ELEN90091 - Semiconductor Devices

Semiconductor basics

Semiconductors are materials whose conductivity can vary over a wide range. Sit between insulators and conductors. Characterised by the bandgap energy.

Semiconductors are dictated by the arrangement of electrons in an atom. There are elemental semiconductors (Si, Ge) and compound semiconductors (GaAs, MCT-Hg_xCd_1-xTe). Clustered around group 4 of the periodic table. Semiconductors can be crystalline, polycrystalline or amorphous, and can vary properties depending on the structure. Different crystalline materials can form different crystal configurations.

Atoms

Atoms are small $\sim 10^{-10}\text{m}(1\text{\AA})$. Composed of positive protons and negative electrons. Due to small size, classical modelling no longer works. Have a complex model, with a densely packed nucleus (protons and neutrons) occupying a small amount of space (1/100000) in the atom but containing most of the mass (>99%). Electrons occupy probabilistic clouds around the nucleus, known as an orbital. The cloud takes on discrete arrangements/energies due to the electron's quantum nature. Only two electrons with different spins can occupy an orbital (Pauli exclusion principle). The orbitals are limited to:

s = 2 electrons
p = 6 electrons
d = 10 electrons

Electrons are filled from the lowest to the highest orbital according to the Aufbau principle (many exceptions). The atom's behaviour is mostly dictated by its valence electrons. Electrons can jump between orbitals if they have enough energy.

Electrons are described by a set of 4 quantum numbers, and no two electrons can share the same set of numbers.

Principle Quantum number (n) The shell number or energy level the electron sits in
Orbital (Azimuthal) number (l) The orbital type the electron is in, s=0, p=1, …
Magnetic quantum number (m or $m_l$) Distinguishes the orbitals within a subshell, like in the 3 p subshells
Spin (s or $m_s$) Describes the spin state of the electron, fermions (like electrons) have +1/2 or -1/2

Primary bonding (covalent, ionic, metallic) is an interaction of the valence electrons, and the core electrons do not participate. Reconfiguration of the electron distribution leads to a lower potential energy, forming a bond releases energy and breaking a bond absorbs energy. Ionic, metallic and covalent are the three extremes of bonding. With few exceptions, atoms bond to satisfy the octet rule, to gain 8 electrons in their valence shell. Electronegativity and its difference determine the ionic, metallic and covalent character of a bond. Electronegativity is a measure of the ability of an atom to attract shared electrons. Electronegativity is approximately inversely proportional to the valence electron's orbital energy. Ionic bonds involves transferring an electron from one atom to another. Covalent bonds have equal sharing of an electron between two atoms. Metallic bonding is the delocalisation and sharing of electrons across the crystal structure.

Band gaps

Bringing two atoms close together reduces the potential barrier between the atoms. The Pauli exclusion principle causes a split in the energy level of the valence electrons. As more atoms are added, the splitting causes more levels, approaching a continuity of energy levels in a band. Silicon is more complex, as the valence orbitals hybridise as we bring them closer together. Moving the atoms closer together causes the s and p orbital energies to overlap and $sp^3$ hybridise, which also the splitting of energy levels. At the distance where silicon atoms sit in crystals, there is a gap in energy where no electrons are, the band gap. The splitting results in a completely filled inner band and a completely empty outer band. If the atoms got closer, the inner shells would also split, although this would require extra energy. The hybridisation causes a new orbital shape, forming a tetrahedron.

By convention, the lower band is the valence band, and the upper band is the conduction band. The valence band is completely filled and the conduction band is completely empty. The difference in energy between the top of the valence ($E_v$) and bottom of the conduction ($E_c$) bands is the band gap ($E_g$). No electrons can have energy in the band gap space. Insulators have a large band gap, semiconductors have a moderate band gap and metals have partially filled bands that allow conduction through them. As temperature changes, the distance between atoms can change, altering the band gap energy.

The band gap depends on the periodic structure of the crystal. Hence, waves propagating in difference directions in a crystal can experience different band features. The band gap depends on the value of the propagation constant ($k$) which is implicitly linked to momentum through: $$p=mv=\hbar k$$ Direct semiconductors are those in which the valence band maximum and conduction band minimum are aligned in k-space, at $k=0$. Indirect semiconductors have a misalignment in the maxima and minuma. This has important consequences for light absorption and emission. Varying the composition can transition the material from a direct to indirect band gap.

Electrons and holes

An intrinsic semiconductor is one with no impurities, such as a silicon lattice without any other elements. At 0K the electrons completely fill the valence band. At room temperature, the average kinetic energy of the electrons is $kT=0.026meV$. However some electrons will have much more energy than this, and will be capable of transitioning to the conduction band, resulting in the generation of the electron hole pair (EHP).

Electrons are negatively charged and their concentration is normally labelled as “n”. Holes are the absence of electrons, and can be thought as a separate particle. Its concentration is labelled as “p”. In an intrinsic semiconductor, all electrons in CB and holes in VB are from thermal generation and have equal concentrations. $n_i$ is known is the intrinsic carrier concentration. $$n=p=n_i$$ Under constant conditions, the thermal generation rate $g_i$ will be balanced by an equal and opposite recombination rate $r_i$, resulting in $$g_i=r_i$$

Doping

Extrinsic semiconductors are those in which an impurity atom (dopant) is incorporated into the crystal lattice and is the most common way to modulate the semiconductor's conductivity. Trivalent elements (B, Ga, Al) have three valence electrons and when substituted for a silicon atom introduce extra holes into the VB, one hole for each atom. These are known as p-type regions. Pentavalent elements (P, As, Sb) have five valence electrons and introduce an extra electron into the CB, one electron for each atom. These are known as n-type regions. Interstitial doping can also be done with smaller atoms (e.g. Li), and sit in the crystal voids.

n-type doping introduces a localised electronic state $E_d$ close to the CB, and above 0K an electron gets donated to the CB. As $E_d$ is close to $E_c$, the process is efficient and has a high ionisation probability. The remaining atom (donor) is positively charged, but the crystal remains neutral. Donor concentrations vary up to $\sim5*10^{20}cm^3$, limited by the solid solubility limits, making the dopant about 1% of the lattice. The electron concentration is now $n=n_i+N_d$. The same happens for p-type doping, with a level near the valence band and the ion being an acceptor. The hole concentration becomes $p=n_i+N_a$.

A doped semiconductor has the carriers take on special names. The carrier associated with the dopant is the majority carrier and the other is the minority carrier. Compensated doping is where a region contains both kinds of dopant (e.g. npn junction). The resulting concentration depends on the difference between the types present. $$n_0\approx N_d-N_a$$

Energy, momentum and mass

The E-k diagram for a free electron forms a parabola, which is proportional to the mass. $$\frac{d^2E}{dk^2}=\frac{\hbar^2}{m}$$ We can use the band gap E-k diagram to approximate mass of an electron as its “effective mass” ($m^*$). The effective mass will often be less than the free electron mass $m_0$ and are often written as a fraction of $m_0$. This can also be applied to the mass of holes. As effective mass is an approximation, there are different effective masses used for different calculations.

Fermi-Dirac statistics

Fermi-Dirac statistics model fermions (like electrons) and their behaviour. The distribution function is: $$f(E)=\frac{1}{1+e^{(E-E_F)/kT}}$$ Where $k=8.62*10^{-5}eV/K=1.38*10^{-23}J/K$. $E_f$ is the Fermi energy, defined as the point where there is a 50% chance of occupancy. The distribution is symmetrical about $E_f$, and spreads out as temperature increases. $E_f$ is roughly in the middle of the band gap. There is a small probability of electrons in the CB and approximately the same probability of holes in the VB. For most situations, Maxwell-Boltzmann statistics are a good enough approximation, and are used instead due to ease of use.

