Semiconductor Devices 1

Crystal and electronic bands

Periodic atomic interactions turn discrete atomic levels into allowed bands separated by gaps. Electrons near a band extremum respond as if they had an effective mass describing band curvature. A hole describes the missing occupation of an otherwise nearly filled valence band.

The density of states counts available levels; the Fermi–Dirac distribution supplies occupancy. A direct gap connects extrema at the same crystal momentum, while an indirect transition needs momentum assistance. Draw energy vertically and state whether the horizontal axis is position or wavevector.

Build the physical picture

Start with one isolated atom: its electrons occupy discrete allowed energies. Bringing many atoms together couples their states and splits each atomic level into many closely spaced levels. A macroscopic crystal therefore has allowed bands separated by forbidden gaps. A band diagram is an energy map of these states, not a drawing of the physical positions of individual electrons.

A completely filled band cannot normally carry a net small-field current because changes in opposite-moving occupied states cancel. Partially occupied conduction-band states and empty states in the valence band can respond. The hole description tracks a missing valence electron as a positive mobile carrier; it is not an extra proton moving through the lattice.

From mechanism to a usable model

The dispersion E(k) connects energy to crystal wavevector. Its slope determines group velocity, and its curvature determines effective mass near a band extremum. A flat band means small velocity for a given wavevector change and often a large effective mass. Real silicon has multiple conduction-band valleys, so one scalar mass is an approximation.

In an indirect-gap material, the valence-band maximum and conduction-band minimum occur at different wavevectors. An optical transition must conserve both energy and momentum, typically involving a phonon as well as the photon. This explains why the bandgap alone does not predict light-emission efficiency.

1/m*=(1/ℏ²)d²E/dk²

Model assumptions: One-dimensional curvature near a parabolic extremum; real effective mass can be anisotropic.

Reading a band diagram

Identify the energy axis and the conduction/valence band edges; spatial band diagrams have position on the horizontal axis.

Mark allowed states separately from their occupation. A state can exist without being filled.

Compare band-edge curvature or extrema before assigning mass or a direct/indirect gap.

Mistake to avoid: Bandgap is not the energy of every conduction electron, and drawing bands bending does not mean the silicon mechanically bends.

Worked problem: If band curvature doubles, how does effective mass change?

It halves in this local model. This is a dynamical response parameter, not a literal change in electron rest mass.

Self-test: Why can two diagrams both labelled “band diagram” have different horizontal axes?

E(k) describes dispersion; E(x) describes spatial band edges and electrostatic bending.

A photon has energy 1.55 eV. Is that enough information to predict strong absorption in a semiconductor with a 1.1 eV gap?

It is energetically sufficient for a band-to-band transition, but transition matrix elements, momentum conservation, thickness and absorption coefficient still determine absorption. An indirect gap may require phonon participation.

Carrier statistics

At thermal equilibrium one Fermi level determines occupation. In a nondegenerate semiconductor, carrier concentrations follow exponential distances between that level and the band edges. Donors and acceptors influence charge neutrality, but incomplete ionization and compensation can change the simple approximation.

Use charge neutrality together with mass action rather than assigning n and p independently. Far from equilibrium, separate electron and hole quasi-Fermi levels may be needed.

Build the physical picture

Separate two questions: how many states exist at an energy, and what fraction are occupied? Density of states answers the first; the Fermi–Dirac function f(E)=1/[1+exp((E−EF)/kT)] answers the second. EF is a chemical-potential parameter. At equilibrium it is spatially constant even when band edges vary with position.

For nondegenerate material, the conduction-band edge lies several kT above EF. The Fermi function there can be approximated by an exponential. Integrating the conduction-band states gives n=Nc exp[−(Ec−EF)/kT]; an analogous expression gives p=Nv exp[−(EF−Ev)/kT]. Their product becomes ni² in thermal equilibrium.

From mechanism to a usable model

Doping sets charge neutrality, not an unconditional equality n=ND. Donors and acceptors can compensate, may not be completely ionized at low temperature, and coexist with thermally generated carriers. With complete ionization, n−p=ND−NA. Combine this with np=ni² rather than discarding minority carriers automatically.

Out of equilibrium, electron and hole populations may need separate quasi-Fermi levels. Applying equilibrium mass action to an illuminated or strongly injected sample can therefore produce the wrong result. Temperature also changes ni, mobility and ionization; treating ni as a universal constant is only a chosen operating-point approximation.

np=nᵢ²; n≈N_D−N_A; p=nᵢ²/n

Model assumptions: Thermal equilibrium, nondegenerate statistics and complete ionization with dominant net donors for the approximate n.

Solve compensated nondegenerate material

Let ΔN=ND−NA and impose n−p=ΔN.

Use p=ni²/n to obtain n²−ΔN n−ni²=0.

Choose n=[ΔN+√(ΔN²+4ni²)]/2, then p=ni²/n; check the ionization and nondegenerate assumptions.

Mistake to avoid: The Fermi level is not necessarily an occupied electron energy, and minority concentration is not zero simply because donors dominate.

Worked problem: For nᵢ=10¹⁰ cm⁻³ and n=10¹⁶ cm⁻³, find p.

p=10²⁰/10¹⁶=10⁴ cm⁻³. Majority doping greatly suppresses equilibrium minority concentration.

