n} ( 0 )}{L _{P} + w}

I = \frac{q A D _{P} n _{i}^{2}}{N _{D} ( L _{P} + x _{b} - x _{n} )} (e^{q V_{A} / k T} - 1), x_{n} = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}

Checks: if $x_{b} - x_{n} \to 0$ the result reduces to the ideal wide-base result $q A D_{P} n_{i}^{2} / (N_{D} L_{P})$ ; if $L_{P} \to 0$ it reduces to the 6.22 result. The diode behaves like a wide-base diode whose effective base is lengthened by the lossless layer of width $w = x_{b} - x_{n}$ : the lossy layer’s recombination is the sink, and the no-loss layer acts as a series “transport resistance”.

Problem 6.24 Denuded zone (multiple considerations)

Statement. Denuding reduces the R-G center concentration in the near-surface region of devices. A planar Si $p^{+}$ - $n$ step junction at 300 K has $N_{D} = 1 0^{16} cm^{- 3}$ . Denuding has created a reduced R-G center concentration $N_{T 1}$ in the n-side region from $x = 0$ to $x = x_{c} = 2 μ m$ (Fig. P6.24); in the $x > x_{c}$ region the concentration is $N_{T 2} = 10 N_{T 1}$ .

(a) Junction built-in voltage? (b) Diode breakdown voltage? (c) Applied voltage needed to expand the n-side depletion edge to $x = x_{c}$ ? (d) If the minority carrier lifetimes are $τ_{p 1}$ and $τ_{p 2}$ in the near-junction ( $x < x_{c}$ ) and ( $x > x_{c}$ ) regions, what is the $τ_{p 1} / τ_{p 2}$ ratio? (e) Establish an expression for the diffusion current $I_{DIFF}$ as a function of $V_{A}$ . To simplify, assume $L_{P 1} > x_{c}$ and the n-side contact is many $L_{P 2}$ from the depletion edge. Sketch the expected $I_{DIFF}$ vs. $V_{A}$ when reverse biased. (f) Assuming $τ_{01} \propto τ_{p 1}$ for $0 \leq x \leq x_{c}$ and $τ_{02} \propto τ_{p 2}$ for $x > x_{c}$ , establish an expression or expressions for $I_{R-G}$ vs. $V_{A}$ when reverse biased more than a few $k T / q$ . Sketch.

Solution.

(a) $p^{+}$ - $n$ : $V_{bi} \approx \frac{E _{G}}{2 q} + \frac{k T}{q} ln \frac{N _{D}}{n _{i}} = 0.56 + 0.0259 ln 1 0^{6} = 0.56 + 0.358 = 0.92$ V.

(b) Breakdown is set by the n-side doping (the denuding changes only R-G centers, not the doping): from Fig. 6.11 for $N_{D} = 1 0^{16}$ , $V_{B R} \approx 55$ V.

(c) $x_{n} = x_{c}$ when

V_{A} = V_{bi} - \frac{q N _{D} x _{c}^{2}}{2 K _{S} ε _{0}} = 0.92 - \frac{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} ) ( 2 \times 1 0 ^{- 4} ) ^{2}}{2 ( 11.8 ) ( 8.85 \times 1 0 ^{- 14} )} = 0.92 - 30.6 = - 29.7 V

(the same computation as Exercise 6.10). Since $29.7 < 55$ V, this is reached before breakdown.

(d) The lifetime is inversely proportional to the R-G center concentration ( $τ = 1/ (c N_{T})$ for the same type of centers): $τ_{p 1} / τ_{p 2} = N_{T 2} / N_{T 1} = 10$ . The denuded region has a ten times longer lifetime.

(e) Diffusion current. Take the n-side with $x^{'} = x - x_{n}$ and let $w \equiv x_{c} - x_{n}$ be the remaining width of the denuded layer (valid for $x_{n} \leq x_{c}$ , i.e. $V_{A} \geq - 29.7$ V). Because $L_{P 1} > x_{c} \geq w$ , recombination in region 1 is negligible: $Δ p_{n}^{''} \approx 0$ (linear profile, constant $J_{P}$ ). In region 2 ( $x > x_{c}$ , contact far away) $Δ p_{n} = Δ_{c} e^{- (x - x_{c}) / L_{P 2}}$ , $J_{P} (x_{c}) = q D_{P} Δ_{c} / L_{P 2}$ . This is exactly the problem solved in 6.23 with $L_{P} \to L_{P 2}$ and $w = x_{c} - x_{n}$ :

