part08

Chapter 9 — Optoelectronic Diodes

Why this chapter is fun. Until now, electrons and holes were pushed around by voltages. Now we let light into the story. Light can create carriers (photodiodes, solar cells), and carriers can create light (LEDs). A pn junction is a two-way translator between photons and electrons. Think of a semiconductor as a currency exchange booth where photons and electrons trade at a fixed rate: the band gap $E_G$.

9.1 Introduction

Photodevices fall into three functional categories:

Category	Energy conversion	Purpose	Examples
Photodetector	light $\to$ electrical	detect or measure information carried by light	pn photodiode, p-i-n, avalanche photodiode
Solar cell	light $\to$ electrical	produce electrical power	pn junction solar cell
Light emitter	electrical $\to$ light	produce light	LED, laser diode

This chapter surveys one family from each category. Applications are everywhere: barcode readers, disc readers and laser printers (LED or laser-diode sources); circuit isolators, intruder alarms and remote controls (emitter plus detector pairs); fiber-optic telecommunication (laser diode transmits, photodiode receives); calculators, chargers and satellites (solar cells).

Wavelength and photon energy

Visible light spans roughly $0.4\mu$m (violet) to $0.7\mu$m (red). Figure 9.1 shows (top) the relative response of the human eye, peaking in the green near $0.55\mu$m; (middle) the colour bands violet, blue, green, yellow, red, with ultraviolet below $0.4\mu$m and infrared above $0.7\mu$m; (bottom) the wavelengths ${\lambda }_G$ at which the photon energy equals the 300 K band gap of 6H-SiC, ZnSe, GaP, AlAs, GaAs and InP.

A photon of frequency $\nu$ has $E_{ph}=h\nu$. With $c=\lambda \nu$:

$\lambda =\frac{c}{\nu }=\frac{hc}{h\nu }=\frac{hc}{E_{ph}}(9.1)$

With $\lambda$ in $\mu$m and $E_{ph}$ in eV:

$\lambda [\mu \text{m}]=\frac{1.24}{E_{ph}[\text{eV}]}(9.2)$

Check: $hc=(6.626\times 10^{-34})(3\times 10^8)=1.988\times 10^{-25}$ J m, and dividing by $1.602\times 10^{-19}$ J/eV gives $1.24$ eV $\mu$m.

The wavelength where $E_{ph}=E_G$ is

${\lambda }_G=\frac{1.24}{E_G}$

Photons with $\lambda <{\lambda }_G$ can be absorbed and create pairs; photons with $\lambda >{\lambda }_G$ pass through the crystal like glass.

Examples: Si, $E_G=1.12$ eV gives ${\lambda }_G=1.107\mu$m. GaAs, $E_G=1.42$ eV gives $0.873\mu$m. Visible light $0.4<\lambda <0.7\mu$m corresponds to $1.77<E_{ph}<3.10$ eV.

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9.2 Photodiodes

9.2.1 pn Junction Photodiodes

A pn junction photodiode is a pn diode fabricated and encapsulated so light can reach the neighbourhood of the metallurgical junction.

Rain-and-gutter picture. The depletion region is a steep roof with a gutter (the built-in field). Light creates raindrops (electron-hole pairs) everywhere. Drops landing on the roof slide into the gutter immediately. Drops landing on flat ground within a diffusion length of the roof wander and often reach it. Drops farther away soak into the ground (recombine) and are lost.

Minority carriers created in the quasineutral regions within roughly a diffusion length of the depletion edges ($x=-x_p$ on the p-side, $x=x_n$ on the n-side) live long enough to diffuse to the depletion region. These, plus carriers generated inside the depletion region, are swept by the field to the other side, adding a reverse-going current. If the photogeneration rate $G_L$ is uniform, the added current is $-q$ times the pairs generated per second in the volume $A(L_N+W+L_P)$:

$I=I_{dark}+I_L(9.3)$

$I_L=-qA(L_N+W+L_P)G_L(9.4)$

A modified derivation of the ideal-diode and R-G current relations with $G_L\ne 0$ confirms this (book Problem 6.8).

Consequences of (9.4).

1. $W$ is typically small compared with $L_N+L_P$, so $I_L$ is essentially independent of bias.
2. The light-on $I$–$V$ curve is the dark curve translated straight down the current axis.
3. $I_L\propto G_L$, so the shift is proportional to light intensity (Fig. 9.3 shows parallel curves for $G_L=0,G_0,2G_0,4G_0$).

Combined photodiode equation:

$I=I_0\left(e^{q{V}_A/kT}-1\right)-qA(L_N+W+L_P)G_L$

Spectral response

Spectral response is how $I_L$ varies with wavelength. In Fig. 9.4 (Si, constant incident power) the response rises from near zero at $0.4\mu$m to a maximum near $0.8$–$0.9\mu$m, then falls sharply to near zero beyond about $1.1\mu$m.

Long-wavelength cutoff. Photons with $E_{ph}<E_G$ cannot lift an electron across the gap, so the crystal is transparent. The cutoff is ${\lambda }_G=1.24/E_G$; for Si ($1.12$ eV) that is about $1.1\mu$m, matching the figure.

Short-wavelength decline: two reasons.

1. Fixed power means fewer photons. The photon flux is $\Phi =P_{opt}/(h\nu )$. As $\lambda$ decreases, $h\nu$ rises, so fewer photons arrive per second.
2. Surface recombination. Intensity decays as $e^{-\alpha x}$ and the mean penetration depth is $1/\alpha$. As $\lambda$ drops below ${\lambda }_G$, $\alpha$ rises rapidly (Fig. 3.20), so light is absorbed nearer the surface, where recombination is strong, and more carriers recombine before reaching the depletion region.

Frequency response

Photogenerated minority carriers outside the depletion region must diffuse to it before current appears externally. Diffusion is slow, so a standard pn photodiode reaches at best tens of MHz.

9.2.2 p-i-n and Avalanche Photodiodes

p-i-n photodiodes

A p-i-n diode has a lightly doped (“intrinsic”) i-region between heavily doped $p^{+}$ and $n^{+}$ regions. In the photodiode (Fig. 9.5a) the surface metal has an opening for light, an antireflection coating covers the surface, the top region is very thin to minimize absorption there, and the i-width is tailored to the desired response.

Band diagram (Fig. 9.5b).

• Low doping means the i-layer is fully depleted at zero or small reverse bias.
• Heavy doping makes the depletion widths in the outer regions tiny.
• The total depletion width is effectively $W_I$, independent of reverse bias.
• Little charge in the i-region gives $d\mathcal{E}/dx\approx 0$: a roughly constant field and straight, tilted bands.
• Heavy outer doping gives short minority-carrier diffusion lengths there, so most photocurrent comes from carriers generated in the central depletion region.

Two advantages.

1. Tuned wavelength response: choose $W_I\approx 1/\alpha$ at the wavelength of interest.
2. High speed: carriers are collected by the large i-field at saturation velocity rather than by diffusion:

$f_{max}\approx \frac{1}{\text{transit time across }{W}_I}=\frac{v_{sat}}{W_I}(9.5)$

Example. $v_{sat}\approx 10^7$ cm/s, $W_I=5\mu \text{m}=5\times 10^{-4}$ cm:

$t_{tr}=\frac{5\times 10^{-4}}{10^7}=5\times 10^{-11}\text{s}\Rightarrow f_{max}\approx 20\text{GHz}$

$W_I$ cannot be shrunk arbitrarily: the junction capacitance $C=\epsilon A/W_I$ grows as $W_I$ falls, and the $R_{S}C$ time constant (with series resistance $R_S$) eventually limits speed. Transit time wants small $W_I$; RC wants large $W_I$.

Fiber-optic use. Pre-1990 silica fiber systems operated at $1.3\mu$m where chromatic dispersion (different wavelengths travelling at slightly different speeds, smearing pulses) is minimal, suitable for multiwavelength sources. Advanced high-bit-rate systems use single-wavelength sources at $1.55\mu$m, where fiber loss is minimal. Both exceed Si’s $1.1\mu$m cutoff, so another material is needed.

Material of choice: ${\text{In}}_{0.53}{\text{Ga}}_{0.47}\text{As}$ on InP. From Fig. 9.6 (band gap versus lattice constant; a line joins two compounds, and the alloy lattice constant is linear in composition), it has $E_G\approx 0.75$ eV:

${\lambda }_G=\frac{1.24}{0.75}=1.65\mu \text{m}$

and is lattice matched to InP (same lattice constant), permitting quality layers on commercial InP substrates.

Interpolation example (Fig. 9.6 caption). ${\text{In}}_{0.8}{\text{Ga}}_{0.2}$P, with InP at $5.87$ Å and GaP at $5.45$ Å:

$a=5.87-0.2(5.87-5.45)=5.79\text{A˚}$

and $E_G\approx 1.5$ eV from the InP–GaP line. Solid lines mark direct alloys, dashed lines indirect.

InGaAs p-i-n cross section (Fig. 9.7), bottom to top: $n^{+}$-InP substrate; $n$-InP buffer (reduces defects in the InGaAs); $n^{-}$ ${\text{In}}_{0.53}{\text{Ga}}_{0.47}$As i-layer (absorber); $p^{+}$-InP window (wide gap, transmits light to the absorber); $p^{+}$-InGaAsP cap (reduces surface dark current); silicon nitride (protects the surface, minimizes surface recombination). Light enters from the top.

Avalanche photodiodes (APD)

APDs are specially built p-i-n, pn or metal-semiconductor photodiodes operated near avalanche breakdown. The standard Si APD (Fig. 9.8) has an $n^{+}$ top region with antireflection coating, a guard ring around the junction periphery, $p$ and $p^{+}$ regions, a passivation coating, and $n$ and $p$ contacts. Junction curvature causes early breakdown at the periphery (Chapter 6); the guard ring prevents this. Uniform, low-defect material is needed for uniform breakdown across the junction.

Advantage: gain without proportionally amplified noise. Normally amplification boosts signal and noise and adds amplifier noise, lowering S/N. In an APD, avalanche multiplication amplifies the photo-signal without amplifying the typically dominant receiver circuit noise, so S/N improves until avalanche noise becomes comparable to circuit noise. InGaAs-on-InP and Ge APDs are alternatives to p-i-n diodes in fiber telecommunications.