Self-test: What fails when the Fermi level approaches or enters a band?

The Boltzmann approximation breaks down; use Fermi–Dirac integrals and appropriate band/ionization models.

At 300 K use ni=10¹⁰ cm⁻³, ND=10¹⁶ and NA=9×10¹⁵ cm⁻³. Estimate n and p.

Net donors are 10¹⁵ cm⁻³, far above ni. Thus n≈10¹⁵ cm⁻³ and p≈10⁵ cm⁻³. Using ND alone would overestimate n tenfold. Both values have concentration units.

Carrier transport

An electric field drives drift; a concentration gradient drives diffusion. Electron particle motion opposes the field, but conventional electron current points with it because electrons carry negative charge. Hole diffusion current has the opposite gradient sign from electron diffusion current.

At equilibrium nonzero drift and diffusion components can cancel. Mobility relates low-field drift velocity to field; high fields can invalidate the linear law through velocity saturation. Keep cm-based or SI units consistent throughout.

Build the physical picture

Drift is a biased response to an electric field superimposed on random thermal motion. Electrons drift opposite to the field, yet their conventional current points along it because their charge is negative. Holes drift with the field and contribute conventional current in the same direction. Mobility connects low-field drift speed to field: |vdrift|=μ|E|.

Diffusion requires no applied field. A population gradient produces more carriers crossing from the crowded side than returning from the dilute side. Particle flux and conventional current have different signs for electrons. In one dimension, Jn=qμn nE+qDn dn/dx and Jp=qμp pE−qDp dp/dx.

From mechanism to a usable model

The Einstein relation D/μ=kT/q links diffusion and mobility for nondegenerate carriers near thermal equilibrium. It explains how drift and diffusion can cancel in equilibrium even though each mechanism remains present. Their cancellation is a zero-net-current statement, not an absence of carrier motion.

At high fields, drift speed need not keep increasing linearly. Scattering, heating and velocity saturation limit the simple mobility model. Resistivity also depends on both carrier types: σ=q(nμn+pμp), and ρ=1/σ. Keep centimetre-based and metre-based mobility/concentration units consistent.

J_n=qμ_n nE+qD_n∇n; J_p=qμ_p pE−qD_p∇p; D/μ=kT/q

Model assumptions: Low-field drift-diffusion and nondegenerate thermal equilibrium for Einstein relation.

Calculate low-field conductivity

Convert n, p and mobilities into one consistent unit system.

Calculate each positive conductivity contribution qnμn and qpμp.

Sum them before taking the reciprocal; then use J=σE only if the field is in the low-field regime.

Mistake to avoid: Thermal speed is not drift speed. Fast random motion can coexist with a much smaller net drift and zero average current.

Worked problem: At 300 K, μ_n=1350 cm²/(V·s). Find D_n.

V_T≈0.02585 V. D_n=1350×0.02585≈34.9 cm²/s. In SI the same mobility is 0.135 m²/(V·s).

Self-test: Can current be zero while a concentration gradient exists?

Yes; an equilibrium field can make drift cancel diffusion.

For n=10¹⁶ cm⁻³, μn=1000 cm²/(V s), negligible holes, find σ and ρ.

σ=1.602×10⁻¹⁹×10¹⁶×1000=1.602 S/cm, giving ρ≈0.624 Ω cm. In SI, those values are 160.2 S/m and 0.00624 Ω m.

Semiconductor equations

Poisson’s equation links electrostatic potential to charge density. Continuity equations conserve carriers while allowing generation and recombination. Current laws connect carrier flux to field and concentration gradients. Together they form a coupled system with contacts and surfaces supplying boundary conditions.

Solving only one equation while holding incompatible quantities fixed can produce an unphysical answer. At equilibrium, charge and carrier distributions must also be consistent with one electrochemical reference.

Build the physical picture

Device equations couple electrostatics to conservation. Poisson’s equation tells you how net charge bends potential: d²ψ/dx²=−ρ/ε. Net charge includes mobile electrons and holes and ionized dopants. With E=−dψ/dx, integrating a known charge density gives field and then potential, but integration constants require boundary conditions.

Continuity says a carrier population changes because carriers enter or leave and because generation or recombination occurs. For electrons, ∂n/∂t=(1/q)∂Jn/∂x+G−R; for holes the current-divergence term has the opposite sign. These signs follow from their opposite charges and conventional-current definitions.

From mechanism to a usable model

A calculation is incomplete without contacts and initial conditions. An ohmic contact can set a population or quasi-Fermi level; an insulating boundary can impose zero normal current. A steady-state calculation sets time derivatives to zero but does not automatically set spatial gradients or current to zero.

Useful approximations remove only justified terms. In a quasi-neutral region, net charge is small and Poisson may simplify. In a depletion region, mobile charge may be neglected. In a thin neutral base, recombination may be small. Never use all three simplifications throughout a device without checking where each applies.

∇·(ε∇ψ)=−ρ; ∂n/∂t=(1/q)∇·J_n+G−R

Model assumptions: Electron continuity sign convention for conventional current; positive holes use −∇·J_p/q.

Construct a one-dimensional boundary-value problem

Write geometry, material parameters, doping profile and contact conditions before equations.