Adding dopants to the material shifts $E_f$, with n-type shifting upwards (towards CB) and p-type downwards (towards VB). This increases the number of electrons or holes being in the other band. The difference $E_c-E_f$ can be used to find the number of electrons in the VB, and $E_f-E_v$ can be used for holes in the VB.

The CB and VB are quasi-continuous in states, so the density of states can be calculated which can be used with the probability the state will be occupied.

The electron concentration can be found from the product of the Fermi-Dirac distribution and the density of states, from the CB upwards. $$n_0=\int_{E_C}^\infty f(E)N(E)dE$$ The density of states doesn't vary with doping, unlike the Fermi-Dirac distribution, which leads to a change in the carrier concentration.

Using the Boltzmann approximation, we can approximate all the states at the CB as a single density of states at the band edge $E_c$: $$N_c=2\left(\frac{2\pi m^*_nkT}{h^2}\right)^\frac{3}{2}$$ This gives the electron concentration as: $$n_0=N_cf(E_c)$$ Assuming that $E_f$ is not too close to $E_c$ (more than a few $kT=0.026V$), we can simplify the Fermi-Dirac to the Boltzmann: $$f(E_c)=\frac{1}{1+e^{E_c-E_f)/kT}}\approx e^{-(E_c-E_f)/kT}$$ Which gives: $$n_0=N_ce^{-(E_c-E_f)/kT}$$ The law of mass action states that the product of electron and hole concentrations is a constant: $$n_0p_0=n_i^2$$ This allows is to find the electron concentration as; $$n_0=n_ie^{-(E_i-E_f)/kT}$$ Where $E_i$ is the equilibrium Fermi energy (intrinsic energy level), where the undoped equilibrium would be. $E_i$ is approximately half way between the CB and VB.

Carrier drift

Drift is the motions of electrons and holes due to the application of an electric field. In the absence of an electric field there is no net velocity of an electron, despite a high thermal velocity. When an electric field is applied, the same velocity and collisions happen, but there is a net direction to the electron's movement, due to its interaction with the field.

Mobility ($\mu$) is the proportionality between the applied field and the resultant velocity of the carriers. $$\langle v_x\rangle=\frac{\langle p_x\rangle}{m_n^*}}=-\frac{q\overline{t}}{m_n^*}\varepsilon_x$$ Here the effective mass is the conductivity effective mass. The mobility is the constant relating the steady state velocity and the electric field. $$\mu_n=-\frac{\langle v_x\rangle}{\varepsilon_x}=\frac{q\overline{t}}{m_n^*}$$

The current density due to an electric field is the number of carriers passing through a given area over a given time multiplied by the charge of those carriers. $$J_x=-qn\langle v_x\rangle=qn\mu_n\varepsilon_x=\sigma\varepsilon_x$$ Where $\sigma=qn\mu_n$. The direction of electron and hole currents are in the same direction, despite their mobility having opposite signs. Hence, we can find the net current: $$J_x=q(n\mu_n+p\mu_p)\varepsilon_x=\sigma\varepsilon_x$$ The carrier concentration and mobility product determines which carrier contributes the most to the current, through conductivity. If the semiconductor is behaving Ohmically, we can relate resistivity to the resistance. $$R=\frac{\rho L}{wt}=\frac{L}{wt}\frac{1}{\sigma}$$

Applying an electric field to a semiconductor slopes the energy bands, with the side close to the positive of the field being lower than the negative side. This causes the electrons to fall down the field and holes to rise. The slope is proportional to the magnitude of the electric field. $$-q\varepsilon_x=\frac{dE_c}{dx}$$

The mobility of carriers are affected by scattering, such as lattice and impurity scattering. Lattice scattering is from lattice vibrations (phonons) and hence has a strong temperature dependence. Lattice scattering has a negative effect on mobility at elevated temperatures, with an approximate dependence of $T^{-3/2}$. Impurity scattering occurs due to interaction between carriers and the charged dopant ions. It is more pronounced when the carrier's momentum is low, such as at low temperatures. Its dependence is approximately $T^{3/2}$ and also is more pronounced at high carrier concentrations. The total mobility is dominated by whichever mechanism is dominating, so they add inversely. $$\frac{1}{\mu}=\frac{1}{\mu_1}+\frac{1}{\mu_2}$$

Under a high electric field, the velocity of carriers ceases to be linearly related to the field strength. Eventually, we hit the thermal velocity limit $\sim10^7cm/s$. At this point, the added energy form the larger field is transferred to the lattice rather than the carriers.

Light absorption and Luminescence

The energy of a photon is related to its wavelength. $$E_{ph}=\frac{hc}{\lambda}$$ $$E_{ph}(eV)=\frac{1240}{\lambda(nm)}$$ In general, if the photon energy is less than the band gap, an electron cannot be promoted and the photon is not absorbed. If the energy is approximately the same, the photon will be absorbed and an electron hole pair will be generated. If the photon energy is greater, the photon will be absorbed, exciting the electron to a higher state in the CB, where after a short time the electron will relax back to the edge of the CB through thermalisation (lattice scattering).

The measurement of absorption can be done by measuring the amount of light that passes through a semiconductor. $$I_t=I_0e^{-\alpha l}$$ Where $I_t$ is the light transmitted, $I_0$ is the light incident on the sample, $\alpha$ is the absorption coefficient of the semiconductor and $t$ is the thickness of the semiconductor.

In direct band gap semiconductors, the bands are aligned in k space, so the photon directly generates an electron hole pair. The absorption occurs according to the following parabolic band relationship: $$hv=E_C+\frac{p_e^2}{2m_e^*}-E_V+\frac{p_h^2}{2m_h^2}$$ The absorption coefficient is proportional to the square root of the difference between the incoming light and the band gap. $$\alpha(hv)\propto(hv-E_G)^\frac{1}{2}$$

In indirect semiconductors, phonons are required due to the misalignment between the CB and VB and are required for the extra momentum. The absorption is now dependent on the availability of the photon and phonon, giving a reduced value. $$\alpha(hv)\propto(hv-E_G\pm E_P)^2$$ This weak absorption can be overcome by more practical factors, such as manufacturing. The direct band gap still exists in indirect materials, but at a higher energy. The slope of the absorption coefficient against wavelength can reveal whether a material is direct or indirect, with direct having a steeper curve.

Adding light stimulates the creation of electron hole pairs, meaning the law of mass action no longer applies. $$np>>n_i^2$$ This increases the generation rate: $$g_{total}=g_i(T)+g_L$$ After some point of constant illumination, the carrier concentration will reach steady state, where the recombination rate it equal to the generation rate. This recombination is proportional to the np product.

Recombination is the annihilation of electron hole pairs, and there are three main types.

Radiative/direct/band-to-band recombination This is the direct relaxation of an electron in the CB to the VB, releasing a photon of energy equal to the band gap in a process known as photoluminescence
Shockley-Read-Hall/Indirect/trap-assisted recombination This is the relaxation of an electron from the CB to the VB through an electronic state (trap) within the band gap $E_t$, whose presence can greatly increase the probability of recombination and cause the energy given off to be as heat
Auger recombination This is a three particle process where two electrons or holes collide, leading to an excitation of one and a relaxation of the other

Photoluminescence is when a semiconductor has excess carriers generated from the absorption of photons and emits photons, relaxing EHPs. Cathodolumiscence is from excess carriers from high energy electron bombardment. The high energy electrons excite multiple sets of EHP as they lose energy. Electroluminescence is from the application of an external bias generating extra carriers and emitting light.