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9.3 Solar Cells

9.3.1 Solar Cell Basics

Most solar cells are large-area pn photodiodes. Photodiodes are designed for spectral or time response; solar cells are designed to minimize energy losses. The $I$–$V$ form is that of Fig. 9.3.

Power is delivered when the operating point is in the fourth quadrant ($I<0$, $V>0$, so $P=IV<0$ means the device supplies power), e.g. by putting a resistor in series with the illuminated cell. Usually only this quadrant is plotted with the $-I$ axis up (Fig. 9.9).

Parameters (Fig. 9.9):

• $V_{OC}$: open-circuit voltage, the maximum voltage.
• $I_{SC}$: short-circuit current, the maximum current.
• $V_m,I_m$: the maximum-power operating point (corner of the “maximum power rectangle”).

From $I=0$ in the photodiode equation, with $I_L^{\prime }=|I_L|$:

$0=I_0\left(e^{q{V}_{OC}/kT}-1\right)-I_L^{\prime }\Rightarrow V_{OC}=\frac{kT}{q}\ln \left(1+\frac{I_L^{\prime }}{I_0}\right),I_{SC}=I_L^{\prime }$

$P_{max}=I_m{V}_m<I_{SC}{V}_{OC}$

Fill factor:

$FF=\frac{P_{max}}{I_{SC}{V}_{OC}}=\frac{I_m{V}_m}{I_{SC}{V}_{OC}}(9.6)$

$FF<1$ always; it measures how square the curve is.

Power conversion efficiency:

$\eta =\frac{P_{max}}{P_{in}}=\frac{FF{I}_{SC}{V}_{OC}}{P_{in}}(9.7)$

9.3.2 Efficiency Considerations

Higher efficiency means less area and cost per electrical output.

Solar spectrum (Fig. 9.10, irradiance in W/m${}^2$-$\mu$m versus wavelength 0–3 $\mu$m).

• AM0 (air mass zero): just outside the atmosphere, $135{\text{mW/cm}}^2$; relevant for satellites.
• AM1.5: normalized to $100{\text{mW/cm}}^2$; average terrestrial U.S. conditions (shows atmospheric absorption dips).
• Most power is visible, with a long infrared tail.

Two intrinsic losses and the optimum gap.

1. Sub-gap photons wasted: the spectral area beyond ${\lambda }_G$ cannot be absorbed: about 20% of incident energy in Si and about 35% in GaAs.
2. Excess photon energy becomes heat: a photon with $E_{ph}>E_G$ yields only $E_G$ usefully; the excess $E_{ph}-E_G$ thermalizes. About 40% of absorbed energy is lost this way in Si, about 30% in GaAs.

Raising $E_G$ reduces loss 2 but increases loss 1, so an optimum band gap exists. Si and GaAs, the most developed semiconductors, both lie very close to it.

Device-related losses (Si cell, Fig. 9.11: $n$ top layer on $p$-Si, $p^{+}$ under the rear contact, oxides top and bottom, metal fingers, inverted pyramids).

Loss	Remedy
Shadowing versus series resistance	Narrow metal fingers connected along one edge. Fingers far apart give long lateral current paths in the thin $n$-layer and high $R_S$ (even a few ohms hurt efficiency); fingers too wide or dense shadow the cell. Size and spacing are optimized. Very narrow fingers also raise metallization and contact resistance
Reflection ($\approx 30\%$ at bare planar Si, normal incidence)	Texturing by anisotropic etching (etch removing Si preferentially along certain crystal planes) forms inverted pyramids so light strikes the surface two or more times before escaping; plus an antireflection coating of intermediate refractive index (${\text{SiO}}_2$ or an optimized deposited film). Net reflection below 1%
Incomplete absorption of long wavelengths	Bottom oxide plus metal form a mirror; light is trapped and bounces repeatedly, dramatically enhancing absorption near ${\lambda }_G$
Recombination	Very long minority-carrier lifetimes (diffusion length greater than cell thickness) and oxidized top and bottom surfaces (low surface recombination) let carriers born almost anywhere reach the depletion region and be swept across

9.3.3 Solar Cell Technology

Three categories: thin-film, single-crystal, concentrator.

• Thin-film: deposited amorphous or polycrystalline absorber. Amorphous Si (a-Si) is commercial; CdTe and ${\text{CuInSe}}_2$ are in development. Cheap, large area (modules to $1.2{\text{m}}^2$). a-Si is vapor-deposited, about $1\mu$m thick, on tin-oxide-coated glass (transparent front contact), doped during deposition into a p-i-n structure, with Al or Ag back contact. Lab efficiency about 13%, commercial often half that; output falls 10%–15% in the first year because light breaks passivating bonds, creating traps and recombination. CdTe reached roughly 15% in the lab (OCR garbled) and 11% in the field.
• Single-crystal: almost all Si or GaAs. GaAs cells often contain AlGaAs or other alloy layers; active GaAs is epitaxially deposited on GaAs, Ge or Si.
• Concentrator: for 100 suns or more. A mirror or lens costs less than equivalent cell area; the higher current and voltage also raise efficiency (Fig. 9.12). Stacking (for example Si or GaSb (${\lambda }_G=1.7\mu$m) beneath GaAs to absorb what GaAs passes) is expected to exceed 35%. Concerns: heat dissipation, elevated temperature, high current density.

Figure 9.12 (lab efficiency versus year, 1978–1992): concentrators (GaAs or Si) top at about 32%; flat-plate single-crystal Si about 23%; flat-plate thin-film about 16%; all rise with time.

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9.4 LEDs

9.4.1 General Overview

The mechanism as a story. Forward bias lowers the hill: electrons flood from n to p and holes from p to n. The injected minority carriers meet majority carriers and recombine. What happens to the released energy?

• In an indirect semiconductor (Si, Ge) electron and hole crystal momenta differ greatly (E-k diagrams, Subsection 3.3.2). A photon carries almost no momentum, so band-to-band recombination is hard; recombination goes through R-G centres and the energy becomes heat.
• In a direct semiconductor (GaAs) the momenta match, so much of the injected carriers recombine band to band emitting photons (Fig. 9.13: forward-biased band diagram with split quasi-Fermi levels $E_{Fn}$, $E_{Fp}$, electrons dropping from $E_C$ to $E_V$).

Emission spectrum. Carrier distributions peak about $kT/2$ from band edges in a nondegenerate semiconductor, so photon energies spread by about $kT$ with a peak slightly above $E_G$. If photons come from a centre just below $E_C$ or from excitons (electrostatically bound electron-hole pairs acting as a unit), the peak can lie slightly below $E_G$. In any case ${\lambda }_{peak}\approx {\lambda }_G=1.24/E_G$. For visible light, $0.4<{\lambda }_G<0.7\mu$m, i.e. $1.77<E_G<3.10$ eV.

Three requirements for a visible LED material: direct gap; $E_G$ between about 1.7 and 3.1 eV; pn junctions can be formed. Candidates:

• Si, Ge: indirect, wrong gap.
• GaAs: direct but gap too small.
• GaP, AlAs: right gap, indirect.
• SiC: right gap, indirect.
• II-VI compounds (ZnSe): direct, right gap, but native defects compensate dopants, so junctions were impossible until recently.

Hence alloys and special “light-enhancing” centres are used.

9.4.2 Commercial LEDs

Table 9.1: Characteristics of commercial LEDs (adapted from Craford)

Material	Colour	Peak $\lambda$ ($\mu$m)	External efficiency $\eta$ (%)	Luminous performance (lm/W)
Established
${\text{GaAs}}_{0.6}{\text{P}}_{0.4}$	Red	0.650	0.2	0.15
${\text{GaAs}}_{0.35}{\text{P}}_{0.65}$:N	Orange-red	0.630	0.7	1
${\text{GaAs}}_{0.14}{\text{P}}_{0.86}$:N	Yellow	0.585	0.2	1
GaP:N	Green	0.565	0.4	2.5
GaP:Zn-O	Red	0.700	2	0.40
Recent additions
AlGaAs	Red	0.650	4–16	2–8
AlInGaP	Orange	0.620	6	20
AlInGaP	Yellow	0.585	5	20
AlInGaP	Green	0.570	1	6
SiC	Blue	0.470	0.02	0.04
GaN	Blue	0.450	2	0.6

Luminous performance is lumens per watt of electrical input (radiant watts times relative eye response, Fig. 9.1). External efficiency is $\eta =P_{photo,out}/P_{elec,in}$.

${\text{GaAs}}_{0.6}{\text{P}}_{0.4}$

GaAs is direct but infrared; GaP is visible but indirect. The alloy ${\text{GaAs}}_{1-x}{\text{P}}_x$ is direct with a light-producing gap for $0.28\le x\le 0.45$ (direct-indirect transition at $x=0.45$, Fig. 9.6). Why $x=0.4$? From $x=0.28$ to $0.45$, $\eta$ falls about tenfold, but the wavelength shortens and eye response rises about 50 times; the product (efficiency $\times$ eye response) peaks at $x=0.4$, giving the brightest-looking device. These were the first mass-marketed solid-state lamps (early-1970s watches and calculators), later replaced by liquid crystal displays (lower battery drain); today they are indoor indicator lamps.

Structure (Fig. 9.14a). GaAs and ${\text{GaAs}}_{0.6}{\text{P}}_{0.4}$ have very different lattice constants, so direct growth would be defective. Process: (1) graded layer $x:0\to 0.4$ on the GaAs substrate; (2) n-type ${\text{GaAs}}_{0.6}{\text{P}}_{0.4}$ active layer; (3) second graded layer $x:0.4\to 0.6$ (wider gap, light passes without reabsorption); (4) Zn diffused to form the p-side. The back n-contact is segmented to enhance bottom-surface reflection; ${\text{Si}}_3{\text{N}}_4$ covers the top.

${\text{GaAs}}_{0.35}{\text{P}}_{0.65}$:N, ${\text{GaAs}}_{0.14}{\text{P}}_{0.86}$:N, GaP:N

Orange-red, yellow and green LEDs use $x=0.65$ to $1$, technically indirect; nitrogen doping makes them pseudo-direct.

