The Physics of Infrared Detectors (with a special emphasis on MID-IR detectors) Massimo Robberto (ESA STScI) Outline

Semiconductors as crystals Energy bands Intrinsic semiconductors Light on intrinsic semiconductors Extrinsic Semiconductors Impurity band conductors (IBC) Periodic Table Atomic structure: (four 2s2p) covalent bonds per atom Crystal Structure: 2-d

simple cubic Bravais lattice In two dimensions, there are five distinct Bravais lattices R ma nb pc Crystal Structure: 3-d

In three dimensions, there are fourteen distinct Bravais lattices A large number of semiconductors are cubic (a1=a2=a3= a lattice constant) simple cubic body-centered cubic face centered cubic Diamond Structure

Diamond Structure Diamond Structure Diamond Structure Diamond Structure Diamond Structure Diamond Structure Diamond Structure

Diamond Structure Two interpenetrating face-centered cubic lattices Diamond lattice: All same atoms (e.g. Si) Zincblende lattice: different atoms in each sublattice (e.g.CdTe, GaAs) Lattice planes in crystals:

Miller indexes Silicon and Germanium break on {1,1,1} planes never confuse the spacing between lattice planes with the spacing between crystal planes Example The surface density of Silicon is 11.8 1014 atoms/cm 2 on {111} 9.6 1014 atoms/cm 2 on {110} 6.8 1014 atoms/cm 2 on {100}

Wigner-Seitz cell The smallest (primitive) cell which displays the full symmetry of the lattice is the Wigner-Seitz cell. Construction method: surfaces passing through the middle points to the nearest lattice points In 3-d think to a polyhedron Reciprocal lattice g ha kb lc

The Bravais lattice after Fourier transform real space normals to the planes(vectors) spacing between planes reciprocal lattice points 1/distance between points (actually, 2/distance) (distance, wavelength) Bravais cell

2k (momentum, wave number) Wigner-Seitz cell Reciprocal lattice Bragg conditions When a wave impinges on a crystal - and it doesn't matter if it is an electromagnetic wave, e.g. X-rays, or an electron, or neutron "wave" - it will be reflected at a particular set of lattice planes {hkl} characterized by its reciprocal lattice vector g only if the so-called Bragg condition is met

k k g If the Bragg condition is not met, the incoming wave just moves through the lattice and emerges on the other side of the crystal (neglecting absorption) Elastic scattering k k For a given k, The Bragg conditions is met on surfaces normal to particular g. These surfaces define cells in the k-space, called

Brillouin zones Brillouin Zone construction All wave vectors that end on a BZ, will fulfill the Bragg condition and thus are diffracted. Wave vectors completely in the interior of the 1. BZ, or in between any two BZs, will never get diffracted; they move pretty much as if the potential would be constant, i.e. they behave very close to the solutions of the free electron gas. Brillouin zone The Brillouin zone is defined in the reciprocal lattice. First BZ

Second BZ Third BZ In 3-d think to a nested set of polyhedra The first BZ is the volume enclosed within a Wigner-Seitz cell in the k-space. WS zone and BZ Lattice Real Space Lattice K-space bcc WS cell

Bcc BZ (fcc lattice in K-space) fcc WS cell fcc BZ (bcc lattice in K-space) The WS cell of bcc lattice in real space transforms to a Brillouin zone in a fcc lattice in reciprocal space while the WS cell of a fcc lattice transforms to a Brillouin zone of a bcc lattice in reciprocal space. Bcc: body-centered cubic; fcc: face-centered cubic

Brillouin zone of Silicon Points of high-symmetry on the Brillouin zone have specific importance. The most important point for optoelectronic devices is the center at k = 0, known as the gamma point . Note the points , X, W, K, Wave vectors near or at a BZ - let's call them kBZ electrons - feel the periodic potential of the crystal while the others do not. E.g., they are diffracted. ENERGY GAP in CRYSTALS

On the BZ it is k=-k: these are two standing waves described by ~eikr and e-ikr Their combination can be symmetric eikr+ e-ikr ~ cos(kr) or antisymmetric eikr- e-ikr ~ sin(kr) The probability density have different values at each point 2 different values of the energy, varying with k on the BZ: ENERGY GAP. E(k BZ ) =

k 2m 2 U (g) Energy bands F. Block solved the Shroedinger equation for an electron in the lattice: 2 2

2m V (r ) k (r ) Ek k (r ) If V(r) is periodic with periodicity of the lattice, then the wave Function is a plane wave (free electron) with periodic modulation k (r ) eik rU n (k , r ) Block function k is a wave vector in the reciprocal lattice, Un(k,r) is periodic in r, i.e. U(r+R)=U(r), and n is the band index. -For a given n, it is sufficient to use ks in the primitive cell of the reciprocal lattice (Brillouin zone). The rest is redundant! From:

Band Structure and Blochs Theorem k (r ) eik rU n (k , r ) Block function It is also: k g (r ) ei ( K g )rU n (k g , r ) k (r ) Then: for dispersion curves E (k g ) E (k ) that have a different origin There are many energy values for

one given k. In particular, all possible energy values are contained within the first Brillouin zone (between -1/2g1 and +1/2g1 in the picture). Band Structure and Blochs Theorem reduced representation of the band diagram Every energy branch in principle should carry an index denoting the band (often omitted) Energy functions of a periodic potential

The electron at k1can go to the upper band if someone gives him 1. E > bandgap, AND 2. k2=kl+g THIS IS THE BRAGGS LAW FOR INELASTIC SCATTERING Band diagram of Silicon Si has a band gap of about 1 eV. Si is an indirect semiconductor because the maximum of the valence band (at ) does not coincide with the minimum of the conduction band (to the left of X).