In cases of moderately doped semiconductors, the minority carrier has a concentration approximately equal to the injected carriers. $$n\approx\Delta n$$ Low injection is when $\Delta n < N_A$ and high injection is when $\Delta n>N_A$. Recombination is always occurring, but adding excess carriers causes an increase in the recombination rate. Steady state is achieved when the generation and recombination rates are equal. $$\frac{dp}{dt}=g+r=0$$ The average time that the excess carriers remain in their excited state is the excess carrier lifetime $\tau$. The recombination rate is equal to the number of excess carriers divided by their lifetime. $$r=\frac{\Delta n}{\tau}$$ This is linked by the behaviour of many systems involving populations. $\tau$ is typically between $1\mu s$ and $100ms$ in silicon, and is typically shorter in other materials. Given the generation and recombination are linked in steady state, the generation rate can be linked to the lifetime. $$\Delta n=g\tau$$

Radiative recombination is a spontaneous process, where both an electron and a hole need to be present, so the rate is proportional to the pn product. $$r_{rad}=Bnp$$ The $B$ value depends on the material and the temperature.

Shockley Read Hall (SRH) recombination is facilitated through a energy state deep within a semiconductor's band gap. The energy state introduced a stepping stone to allow recombination, with the most effective energy state being near the mid band gap. This is typically caused by metal impurities or surface dangling bonds. Traps can either be donors or acceptors. We can use Fermi-Dirac statistics to find the probability that a trap will be occupied. $$p_t=N_t(1-f_t)$$ The hole concentration is related to the trap concentration and the fermi function at trap energy. The rate of transition is proportional to the electron and hole concentration in the trap. $$r_a=Bnp_t$$ There are 4 different rates for transitions, however commonly one is the limiting factor. $$R_{SRH}=An$$

Auger recombination is a three particle process where one carrier loses energy in a collision, with the other carrier eventually losing the gained energy through thermalisation. It is normally only significant in non-equilibrium conditions with high carrier density. The eeh process involves two electrons and a hole. $$r_{avg_{eeh}}=c_nn^2p$$ The ehh process is two holes colliding. $$r_{avg_{ehh}}=c_pp^2n$$ Hence the total rate can be found from both acting together. The constants are small compared to the other processes, happening less often than the others.

The total recombination can be found from the sum of the three mechanisms, and can be used to find an effective carrier lifetime. This can be used to find the ABC model, which is an oversimplification and applies only in certain situations. $$r_{total}=An+Bn^2+Cn^3$$ With each term being SRH, radiative and Auger recombination respectively. The effective lifetime is the parallel sum of each lifetime. $$\frac{1}{\tau_{eff}}=\frac{1}{\tau_{rad}}+\frac{1}{\tau_{auger}}+\frac{1}{\tau_{SRH,bulk}}+\frac{1}{\tau_{SRH,surf}}$$

The carrier lifetime can be modelled as: $$\tau=\frac{\Delta n}{g+\frac{dn}{dt}}$$ In steady state, $\frac{dn}{dt}\to0$ and in transient $g\to0$, allowing the lifetime to be measured. A change in conductivity can be measured and is related to the number of carriers in the semiconductor.

When we apply external stimulus, the Fermi level of the semiconductor splits into two levels, $E_{FN}$ and $E_{FP}$, one for each carrier. The spacing between the two quasi-Fermi levels represents the degree of departure from equilibrium.

Diffusion

Diffusion is a process based on the random motion of particles due to random thermal motion. There is no interaction between particles and keeps occurring until the distribution is constant. Diffusivity is a measure of how carriers in a material respond to a gradient in carrier concentration. The flux of carriers through a point is: $$\Phi_{n(0)}=-v_{th}l\frac{dn}{dx}=-v_{th}^2\tau\frac{dn(x)}{dx}=-D_n\frac{dn(x)}{dx}$$ Fick's law states that the flux of particles is proportional to the concentration gradient. Carriers diffuse from the region of high concentration to the one of low concentration. We can also write diffusivity as: $$D_n=\left[\frac{kT}{m_n^*}\right]\left[\frac{\mu_nm_n^*}{q}\right]=v_{th}^2\tau$$ The Einstein relation is: $$D_n=\frac{kT}{q}\mu_n$$

The diffusion of charge carriers creates a current. Hence current can be due to an electric field or a concentration gradient. The carrier flux for diffusion is given by: $$\Phi_n=-D_n\frac{dn(x)}{dx}$$ From this we can get current: $$J_{n,diff}=-(-q)D_n\frac{dn(x)}{dx}$$ The current with the electric field is: $$J_n(x)=q\mu_nn(x)\varepsilon(x)+qD_n\frac{dn(x)}{dx}$$ The total current is then: $$J_{total}(x)=J_p(x)+J_n(x)$$

The total current flowing is due drift and diffusion is linked to the gradient of the quasi Fermi levels. In equilibrium the quasi Fermi energy has to be flat. The total current is; $$J_n(x)=q\mu_nn(x)\varepsilon(x)+qD_n\frac{dn(x)}{dx}=q\mu_nn(x)\varepsilon(x)+\mu_nn(x)\left[\frac{dF_n}{dx}-\frac{dE_i}{dx}\right]=q\mu_nn(x)\frac{dF_n}{dx}$$ Thus a gradient in the quasi Fermi level indicates a current must be flowing.

The continuity equation states that the number of carriers are conserved and are affected by:

Carriers flowing in
Carriers flowing out
Carriers being generated
Carriers recombining

$$\frac{dn(x,t)}{dt}=\frac{d\Delta n}{dt}=-\frac{1}{q}J_n(x)-\left(-\frac{1}{q}J_n(x+\Delta x)\right)+g_L(x)\Delta x-r(x)\Delta x$$ The diffusion length is: $$D_p\tau_{eff}=L_p^2$$ The diffusion length is the distance at which the excess hole distribution is reduced to $1/e$ of its value.

Junctions

Junctions form the basic building blocks of electronic devices. They form on the intersection of two different materials or two disparate regions of a semiconductor with different characteristics.

PN junctions

An abrupt junction is a junction that is formed by a immediate transition from a p type to n type material. The high concentration of carriers promotes the diffusion flow into the other material (holes into n type and electrons into p type). For every carrier that moves, there is an ion that is left behind, leaving a charge and establishing an electric field at the junction. The electric field tries to cause a current counter to the diffusion currents. The drift current will oppose the diffusion current and a balance will be reached when there is no net current. $$J_{drift}+J_{diff}=0$$ The width of the region with uncovered dopants is the transition, depletion or space-charge region.

The bands are bended over the transition region, representing the electric field. Outside the region, the bands are flat. Diffusion is motivated by the concentrations and the drift is motivated by the potential barrier $qV_0$ at the junction. The fermi level is flat at the junction, so there is no net current. The energy between the sides is the contact potential $V_0$. $$V_0=\frac{kT}{q}\ln\frac{p_p}{p_n}=\frac{kT}{q}\ln\frac{N_aN_d}{n_i^2}$$ $$\frac{p_p}{p_n}=e^{qV_0/kT}$$

In a junction there are a balanced set of charges to maintain net charge neutrality.