Isoelectronic trap. N is a Column V element like As and P, so it replaces a Column V atom and adds no free electron or hole. It instead creates an electron-trapping level about $0.1$ eV below $E_C$. Being formed from an element with the same (iso) valence structure as the atom replaced, it is an isoelectronic trap. A trapped electron is highly localized, so by $\Delta x\Delta p\gtrsim \hslash /2$ it has a large spread in momentum that includes the momentum of valence-band holes, giving a greatly enhanced probability of a photon-emitting transition. The trap is a conduit for relatively efficient radiative recombination.

These devices resemble the ${\text{GaAs}}_{0.6}{\text{P}}_{0.4}$ LED but are built on GaP substrates. The GaP:N diode (Fig. 9.14b) needs no substrate grading layer, and both n and p regions are epitaxial. Photons originate from the trap ($E_{ph}<E_G$), so the epitaxial layers and the GaP substrate are transparent; even downward light reflects and may escape, giving higher efficiency and brightness. With GaP:Zn-O they hold the largest LED market share.

GaP:Zn-O

Zn alone is an acceptor and O alone a donor in GaP. Added in about equal numbers and annealed, Zn on a Ga site and O on an adjacent P site form one isoelectronic trap complex, deeper than N: about $0.3$ eV below $E_C$. Emission is red, ${\lambda }_{peak}=0.700\mu$m. Fabrication parallels GaP:N.

AlGaAs

Introduced in the late 1980s as the “high-brightness” red LED, up to 20 times more luminous than GaAsP:N and GaP:Zn-O red LEDs. Uses: automobile spoiler (third) brake lights (about 10% of the market) and athletic-shoe heel lights. Structure (Fig. 9.14c): a narrow active layer ($E_G\le 1.9$ eV) between slightly wider-gap confining layers (minimize reabsorption, maximize extraction of randomly directed light); in the brightest devices a thick (100–200 $\mu$m) epitaxial layer is grown below the lower confining layer and the GaAs substrate removed, so $E_G>E_{ph}$ below the emitter too. AlGaAs with $E_G<2$ eV is direct (Fig. 9.6) and all AlGaAs alloys are lattice matched to GaAs. Lattice-matched perfection (minimal non-radiative recombination) plus direct recombination (about an order of magnitude more efficient than via isoelectronic traps) give high efficiency and brightness.

AlInGaP

On sale from 1992; three Column III elements plus one Column V element. Already beats the established orange, yellow and green devices and may become the dominant high-performance technology from red through green. Structure (Fig. 9.14d): confining and active layers lattice matched to GaAs, direct recombination; colour set by active-layer composition. GaP is the top contact/window because its conductivity far exceeds that of p-type AlInGaP.

SiC

Electroluminescence from SiC (surely with built-in pn junctions) was reported in 1907. SiC LEDs were intensely researched in the 1960s and briefly sold in the early 1970s, then offered again in 1990 after a substrate-quality breakthrough, given the need for a blue LED. Efficiency improved only to 0.02% because operation still relies on low-probability radiative recombination in an indirect material.

GaN

Direct III-V with $E_G=3.36$ eV. The blue GaN LED was first sold in April 1994, developed by a chemical company thanks to a breakthrough in forming pn junctions in GaN and its alloys. Light is actually produced in an $E_G=2.75$ eV InGaN film sandwiched between wider-gap AlGaN layers. Expected to overtake SiC; it supplies the missing colour for full-colour LED outdoor video displays at reasonable power.

9.4.3 LED Packaging and Photon Extraction

Package (Fig. 9.15). The chip, about $250\mu$m on a side, sits in a reflecting cup because a portion of the light leaves from the chip sides. One lead connects to the cup; a wire bond joins the chip’s top contact to the other lead. An epoxy dome holds the leadframe and helps photon extraction.

Extraction is harder than generation.

• Wide-gap overlayers act as windows.
• N- and Zn-O-doped LEDs need no window since $E_{ph}$ is at least $0.1$ eV below the host gap.
• Ideally $E_G>E_{ph}$ also below the emitting layer, since half the light starts downward; the light can then bounce around with minimal reabsorption and eventually hit the surface.

Optics at the surface (Fig. 9.16). At normal incidence the amplitude reflection coefficient is $r=(n_1-n_2)/(n_1+n_2)$, so the transmitted power fraction is

$T_N=1-r^2=\frac{(n_1+n_2{)}^2-(n_1-n_2{)}^2}{(n_1+n_2{)}^2}=\frac{4n_1{n}_2}{(n_1+n_2{)}^2}$

Dividing top and bottom by $n_2^2$ with $\bar{n}=n_1/n_2$:

$T_N=\frac{4\bar{n}}{(1+\bar{n}{)}^2}(9.8)$

For $n_1>n_2$ (semiconductor into another medium, so $\bar{n}>1$), light beyond the critical angle undergoes total internal reflection ($T=0$). Snell’s law $n_1\sin {\theta }_c=n_2\sin 90^{\circ }$ gives

$\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1}{\bar{n}}(9.9)$

Overall transmission (9.10). Assume isotropic, randomly polarized internal light, only the upward half able to exit. Flux through a surface element from angle $\theta$ is proportional to $\cos \theta \sin \theta d\theta$. The fraction of upward flux within the escape cone:

$\frac{{\int }_0^{{\theta }_c}\cos \theta \sin \theta d\theta }{{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta }=\frac{\frac{1}{2}{\sin }^2{\theta }_c}{\frac{1}{2}}={\sin }^2{\theta }_c$

Inside the (small) cone the transmission is close to $T_N$; multiply by $\frac{1}{2}$ for the upward half:

$T\approx \frac{T_N{\sin }^2{\theta }_c}{2}=\frac{1}{2}\cdot \frac{4\bar{n}}{(1+\bar{n}{)}^2}\cdot \frac{1}{{\bar{n}}^2}=\frac{2}{\bar{n}(1+\bar{n}{)}^2}(9.10)$

Example: GaP into air. $n_1=3.4$, $n_2=1$, $\bar{n}=3.4$:

$T_N=\frac{4(3.4)}{(4.4{)}^2}=\frac{13.6}{19.36}=0.70$

${\theta }_c={\sin }^{-1}\frac{1}{3.4}=17.1^{\circ }$

$T=\frac{2}{3.4(4.4{)}^2}=\frac{2}{65.8}=0.030(3.0\%)$

Most light is trapped by total internal reflection. Epoxy (index between air and semiconductor) widens the escape cone at the chip interface, and the conical dome lets much light reach the epoxy-air surface within its critical angle.

Chapter 9 Recap (study aid)

The OCR slice has no formal Chapter 9 summary or problem set, so this recap stands in.

1. ${\lambda }_G=1.24/E_G$ separates absorbing from transparent behaviour.
2. Photodiode: $I=I_{dark}+I_L$, $I_L=-qA(L_N+W+L_P)G_L$; a bias-independent downward shift proportional to light intensity.
3. p-i-n: wide constant-field depletion region, tailored response, $f_{max}\approx v_{sat}/W_I$ (RC limited at small $W_I$). APD: multiplication gain with improved S/N.
4. Solar cell: fourth-quadrant operation, $FF=P_{max}/(I_{SC}{V}_{OC})$, $\eta =FF{I}_{SC}{V}_{OC}/P_{in}$; losses from sub-gap photons, excess photon energy, shadowing, reflection, incomplete absorption and recombination.
5. LED: direct gap, 1.7–3.1 eV, junction-capable; achieved through alloys and isoelectronic traps. Extraction limited by total internal reflection, $T=2/[\bar{n}(1+\bar{n}{)}^2]$.

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Chapter 10 — BJT Fundamentals

The big idea. A diode has one junction and two terminals. Put two junctions so close together that carriers can run from one to the other before they have time to die, and you get a transistor. Picture a relay race: the emitter hands the baton (carriers) to the base, which is so short that nearly every runner crosses it, and the collector catches them. A tiny trickle of effort in the base lane controls a big stream in the main lane. That is amplification.

Chapters 5–8 analysed the pn diode. This chapter begins the bipolar junction transistor (BJT): terms and symbols, qualitative operation, key relations; the knowledge base for the quantitative analysis in the next chapter.

10.1 Terminology

A BJT has three adjoining, alternately doped regions, the middle one being very narrow compared with the minority-carrier diffusion length in it. pnp and npn types are shown in Fig. 10.1. The narrow centre is the base; the outer regions are the emitter and collector. They are not interchangeable: the emitter has a different geometry and is typically more heavily doped than the collector, so swapping terminals changes the characteristics substantially.

Symbols and polarities (Fig. 10.2). For the pnp: $I_E$ flows into the emitter, $I_B$ and $I_C$ flow out; $V_{EB}$, $V_{EC}$, $V_{CB}$ are voltage drops. The ”+” and ”−” signs are redundant with the double subscripts: the first letter is the terminal assumed more positive, so $V_{EB}$ is the drop from emitter (+) to base (−). Some chosen current directions are contrary to the convention “into a terminal is positive”, but with these choices all terminal currents are positive in standard amplifying operation, which keeps the physical discussion uncomplicated. (For npn, $I_C$ and $I_B$ flow in and $I_E$ out, with $V_{BE}$, $V_{CE}$, $V_{BC}$.)

Only two currents and two voltages are independent. Current into the device equals current out (Kirchhoff’s current law) and voltage drops around a closed loop sum to zero (Kirchhoff’s voltage law):

$I_E=I_B+I_C(10.1)$

$V_{EB}+V_{BC}+V_{CE}=0(V_{CE}=-V_{EC})(10.2)$

If two currents (or voltages) are known, the third follows.