Direct and Indirect Semiconductors Simplified band diagram Conduction band Eg bandgap Valence band 1.24 co Eg (eV )

Fermi Energy and Carrier Concentration The number (or density) of something is given by the density of available places times the probability of occupation. Density of electrons in the energy interval E, E + E = density of states probability for occupancy energy interval dN D( E ) f ( E , T ) dE Density of states A free (or under constant potential) particle in a rectangular box has:

1) only kinetic energy 2k 2 Ek 2m 2) k is discrete (stationary waves) kx, y,z 2 nx , y , z L

ky kx Therefore, the energy is quantized 2 2 2 2 2 2 Ek

n x n y nz 2m L kz Density of states Ds In phase space a surface of constant energy is a sphere. The volume is 4

V k3 3 Any "state", i.e. solution of the Schroedinger equation with a specific k, occupies with 2 electrons the volume 3 2 Vk L The number of cubes fitting inside the sphere at energy E thus is the number of all energy levels up to E: Vk k 3 L3

N s 2 2 V 3 Finally: 1 dN s 1 2m Ds 3 2 2 L dE 2 3/ 2

E1/ 2 Density of states Ds 1 2m Ds 2 2 2 3/ 2 E1/ 2

At 0 K one can place electrons up to the Fermi level EF Probability of occupancy Fermi distribution 1 f(E,T)= e E-E F kT

1 Fermi level E EF >> kT Intrinsic Semiconductors Si HgCdTe InSb The concentration of electrons

in the conduction band is E ne D ( E ) f ( E , T )dE Ec N e eff AT

e 3/ 2 e E EF KT Eg

2 KT N e eff 2 mekT 4 2 h

3/ 2 effective density of states at the band edge With A=4.831021 electrons m-3 K-3/2. The nr. of electrons in the conduction band depends on Eg and T. In Si ne doubles for ~8 degree rise in temperature Same for the holes in the valence band: ne = nh = ni

INTRINSIC CONCENTRATION And the mass action law is 3 2 kT n n n 4 2 e h i2 e

p Eg kT m e

m h 3/ 2 The bandgap energy and the effective mass depend on T Eg(300 K) = 1.1242 eV for Silicon:

Eg(0 K) = 1.700 eV The CONDUCTIVITY e e n e h n h decreases with Eg. Leakage currents are lower in large Eg materials. The mobility is the proportionality constant between the average drift velocity vD of carriers in the presence of an electrical field E: v D E Mobility depends on the average time between scattering processes

e s m Radiation on intrinsic semiconductors at low T the conduction band is empty. Low intrinsic

conductivity radiation with h>EEg creates electron-hole pairs: ne=nh both electrons and holes contribute to the photocurrent, depending on mobilities The conductivity changes: PHOTOCONDUCTOR e e ne h n h e n e h n h Where ne,h = quantum efficiency) photon flux) all atoms count: absorption occurs in thin layer (~10micron) Extrinsic semicondictors

Semiconductor Doped semiconductors acceptors donors N-type and p-type Ruby 4.35ct avg 10mm red serengeti ruby, ACNTV

PRICE $74.95 @ www.jewelrytelevision.com The ruby mineral (corundum) is aluminum oxide with a small amount (about 0.05%) of chromium which gives it its characteristic pink or red color by absorbing green and blue light. ne N e eff f ( Ec , EF , T ) N D N D 1 f ( E A , EF , T ) n h N h eff 1 f ( EV , EF , T )

N A N A f ( E A , EF , T ) e h n N n ND Charge neutrality A

gives EF and therefore the concentrations Extrinsic semiconductors Normally N D N A: Dopant or majority vs. residual impurities Ex: Si:Ga (III group) is p-type N D N A Si:As (V group) is n-type N D N A There are shallow- and deep-level impurities shallow-level impurities increase the response

Atoms of dopant and residual impurities do not interact with each other. p-type: at low T the conduction band is empty; donor impurities have lost their electrons and are ionized, acceptor dopant has electrons from valence band and impurities; valence band has free holes. Opposite requirements To move away and collect charges we need an electric field: dV Ex dx To have a good drop of potential, no current must flow:

Detector material must have high impedence (I.e. low conductivity) On the other hand, we want high concentration of absorbers: this goes in the direction of high conductivity Impurity band conduction The donor band (As in Si:As ) can be heavily doped Photons are absorbed in thin layer (smaller volume) providing higher QE higher radiation immunity lower applied bias better uniformity

faster response High dopant concentration creates a band ~1meV wide, thus the cutoff increasees from 24 to 28m. However: The donor band becomes a conduction band: Impurity Band Conduction Impurity band conduction The donor band (As in Si:As ) is heavily doped Electric field is applied (bias) and the free charge carriers are driven out from the IR active region (depletion): High resistance and high electric field Transparent contact (V+)

Blocking layer (intrinsic Si) h+ e- IR active region (heavily As doped) Substrate (heavily doped n-type) at ground Problem: charges hop in the impurity conduction band; If they are sensed there is an extra dark current: need for a blocking layer (BIB) hole

Photo-electrons and relative holes are collected Si:As Engeneering Dopant As is n-type. P-type impurities (e.g. B) are a potential problem They are neutralized by As, leaving them as negative charge centers in the depletion region These NA charges create an electric field that limits the extention of the depletion region w

Poisson equation for the ionized impurities: dEx eN A dx k0 0 k 0 0 -tB 2k 2

w 0 0 Vb t B eN A 1/ 2 tB The width of the depletion region depends on the bias and on the impurity concentration Assuming NA = 1012 cm-3 and Vb = 1V it is w=32m An acceptable arsenic concentration is ND=3 x 1017 cm-3.

For arsenic in silicon, the absorption cross section is Si:As= 2.2 x 10-15 cm2 the absorption length is l=1/ ND Si:As = 15m. Since l