Choose unknowns ψ(x), n(x), p(x); connect them through Poisson, continuity and drift–diffusion.

Check charge/current conservation and limiting cases after solving; a numerical curve is not validation by itself.

Mistake to avoid: Steady state means time-independent populations, not necessarily equilibrium. A biased diode can have stationary nonzero current.

Worked problem: In a spatially uniform region with G=2×10²⁰ and R=1.5×10²⁰ cm⁻³s⁻¹, find dn/dt.

With current divergence zero, dn/dt=5×10¹⁹ cm⁻³s⁻¹. Constant excess generation cannot sustain a steady state without changing recombination or outflow.

Self-test: Why specify contact boundary conditions?

They fix potentials and carrier injection/removal, selecting the actual solution.

If generation and recombination balance everywhere in a steady neutral region, what does electron continuity imply?

With ∂n/∂t=0 and G−R=0, ∂Jn/∂x=0. Electron current is spatially constant; its value is set by boundaries and need not be zero.

Generation and recombination

Generation creates electron–hole pairs; recombination removes them. Traps mediate Shockley–Read–Hall processes, radiative transitions emit photons, and Auger recombination transfers energy to another carrier. Relative importance depends on material, carrier density and defects.

After a small excitation, a simple lifetime model gives exponential decay. This lifetime can reflect bulk and surface contributions; a measured decay is not automatically one microscopic mechanism. Diffusion length combines lifetime and diffusivity.

Build the physical picture

Generation promotes an electron into the conduction band and leaves a hole; recombination removes an electron–hole pair. Thermal equilibrium has generation and recombination balancing statistically, not both stopping. An optical pulse creates excess populations above the equilibrium values. In low injection, the minority population changes substantially while the majority population changes only slightly relative to its large background.

Shockley–Read–Hall recombination uses defect states, radiative recombination emits photons, and Auger recombination transfers energy to another carrier. Their concentration dependences differ, so the same measured lifetime cannot describe every injection level.

From mechanism to a usable model

For a uniform small excess population with a constant effective lifetime, dΔn/dt=−Δn/τ gives exponential decay. If carriers also diffuse, the continuity equation contains D d²Δn/dx². The characteristic distance traveled before recombination is the diffusion length L=√(Dτ).

Surface recombination adds a boundary loss described by a surface recombination velocity, with units of length per time. Thin samples can be surface-dominated even when their bulk material is good. A measured decay may therefore combine bulk defects, surfaces, trapping and transport; fitting one exponential does not prove one microscopic mechanism.

Δn(t)=Δn₀exp(−t/τ); L_n=√(D_nτ)

Model assumptions: Uniform low-level excess population and constant lifetime; coupled transport changes spatial response.

Relate a transient to a length scale

Determine whether a uniform single-lifetime approximation is credible.

Read τ from the time for excess concentration to fall to 1/e of its initial value, not to zero.

Compute L=√(Dτ); compare L with device thickness to judge collection or recombination losses.

Mistake to avoid: A longer lifetime is not always better for every application: charge storage can slow switching while benefiting photocarrier collection.

Worked problem: For D=25 cm²/s and τ=1 µs, find diffusion length.

L=√(25×10⁻⁶)=0.005 cm=50 µm. Larger lifetime allows carriers to travel farther before recombination.

Self-test: What fraction remains after three lifetimes?

exp(−3)≈4.98% under the exponential model.

For D=25 cm²/s and τ=1 µs, calculate diffusion length.

Dτ=25×10⁻⁶ cm², so L=0.005 cm=50 µm. The square root restores length units. A 5 µm region is much shorter than L; a 500 µm region is not.

Metal–semiconductor contacts

Joining a metal and semiconductor aligns electrochemical potentials and can bend semiconductor bands. A rectifying Schottky contact has a barrier that limits carrier injection. An ohmic contact is approximately linear over its intended range, often achieved through suitable material and heavy doping that narrows the barrier for tunnelling.

Interface states and chemical reactions can pin barrier behaviour, so work-function difference alone may not predict reality. Contact resistance can dominate a small device even when its channel is conductive.

Build the physical picture

A metal and a semiconductor have different work functions before contact. When connected, charge redistributes until their equilibrium Fermi levels align. The semiconductor bands bend near the interface, forming accumulation or depletion depending on material and doping. For an ideal n-type Schottky interface, the electron barrier height is approximately ΦBn=ΦM−χ.

The barrier height and the built-in potential are different quantities. The barrier refers to injection from the metal relative to the conduction-band edge; the semiconductor potential drop also depends on where its bulk Fermi level lies. Draw these energies before assigning a numerical voltage.

From mechanism to a usable model

Real contacts are not determined by work functions alone. Interface states, interfacial oxides, reaction products and dipoles can pin the Fermi level or modify the barrier. Heavy semiconductor doping narrows the depletion region and enables tunnelling, which can make a contact nearly ohmic without eliminating its microscopic barrier.

An ohmic contact is a low-resistance approximately nonrectifying interface over its operating range. That is a measured electrical property, not simply “metal touches silicon.” Contact resistance matters strongly as device dimensions shrink because the interface contribution can dominate the channel resistance.