Circuit configurations (Fig. 10.3). One lead must be common to input and output; the name of that lead names the configuration:

Configuration	Common lead	Input	Output	Notes
Common base	base	$I_E$, $V_{EB}$	$I_C$, $V_{CB}$	occasional use
Common emitter	emitter	$I_B$, $V_{EB}$	$I_C$, $V_{EC}$	most widely used
Common collector	collector	$I_B$, $V_{CB}$	$I_E$, $V_{EC}$	seldom used, neglected from here on

Figure 10.4 shows idealized output characteristics of a pnp. (a) Common base: $I_C$ (0–5 mA) versus $-V_{CB}$ (0–60 V… as plotted, nearly flat lines at $I_C\approx I_E$ for $I_E=0,1,2,3,4$ mA). (b) Common emitter: $I_C$ (0–16 mA) versus $V_{EC}$ (0–20 V) for $I_B=0,10,20,30,40\mu$A, rising lines with a modest upward slope.

Biasing modes (Table 10.1, Fig. 10.5). Each junction is forward (F) or reverse (R) biased: four combinations.

Mode	E-B junction	C-B junction	Typical use
Saturation	Forward	Forward	switch “on”: high current, low voltage
Active	Forward	Reverse	linear amplifiers; largest gain, least distortion
Inverted	Reverse	Forward	roles of emitter and collector swapped
Cutoff	Reverse	Reverse	switch “off”: low current, high voltage

In digital circuits the low-voltage and high-voltage states of saturation and cutoff correspond to logic levels. Figure 10.5(a) is a plane with axes $V_{EB}$ (pnp) or $V_{BE}$ (npn) and $V_{CB}$ (pnp) or $V_{BC}$ (npn), the four quadrants being the four modes; Fig. 10.5(b) marks the regions of the common-emitter output characteristics: cutoff at $I_B\le 0$ (along the $V_{EC}$ axis), active in the flat-ish region, saturation at small $V_{EC}$, and the inverted-active region for negative $V_{EC}$.

The development mostly uses a pnp because it is a convenient vehicle for the concepts; the npn is easily obtained by swapping carrier types and polarities.

10.2 Fabrication

An idealized pseudo-one-dimensional model (Fig. 10.1) is used for qualitative and first-order quantitative work. Fabrication extends diode fabrication (Chapter 4); the new feature is two pn junctions in close proximity.

Discrete double-diffused pnp (Fig. 10.6a). Start with a heavily doped $p^{+}$ wafer (mechanical support, low-resistance path to the bottom collector contact). Grow a high-resistivity (lightly doped) $p$ epilayer 5–10 $\mu$m thick; this contains the transistor proper (it becomes the collector). Grow an oxide, open windows, diffuse phosphorus or arsenic to form the n-type base. Repeat with boron to form the $p^{+}$ emitter. Deposit aluminium and pattern the emitter and base contacts. (Dimensions in the figure are not to scale; the lateral scale should be expanded about 50 times for a true picture: the active region is about $5\mu$m deep while the wafer is about $200\mu$m thick.)

IC npn (Fig. 10.6b). Diffuse p-type then n-type into a high-resistivity n-type epilayer. The epilayer has doping opposite to the substrate, and diode isolation is completed by a deep p-type diffusion around the transistor. The most negative circuit potential is applied to the p-substrate, so the n-epi region is isolated by a reverse-biased pn junction from other devices. Advanced designs replace the lateral junctions with silicon dioxide isolation. The $n^{+}$ buried layer, diffused before the epilayer is grown, gives a low-resistance path between the active collector region and the top-side collector contact.

10.3 Electrostatics

Under equilibrium and normal operating conditions, the BJT is electrostatically just two independent pn junctions. Chapter 5 results (built-in potential, charge density, field, potential, depletion width) apply separately to the E-B and C-B junctions. For a standard uniformly doped pnp with $N_{AE}\ge N_{DB}>N_{AC}$ the equilibrium band diagram and variables are as in Fig. 10.7 (depletion approximation).

Reading Fig. 10.7.

• Almost all of the E-B depletion width $W_{EB}$ lies in the base; most of the C-B depletion width $W_{CB}$ lies in the collector.
• $W_{CB}>W_{EB}$ because the lightly doped side of the C-B junction (collector) is more lightly doped than that of the E-B junction (base).
• $W_B$ is the total (metallurgical) base width; $W$ is the quasineutral base width, the undepleted part:

$W=W_B-x_{nEB}-x_{nCB}(10.3)$

where $x_{nEB}$, $x_{nCB}$ are the parts of the E-B and C-B depletion widths lying in the n-type base. This $W$ is not the diode depletion width of earlier chapters.

Useful formulas (from Chapter 5, applied per junction, with $N_l$ and $N_r$ the doping on either side):

$V_{bi}=\frac{kT}{q}\ln \frac{N_A{N}_D}{n_i^2},W_{dep}=\sqrt{\frac{2\epsilon V_{bi}}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)}$

$x_n=W_{dep}\frac{N_A}{N_A+N_D},x_p=W_{dep}\frac{N_D}{N_A+N_D},|\mathcal{E}{|}_{max}=\frac{q{N}_D{x}_n}{\epsilon }=\frac{q{N}_A{x}_p}{\epsilon }=\frac{2V}{W_{dep}}$

For a general bias $V$ across a junction (forward positive): replace $V_{bi}$ by $V_{bi}-V$.

Exercise 10.1

Problem. Draw the energy band diagram of (a) an npn transistor in equilibrium; (b) a pnp transistor under active-mode biasing. Assume uniform doping and the standard profile: emitter doping > base doping > collector doping.

Solution.

(a) An npn is the complement of a pnp, so its diagram is the pnp diagram (Fig. 10.7b) turned upside down (equivalently, combine the equilibrium diagrams of an np and a pn junction). Features:

• $E_F$ is a single flat line.
• Emitter (n${}^{+}$): $E_C$ very close to $E_F$ (just above it, or even at it).
• Base (p): $E_F$ close to $E_V$, so $E_C$ is far above $E_F$; $E_C$ in the base is higher than in the emitter and collector: a hill in $E_C$ for electrons.
• Collector (n, lightest doping): $E_F$ sits between $E_i$ and $E_C$, farther below $E_C$ than in the emitter.
• The emitter-base hill is $q{V}_{bi,EB}$ and the base-collector hill $q{V}_{bi,CB}$, with $V_{bi,EB}>V_{bi,CB}$ because the emitter is more heavily doped than the collector. The E-B depletion region is narrower than the C-B one and penetrates mostly into the base; the C-B depletion region penetrates mostly into the collector.

(b) Active mode: E-B forward biased, C-B reverse biased. Starting from Fig. 10.7(b):

• Forward bias on E-B raises the base-side Fermi level (electron energy) relative to the emitter-side one by $q{V}_{EB}$; both the E-B depletion width and potential hill are reduced.
• Reverse bias on C-B lowers the base-side Fermi level relative to the collector side by $q|V_{CB}|$; the C-B depletion width and potential hill increase.
• The result: $E_F$ is not flat but separated in three parts: emitter at the lowest electron energy, collector at the highest, with the base level between them (for pnp, $E_F$ steps up from emitter to base by $q{V}_{EB}$ and from base to collector by $q|V_{CB}|$). The hole potential energy hill for holes in the valence band is low at the E-B junction and the holes injected into the base roll down the large drop at the C-B junction.

Exercise 10.2 (computer): equilibrium band diagram of a BJT

Problem. Write a MATLAB program that draws the equilibrium energy band diagram of a nondegenerately doped Si pnp or npn BJT with uniformly doped regions at room temperature.

Logic of the solution. The book’s file BJT_Eband (code in Appendix M) uses the defaults $N_{AE}=10^{18}$, $N_{DB}=10^{16}$, $N_{AC}=10^{15}{\text{cm}}^{-3}$, $W_B=1\mu$m (doping values input with n-type negative). The algorithm is:

1. For each region compute $E_i-E_F=\pm kT\ln (|N|/n_i)$ (positive for p-type, negative for n-type).
2. At each of the two junctions, $V_{bi}=|(E_i-E_F{)}_{left}-(E_i-E_F{)}_{right}|/q$; compute the depletion width and its split $x_l,x_r$.
3. Inside the depletion region of a junction, the band-edge shift is parabolic: on the left side it moves by the fraction $x_l/W_{dep}$ of the total change as $(s/x_l{)}^2$; on the right side the remaining fraction approaches the bulk value as $((W_{dep}-s)/x_r{)}^2$.
4. Build $E_i(x)$, then $E_C=E_i+E_G/2$, $E_V=E_i-E_G/2$ (approximately, ignoring the small difference between $E_i$ and midgap), with $E_F=0$ as reference, and plot.

% BJT equilibrium band diagram (uniformly doped, depletion approximation)
% Doping convention: p-type positive, n-type negative (cm^-3)
clear; clc;
DOPING = [1e18  -1e16  1e15];   % [emitter base collector]  pnp example
WIDTH  = 1e-4;                   % total base width WB in cm (1 um)
T  = 300;  kT = 8.617e-5*T;      % eV
ni = 1e10; EG = 1.12; q = 1.602e-19;
eps = 11.8*8.854e-14;            % F/cm
 
% step 1: Ei - EF in each region (eV)
EiEF = sign(DOPING).*kT.*log(abs(DOPING)/ni);
 
% step 2: junction parameters
for j = 1:2
    Nl = abs(DOPING(j)); Nr = abs(DOPING(j+1));
    Vbi = abs(EiEF(j) - EiEF(j+1));                  % volts
    Wd(j) = sqrt(2*eps*Vbi/q*(1/Nl + 1/Nr));         % cm
    xl(j) = Wd(j)*Nr/(Nl+Nr);                        % left depletion edge
    xr(j) = Wd(j)*Nl/(Nl+Nr);                        % right depletion edge
end
 
% positions: junction 1 at x=0, junction 2 at x=WIDTH
xc = [0 WIDTH];
x  = linspace(-2*xl(1)-0.5e-4, WIDTH + 2*xr(2)+3e-4, 4000);
Ei = zeros(size(x));
for k = 1:numel(x)
    xx = x(k);
    % default: bulk value of the region
    if xx < WIDTH/2, Ei(k) = EiEF(1)*(xx<0) + EiEF(2)*(xx>=0);
    else,            Ei(k) = EiEF(2)*(xx<=WIDTH) + EiEF(3)*(xx>WIDTH); end
    for j = 1:2
        s = xx - (xc(j) - xl(j));                    % distance from left edge
        D = EiEF(j+1) - EiEF(j);                     % total change of Ei
        if s >= 0 && s <= xl(j)
            Ei(k) = EiEF(j) + D*(xl(j)/Wd(j))*(s/xl(j))^2;
        elseif s > xl(j) && s <= Wd(j)
            Ei(k) = EiEF(j+1) - D*(xr(j)/Wd(j))*((Wd(j)-s)/xr(j))^2;
        end
    end
end
Ec = Ei + EG/2; Ev = Ei - EG/2;                      % EF = 0 reference
% sign: Ei-EF>0 means Ei above EF, i.e. p-type has Ei above EF
plot(x*1e4, Ec, 'b', x*1e4, Ei, 'k--', x*1e4, Ev, 'r', ...
     x*1e4, zeros(size(x)), 'g');
xlabel('x (\mum)'); ylabel('Energy (eV)');
legend('E_C','E_i','E_V','E_F'); title('Equilibrium BJT band diagram');

Sign check. For a p-type region $E_F$ lies below $E_i$, so $E_i-E_F>0$; the code uses $+kT\ln (N_A/n_i)$ for positive $N$, which is correct. The same code draws an npn by entering e.g. DOPING = [-1e18 1e16 -1e15].