The electric field can be calculated using Poisson's equation. $$\frac{d\varepsilon(x)}{dx}=\frac{q}{\epsilon}(p0n+N_d^+-N_a^-)$$ Which during the transition region becomes: $$\frac{d\varepsilon}{dx}=\begin{cases}-\frac{q}{\epsilon}N_a,&-x_{p0}<x<0\\\frac{q}{\epsilon}N_d,&0<x<x_{n0}\end{cases}$$ The maximum field exists at the metallurgical junction, and has a value of: $$\varepsilon_0=-\frac{q}{\epsilon}N_dx_{n0}=-\frac{q}{\epsilon}N_ax_{p0}$$ Similarly the contact potential can be found by integrating the electric field to be: $$V_0=\frac{1}{2}\frac{q}{\epsilon}\frac{N_aN_d}{N_a+N_d}W^2$$ The transition region will extend into the side with the lowest doping. $$x_{p0}=\frac{W}{1+N_a/N_d}$$ The width is important when applying external biases.

Junctions under bias

When biased, the pn junction will behave differently depending on the bias direction. This is called rectification. Under equilibrium, there is no net current as drift and diffusion currents are equal.

Under forward bias (p side positive), the bias opposes the contact potential $V_0$ and the built-in electric field, reducing them. The new potential becomes $V=V_0-V_f$. The diffusion current increases (smaller barrier to block diffusion) and the width of the transition region decreases. Reverse bias (p side negative), the bias is in the same direction as the electric field and contact potential. The new potential becomes $V=V_0+V_r$. Diffusion current decreases (larger barrier) and the depletion width increases. The difference between the Fermi levels in the neutral regions is approximately the $V$ applied. The electron and hole diffusion currents are in the same direction (p to n) and are both dependent on the size of the barrier. The electron and hole drift currents are in the same direction (n to p) and are limited by the number of carriers, not the size of the barrier. The current under bias is: $$I=I_0(e^{qV/kT}-1)$$ $I_0$ has many names, and represents the diffusion and drift currents flowing in equilibrium.

We can characterise the excess minority carrier concentration on either side of the junction as: $$\Delta p_n=p(x_{n0})-p_n=p_n(e^{qV/kT}-1)$$ From the number of minority carriers on the edge of the boundaries in low-injection conditions, we can find $I_0$. $$I_0=qA\left(\frac{D_p}{L_p}p_n+\frac{D_n}{L_n}n_p\right)$$ $I_0$ is dependent on factors that are theoretically independent of voltage.

Under forward bias, the Fermi levels split, giving: $$pn=n_i^2e^{(F_n-F_p)/kT}=n_i^2e^{qV/kT}$$ The energy difference between the Fermi levels is $qV$. Under bias, the minority carrier's Fermi level changes, the majority carrier doesn't vary much. The large slop in the Fermi level means current is flowing, and the minority carriers eventually recombine. The majority carriers move to replenish the recombination, completing the circuit. Eventually there is negligible minority injection, where the Fermi levels remerge.

In reverse bias, there are negative minority carriers being injected, meaning minority carriers are being taken from the surrounding region. Again, there is a split in the quasi-Fermi energies, with the split being equal to the bias applied. The Fermi levels may be past the minority band, meaning a very small amount of carriers may be present. The effective $pn$ is 0, so almost no charge flows ($I_O$ only, which is very small).

Reverse breakdown and rectifier diodes

Reverse breakdown occurs when certain pn junctions reach a critical reverse voltage. At this point the reverse current rapidly increases. It can be a non-destructive processm provided $I$ is limited. Can be useful for the detection of small signals, voltage regulation, etc. It commonly occurs due to two mechanisms:

Zener effect Typically at low biases, quantum mechanical tunnelling through the bandgap occurs, and is uncommon
Avalanche breakdown Typically at moderate to high biases, where carriers have enough kinetic energy under high fields to generate new EHPs

The Zener effect occurs due to quantum mechanical tunnelling, where the probability of an electron being on the wrong side of a barrier is nonzero. It only occurs when the CB and VB on either side of the junction are past each other, and when the width of the junction is small (heavy doping). The barrier between the n and p regions is small enough to allow for tunnelling between the regions. An electron tunnelling from the p side to the n side produces a reverse current from n to p. It increases with doping (sharp junction) and the level of reverse bias (steeper transition).

Avalanche breakdown occurs in lighter doped junctions where tunnelling (and thus the Zener effect) is negligible. At high reverse bias the carriers can be accelerated to very high kinetic energy. Impact ionisation can occur where carriers with high energy can excite new EHPs during normal scattering events. We get carrier multiplication, where one carrier can eventuate in many carriers moving. $$M_n=\frac{n_{out}}{n_{in}}=\frac{1}{1-P}$$ Where $P$ is the probability of a collision. $P$ can be hard to find, so an empirical relationship is used instead. $$M=\frac{1}{1-(V/V_{br})^n}$$ $n$ varies by material and $V_{br}$ is the breakdown voltage. $V_{br}$ increases with the $E_g$ of the material and decreases with doping.

Diodes can generally be modelled by a simplified piecewise model. The control characteristics of the rectifier can be modified by:

Increasing the bandgap
- Decreases $n_i$, decreases $I_0$
- Decreases sensitivity to temperature
- Increases $V_{br}$
- Increases offset voltage
Increased doping decreases $V_{br}$ and R
Geometry can be used to reduce $R$ for lowly doped samples

Transient and AC behaviour of junctions

After an abrupt change in junction current, it will take time for the charges to respond. We can write a time dependent continuity equation: $$i(t)=\frac{Q_p(t)}{\tau_p}+\frac{dQ_p(t)}{dt}$$ The first term accounts for the recombination of charges, and the second accounts for build up or decay of charges (zero in steady state). For a sudden turn off of current, the charge over time becomes: $$Q(t)=I\tau_pe^{t/\tau_p}$$ The voltage over the junction will persist until the charge becomes 0. We can approximate the voltage response to an instantaneous change in current by monitoring the excess carrier concentration at $x_n=0$. The voltage is a function of $\Delta p(x,t)$. Ignoring the zero gradient at $x=0$ and redistribution due to diffusion, we can approximate the decay at every point $x$ as: $$\delta p(x_n,t)=\Delta p(t)e^{-x_n/L_p}$$ $$Q_p(t)=qAL_p\Delta p_n(t)$$ This can help find $v(t)$: $$v(t)=\frac{kT}{q}\ln\left(\frac{I\tau_p}{qAL_pp_n}e^{-t/\tau_p}+1\right)$$ The delay can be reduced by introducing traps (reducing $\tau_p$) or by reducing the width of the $n$ region (reducing $Q_p$)

When switching from forward to reverse bias, there is a time where the diode is still in forward bias, resulting in a large current compared to the saturation current. The current then approaches the saturation current, with the increased negative current acting as a transient. The voltage over the diode also starts low (forward bias) and is maintained until the injection reaches 0, and then begins increasing until it reaches the full source over the diode (reverse bias). The storage delay time $t_{sd}$ is the time at which the voltage across the junction reaches zero. $$t_{sd}=\tau_p\left[\text{erf}^{-1}\left(\frac{I_f}{I_f+I_r}\right)\right]^2$$ In most applications, a short $t_{sd}$ is desirable, and can be reduced by introducing traps (reducing $\tau_p$) and/or reducing the width of the region (reducing $Q_p$). Making the lightly doped region shorter than $L_p$ (narrow base diode) can cause the holes to flow to the contact and help charge to be dissipated quickly.

The pn junction has a voltage dependent capacitance and is dominated by different mechanisms in forward and reverse bias.