Ideal n-type Schottky barrier φ_Bn=φ_m−χ

Model assumptions: Schottky–Mott approximation, energies expressed consistently; ignores interface pinning and reactions.

Sketch equilibrium band alignment

Use one common energy reference and align EF after contact.

Locate Ec at the interface and far into the semiconductor; distinguish barrier height from band bending.

Consider interface chemistry and depletion width before deciding whether thermionic emission or tunnelling dominates.

Mistake to avoid: A low apparent two-terminal resistance does not alone isolate contact resistance; the semiconductor and leads also contribute.

Worked problem: For φ_m=4.8 eV and χ=4.05 eV, estimate φ_Bn.

0.75 eV. This is an ideal estimate, not a measured guarantee for a fabricated interface.

Self-test: How can heavy doping improve an ohmic contact?

It narrows the depletion barrier and permits stronger tunnelling, subject to interface chemistry.

Why can increased doping improve an ohmic contact even when the ideal barrier height is unchanged?

The depletion barrier becomes thinner, increasing tunnelling probability. Current can then flow with less rectification and lower contact resistivity; the barrier need not vanish.

Contact transport and dynamics

Thermionic emission sends carriers over a barrier; tunnelling sends them through it. Temperature and barrier width distinguish regimes. Applied voltage alters injection and space charge, creating non-linear current and voltage-dependent capacitance.

Series resistance can flatten a measured exponential at large current. Image-force lowering, interface states and nonuniform barriers can change extracted parameters. Use temperature-dependent and small-signal measurements to constrain a model.

Build the physical picture

Thermionic emission selects carriers with sufficient energy to cross a barrier. For a simple Schottky model, saturation current scales approximately as area×T² exp[−qΦB/(kT)]. Tunnelling instead depends on barrier shape and width, while thermionic-field emission combines thermal excitation and tunnelling. Their relative importance changes with temperature and doping.

The contact also stores electrostatic charge. Small changes of reverse bias change the depletion width and thus capacitance. This charge response is separate from the DC current mechanism, so a contact can rectify current and still behave as a voltage-dependent capacitor.

From mechanism to a usable model

For uniform doping, depletion capacitance per area is ε/W and W grows roughly as the square root of the reverse junction potential. A plot of 1/C² against bias can estimate doping under specific assumptions. Use total capacitance and device area consistently, and identify series resistance or interface-state dispersion before extracting parameters.

A measured barrier from I–V can differ from C–V because the methods weight nonuniformity differently. Current can preferentially cross low-barrier patches, while capacitance samples a broader area. Disagreement is a clue about the interface, not an automatic reason to average the values.

J≈A**T²exp(−φ_B/V_T)[exp(V/(ηV_T))−1]

Model assumptions: Ideal thermionic-emission model, φ_B in volts numerically equivalent to eV per elementary charge; η is ideality factor.

Separate conduction and charge response

Measure I–V with current compliance and record temperature.

Measure small-signal C–V over frequency to look for dispersion and series-resistance artifacts.

Fit only a valid regime and report area, bias convention and assumed uniformity with any extracted parameter.

Mistake to avoid: A straight semilog current segment alone is not proof that every part of the interface shares one barrier height.

Worked problem: With η=1 at 300 K, estimate voltage increase for a tenfold forward current rise in the exponential regime.

ΔV=V_T ln10≈59.5 mV. Series resistance and other mechanisms can change this slope.

Self-test: Why is a barrier-fit result sensitive to temperature?

Thermionic emission depends exponentially on barrier divided by thermal energy.

At constant temperature, what factor does ideal thermionic saturation current acquire if barrier energy increases by 0.1 eV at 300 K?

It is multiplied by exp(−0.1/0.02585)≈0.0209: roughly a 48-fold decrease, assuming unchanged prefactor, area and transport mechanism.

PN-junction electrostatics

Initially electrons diffuse from n to p and holes from p to n. Ionized dopants left near the interface create a space-charge field that opposes further majority diffusion. At equilibrium the built-in potential balances these tendencies, with a constant Fermi level.

Under the depletion approximation, mobile carriers are neglected inside the space-charge region. Charge neutrality gives N_A x_p=N_D x_n, so most width lies on the lightly doped side. Reverse bias increases depletion width and lowers junction capacitance.

Build the physical picture

At the junction, majority carriers initially diffuse down concentration gradients. Electrons leaving the n side uncover positively charged ionized donors; holes leaving the p side uncover negatively charged ionized acceptors. These dopants remain in the lattice. The resulting electric field points from n to p and produces drift opposing the initial diffusion.

Equilibrium is reached when the Fermi level is constant. The depletion region contains very few mobile carriers compared with fixed dopant charge, but it is not literally empty silicon. Its electrostatic potential is an internal balance, not a free energy source extractable by attaching a voltmeter.

From mechanism to a usable model

Inside an abrupt uniformly doped depletion region, charge density is piecewise constant. Poisson therefore gives a field that is linear on either side, and integrating that field gives parabolic potential segments. Requiring the field to vanish at both depletion edges implies NA xp=ND xn. The lightly doped side needs more width to supply the same total charge.