10.4 Introductory Operational Considerations

Consider a pnp in active mode and watch the holes in and near the base (Fig. 10.8). The main activity at the forward-biased E-B junction is majority-carrier injection across the junction into the opposite-side quasineutral regions. The $p^{+}$-n nature means many more holes are injected from emitter into base than electrons from base into emitter.

The key to transistor action is what happens to carriers injected into the base. If $W$ were much larger than a diffusion length, the injected holes would recombine in the n-base and there would be no interaction between junctions: just two back-to-back diodes. But the BJT’s base is narrow compared with a minority-carrier diffusion length, so the vast majority of injected holes diffuse completely across the quasineutral base and enter the C-B depletion region, where the accelerating field sweeps them into the collector. The narrow base thus couples the E-B and C-B currents.

The names make sense: the emitter emits carriers into the base; the collector (the reverse-biased side of the C-B junction) collects them after they cross the control (base) region.

Currents (Fig. 10.9), neglecting R-G currents in the depletion regions initially:

• $I_{Ep}$: hole diffusion current across the E-B junction (holes injected into the base).
• $I_{Cp}$: hole diffusion current across the C-B junction, almost entirely holes that crossed the base. Very few are lost by recombination in a well-made transistor, so $I_{Cp}\approx I_{Ep}$.
• $I_{En}$: electron current injected from base into emitter (the minority-carrier diffusion current on the heavily doped side).
• $I_{Cn}$: electrons that are minority carriers in the collector and wander into the C-B depletion region to be swept into the base (a small reverse-bias current).

$I_E=I_{Ep}+I_{En}(10.4)$

$I_C=I_{Cp}+I_{Cn}(10.5)$

Since both electron components are small compared with the hole components and $I_{Cp}\approx I_{Ep}$, it follows that $I_C\approx I_E$, a well-known property. $I_B=I_E-I_C$ is small compared with both. Inspection of Fig. 10.9 agrees: the three components of base current are $I_{B1}=I_{En}$ (electrons injected into the emitter, resupplied through the base lead), $I_{B2}$ (electrons that replace those lost by recombination with injected holes in the base) and $I_{B3}=-I_{Cn}$ (the reverse current into the collector, flowing opposite): all small.

Amplification (Fig. 10.10). In common emitter, output current is $I_C$, input is $I_B$, and the d.c. gain is $I_C/I_B$. $I_B$ (an electron current in a pnp) and $I_C$ (predominantly hole current) are tied together by the E-B junction: raising $I_B$ proportionally raises $I_{Ep}$ and hence $I_C$. Control of large $I_C$ by small $I_B$ is possible because the coupled-junction arrangement splits the small electron current and the large hole current crossing the E-B junction into two separate current loops: the hole loop passes through emitter, base, collector; the electron loop closes through the base lead and emitter.

10.5 Performance Parameters

Parameters for amplifier performance: emitter efficiency and base transport factor (internal operation), and d.c. current gains (external operation).

Emitter efficiency

If the hole current injected across E-B increased while total emitter current stayed constant, the electron current would drop and overall gain would increase. Treating $I_E$ as emitter input and $I_{Ep}$ as the useful output:

$\gamma =\frac{I_{Ep}}{I_E}=\frac{I_{Ep}}{I_{Ep}+I_{En}}\text{(pnp)}(10.6)$

$0\le \gamma \le 1$; gain is maximized by making $\gamma$ as close to 1 as possible. In words: the fraction of emitter current that is carried by the useful (majority-of-emitter) carriers injected into the base.

Base transport factor

The fraction of the minority carriers injected into the base that successfully cross the quasineutral base and enter the collector. Number injected $\propto I_{Ep}$; number reaching the collector $\propto I_{Cp}$:

${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}\text{(pnp)}(10.7)$

$0\le {\alpha }_T\le 1$; recombination in the base degrades it. Again maximum amplification needs ${\alpha }_T\to 1$.

Common-base d.c. current gain

In common base, the active-mode output characteristics (Fig. 10.4a, $-V_{CB}>0$) follow

$I_C={\alpha }_{dc}{I}_E+I_{CBO}(10.8)$

with ${\alpha }_{dc}$ the common-base d.c. current gain and $I_{CBO}$ the collector current when $I_E=0$ (“C-B, emitter open”). Using (10.6) and (10.7):

$I_{Ep}=\gamma I_E(10.9a)$

$I_{Cp}={\alpha }_T{I}_{Ep}={\alpha }_T\gamma I_E(10.9b)$

(Eq. 10.9 in the book collects these relations.) Then

$I_C=I_{Cp}+I_{Cn}=\gamma {\alpha }_T{I}_E+I_{Cn}(10.10)$

Comparing (10.8) with (10.10):

${\alpha }_{dc}=\gamma {\alpha }_T(10.11)$

$I_{CBO}=I_{Cn}(10.12)$

(10.11) links the external gain to the internal parameters, and $0\le {\alpha }_{dc}\le 1$.

Common-emitter d.c. current gain

In common emitter, active-mode characteristics (Figs. 10.4b, 10.5b) follow

$I_C={\beta }_{dc}{I}_B+I_{CEO}(10.13)$

with $I_{CEO}$ the collector current when $I_B=0$. A second relation comes from substituting $I_E=I_C+I_B$ into (10.8):

$I_C={\alpha }_{dc}(I_C+I_B)+I_{CBO}(10.14)$

Collect $I_C$ terms:

$I_C(1-{\alpha }_{dc})={\alpha }_{dc}{I}_B+I_{CBO}$

$I_C=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}{I}_B+\frac{I_{CBO}}{1-{\alpha }_{dc}}(10.15)$

Comparing (10.13) and (10.15):

${\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}(10.16)$

$I_{CEO}=\frac{I_{CBO}}{1-{\alpha }_{dc}}=({\beta }_{dc}+1)I_{CBO}(10.17)$

and, if $I_{CEO}\ll I_C$ at the operating point,

${\beta }_{dc}\approx \frac{I_C}{I_B}(10.18)$

Since ${\alpha }_{dc}$ is typically near unity, ${\beta }_{dc}\gg 1$; this agrees with (10.18) since $I_C\gg I_B$.

Intuition for $\beta$. If ${\alpha }_{dc}=0.99$, then $\beta =0.99/0.01=99$. If ${\alpha }_{dc}$ improves from 0.99 to 0.995, $\beta$ doubles to 199. The gain depends on how close to 1 ${\alpha }_{dc}$ gets, the “loss fraction” $1-{\alpha }_{dc}$ that must leak out through the base.

10.6 Summary

This chapter covered the basics of the BJT: terminal current and voltage symbols, polarities and relations; circuit configurations and biasing modes and the general form of the output characteristics; fabrication, to relate the real structure to the one-dimensional model used later; electrostatics (the BJT is just two independent pn junctions in equilibrium and under normal operation); the carrier activity and currents in active mode that explain how the device works; and the performance parameters, namely emitter efficiency $\gamma$, base transport factor ${\alpha }_T$ and the d.c. current gains ${\alpha }_{dc}=\gamma {\alpha }_T$ and ${\beta }_{dc}={\alpha }_{dc}/(1-{\alpha }_{dc})$, interrelated through (10.11)–(10.18).

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Chapter 10 Problems

Problem information table (from the book).

Problem	After section	Difficulty	Points	Topic
10.1	10.6	–	10	Quick quiz
10.2	10.1	1	4	Mode tables
10.3	10.3	2–3	22	pnp equilibrium electrostatics
10.4	10.3	2–3	22	npn equilibrium electrostatics
10.5	10.3	2–3	10	pnp active-mode electrostatics
10.6	10.3	2	8	npn active-mode electrostatics
10.7	10.4	3	12	pnp carrier activity/currents
10.8	10.4	3	12	npn carrier activity/currents
10.9	10.5	2	11	pnp parameter calculations
10.10	10.5	2	11	npn parameter calculations
10.11	10.5	3	10	$I_{CBO}$ thought problem

Constants used: $kT/q=0.0259$ V, $n_i=10^{10}{\text{cm}}^{-3}$, ${\epsilon }_{Si}=11.8\times 8.854\times 10^{-14}=1.045\times 10^{-12}$ F/cm, $q=1.6\times 10^{-19}$ C.

Problem 10.1 Quick quiz

Problem. Answer concisely: (a) name of the circuit configuration where the input current is… and the output current is…; (b) name of the configuration where $I_C$ is the output current and $V_{E?}$ is the output voltage; (c) four biasing modes; (d) purpose of the buried layer in IC BJTs; (e) relative sizes of $N_{AE}$, $N_{DB}$, $N_{AC}$ in a standard pnp; (f) standard symbols for diode depletion width and BJT quasineutral base width; (g) precise definition of “narrow” base; (h) why the base must be narrow; (i) emitter efficiency in words; (j) base transport factor in words.

Solution. (The OCR of (a) and (b) is garbled. Assumption: (a) input current $I_B$, output $I_C$; (b) output voltage $V_{EC}$. Both then describe the common emitter; if (a) instead meant input $I_E$, output $I_C$, the answer is common base.)