I_{DIFF} = \frac{q A D _{P} n _{i}^{2}}{N _{D} ( L _{P 2} + x _{c} - x _{n} )} (e^{q V_{A} / k T} - 1), V_{A} \geq - 29.7 V

with $x_{n} = [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}$ and $L_{P 2} = D_{P} τ_{p 2} = L_{P 1} / 10$ . For reverse bias beyond $- 29.7$ V, the depletion edge has moved into region 2 ( $x_{n} > x_{c}$ ); the n-side quasineutral region is now entirely region 2 with a wide base, and

I_{DIFF} \to - \frac{q A D _{P} n _{i}^{2}}{N _{D} L _{P 2}} \equiv - I_{0}^{(2)} (V_{A} \leq - 29.7 V)

Sketch (reverse bias). Starting at zero, the magnitude of $I_{DIFF}$ at $V_{A} = 0^{-}$ is $\frac{q A D _{P} n _{i}^{2}}{N _{D} ( L _{P 2} + x _{c} - x _{n} ( 0 ))}$ ( $≪$ saturation scale), then it increases gradually (concave up) as $x_{n} \to x_{c}$ because the denominator shrinks from $L_{P 2} + w (0)$ to $L_{P 2}$ , and for $V_{A} < - 29.7$ V it is constant at $- I_{0}^{(2)}$ (saturated) until breakdown at $- 55$ V. The total variation is modest: the magnitude goes from $1/ (L_{P 2} + w (0))$ to $1/ L_{P 2}$ . (Numerically, $w (0) = x_{c} - x_{n} (0) = 2 - 2 (11.8) (8.85 \times 1 0^{- 14}) (0.92) /1.6 \times 1 0^{- 3} μ m = 2 - 0.35 = 1.65 μ m$ ; the ratio is $1 + 1.65 μ m / L_{P 2}$ , which is $\approx 1$ if $L_{P 2} ≫ 1.65 μ m$ , as is typical, $L_{P 2} \sim 10$ - $30 μ m$ .)

(f) R-G current, reverse biased. For $V_{A}$ beyond a few $k T / q$ reverse, $∣ I_{R-G} ∣ = q A n_{i} \int d x / (2 τ_{0} (x))$ over the depletion region (Problem 6.17 approach). Since $τ_{01} \propto τ_{p 1}$ and $τ_{02} \propto τ_{p 2}$ , $τ_{01} = 10 τ_{02}$ . The depletion region lies on the n-side (width $W = x_{n}$ ):

Case 1: $x_{n} \leq x_{c}$ ( $0 \geq V_{A} \geq - 29.7$ V):

I_{R-G} = - \frac{q A n _{i} x _{n}}{2 τ _{01}} = - \frac{q A n _{i}}{2 τ _{01}} [\frac{2 K _{S} ε _{0}}{q N _{D}} (V_{bi} - V_{A})]^{1/2}

Case 2: $x_{n} > x_{c}$ ( $V_{A} < - 29.7$ V):

I_{R-G} = - q A n_{i} [\frac{x _{c}}{2 τ _{01}} + \frac{x _{n} - x _{c}}{2 τ _{02}}] = - \frac{q A n _{i}}{2 τ _{02}} [\frac{x _{c}}{10} + (x_{n} - x_{c})]

Sketch. $∣ I_{R-G} ∣$ vs. $∣ V_{A} ∣$ : $\propto V_{bi} + ∣ V_{A} ∣$ up to $∣ V_{A} ∣ = 29.7$ V (where $∣ I_{R-G} ∣ = q A n_{i} x_{c} /2 τ_{01}$ ), then a kink: beyond 29.7 V the current continues to increase as $V_{bi} + ∣ V_{A} ∣$ , but with a tenfold larger coefficient for the added width (the incremental generation per unit width is ten times larger in the not-denuded layer), until avalanche breakdown at 55 V (at which $x_{n} = 2.7 μ m$ ). Between $x_{c} = 2.0 μ m$ and $2.7 μ m$ the added current is $(q A n_{i} /2 τ_{02}) (0.7 μ m)$ , i.e. the current at 55 V is $[0.2 + 0.7] / [0.2] = 4.5 \times$ larger than the current at 29.7 V.