Reverse bias increases the total potential barrier from Vbi to Vbi+|VR|. Width grows as the square root of that potential, so capacitance falls. Forward bias reduces the barrier; as high injection or a very small barrier is reached, the depletion approximation becomes unreliable.

V_bi=V_T ln(N_A N_D/nᵢ²); W=√[(2ε/q)(1/N_A+1/N_D)(V_bi−V)]

Model assumptions: Abrupt junction, complete ionization, nondegenerate statistics; V is applied forward bias and depletion model requires V_bi−V>0.

Derive depletion width rather than memorize it

Integrate dE/dx=ρ/ε on each side using E=0 at the outer edges.

Match field at the metallurgical junction to obtain NA xp=ND xn.

Integrate the triangular field magnitude: potential drop is Emax W/2. Substitute W=xp+xn to obtain the depletion-width formula.

Mistake to avoid: Depletion width is not distributed equally unless the doping magnitudes are equal. Width and charge density compensate one another.

Worked problem: At 300 K with N_A=N_D=10¹⁶ cm⁻³ and nᵢ=10¹⁰ cm⁻³, find V_bi.

V_bi=0.02585 ln(10¹²)≈0.714 V. For silicon ε≈11.7ε₀, W≈0.430 µm at zero bias.

Self-test: Can a voltmeter directly read V_bi between ordinary external contacts at equilibrium?

No; other contact potentials enter the full circuit, and an equilibrium built-in field does not provide a power source.

For NA=10¹⁵ and ND=10¹⁷ cm⁻³, what fractions of W lie on each side?

xp/W=ND/(NA+ND)≈0.9901 and xn/W≈0.0099. About 99% lies in the lightly doped p side, and NA xp=ND xn still holds.

Diode operation

Forward bias reduces the junction barrier and injects minority carriers. Their transport and recombination produce current. Reverse bias normally permits a small leakage current until breakdown mechanisms become important. Breakdown can be avalanche or tunnelling dominated depending on junction structure.

The ideal diode law excludes many practical effects, including series resistance, generation current and high injection. A diode has no exact universal 0.7 V switching boundary. Dynamic charge storage can delay switching after forward conduction.

Build the physical picture

Forward bias lowers the junction barrier and raises minority-carrier concentrations at the depletion edges. Electrons injected into the p region and holes injected into the n region then diffuse and recombine. In the ideal low-injection long-region model, this diffusion current gives the exponential diode law. Reverse bias suppresses injection and leaves a small current associated with minority supply.

The depletion region is not a resistor whose resistance simply decreases. Changing bias changes populations and transport boundary conditions. This is why a constant 0.7 V drop is a rough circuit approximation, not the defining physics of a silicon diode.

From mechanism to a usable model

The small-signal resistance follows the local slope: rd=(dI/dV)⁻¹≈nVT/I at significant forward current. Diffusion capacitance comes from stored minority charge, while depletion capacitance comes from changing space charge. Both affect switching.

Reverse breakdown can involve avalanche multiplication or tunnelling. At large forward current, series resistance makes terminal voltage include I Rs, and self-heating changes parameters. Choose a model for the operating region rather than extending the ideal exponential across all voltages and currents.

I=I_S[exp(V/(ηV_T))−1]

Model assumptions: Idealized exponential region; η=1 for ideal diffusion current, with real mechanisms changing ideality.

Move from large-signal law to incremental response

Choose the DC operating point and evaluate current there.

Differentiate I=IS[exp(V/(nVT))−1] to obtain local conductance.

Use ΔI≈ΔV/rd only for a sufficiently small perturbation; add capacitances for time-dependent analysis.

Mistake to avoid: A diode is not “off until 0.7 V.” There is no universal abrupt turn-on voltage independent of current, area and temperature.

Worked problem: For I_S=1 nA, η=1 and V=0.3 V at 300 K, estimate current.

I≈10⁻⁹[exp(11.61)−1]≈0.110 mA. A larger current may require including series resistance.

Self-test: Why does forward voltage change with current?

Injection is exponential rather than an ideal fixed-drop switch; real series resistance adds further voltage.

At 300 K, n=1 and I=1 mA, estimate rd and the current change for a 1 mV perturbation.

rd≈25.85 Ω. A 1 mV small perturbation gives ΔI≈38.7 µA. This is an incremental prediction around 1 mA, not a replacement for the full nonlinear I–V curve.

Solar cells

Light generates carriers that can be separated and collected by device fields and selective contacts. The resulting photocurrent opposes the dark diode current under the generator convention. Short-circuit current occurs at zero voltage; open-circuit voltage occurs at zero external current.

Maximum power is below V_oc I_sc because no point simultaneously achieves both. Fill factor compares actual maximum power with that rectangle. Series resistance, shunts and recombination reduce output.

Build the physical picture

Light absorbed above an allowed transition energy generates electron–hole pairs. A solar cell must separate and collect those carriers before they recombine. The depletion field helps near the junction, while diffusion carries photogenerated minority carriers from nearby neutral regions. Collection therefore depends on absorption depth, diffusion length and surface losses.

A useful terminal model is a photogenerated current source in parallel with a diode, plus series and shunt resistances. Under the usual power-delivery convention, I=Iph−IS[exp(V/(nVT))−1]. Short circuit probes collection at V=0; open circuit establishes a voltage that balances photocurrent against recombination-related current.