(a) Input $I_B$, output $I_C$ with the emitter common: common emitter (if input $I_E$: common base).

(b) $I_C$ output, output voltage $V_{EC}$ (collector-to-emitter): common emitter. (An output voltage $V_{CB}$ would signal common base.)

(c) Saturation, active (forward active), inverted, cutoff.

(d) The $n^{+}$ buried layer provides a low-resistance path between the active collector region and the top-side collector contact.

(e) $N_{AE}\ge N_{DB}>N_{AC}$ (emitter most heavily doped, collector least).

(f) $W$ for the pn diode depletion width; $W$ also denotes the BJT quasineutral base width (with $W_B$ the total metallurgical base width, $W=W_B-x_{nEB}-x_{nCB}$ for a pnp).

(g) “Narrow” means the quasineutral base width is much smaller than the minority-carrier diffusion length in the base: $W\ll L_{min}$ ($L_P$ for a pnp, $L_N$ for an npn).

(h) A narrow base lets nearly all carriers injected from the emitter diffuse across and be swept into the collector instead of recombining; this couples the two junctions (transistor action) and keeps $I_B$ small and the gain large.

(i) Emitter efficiency: the fraction of the total emitter current that is carried by the carriers injected into the base (the useful part).

(j) Base transport factor: the fraction of the minority carriers injected into the base that reach the collector without recombining.

Problem 10.2 Mode tables

Problem. Give the polarities (+ or −) of the input and output junction voltages in each of the four biasing modes: (a) pnp with $V_{EB}$, $V_{CB}$; (b) npn with $V_{BE}$, $V_{BC}$.

Solution. A junction is forward biased when its p-side is more positive than its n-side.

pnp. E-B: emitter (p) minus base (n) is $V_{EB}$; forward means $V_{EB}>0$. C-B: collector (p) minus base (n) is $V_{CB}$; forward means $V_{CB}>0$, reverse means $V_{CB}<0$.

Mode	$V_{EB}$	$V_{CB}$
Active	+	−
Inverted	−	+
Saturation	+	+
Cutoff	−	−

npn. E-B: base (p) minus emitter (n) is $V_{BE}$; forward means $V_{BE}>0$. C-B: base (p) minus collector (n) is $V_{BC}$; forward means $V_{BC}>0$.

Mode	$V_{BE}$	$V_{BC}$
Active	+	−
Inverted	−	+
Saturation	+	+
Cutoff	−	−

Problem 10.3 Equilibrium electrostatics of a pnp

Problem. A Si pnp BJT with $N_{AE}=5\times 10^{17}$, $N_{DB}=10^{15}$ (OCR reads "$10^5$"; taken as $10^{15}$), $N_{AC}=10^{14}{\text{cm}}^{-3}$, $W_B=3\mu$m at room temperature in equilibrium. (a) Sketch the band diagram with the Fermi level in the three regions. (b) Sketch (i) electrostatic potential with $V=0$ in the emitter region, (ii) the electric field, (iii) the charge density versus position. (c) Calculate the net potential difference between collector and emitter. (d) Find the quasineutral base width. (e) Find the maximum electric fields in the E-B and C-B depletion regions.

Solution.

Fermi level positions. For p-type $E_i-E_F=kT\ln (N_A/n_i)$; for n-type $E_F-E_i=kT\ln (N_D/n_i)$.

$E_i-E_F{|}_E=0.0259\ln \frac{5\times 10^{17}}{10^{10}}=0.0259\ln (5\times 10^7)=0.0259(17.73)=0.459\text{eV}$

$E_F-E_i{|}_B=0.0259\ln \frac{10^{15}}{10^{10}}=0.0259(11.51)=0.298\text{eV}$

$E_i-E_F{|}_C=0.0259\ln \frac{10^{14}}{10^{10}}=0.0259(9.21)=0.239\text{eV}$

(a) Band diagram: a single flat $E_F$. In the emitter $E_F$ is $0.459$ eV below $E_i$ (so $E_F$ is $\approx 0.10$ eV above $E_V$: $E_i-E_V\approx 0.56$, and $0.56-0.459=0.10$); in the base $E_F$ is $0.298$ eV above $E_i$ ($0.26$ eV below $E_C$); in the collector $E_F$ is $0.239$ eV below $E_i$ ($0.32$ eV above $E_V$). The bands bend: $E_C$, $E_i$, $E_V$ in the base lie lower than in the emitter and collector… in electron energy terms the bands in the base lie lower in electron energy than in the emitter (a valley for electrons, a hill for holes): going emitter to base, $E_i$ falls by $0.757$ eV; going base to collector it rises by $0.537$ eV, a net fall of $0.221$ eV from emitter to collector. The depletion regions are narrow at E-B (mostly on the base side) and wide at C-B (mostly on the collector side).

(c) Built-in potentials:

$V_{bi,EB}=0.459+0.298=0.757\text{V}{V}_{bi,CB}=0.298+0.239=0.537\text{V}$

The potential is $V=-E_i/q$ (relative). Going emitter to collector the potential first rises by $V_{bi,EB}$ into the base and then falls by $V_{bi,CB}$ into the collector. Net:

$V_C-V_E=V_{bi,EB}-V_{bi,CB}=0.757-0.537=0.221\text{V}$

(The collector is $0.221$ V higher than the emitter in electrostatic potential; equivalently ${\varphi }_E-{\varphi }_C$ computed from the doping: $-0.459+0.239$ in terms of $E_i-E_F$ gives the same magnitude.)

(b) With $V=0$ in the emitter bulk: $V=0$ for $x$ in emitter bulk, rising in an S-shaped (parabolic) step of height $0.757$ V across the E-B depletion region (almost all in the base side, width about $1\mu$m) to $+0.757$ V in the quasineutral base, then falling in a step of $0.537$ V across the C-B depletion region (mostly in the collector, width about $2.8\mu$m) to $+0.221$ V in the collector bulk.

Field $\mathcal{E}=-dV/dx$: zero in neutral regions; in the E-B depletion region a negative (pointing from n-base toward p-emitter… pointing from base to emitter) triangular profile, peak at the metallurgical junction; same direction triangle at the C-B junction but wider and of smaller peak (see (e)). Both triangles have the same sign since both junctions are pn junctions oriented with n-side in the middle.

Charge density $\rho$ (depletion approximation): E-B: $-q{N}_{AE}$ over $x_{pEB}$ (very thin, in the emitter), $+q{N}_{DB}$ over $x_{nEB}$ (in the base); C-B: $+q{N}_{DB}$ over $x_{nCB}$ (in the base), $-q{N}_{AC}$ over $x_{pCB}$ (long, in the collector); zero elsewhere.

(d) E-B depletion:

$W_{EB}=\sqrt{\frac{2\epsilon V_{bi}}{q}\left(\frac{1}{N_{AE}}+\frac{1}{N_{DB}}\right)}$

$\frac{2\epsilon V_{bi}}{q}=\frac{2(1.045\times 10^{-12})(0.757)}{1.6\times 10^{-19}}=9.89\times 10^6{\text{cm}}^{-1}{\text{ (units V cm/C... in cm}}^{-3}\cdot {\text{cm}}^2)$

$\frac{1}{N_{AE}}+\frac{1}{N_{DB}}=2\times 10^{-18}+10^{-15}=1.002\times 10^{-15}{\text{cm}}^3$

$W_{EB}=\sqrt{(9.89\times 10^6)(1.002\times 10^{-15})}=\sqrt{9.91\times 10^{-9}}=9.96\times 10^{-5}\text{cm}=0.996\mu \text{m}$

$x_{nEB}=W_{EB}\frac{N_{AE}}{N_{AE}+N_{DB}}=0.996\left(\frac{5\times 10^{17}}{5.01\times 10^{17}}\right)=0.994\mu \text{m}$

C-B depletion:

$\frac{2\epsilon V_{bi}}{q}=\frac{2(1.045\times 10^{-12})(0.537)}{1.6\times 10^{-19}}=7.01\times 10^6$

$\frac{1}{N_{AC}}+\frac{1}{N_{DB}}=10^{-14}+10^{-15}=1.1\times 10^{-14}$

$W_{CB}=\sqrt{(7.01\times 10^6)(1.1\times 10^{-14})}=\sqrt{7.71\times 10^{-8}}=2.78\times 10^{-4}\text{cm}=2.78\mu \text{m}$

$x_{nCB}=W_{CB}\frac{N_{AC}}{N_{AC}+N_{DB}}=2.78\left(\frac{10^{14}}{1.1\times 10^{15}}\right)=0.252\mu \text{m}$

$x_{pCB}=2.78-0.252=2.525\mu \text{m}$

Quasineutral base width, by (10.3):

$W=W_B-x_{nEB}-x_{nCB}=3-0.994-0.252=1.75\mu \text{m}$

(e) Peak fields $|\mathcal{E}{|}_{max}=q{N}_D{x}_n/\epsilon$ on the n-side:

$|\mathcal{E}{|}_{max,EB}=\frac{(1.6\times 10^{-19})(10^{15})(0.994\times 10^{-4})}{1.045\times 10^{-12}}=1.52\times 10^4\text{V/cm}$

$|\mathcal{E}{|}_{max,CB}=\frac{(1.6\times 10^{-19})(10^{15})(0.252\times 10^{-4})}{1.045\times 10^{-12}}=3.86\times 10^3\text{V/cm}$

Check with $2V_{bi}/W$: $2(0.757)/(0.996\times 10^{-4})=1.52\times 10^4$ and $2(0.537)/(2.78\times 10^{-4})=3.86\times 10^3$ V/cm.

Problem 10.4 Equilibrium electrostatics of an npn

Problem. A Si npn with $N_{DE}=10^{18}$, $N_{AB}=10^{16}$, $N_{DC}=10^{1?}{\text{cm}}^{-3}$ (OCR shows "$10^1$"; taken as $10^{14}$ by analogy with 10.3), $W_B=2\mu$m in equilibrium at room temperature. Parts (a)–(e) as in 10.3, with $V=0$ in the emitter.

Solution.

Fermi levels.