The junction capacitance This is due to the dipole formed from the fixed dopant ions, and is prominent under reverse bias
The charge storage capacitance This is due to the charge storage effects and is significant under forward bias

As the diode is variable, it can be used as a tunable capacitor (a varactor). The width of the transition region is: $$W=\left[\frac{2\epsilon(V_0-V)}{q}\left(\frac{N_a+N_d}{N_aN_d}\right)\right]^{1/2}$$ Similarly the charge can be related to width: $$Q=qAx_{p0}N_a=qAn_{n0}N_d$$ This gives the capacitance as: $$C_j=\epsilon A\left[\frac{q}{2\epsilon(V_0-V)}\frac{N_dN_a}{N_d+N_a}\right]^{1/2}=\frac{\epsilon A}{W}$$ For an asymmetric junction, the lightly doped side will dominate the capacitance, leading to: $$C_j=\epsilon A\left[\frac{q}{2\epsilon(V_0-V)}N_d\right]^{1/2}$$

Charge storage capacitance is dominant in reverse bias and is a complex mechanism. $$C_s=\frac{1}{3}\frac{q^2}{kT}Ace^{qV/kT}$$ Where $c$ is the n-side length.

Non-ideality in diode modelling

When a junction is forward biased, the transition region contains excess n and p, which move through the junction. Unless $W$ is small compared to $L_p$ and $L_n$, recombination will take place in $W$, being proportional to: $$\propto n_ie^{qV/2kT}$$ In the neutral region, recombination is: $$\propto n_i^2e^{qV/kT}$$ To compensate, we can add an ideality factor to the ideal diode equation. $$I=I_0^{'}(e^{qV/nkT}-1)$$ Recombination only in the neutral region has $n=1$ and only in transition $n=2$, typically it is somewhere in between. $$\frac{I(\text{recombination in neutral})}{I(\text{recombination in tranition})}\propto\frac{n_ie^{qV/kT}}{n_ie^{qV/2kT}}\propto n_ie^{qV/2kT}$$ Departures are more likely for large bandgaps,low $n_i$ (low temperatures), low voltage.

Generation can occur in the transition region in reverse bias, increasing $I_0$. $R_n$ and $R_p$ can be greatly reduced within $W$ because of the very small $n$ and $p$ concentrations under reverse-bias. The generation is not dependent on $n$ and $p$ concentrations. Normally, the generation and recombination are balanced, but in reverse bias the carriers are swept out before recombination, resulting in net generation. Generation in $W$ increases es with $W$ and reverse bias.

We have assumed that majority carrier concentrations are not affected by forward bias, however this condition is breached at moderate forward biases. Under high injection conditions, carrier concentrations are comparable. $$pn=n_i^2e^{qV/kT}\implies p_n=n_ie^{qV/2kT}$$ And the current increases too. $$\propto e^{qV/2kT}$$

The assumption that the potential only falls across the junction is only true for modest values of forward bias. At moderate biases a significant amount of the potential can fall across the neutral regions, introducting effective series resistance. $$I=I_0\exp\left[\frac{q(V+IR_s)}{nkT}\right]-\frac{V-IR_s}{R_{sh}}$$ Series resistance can also occur from the metal electrode connected to either side of the pn junction. Shunt resistance can also play a role in resistive losses; the alternative pathway can be caused by fabrication faults. Series resistance effects are the most common and can cause a flattening/saturation of the current at high forward biases.

In reality junctions aren't abrupt and may have a doping profile. To deal with this profile, we can write a function for the graded transition. $$N_d-N_a=Gx$$ In a graded junction the usual depletion approximation is accurate. If the grade is small, the carrier concentrations (p-n) is important. The regions outside the transition region still have some doping, and can more accurately be described as quasi neutral. This makes it difficult to calculate analytically, requiring simulation to get a good model.

Metal-semiconductor junctions and heterojunctions

The vacuum level $E_{vac}$ can be used as a reference to compare the energies of different materials. The vacuum level is the energy of a free stationary electron that is outside of a material in a perfect vacuum. The work function $\Phi_m, \Phi_s$ is the distance between the $E_F$ and the $E_{vac}$, incorporating chemical and electrical potential. The electron affinity $\chi$ is the distance between the conduction band and the vacuum. The ionisation energy $I$ is the distance between the valence band and the vacuum.

A metal's work function behaves similarly to a heavily doped semiconductor. Hence the contact potential will be the difference between the two work functions. $$V_0=\Phi_m-\Phi_s$$ When the metal and semiconductor join, majority carriers from the semiconductor will diffuse into the metal. Approximately all of the potential difference will fall over the semiconductor. In the other direction a potential barrier forms $\Phi_B$ and prevents injection from the metal into the semiconductor conduction band. $$\Phi_B=\Phi_m-\chi$$ We can use the same equations from modelling a pn junction, using a very highly doped semiconductor. This is the same as for homojunctions, except bending can occur in both materials in the homo case.

Typically if $\Phi_m>\Phi_s$ on an n type surface, it will behave differently in a rectifying manner. This results in the same diode equation to the pn junction, with a different $I_0$. In both directions, the flow of carriers from the metal to the semiconductor is regulated by $\Phi_B$. $I_0$ should depend on the size of the barrier, and is ideally unaffected by the bias voltage. $$I_0\propto e^{-q\Phi_B/kT}$$

When $\Phi_B>\Phi_s$ on a p type surface, current will flow in both directions, exhibiting ohmic behaviour. In this case, the barrier is small and easily overcome. Ideal metal-semiconductor contacts are ohmic when the charge induced in the semiconductor in aligning the Fermi levels is provided by the majority carriers. No depletion region occurs in the semiconductor. As long as the barrier is small, the barrier can be small enough to be easily overcome at room temperature. Alternatively if the barrier is doped highly enough, the junction is narrow enough to be quantum tunnelled through.

Transistors

Field Effect Transistors

A FET is a majority carrier device, often called a unipolar transistor. BJTs operate by injection and collection of minority carriers. Both types of transistors are three terminal devices, with the current between two is controlled by the third. Transistors are mainly used for amplification and switching.

Junction FET (JFET)

In a JFET, the voltage-variable depletion region width is used to control the effective cross-sectional area of a conducting channel. It consists of a channel sandwiched between two oppositely doped semiconductors. The gate is connected to the surrounding semiconductors and by varying the applied voltage, the width of the transition region can vary, altering the effective width of the channel. As a larger reverse bias is applied to the gate, the effective channel resistance increases. As the transition region is desired to be in the channel, the gates should be higher doped than the channel, which also causes low gate resistance.

Initially with low voltage, the channel doesn't vary much (effectively the same width all the way through) and the drain current and voltage are linearly related. As the voltage is increases, the channel beings to narrow closer to the drain due to the reverse biasing near the drain. This increases the resistance with the voltage, having non-linear IV relationship. Eventually, if enough voltage is applied, the channel is pinched-off when the two depletion regions meet. Ideally this cause a flat IV relationship, where I doesn't increase with increased V. We say its saturated at the pinch-off value, where $\frac{dV_D}{dI_D}$ is very high. Altering $V_G$ affects where the pinch-off voltage occurs. When $V_G<0$, the depletion regions are larger, the effective channel is smaller and the channel resistance is higher.