From mechanism to a usable model

At open circuit, no external current flows, but generation and internal recombination continue. The relation Voc≈nVT ln(Iph/IS+1) shows logarithmic rather than linear growth with illumination. Maximum power lies between short and open circuit; neither endpoint delivers power because one has zero voltage and the other zero current.

Efficiency divides maximum electrical power by incident optical power. Fill factor compares maximum power with Voc Isc. Report irradiance, illuminated area, temperature and spectral conditions; efficiencies from different conditions are not automatically comparable.

I=I_ph−I_S[exp(V/(ηV_T))−1]; FF=P_max/(V_oc I_sc); efficiency=P_max/P_in

Model assumptions: Single-diode model without resistance for the current expression; defined illuminated area and incident power.

Find the relevant photovoltaic quantities

Read Isc at zero terminal voltage and Voc at zero terminal current using a consistent sign convention.

Calculate V I over the power-producing branch and locate its maximum.

Compute FF=Pmax/(Voc Isc) and η=Pmax/Pin; identify contact or shunt losses from curve shape.

Mistake to avoid: The built-in potential is not identical to Voc, and a large absorption coefficient alone does not guarantee efficient charge collection.

Worked problem: V_oc=0.6 V, I_sc=30 mA and P_max=13.5 mW. Find FF.

FF=13.5/(0.6×30)=0.75. If incident power is 100 mW, efficiency=13.5%.

Self-test: Why does current alone not identify maximum power?

Power is VI; output voltage changes with the load, so the product must be evaluated.

For Voc=0.6 V, Isc=20 mA and Pmax=8 mW, find fill factor.

Voc Isc=12 mW, so FF=8/12≈0.667. Efficiency cannot be found until incident optical power is given.

Bipolar junction transistors

In forward-active NPN operation, the emitter–base junction injects electrons into a thin p-type base. Most traverse the base and are collected by the reverse-biased base–collector junction. Base current accounts for recombination and injection inefficiencies; it is not the current that simply passes through a mechanical valve.

Cutoff, forward active, saturation and reverse active follow the bias of both junctions. Saturation forward-biases both and stores charge. PNP devices reverse carrier and voltage polarities.

Build the physical picture

For an NPN transistor in forward-active mode, the emitter–base junction is forward biased and the collector–base junction reverse biased. Electrons injected from the heavily doped emitter diffuse across the thin p-type base. Most reach the collector depletion field and are swept into the collector. Base current accounts for recombination and less efficient injection in the other direction.

The collector supply provides output power; a small base current controls a larger carrier flow but does not create energy. A transistor is not just two independent back-to-back diodes: shared-base transport couples its junctions, which independent diodes do not reproduce.

From mechanism to a usable model

Define α=IC/IE and β=IC/IB in the stated operating regime. Since IE=IC+IB, β=α/(1−α). A small change of α near one causes a large change of β. In forward active operation, IC depends exponentially on VBE in the simplest model, and gm=IC/VT.

When both junctions become forward biased, the transistor is in saturation and the active-mode relation IC=βIB no longer reliably sets collector current. Early effect, high injection and base resistance also alter characteristics. Identify the bias of each junction before choosing equations.

I_C≈I_S exp(V_BE/V_T); β=I_C/I_B; g_m=I_C/V_T

Model assumptions: Forward active, low injection, negligible Early effect for the first expression.

Diagnose BJT operating mode

For NPN, compare emitter, base and collector potentials to determine each junction bias.

Use forward-active transport equations only when the emitter injects and the collector junction remains reverse biased.

Check terminal-current conservation and load limits; if the active-mode prediction is impossible, re-evaluate the mode.

Mistake to avoid: β is not a precise universal design constant. It depends on device, current, temperature and operating mode.

Worked problem: For I_C=1 mA and β=100, find I_B and g_m at 300 K.

I_B=10 µA; g_m=0.001/0.02585≈38.7 mS. β varies with operating point and device.

Self-test: Why must the base be thin for useful transistor action?

Injected carriers need to reach the collector before substantial base recombination.

If α=0.99, find β. If collector current is 1 mA at 300 K, find gm.

β=0.99/0.01=99. gm=0.001/0.02585≈0.0387 S. The first is dimensionless; the second is a slope in amperes per volt.

BJT dynamic behaviour

Small-signal models linearize the transistor around a bias point. Transconductance links small base–emitter voltage changes to collector current, while input resistance relates base current and voltage. Junction and diffusion capacitances store charge and limit high-frequency response.

Deep saturation can create long storage delay when switching off. Small-signal frequency limits and large-signal switching delay answer different questions. Always state bias and amplitude before applying a linear model.

Build the physical picture

A BJT takes time to move and remove stored minority charge. In forward active operation, base transit and emitter/collector charging contribute delay. Driving a device into saturation can store additional charge that must be removed before it turns off, producing a storage interval even after the input changes.

Small-signal hybrid-π parameters linearize around a DC operating point: gm=IC/VT, rπ=β/gm, and capacitances Cπ and Cμ describe charge response. These are local parameters, not one fixed model for cutoff, active operation and deep saturation.