$E_F-E_i{|}_E=0.0259\ln (10^8)=0.0259(18.42)=0.477\text{eV}$

$E_i-E_F{|}_B=0.0259\ln (10^6)=0.0259(13.82)=0.358\text{eV}$

$E_F-E_i{|}_C=0.0259\ln (10^4)=0.0259(9.21)=0.239\text{eV}$

(a) Flat $E_F$. Emitter: $E_F$ $0.477$ eV above $E_i$ (about $0.08$ eV below $E_C$). Base: $E_F$ $0.358$ eV below $E_i$ ($0.20$ eV above $E_V$). Collector: $E_F$ $0.239$ eV above $E_i$. The base is a potential hill for electrons: $E_C$ in the base is higher than in the emitter by $0.835$ eV and higher than in the collector by $0.597$ eV; the E-B depletion region is narrow (mostly in the base), C-B wide (mostly in the collector). The diagram is the pnp diagram of 10.3 flipped upside down.

(c) Built-in potentials:

$V_{bi,EB}=0.477+0.358=0.835\text{V}{V}_{bi,CB}=0.358+0.239=0.597\text{V}$

$V=-E_i/q$ with $V=0$ in the emitter: the base potential is $0.835$ V lower (electrons see a hill), so $V_B=-0.835$ V, and the collector rises by $0.597$ V from the base:

$V_C-V_E=-V_{bi,EB}+V_{bi,CB}=-0.835+0.597=-0.238\text{V}$

(b) Potential profile: $0$ in emitter, a parabolic step down to $-0.835$ V across the thin E-B depletion region, flat at $-0.835$ V in the base, a step up of $0.597$ V across the wide C-B depletion region to $-0.238$ V in the collector. Field: triangular, pointing (from n to p side) in the sense opposite to the pnp case, i.e. the field points from the n-side toward the p-side, namely from emitter toward base at E-B and from collector toward base at C-B; peaks at the junctions. Charge: E-B: $+q{N}_{DE}$ in a very thin layer ($x_{nEB}$) in the emitter, $-q{N}_{AB}$ over $x_{pEB}$ in the base; C-B: $-q{N}_{AB}$ over $x_{pCB}$ in the base (very thin), $+q{N}_{DC}$ over $x_{nCB}$ (wide) in the collector.

(d) E-B junction:

$\frac{2\epsilon V_{bi}}{q}=\frac{2(1.045\times 10^{-12})(0.835)}{1.6\times 10^{-19}}=1.091\times 10^7$

$\frac{1}{N_{DE}}+\frac{1}{N_{AB}}=10^{-18}+10^{-16}=1.01\times 10^{-16}$

$W_{EB}=\sqrt{(1.091\times 10^7)(1.01\times 10^{-16})}=\sqrt{1.10\times 10^{-9}}=3.32\times 10^{-5}\text{cm}=0.332\mu \text{m}$

$x_{pEB}=W_{EB}\frac{N_{DE}}{N_{DE}+N_{AB}}=0.332\left(\frac{10^{18}}{1.01\times 10^{18}}\right)=0.329\mu \text{m}$

C-B junction:

$\frac{2\epsilon V_{bi}}{q}=\frac{2(1.045\times 10^{-12})(0.597)}{1.6\times 10^{-19}}=7.80\times 10^6$

$\frac{1}{N_{AB}}+\frac{1}{N_{DC}}=10^{-16}+10^{-14}=1.01\times 10^{-14}$

$W_{CB}=\sqrt{(7.80\times 10^6)(1.01\times 10^{-14})}=\sqrt{7.88\times 10^{-8}}=2.81\times 10^{-4}\text{cm}=2.81\mu \text{m}$

$x_{pCB}=W_{CB}\frac{N_{DC}}{N_{DC}+N_{AB}}=2.81\left(\frac{10^{14}}{1.01\times 10^{16}}\right)=0.0278\mu \text{m}$

$W=W_B-x_{pEB}-x_{pCB}=2-0.329-0.028=1.64\mu \text{m}$

(e) Peak fields:

$|\mathcal{E}{|}_{max,EB}=\frac{q{N}_{AB}{x}_{pEB}}{\epsilon }=\frac{(1.6\times 10^{-19})(10^{16})(0.329\times 10^{-4})}{1.045\times 10^{-12}}=5.04\times 10^4\text{V/cm}$

$|\mathcal{E}{|}_{max,CB}=\frac{(1.6\times 10^{-19})(10^{16})(0.0278\times 10^{-4})}{1.045\times 10^{-12}}=4.26\times 10^3\text{V/cm}$

Checks: $2(0.835)/(3.32\times 10^{-5})=5.03\times 10^4$; $2(0.597)/(2.81\times 10^{-4})=4.25\times 10^3$ V/cm.

Problem 10.5 Active-mode electrostatics of the 10.3 BJT

Problem. Biases $V_{EB}=0.5$ V and $V_{CB}=2$ V are applied to the Problem 10.3 BJT. (a) Sketch the band diagram with $E_F$ in the three regions. (b) Superimpose on the 10.3 sketches the potential, field and charge density.

Assumption. The pnp is meant to be in active mode, so the C-B junction is reverse biased by $2$ V (the sign ”−” of $V_{CB}=-2$ V was lost in the OCR).

Solution.

(a) Fermi level separation: the base-side $E_F$ is raised relative to the emitter by $q{V}_{EB}=0.5$ eV, and the collector $E_F$ is raised relative to the base by $q|V_{CB}|=2$ eV, so the collector $E_F$ lies $2.5$ eV above the emitter $E_F$. The E-B barrier falls from $0.757$ to $0.257$ eV, the C-B barrier rises from $0.537$ to $2.537$ eV.

(b) New depletion widths. E-B (forward): $V_{bi}-V=0.757-0.5=0.257$ V:

$W_{EB}=0.996\sqrt{\frac{0.257}{0.757}}=0.996(0.583)=0.580\mu \text{m}$

$x_{nEB}=0.580(0.998)=0.579\mu \text{m}$

C-B (reverse): $V_{bi}+|V|=0.537+2=2.537$ V:

$W_{CB}=2.78\sqrt{\frac{2.537}{0.537}}=2.78(2.174)=6.04\mu \text{m}$

$x_{nCB}=\frac{6.04}{11}=0.549\mu \text{m},x_{pCB}=5.49\mu \text{m}$

Quasineutral base: $W=3-0.579-0.549=1.87\mu$m (up from $1.75\mu$m; forward bias shrinks the E-B region more than reverse bias grows the C-B one on the base side).

Peak fields:

$|\mathcal{E}{|}_{max,EB}=\frac{2(0.257)}{0.580\times 10^{-4}}=8.9\times 10^3\text{V/cm}(\text{down from }1.52\times 10^4)$

$|\mathcal{E}{|}_{max,CB}=\frac{2(2.537)}{6.04\times 10^{-4}}=8.4\times 10^3\text{V/cm}(\text{up from }3.86\times 10^3)$

Sketches: potential steps become $0.257$ V (E-B) and $2.537$ V (C-B). With emitter bulk at $V=0$, the base bulk is at $+0.257$ V and the collector bulk at $0.257-2.537=-2.28$ V, … check with the biases: the emitter is $0.5$ V above the base and the base $2$ V above the collector (equilibrium offsets are $+0.757$ and $-0.537$ V, so the base sits at $0.757-0.5=0.257$ V and the collector at $0.221-2.5=-2.28$ V), i.e. the collector is $2.5$ V below the emitter and its electron energy is $2.5$ eV higher. Field triangles: E-B narrower and smaller; C-B wider and larger. Charge density rectangles: same heights as in 10.3 ($qN$ values) but E-B width reduced and C-B width increased.

Problem 10.6 Active-mode electrostatics of an npn

Problem. For a typically doped Si npn, sketch the band diagram, potential, field and charge density versus position in active mode.

Solution. (Typical doping: $N_{DE}\gg N_{AB}>N_{DC}$, e.g. $10^{19},10^{17},10^{15}{\text{cm}}^{-3}$.) Active mode: $V_{BE}>0$ (forward), $V_{BC}<0$ (reverse).

• Band diagram: flat Fermi level in each region, with separations $q{V}_{BE}$ (emitter $E_F$ lower… the emitter’s electron energy is higher than the base’s by $q{V}_{BE}$, so $E_F$ in the emitter is $q{V}_{BE}$ above the base $E_F$? For an npn forward-biased E-B junction, the emitter is at lower potential than the base, so emitter electrons have energy raised by $q{V}_{BE}$ relative to the base): $E_{F,E}=E_{F,B}+q{V}_{BE}$, and the collector at higher potential (lower electron energy): $E_{F,C}=E_{F,B}-q|V_{BC}|$. The emitter-base hill is reduced to $q(V_{bi,EB}-V_{BE})$ (electrons spill into the base); the base-collector hill increases to $q(V_{bi,CB}+|V_{BC}|)$, a steep downhill slope for electrons into the collector.
• Potential: a small rise (reduced barrier) at E-B, then a large step at C-B.
• Field: a small triangle at E-B (narrow, mostly in the base), a large wide triangle at C-B (mostly in the collector).
• Charge density: E-B: thin $+q{N}_{DE}$ slab in the emitter, $-q{N}_{AB}$ slab of reduced width in the base; C-B: $-q{N}_{AB}$ slab in the base (thin) and a wide $+q{N}_{DC}$ slab in the collector. The quasineutral base width $W=W_B-x_{pEB}-x_{pCB}$ is reduced by the C-B reverse bias (base-width modulation).

Problem 10.7 Saturation-biased pnp, $I_C>0$

Problem. A pnp is biased in saturation with $I_C>0$. Construct figures like Figs. 10.8 and 10.9 showing carrier activity and diffusion currents.

Solution (verbal description of the figures). Both junctions are forward biased; both barriers are lowered.

Carrier activity (like Fig. 10.8):

• E-B: holes injected from the $p^{+}$ emitter into the n-base (large), and electrons from the base into the emitter (small).
• C-B: now forward biased, so holes are injected from the p-collector into the base (a hole flow in the opposite direction to the emitter’s flow), and electrons are injected from the base into the collector.
• Base: holes from both ends in a narrow region; some recombine with base electrons, which are supplied by the base lead.
• $I_C>0$ (standard direction, hole current leaving the collector) requires the emitter-injected hole current reaching the collector to exceed the collector-injected hole current going the other way.