The pinch-off voltage can be found where the width of one of the junctions is equal to half the channel width $a$ (assuming the device is symmetric). $$W(x=L)=\left[\frac{2\epsilon(-V_{GD})}{qN_D}\right]^{0.5}$$ $$V_P=\frac{qa^2N_d}{2\varepsilon}$$ Here, $a$ is the metallurgical half width, the distance between the material change and halfway through the channel. This assumes the supply voltage is large compared to $V_0$ ($V_P\approx V_P+V_0$). $V_P$ is dependent on the gate and drain voltage. $$V_P=-V_{GD}=-V_G+V_D$$

Where $x$ is the length of the channel, $Z$ is the depth of the channel and $\rho$ is the resistivity of the channel, we can begin to model the transistor IV characteristics. For a volume of neutral channel material: $$Z2h(x)dx$$ Where $h(x)$ is the channel half width, we have a resistance: $$\rho dx/(Z2h(x))$$ As the current doesn't change across the channel, the current is related to the differential voltage change in the element by the conductance of the element. $$I_D=\frac{Z2h(x)}{\rho}\frac{dV_x}{dx}$$ And $h(x)$ is: $$h(x)=a\left[1-\left(\frac{V_x-V_G}{V_P}\right)^{0.5}\right]$$ This is valid if $h(x)$ doesn't vary too abruptly across $x$. All together this gives: $$I_D=G_0V_P\left[\frac{V_D}{V_P}+\frac{2}{3}\left(-\frac{V_G}{V_P}\right)^{3/2}-\frac{2}{3}\left(\frac{V_D-V_G}{V_P}\right)^{3/2}\right]$$ Where $V_G<0$ and $G_0=\frac{2aZ}{\rho L}$. This equation is only valid until pinch-off, where: $$I_{D,sat}=G_0V_P\left[\frac{V_D}{V_P}+\frac{2}{3}\left(-\frac{V_G}{V_P}\right)^{3/2}-\frac{2}{3}\right]=G_0V_P\left[\frac{V_G}{V_P}+\frac{2}{3}\left(-\frac{V_G}{V_P}\right)^{3/2}+\frac{1}{3}\right]$$ The saturation current is greatest when $V_G=0$ and decreases as $V_G$ decreases. The transconductance relates the drain current to the gate voltage by: $$g_{m,sat}=\frac{I_{D,sat}}{V_G}=G_0\left[1-\left(-\frac{V_G}{V_P}\right)^{0.5}\right]$$ $g_m$ is the mutual transconductance and has units siemens or mhos.

This assumes $\rho$ is constant as electron mobility is constant. This may not be true for high fields, due to mobility saturation. Such high fields are common for short channels, with moderate drain voltages resulting in high channel fields. Additionally, the channel length can decrease as drain voltage increases beyond pinch-off. This has more of an effect in short channel devices and cause constant saturation current not to be valid.

The Metal-Semiconductor FET

If instead of using a p-n junction like in a JFET, a Schottky barrier is used, a MESFET is created. This is advantageous for high-speed digital or microwave devices, where the simplicity of the Schottky barrier allows for close geometrical tolerances. InP and GaAs MESFETs can achieve high-speed operation due to their high carrier mobilities and drift velocities compared to Si. Different metals are used for the source and drain (ohmic) compared to the gate (junction).

High Electron Mobility Transistor (HEMT)

The MESFET is compatible with the use of III-V compounds. Heterojunctions can be used to engineer the bandgaps. Different atoms can be fit into the same lattice structure (minimising degradation and cracking) while having different bandgaps. We want to maximise the channel conductance, so the channel is highly doped increasing carrier concentration. Increasing doping also increases scattering by the ionised impurities, leading to degradation of mobility. This is avoided by having a thin undoped well surrounded by highly doped, wider bandgap regions to provide carriers. This causes the carriers to fall into the well, where there is no impurity scattering. Performance is very good at low temperatures, where lattice (phonon) scattering is low. The gate is constructed with the channel along the well.

Only one heterojunction is needed to trap the electrons, as a natural well forms due to band bending. In these devices, the donors are separated by 100$\text{\AA}$ to ensure separation. This allows high electron concentration, in an almost triangular well. They form an almost 2 dimensional electron gas (2-DEG). The mobility is limited by the lattice, rather than impurities, and results in large mobility that improves with lower temperatures. The electron density is very high in the thin layer.

The HEMT has high mobility and maximum electron velocity. Additionally a smooth interface is possible. The high performance means high cutoff frequencies.

Short channel effects

Short channels ($<1\mu m$) can alter the IV characteristics due to the high values of the electric fields ($>10kV/cm$). This causes the IV curve to smooth out, with less distinct a linear and saturation region. The drain current becomes: $$I_D=qnv_sA=qN_dv_sZh$$ Where $h$ varies slowly with $V_G$. Here, the saturated current follows velocity saturation and doesn't require true pinch-off. In the saturated velocity case, $g_m$ is approximately constant.

Additionally, pinch-off can cause a reduction in channel length as the drain voltage is further increased. This is important in short channel transistors. This leads to a slope in the saturated IV characteristics, and is analogous to the Early effect in BJTs.

Metal-Insulator-Semiconductor FET

In a MIS transistor, the channel current is controlled by the voltage at the gate, and is isolated from the channel by an insulator. The most common type is the MOSFET. Applying a voltage to the gate changes the concentration of charges bridging the source and drain, eventually changing it enough to form a current carrying channel, with the carriers being from the minority carriers and source and drain. The threshold voltage is the minimum gate voltage required to induce the channel. For a given $V_G$, there is a $V_G$ where the channel becomes saturated.

Depletion mode transistors have a channel for $V_G=0$, and the applied voltage removes the channel, enhancement mode forms a channel when $V_G-V_T>V_D$. The linear region is where $V_D<V_G-V_T$, where the channel is completely open and changes in $V_D$ have a corresponding change in the current. Saturation occurs when $V_D=V_G-V_T$, where the channel pinches off at the drain contact. Strong saturation is when $V_D>V_G-V_T$ and varying $V_D$ alters the channel length and causes no change in current. The drain current changes slightly with $V_D$, as the channel changes.

MOSFETs are useful in digital circuits, where they are switched on and off from the insulated gate electrode. This insulation causes a high DC input impedance. In Si, NMOS is commonly used due to a higher electron mobility.

The ideal MOS capacitor consists of a metal, oxide and semiconductor being layered together. For convenience, a modified work function is used $q\phi_m$ for the metal oxide interface, being the difference between the metal's Fermi level and the oxide's conduction band. Likewise we can define $q\phi_s$ for the semiconductor. We define $q\phi_F$ as the difference between the Fermi level below the intrinsic level for the semiconductor. If we deposit a negative charge on the metal, an equal net positive charge accumulates on the p type semiconductor as it is operating in the “accumulation” region. As $\phi_m$ and $\phi_s$ don't change, the oxide's conduction band is tilted due to the electric field causing a gradient. Near the interface, the bands of the semiconductor bend to be consistent with the accumulation of charges. Applying a small positive charge to the metal causes the depletion of charges in the semiconductor, as the Fermi level approaches the intrinsic level. If the charge increases further, the Fermi level will cross the intrinsic level, and there will be charge inversion. Strong inversion occurs when the surface is as n-type as the substrate is p-type. $$\phi_s(inv)=2\phi_F=2\frac{kT}{q}\ln{n/n_i}$$ The Debye screening length is the distance in which changes in charge are screened out. $$L_D=\sqrt{\frac{\epsilon_skT}{q^2p_0}}$$