From mechanism to a usable model

Cμ connects output back to input. In an inverting amplifier, the changing collector voltage makes it appear larger at the input through the Miller effect, approximately Cin≈Cπ+Cμ(1−Av) for an appropriate voltage-gain model. This can dominate bandwidth even when the physical capacitor is small.

fT is the frequency where short-circuit current gain falls to unity under a specified small-signal test. It is not a guaranteed usable amplifier bandwidth or maximum switching frequency. Real load, voltage gain, feedback and layout change the response.

g_m=I_C/V_T; r_π=β/g_m; f_T≈g_m/[2π(C_π+C_μ)]

Model assumptions: Simplified hybrid-π estimate; parasitics, bias and transit effects affect actual values.

Build a first bandwidth estimate

Find the bias current, then calculate gm and rπ.

Identify capacitances connected to moving nodes; apply Miller only where its assumptions fit.

Combine effective capacitance with the resistance seen at that node to estimate a pole, and compare with a fuller multi-pole model.

Mistake to avoid: Faster input edges do not necessarily remove saturation storage. The stored charge and available reverse base drive matter.

Worked problem: For g_m=40 mS and β=100, find r_π.

r_π=100/0.04=2500 Ω. With total capacitance 20 pF, the simplified f_T≈318 MHz.

Self-test: Why might a saturated switch turn off more slowly than an active transistor?

Stored excess charge must be removed before collector current can fall rapidly.

For Av=−20 and Cμ=2 pF, estimate its Miller contribution to input capacitance.

Cμ(1−Av)=2×21=42 pF. Add Cπ and parasitics before calculating an input pole; this approximation does not independently predict every pole.

MOS capacitor

A MOS capacitor places a gate and dielectric over a semiconductor. For p-type silicon, negative gate bias accumulates holes. Positive bias first repels holes and creates depletion, then can establish an electron inversion layer. Flatband voltage includes work-function and charge effects.

Measured C–V depends on frequency and whether minority carriers can respond. Oxide and depletion capacitances appear in series in depletion. An ideal electrostatic sketch omits interface traps and leakage; these matter in fabricated structures.

Build the physical picture

A MOS capacitor has a conducting gate, an insulating oxide and a semiconductor substrate. The gate need not inject carriers through the oxide to control surface charge: its field redistributes carriers already supplied through the semiconductor. For p-type silicon, negative gate bias attracts majority holes into accumulation. Positive bias first repels them, leaving ionized acceptors in depletion, then attracts electrons into inversion.

The surface potential describes band bending relative to the bulk. It is not simply the full applied gate voltage because some voltage falls across the oxide. Work-function difference and fixed oxide charge shift flatband voltage, where semiconductor bands are ideally flat.

From mechanism to a usable model

In depletion, the gate-to-bulk small-signal capacitance consists of oxide capacitance and semiconductor depletion capacitance in series. A thicker depletion region reduces Cdep. Around strong inversion, the surface potential is approximately twice the bulk Fermi potential in the conventional ideal model.

Frequency matters because inversion carriers must respond to the AC perturbation. At low frequency, an equilibrium inversion layer can restore capacitance toward Cox. At high frequency, minority response may be too slow, leaving a low series capacitance. A rapid sweep can cause deep depletion; interface traps produce frequency-dependent charge response distinct from ideal minority-carrier behaviour.

C_ox/A=ε_ox/t_ox; 1/C_total=1/C_ox+1/C_dep

Model assumptions: Planar capacitor, neglect fringing and defects; high- versus low-frequency inversion response must be specified.

Derive the series-capacitance relation

Apply a small charge change dQ to the gate; the oxide drop changes by dQ/Cox.

The semiconductor drop changes by dQ/Cdep under the depletion approximation.

Add the drops: dV=dQ(1/Cox+1/Cdep), hence C=dQ/dV. Use either per-area or total capacitances consistently.

Mistake to avoid: A MOS capacitor has no source-to-drain channel current. Its C–V is a charge-response measurement, not a MOSFET output characteristic.

Worked problem: For SiO₂ ε_r=3.9 and t=10 nm, find oxide capacitance per area.

C_ox/A≈3.45×10⁻³ F/m²=0.345 µF/cm². Halving ideal oxide thickness doubles capacitance but can increase leakage risks.

Self-test: Why can high-frequency and low-frequency inversion capacitances differ?

Minority-carrier response to the small AC signal can be too slow at high frequency.

If Cox/A=0.345 µF/cm² and Cdep/A=0.100 µF/cm², find total capacitance per area.

C/A=(0.345×0.100)/(0.345+0.100)≈0.0775 µF/cm², smaller than either series element. Adding them as parallel capacitors would give the wrong result.

MOSFET fundamentals

A MOSFET adds source and drain contacts to an inversion channel controlled by the gate. Drain voltage creates a lateral potential gradient; channel charge varies along the path. In a long-channel model, increasing V_DS eventually pinches off the channel near the drain and produces a saturation region.

Saturation does not mean the transistor is off or that current becomes exactly independent of voltage. Body effect changes threshold, and short-channel effects invalidate simple square-law predictions. Distinguish device threshold from a logic-level guarantee or a low-resistance operating condition.