Appendix: Quick Reference for Chapter 6

Quantity	Expression
Ideal diode equation	$I = I_{0} (e^{q V_{A} / k T} - 1)$
Saturation current	$I_{0} = q A (\frac{D _{N}}{L _{N}} \frac{n _{i}^{2}}{N _{A}} + \frac{D _{P}}{L _{P}} \frac{n _{i}^{2}}{N _{D}})$
Law of the junction	$n p = n_{i}^{2} e^{q V_{A} / k T}$ in the depletion region
Edge boundary conditions	$Δ n_{p} (- x_{p}) = \frac{n _{i}^{2}}{N _{A}} (e^{q V_{A} / k T} - 1)$ , $Δ p_{n} (x_{n}) = \frac{n _{i}^{2}}{N _{D}} (e^{q V_{A} / k T} - 1)$
Minority profile (wide base)	$Δ p_{n} (x^{'}) = Δ p_{n} (0) e^{- x^{'} / L_{P}}$
Multiplication factor	$M=\left[1-(
Reverse R-G current	$I_{R-G} = - \frac{q A n _{i} W}{2 τ _{0}}$
Forward R-G	$\propto e^{q V_{A} /2 k T}$
Series resistance	$V_{J} = V_{A} - I R_{S}$ , $I = I_{0} e^{q V_{J} / k T}$
Narrow-base	$I_{0}^{'} = I_{0} coth (x_{c}^{'} / L_{P}) \to I_{0} L_{P} / x_{c}^{'}$ for $x_{c}^{'} ≪ L_{P}$
Charge control	$d Q_{P} / d t = i_{DIFF} - Q_{P} / τ_{p}$ ; steady state $I_{DIFF} = Q_{P} / τ_{p}$

Vault

Explorer

part06

Chapter 6: pn Junction Diode: I-V Characteristics

6.1 The Ideal Diode Equation

6.1.1 Qualitative Derivation

The “big picture” objection

6.1.2 Quantitative Solution Strategy

General considerations: assumptions

Current relationships available

Quasineutral region considerations

Depletion region considerations

Boundary conditions

”Game plan” summary

6.1.3 Derivation Proper

6.1.4 Examination of Results

Ideal I-V

The saturation current: two big observations

Exercise 6.1

Exercise 6.2 (computer): Ideal I-V versus doping

Carrier currents

Carrier concentrations

Exercise 6.3

Exercise 6.4

6.2 Deviations from the Ideal

6.2.1 Ideal Theory Versus Experiment

6.2.2 Reverse-Bias Breakdown

Avalanching

Exercise 6.5 (computer): the multiplication factor

Zener Process

6.2.3 The R-G Current

Quantitative expression

Total current and which one wins

Exercise 6.6 (computer): I01​ and I02​ for Si and Ge

Exercise 6.7: generalized forward-bias model

6.2.4 VA​→Vbi​ High-Current Phenomena

Series resistance

Exercise 6.8

High-level injection

Exercise 6.9: diagnose the nonidealities

6.3 Special Considerations

6.3.1 Charge Control Approach

6.3.2 Narrow-Base Diode

Current derivation

Limiting cases and punch-through

Exercise 6.10

6.4 Summary and Concluding Comments

Problems (Chapter 6)

Problem 6.1 Ideal Diode Quiz

Problem 6.2 Actual versus ideal characteristic

Problem 6.3 Band diagrams

Problem 6.4 The big picture, forward bias

Problem 6.5 Boundary condition via the “Videri” (JN​=0) approach

Problem 6.6 Typical ideal currents (computer)

Problem 6.7 Ideal I-V versus temperature (computer)

Problem 6.8 Photodiode / solar cell

Problem 6.9 Solar cell characteristic (computer)

Problem 6.10 Reading information from a concentration plot

Problem 6.11 Sketching carrier concentrations

Problem 6.12 Diode temperature sensor

Problem 6.13 Breakdown calculations

Problem 6.14 Avalanche multiplication in an otherwise ideal diode (computer)

Problem 6.15 Breakdown with multiplied R-G current (computer)

Problem 6.16 When does the diffusion current dominate?

Problem 6.17 R-G current with a special region

Problem 6.18 Series resistance computations

Problem 6.19 Extracting I0j​ and nj​ from data

Problem 6.20 Series resistance from the 1N757 data

Problem 6.21 A generalized diode model (computer)

Problem 6.22 A pseudo-narrow-base diode

Problem 6.23 Narrow base with a finite-lifetime lossy layer

Problem 6.24 Denuded zone (multiple considerations)

Appendix: Quick Reference for Chapter 6

Graph View

Table of Contents

Ideal $I$ - $V$

Exercise 6.2 (computer): Ideal $I$ - $V$ versus doping

Exercise 6.6 (computer): $I_{01}$ and $I_{02}$ for Si and Ge

6.2.4 $V_{A} \to V_{bi}$ High-Current Phenomena

Problem 6.5 Boundary condition via the “Videri” ( $J_{N} = 0$ ) approach

Problem 6.7 Ideal $I$ - $V$ versus temperature (computer)

Problem 6.19 Extracting $I_{0 j}$ and $n_{j}$ from data