The inversion region is generally very narrow $<100\text{\AA}$. The applied voltage is mostly over the insulator, with some over the semiconductor's depletion region. The depletion width is: $$W=\left[\frac{2\epsilon_s\phi_s}{qN_a}\right]^{0.5}$$ Once strong inversion is reached, the stronger inversion results in more depletion, rather than a wider region, giving the maximum depletion width as: $$W_\max=2\left[\frac{\epsilon_skT\ln(N_a/n_i)}{q^2N_a}\right]^{0.5}$$ We can use this to find the required voltage for strong inversion: $$V_T=-\frac{Q_d}{C_i}+2\phi_F$$ Where $C_i$ is the capacitance per unit area from the insulator and $Q_d$ is the charge per unit area in the depletion region. $$Q_d=-qN_aW_m=-2(\epsilon_sqN_a\phi_F)^{0.5}$$

The capacitance of the MOSFET is dependent on voltage, as it will vary between accumulation, depletion and inversion. $$C_i=\frac{dQ_s}{d\phi_s}$$ Under strong accumulation or inversion, there are a lot of free charges, so the capacitance is high. Both of these cases resemble a parallel plate capacitor. Under low frequency, inversion is high capacitance, but at high frequency, it remains low capacitance. At high frequencies the channel is unable to form as the substrate's minority carriers are unable to move to the channel. High and low frequency depends on the generation-recombination rate of the substrate. The majority carriers respond much faster than the minority, so this behaviour isn't seen under accumulation. During weak accumulation and depletion, the capacitance drops, until the channel is formed and the capacitance sharply increases.

In real materials, we need to apply a negative voltage to the metal to achieve flat bands. $$V_{FB}=\phi_{ms}$$ Trapped charges in the oxide can affect its performance, including making it remember past biases. This means that an additional component must be added to the flat band voltage: $$V_{FB}=\phi_{ms}-\frac{Q_i}{C_i}$$ Additionally, charges can accumulate on the surface of the oxide, making the threshold voltage: $$V_T=\phi_{ms}-\frac{Q_i}{C_i}-\frac{Q_d}{C_i}+2\phi_F$$ The first two terms achieve flat bands. the third accommodates the charge in the depletion region and the fourth induces inversion.

The gate voltage can be written as: $$V_G=V_{FB}-\frac{Q_S}{C_i}+\phi_s$$ As the induced charge in the semiconductor is mobile and fixed depletion region charges: $$Q_s=Q_n+Q_d$$ We can find the mobile charge to be: $$Q_n=-C_i\left[V_G-\left(V_{FB}-\phi_S-\frac{Q_d}{C_i}\right)\right]$$ When we apply a voltage $V_D$, the voltage in the channel $V_x$ varies with distance in the channel. The mobile charge can then be written as: $$Q_n=-C_i\left[V_G-V_{FB}-2\phi_F-V_x-\frac{1}{C_i}\sqrt{2q\epsilon_sN_a(2\phi_F+V_x)}\right]$$ Neglecting the variation of $Q_d$ with bias, we can simplify to: $$Q_n=-C_i(V_G-V_T-V_x)$$ We can integrate over the charges to find the current. $$I_D=\frac{\overline{\mu}_nZC_i}{L}\left[(V_G-V_T)V_D-\frac{1}{2}V_D^2\right]$$ $k_N$ determines the conductance and transconductance of the MOSFET. $$k_N=\frac{\overline{\mu}_NZC_i}{L}$$ Including the variation in $Q_d(x)$ forms the bulk charge model, and while is generally more accurate, is often not needed. $$I_D=\frac{\overline{\mu}_nZC_i}{L}\left\{\left(V_G-V_{FB}-2\phi_F-\frac{1}{2}V_D\right)V_D-\frac{2}{3}\frac{\sqrt{2\epsilon_sqN_a}}{C_i}\left[(V_D+2\phi_F)^{3/2}-(2\phi_F)^{3/2}\right]\right\}$$

In the linear regime the channel conductance can be found to be: $$g=\frac{\partial I_D}{\partial V_D}\approxeq\frac{Z}{L}\overline{\mu}_nC_i(V_G-V_T)$$ Which assumes $V_D<<V_G-V_T$.

Pinch off occurs approximately at saturation. $$V_D\approx V_G-V_T$$ The saturation current is: $$I_D(sat)\approx\frac{1}{2}\overline{\mu}_nC_i\frac{Z}{L}(V_G-V_T)^2=\frac{1}{2}\overline{\mu}_nC_i\frac{Z}{L}V_D^2(sat)$$ The transconductance is: $$g_m(sat)\approx\frac{Z}{L}\overline{\mu}_nC_i(V_G-V_T)$$

Due to additional scattering, the mobility in the channel is lower than in the bulk semiconductor. Due to the proximity to the semiconductor-oxide interface, there is scattering due to the surface roughness and by Coulomic interactions with the fixed oxide charges.

The saturation current is limited by the charge mobility in a short channel device. $$I_D(sat)\approx ZC_i(V_G-V_T)v_s$$ The saturation current has a linear dependence on the gate voltage, rather than a quadratic as in the long channel model.

It can be important to control the threshold voltage, to ensure it is within our voltage limits and that the transistors work with each other. All the terms that comprise the threshold voltage are controllable. $$V_T=\phi_{ms}-\frac{Q_i}{C_i}-\frac{Q_d}{C_i}+2\phi_F$$ $\phi_{ms}$ can be controlled by that gate conductor material. The substrate determines $\phi_F$. The oxidation process and orientation control $Q_i$. Substrate doping influences $Q_d$. Oxide thickness influence $C_i$.

A low threshold voltage is desirable for the gate of the transistor. The region between transistors (field) is undesirable for transistor behaviour, so a high threshold is desired. A thick oxide is constructed to avoid inversion between devices.

Ion dopants can be added to the channel to reduce the effect of the positive depletion charge. This helps to alter $V_T$.

Bipolar Junction Transistor

Have a situation where carriers are motivated to move over a junction by the injection of carriers from an oppositely biased junction. The junctions are connected by a base, with width $W_b<<L_p$, such that the holes are capable of arriving to the opposite junction. The current from the base is recombined in the base and injected into the emitter. The injected current motivates additional current from the emitter, allowing for amplification controllable by the base current. The current through the collector is proportional to the hole component of the emitter current. $$i_C=Bi_{Ep}$$ $B$ is the constant of proportionality (base transport factor) representing the fraction of injected holes coming from the base to the collector. The emitter injection efficiency is: $$\gamma=\frac{i_{Ep}}{i_{En}+i_{Ep}}$$ We want $B$ and $\gamma$ to be close to unity, meaning that the emitter current is mostly due to holes ($\gamma\approx 1$) and that the injected holes eventually participate in the collector current ($B\approx 1$). The ratio between the emitter and collector currents is: $$\frac{i_C}{I_E}=B\gamma=\alpha$$ $\alpha$ is the current transfer ratio, and represents the emitter-to-collector current amplification. As $\alpha<1$, there is no real amplification. The base current is given by: $$i_B=i_{En}+(1-B)i_{Ep}$$ The ratio between the collector and the base currents is: $$\frac{i_C}{i_B}=\frac{\alpha}{1-\alpha}=\beta$$ $\beta$ is the base-to-collector amplification factor, and can be large. The ratio of excess hole lifetime (transit time) $\tau_t$ to excess electron lifetime $\tau_p$ in the base can also be used to give $\beta$. $$\beta=\frac{\tau_p}{\tau_t}$$ This comes from the number of holes that pass through the base before the excess electron (from space charge neutrality) is recombined.