Diffusion currents (like Fig. 10.9):

$I_E=I_{Ep}+I_{En},I_C=I_{Cp}^{net}+I_{Cn}^{\prime },$

where $I_{Cp}^{net}=I_{Cp}^{(E\to C)}-I_{Cp}^{(C\to E)}$ is the difference between the hole current arriving from the emitter and the hole current injected by the collector, and $I_{Cn}^{\prime }$ is now an electron current flowing from the base into the collector (the opposite sense to the active-mode reverse current, since the C-B junction is forward). The base current has three parts: electrons injected into the emitter, electrons injected into the collector, and electrons replacing recombination. All are larger than in active mode, so $I_B$ is larger and $I_C/I_B$ ($\beta$) is reduced; the hole concentration profile in the base is almost flat (high on both sides) instead of dropping linearly to zero at the collector edge, and the collector’s current is small relative to $I_E$ (a low voltage $V_{EC}$ is the saturation hallmark).

Problem 10.8 Cutoff-biased npn

Problem. An npn is biased into cutoff. Construct figures like Figs. 10.8 and 10.9 showing carrier activity and diffusion currents.

Solution (verbal description of the figures). Both junctions are reverse biased: both barriers are raised (larger than equilibrium), there is no injection.

• Carrier activity: no majority-carrier injection across either junction. The only carriers moving are thermally generated minority carriers near the depletion edges: electrons that are minority carriers in the p-base and drift across the (reverse-biased) E-B and C-B depletion regions into the emitter and collector, and holes that are minority carriers in the emitter and in the collector and are swept into the base. Depletion regions are wider than at equilibrium.
• Diffusion currents: the E-B current is a small reverse-saturation current (electrons from base into emitter, holes from emitter into base); the C-B current is also a small reverse-saturation current (electrons from base into collector, holes from collector into base). All terminal currents are tiny (including generation current in the depletion regions if included), and the minority-carrier concentrations in the base are depleted to near zero on both sides of the base:

$I_E\approx -(I_{En,sat}+I_{Ep,sat}),I_C\approx -(I_{Cn,sat}+I_{Cp,sat}),I_B=I_E-I_C\text{(a small sum)}$

(signs correspond to currents opposite to forward direction). This is the off state: low current and high voltage.

Problem 10.9 Parameter calculations for a pnp

Problem. A pnp BJT has $I_{Ep}=1$ mA, $I_{En}=0.01$ mA, $I_{Cp}=0.98$ mA, $I_{Cn}=0.1\mu$A. Calculate (a) ${\alpha }_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) ${\alpha }_{dc}$ and ${\beta }_{dc}$, (e) $I_{CBO}$ and $I_{CEO}$. (f) If $I_{Cp}$ is increased toward $I_{Ep}$ with all other components fixed, effect on ${\beta }_{dc}$? (g) If $I_{En}$ is increased with all other components fixed, effect on ${\beta }_{dc}$?

Solution.

(a) By (10.7):

${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=\frac{0.98}{1}=0.98$

(b) By (10.6):

$\gamma =\frac{I_{Ep}}{I_{Ep}+I_{En}}=\frac{1}{1+0.01}=0.990$

(c) By (10.4), (10.5), (10.1):

$I_E=I_{Ep}+I_{En}=1+0.01=1.01\text{mA}$

$I_C=I_{Cp}+I_{Cn}=0.98+0.0001=0.9801\text{mA}$

$I_B=I_E-I_C=1.01-0.9801=0.0299\text{mA}=29.9\mu \text{A}$

(d) By (10.11) and (10.16):

${\alpha }_{dc}=\gamma {\alpha }_T=(0.990099)(0.98)=0.9703$

${\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}=\frac{0.9703}{0.0297}=32.7$

(Check: $I_C/I_B=0.9801/0.0299=32.8$, consistent because $I_{CEO}$ is small.)

(e) By (10.12) and (10.17):

$I_{CBO}=I_{Cn}=0.1\mu \text{A}$

$I_{CEO}=\frac{I_{CBO}}{1-{\alpha }_{dc}}=\frac{0.1\mu \text{A}}{0.0297}=3.37\mu \text{A}(=({\beta }_{dc}+1)I_{CBO})$

Check of (10.13): $I_C=\beta I_B+I_{CEO}=(32.67)(0.0299)+0.00337=0.9767+0.0034=0.9801$ mA. ✓.

(f) Raising $I_{Cp}$ raises ${\alpha }_T=I_{Cp}/I_{Ep}$ (fewer injected holes lost to recombination in the base); $\gamma$ is unchanged, so ${\alpha }_{dc}=\gamma {\alpha }_T$ rises, and since ${\beta }_{dc}={\alpha }_{dc}/(1-{\alpha }_{dc})$ is an increasing function of ${\alpha }_{dc}$, ${\beta }_{dc}$ increases. Physically $I_B=I_E-I_C$ falls while $I_C$ rises. (The limit is $I_{Cp}=I_{Ep}$, i.e. ${\alpha }_T=1$, giving ${\alpha }_{dc}=\gamma =0.990$, ${\beta }_{dc}=99.0$.)

(g) Raising $I_{En}$ lowers $\gamma =I_{Ep}/(I_{Ep}+I_{En})$ with ${\alpha }_T$ fixed, so ${\alpha }_{dc}$ falls and ${\beta }_{dc}$ decreases. Physically, more of the emitter current is wasted on carriers that do not go to the collector, and $I_B$ rises.

Problem 10.10 Parameter calculations for an npn

Problem. An npn has $I_{En}=100\mu$A, $I_{Ep}=1\mu$A, $I_{Cn}=99\mu$A, $I_{Cp}=0.1\mu$A. Calculate (a) ${\alpha }_T$, (b) $\gamma$, (c) $I_E$, $I_C$, $I_B$, (d) ${\alpha }_{dc}$, ${\beta }_{dc}$, (e) $I_{CBO}$, $I_{CEO}$. (f) $I_{Cn}$ is increased toward $100\mu$A with all else fixed: effect on ${\beta }_{dc}$? (g) (The OCR is garbled; the item is taken to be: $I_{Ep}$ is increased with all else fixed: effect on ${\beta }_{dc}$?)

For an npn the roles of electrons and holes swap: the useful (injected into base) carriers are electrons, so the definitions are $\gamma =I_{En}/I_E$, ${\alpha }_T=I_{Cn}/I_{En}$, $I_{CBO}=I_{Cp}$.

Solution.

(a)

${\alpha }_T=\frac{I_{Cn}}{I_{En}}=\frac{99}{100}=0.99$

(b)

$\gamma =\frac{I_{En}}{I_{En}+I_{Ep}}=\frac{100}{101}=0.9901$

(c)

$I_E=I_{En}+I_{Ep}=101\mu \text{A}$

$I_C=I_{Cn}+I_{Cp}=99+0.1=99.1\mu \text{A}$

$I_B=I_E-I_C=101-99.1=1.9\mu \text{A}$

(d)

${\alpha }_{dc}=\gamma {\alpha }_T=(0.9901)(0.99)=0.9802$

${\beta }_{dc}=\frac{0.9802}{1-0.9802}=\frac{0.9802}{0.01980}=49.5$

(e)

$I_{CBO}=I_{Cp}=0.1\mu \text{A}$

$I_{CEO}=\frac{I_{CBO}}{1-{\alpha }_{dc}}=\frac{0.1}{0.01980}=5.05\mu \text{A}$

Check: ${\alpha }_{dc}{I}_E+I_{CBO}=(0.9802)(101)+0.1=99.0+0.1=99.1\mu$A ✓.

(f) With $I_{Cn}\to 100\mu$A (equal to $I_{En}$): ${\alpha }_T\to 1$, ${\alpha }_{dc}\to \gamma =0.9901$, and

${\beta }_{dc}\to \frac{0.9901}{0.0099}=100$

so ${\beta }_{dc}$ increases (from $49.5$ to about $100$): fewer injected electrons are lost in the base, so $I_B$ falls.

(g) Increasing $I_{Ep}$ (assumed) lowers $\gamma =I_{En}/(I_{En}+I_{Ep})$, so ${\alpha }_{dc}$ and ${\beta }_{dc}$ decrease: more of the emitter current is carried by holes that never reach the collector and must be supplied through the base lead.

Problem 10.11 Thought problem on $I_{CBO}$

Problem. Explain why $I_{CBO}$ of Eq. (10.12) has no hole component. Why aren’t there contributions associated with minority carrier holes (in the n-base of a pnp) that wander into the C-B depletion region and are swept into the collector?

Solution. $I_{CBO}$ is the collector current with the emitter open, $I_E=0$.

1. In a pnp the holes in the narrow n-base are minority carriers. A hole current in the quasineutral base from emitter to collector would be continuous across the base (the base is narrow, so recombination is negligible): $I_{Cp}\approx I_{Ep}$.
2. With the emitter open, $I_E=I_{Ep}+I_{En}=0$. The E-B junction floats to whatever small bias makes the net emitter current vanish; the emitter’s electron component $I_{En}$ is tiny (the emitter is heavily doped), so $I_{Ep}\approx 0$.
3. Therefore $I_{Cp}=I_{Ep}\approx 0$: any hole leaving the base through the collector would have to be supplied through the emitter, and the open emitter cannot supply it. The hole population in the base simply relaxes to a distribution in which the net hole flow vanishes.
4. The only holes that could contribute independently are those thermally generated inside the narrow base itself. Since the base width $W$ is much less than a diffusion length, the volume (and generation) is tiny compared with the collector region within a diffusion length of the C-B depletion edge, so this contribution is negligible; and by (3) it must be balanced anyway.
5. In contrast, the electrons that are minority carriers in the p-type collector are generated throughout a volume extending about one electron diffusion length $L_N$ into the collector. Those that wander into the C-B depletion region are swept into the base and out the base lead. No emitter is needed to supply them: they are thermally generated in the collector, so this current exists with $I_E=0$.

Hence $I_{CBO}=I_{Cn}$ and carries no hole component.