Build the physical picture

In an NMOS device, positive gate overdrive creates an inversion layer connecting source and drain. Along the channel, potential rises from source toward drain. The local inversion charge is approximately −Cox(VGS−VT−V(x)), so the channel becomes less strongly inverted near the drain as VDS increases.

A lateral field drives carriers through this position-dependent sheet charge. Integrating the gradual-channel relation gives the triode-region equation. At VDS≈VGS−VT, inversion charge at the drain end tends toward zero in the ideal model. Carriers still traverse a high-field drain-end region: pinch-off does not mean zero drain current.

From mechanism to a usable model

In long-channel saturation, the simple square-law model predicts ID=β(VGS−VT)²/2. Transconductance then follows gm=β(VGS−VT)=2ID/(VGS−VT). Body bias changes threshold, channel-length modulation gives finite output conductance, and high-field mobility/velocity effects invalidate the square law for many short-channel devices.

MOSFET “saturation” differs from BJT saturation. The MOS term denotes a channel-current regime useful for amplification; the BJT term denotes both junctions forward biased. For a power switch, low RDS(on) is often achieved in the MOS ohmic/linear regime with adequate gate drive.

Linear: I_D=β[(V_GS−V_T)V_DS−V_DS²/2]; saturation: I_D=β(V_GS−V_T)²/2

Model assumptions: Long-channel gradual-channel model; β=μC_oxW/L, constant mobility, no channel-length modulation.

Obtain the long-channel drain-current equation

Write inversion sheet charge Qinv(x)=−Cox[VGS−VT−V(x)].

Combine charge, drift mobility and lateral potential gradient with a constant channel current.

Integrate V from 0 to VDS over length L to get ID=μCox(W/L)[(VGS−VT)VDS−VDS²/2]. Identify the saturation boundary before using it.

Mistake to avoid: The channel pinching off at its drain end does not disconnect the whole transistor. Nor does a threshold specification guarantee low on-resistance.

Worked problem: For β=1 mA/V², V_GS=2 V, V_T=1 V and V_DS=0.2 V, find I_D.

Since V_DS<1 V overdrive, use linear region: I_D=1×(0.2−0.02)=0.18 mA. At V_DS≥1 V the ideal saturation value is 0.5 mA.

Self-test: Why is threshold voltage inadequate for selecting a power-switch gate drive?

It is defined at a small test current and does not establish the required low on-resistance.

At ID=0.5 mA and overdrive 1 V in ideal saturation, find gm.

gm=2ID/overdrive=1 mS. A sufficiently small 10 mV gate change produces approximately 10 µA current change at that operating point.

Advanced semiconductor devices

As channel length shrinks, drain fields affect source injection and threshold, while velocity saturation changes current scaling. Gate leakage and electrostatic control become central constraints. Multigate structures improve control by surrounding more of the channel.

Heterojunctions use band offsets to confine or separate carriers and can create transport advantages unavailable in one material. Compact models summarize measured device behaviour for circuit design; their parameters are tied to a process and valid operating domain.

Build the physical picture

Scaling strengthens the influence of the drain and interfaces relative to the gate. Drain-induced barrier lowering changes the source injection barrier, while velocity saturation can replace the long-channel square-law dependence. FinFET and gate-all-around structures improve electrostatic control by surrounding more of the channel; they do not abolish contacts, scattering or leakage.

A heterojunction joins materials with different band edges. Conduction- and valence-band offsets can confine carriers, block one carrier species or create a two-dimensional population. Always draw the band alignment: “different bandgaps” alone does not specify which carrier sees a barrier.

From mechanism to a usable model

Conventional subthreshold current is approximately exponential in gate voltage. The subthreshold swing S=dVGS/d(log10 ID) describes the voltage required for a decade change. At 300 K the ideal thermionic factor ln(10)kT/q is about 59.5 mV/decade; semiconductor coupling and interface traps increase the swing in the usual MOS model.

This thermal statement is model-specific, not a law forbidding all alternative injection mechanisms. Device comparisons also need leakage, supply voltage, contact resistance, variability and energy-delay trade-offs. A compact model interpolates measured operating behaviour; its name does not validate extrapolation far beyond the fitted domain.

Thermal subthreshold swing S≈ln(10)(kT/q)(1+C_dep/C_ox)

Model assumptions: Conventional thermionic MOSFET approximation; not a universal bound for all transistor concepts.

Compare scaling proposals fairly

Identify which limitation is addressed: electrostatics, mobility, leakage, contacts or variability.

Compare at matched geometry, temperature, bias and a stated off-current criterion.

Check whether the model includes the mechanism being claimed; use a device model appropriate to the structure.

Mistake to avoid: A smaller physical transistor does not automatically yield a proportional speed increase, because interconnect and contact delays may dominate.

Worked problem: At 300 K with C_dep/C_ox=0.5, estimate swing.

S≈59.5×1.5=89.3 mV/decade. Lower swing gives stronger current change per gate-voltage change.

Self-test: What does drain-induced barrier lowering mean?

Drain voltage reduces the source-channel barrier, increasing off-state injection and changing apparent threshold.

For S=90 mV/decade, what gate-voltage increase gives a 1000-fold subthreshold current rise?

1000 is three decades, so ΔVGS≈3×90=270 mV, within the bias interval where the swing is approximately constant.

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