Under forward biasing (EB forward biased and CB reverse), the following hold. In the base region, the concentration of holes varies almost linearly between the emitter and collector. The concentration at the emitter is $p_ne^{\frac{qv_{EB}}{kT}}$ and the concentration at the collector is $-p_n$. The concentration slopes down from the emitter to collector, i.e. $i_E>i_C$ and $i_E-i_C=i_B$. The collector and emitter regions have excess concentrations exponentially decaying to zero. The emitter current can be found to be: $$I_{Ep}=qA\frac{D_p}{L_p}\left(\Delta p_Ecoth\left(\frac{W_b}{L_p}\right)-\Delta p_Ccsch\left(\frac{W_b}{L_p}\right)\right)$$ Similarly the collector current can be found to be: $$I_{C}=qA\frac{D_p}{L_p}\left(\Delta p_Ecsch\left(\frac{W_b}{L_p}\right)-\Delta p_Ccoth\left(\frac{W_b}{L_p}\right)\right)$$ This gives the base current as: $$I_B=qA\frac{D_p}{L_p}\left[(\Delta p_E+\Delta p_C)\left(coth\left(\frac{W_b}{L_p}\right)-csch\left(\frac{W_b}{L_p}\right)\right)\right]=qA\frac{D_p}{L_p}\left[(\Delta p_E+\Delta p_C)tanh\left(\frac{W_b}{2L_p}\right)\right]$$ We can simplify these equations if the collector is reverse biased, the equilibrium hole concentration is small and $\gamma\approx 1$, then: $$I_E\approx qA\frac{D_d}{L_p}\Delta p_Ecoth\left(\frac{W_b}{L_p}\right)$$ $$I_C\approx qA\frac{D_d}{L_p}\Delta p_Ecsch\left(\frac{W_b}{L_p}\right)$$ $$I_B\approx qA\frac{D_d}{L_p}\Delta p_Etanh\left(\frac{W_b}{L_p}\right)$$ We can further simplify by using the standard series expansions, for which the first terms are accurate for small $W_b/L_p$. This reveals that $I_C$ is only slightly smaller than $I_E$, and gives the base current as: $$I_B\approx\frac{qAW_b\Delta p_E}{2\tau_p}$$ The stored charge is; $$Q_p\approx\frac{1}{2}qA\Delta p_EW_b$$ This charge needs to be replaced every $\tau_p$ seconds. This gives the base current: $$I_B\approx\frac{Q_p}{\tau_p}=\frac{qAW_b\Delta p_E}{2\tau_p}$$ Having light base doping increases $\tau_p$ and emitter injection efficiency. The straight line approximation is accurate for base current, but not collector or emitter. We can also write the emitter injection efficiency in terms of the emitter and base properties: $$\gamma=\left[1+\frac{L_p^nn_n\mu_n^p}{L_n^pp_p\mu_p^n}tanh\left(\frac{W_b}{L_p^n}\right)\right]^{-1}\approx\left[1+\frac{W_bn_n\mu_n^p}{L_n^pp_p\mu_p^n}\right]$$ Where $L_p^n$ is the hole diffusion length in the n-type base region and $\mu_n^p$ is the electron mobility in the p-type emitter region. The base transport factor then becomes: $$B=sech\left(\frac{W_b}{L_p}\right)$$

Considering the generalised biased case, can be done by considering the component of the current from injection and collection in normal mode and in inverted mode. The current in inverted mode is in the opposite direction to normal mode. Together they add to form the current through the base. We can find the normal and inverted mode currents to be: $$I_{EN}=a\Delta p_E$$ $$I_{CN}=b\Delta p_E$$ $$I_{EI}=-b\Delta p_C$$ $$I_{CI}=-a\Delta p_C$$ Where the constants are: $$a=\frac{qAD_p}{L_p}coth\left(\frac{W_b}{L_p}\right)$$ $$b=\frac{qAD_p}{L_p}csch\left(\frac{W_b}{L_p}\right)$$ These can be combined to give the overall collector and emitter currents. Where the transistor can be asymmetric, the normal current can be written as: $$I_{EN}=I_{ES}\left(e^{\frac{qV_{EB}}{kT}}-1\right)$$ Where $I_{ES}$ is the magnitude of the emitter saturation current in normal mode. We can then find the collector current to be: $$I_{CN}=\alpha_NI_{EN}=\alpha_NI_{ES}\left(e^{\frac{qV_{EB}}{kT}}-1\right)$$ $$I_{EI}=\alpha_II_{CI}=-\alpha_II_{CS}\left(e^{\frac{qV_{EB}}{kT}}-1\right)$$ Where each $\alpha$ is the ratios of collected to injected current in each mode. We can again find the superposition of these components to find the Ebers-Moll equations. $$I_E=I_{ES}\left(e^{\frac{qV_{EB}}{kT}}-1\right)-\alpha_II_{CS}\left(e^{\frac{qV_{CB}}{kT}}-1\right)$$ $$I_C=\alpha_NI_{ES}\left(e^{\frac{qV_{EB}}{kT}}-1\right)-I_{CS}\left(e^{\frac{qV_{CB}}{kT}}-1\right)$$ From doping we can find the currents to be: $$I_E=I_{ES}\frac{\Delta p_E}{p_n}-\alpha_II_{CS}\frac{\Delta p_C}{p_n}=\frac{I_{ES}}{p_n}(\Delta p_E-\alpha_N\Delta p_C)$$ $$I_C=\alpha_NI_{ES}\frac{\Delta p_E}{p_n}-I_{CS}\frac{\Delta p_C}{p_n}=\frac{I_{CS}}{p_n}(\alpha_I\Delta p_E-\Delta p_C)$$ We can find these equations to be: $$I_E=Q_N\left(\frac{1}{\tau_{tN}}+\frac{1}{\tau_{pN}}\right)-\frac{Q_I}{\tau_{tl}}$$ $$I_C=\frac{Q_N}{\tau_{tN}}-Q_I\left(\frac{1}{\tau_{tI}}+\frac{1}{\tau_{pI}}\right)$$ We can also find the base current to be: $$I_B=I_E-I_C=-(1-\alpha_N)I_{ES}-(1-\alpha_I)I_{CS}$$

Saturation occurs when the bias across the collector junction is reduced to zero, i.e. $\Delta p_C=0$. In saturation, the transistor approximates the ideal on switch. As the transistor moves deeper into saturation, the collector current stays constant while the base current increases. In saturation charge is stored in the transistor, and exiting saturation takes time as the stored charge needs to be dissipated over a time $t_{sd}$. Once the stored charge is withdrawn, the base current cannot be maintained and it becomes described by the equation above.

As the collector voltage is increased, the base width narrows from the increase in depletion region. This is the Early effect. This causes the saturation current to be slanted rather than flat. The saturation curves intersect at $V_A$, the Early voltage. The length is given as: $$l=\left(\frac{2\epsilon V_{BC}}{qN_d}\right)^{1/2}$$ When the depletion region is the size of the base, punch-through occurs. This means holes are directly swept from the emitter to the collector and transistor action is lost. Avalanche breakdown occurs before punch through. Punch-through couples the junctions of the transistor, allowing for an exponentially increasing current for a small change in CB voltage, limiting the maximum voltage that can be applied to the transistor. In common base the collector current sharply increases at a well-defined breakdown voltage $BV_{CBO}$. For common emitter, there is still an increase at $BV_{CEO}$, but the increase is over a larger range. Also, $VB_{CEO}$ is significantly smaller than $BV_{CBO}$. In the common base case, the collector current is: $$I_C=(\alpha_NI_E+I_{C0})M$$ Where $M$ is the multiplication factor. For the common emitter case: $$I_C=\frac{MI_{C0}}{1-M\alpha_N}$$ Here the collector current increases as $M\alpha_N\to1$ but in the common base case $M\to\infty$ before $BV_{CBO}$ is reached.

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