semiconductors [Fonstad, Sze02, Ghione] Semiconductors - - PowerPoint PPT Presentation

Properties of semiconductors

[Fonstad, Sze02, Ghione]

Semiconductors

 Conducibility:

• Insulators: s<10-8 S/cm
• semiconductors: 10-8<s<103 S/cm, according to doping
• conductors: s >103 S/cm

 conduction is due to free charges, knows as carriers

• electrons
• holes

Semiconductors

 These is the part of

the periodic table of elements we are interested in

Silicon crystal structure

 Two face-centered cubic

structures

 Si: 5.1022 atoms/cm3  GaAs is similar, but with two

different species of atoms

Where do bands come from?

 Atomic Si outer shell is 3s2p2  With N atoms: due to the atoms’ close proximity, the two outer

electronic levels split in N levels, generating two bands of allowed energies

vertical axis: energy of electrons

For a< ao (it’s the real case)

• for T=0
• the valence band

(VB) is completely full

• the conduction

band (CB) is completely empty

• > no conduction

states states states

Inter-atom distance

states states

Where do bands come from?

• for T>0
• some electrons

jump from BV to BC

• VB is almost

completely full, i.e. there are some holes

• CB is almost

completely empty

• conduction is given

by both holes in BV and electrons in BC

states states states

Inter-atom distance

states states

Holes

 Holes correspond to broken covalent bonds  A hole moving is the motion of the broken bond from one

electron to another

hole

The gap

 is the forbidden

band between VB and CB

 its width

depends on

• material
• temperature

 Isolators have a

very large gap (~ 5 eV or more)

 At higher T, it’s

easier for electrons to jump from VB to CB

How many electrons in CB, and holes in VB?



with rn/p(E): number of free electrons/holes per unit of volume and with energy between E and E+dE n/p: density of free electrons/holes in CB/VB

 rn/p(E) depends on

• number of available states per unity of volume and

energy, Nn/p(E)

• probability of occupation of each state, fn/p(E)

Density of states

 For electrons, it’s proportional to

(E-EC)1/2

 for holes, the vertical axis should

be inverted (they are positive -> their potential energy has different sign)

 the two curves are slightly different

due to different m* (see later)

as

Fermi function

 Probability of occupation of an allowed state at anergy E

for electrons: Fermi Dirac function with kB Boltzmann constant, 1.38*10-23 J/K EF Fermi level, the energy for which f(E)=1/2 (for any T)

 For holes, it’s the

complement to one: fp(E)=1-fn(E)

In an intrinsic material

 at thermal equilibrium only we have

Computing n and p

 It is not possible to compute

in a closed form. But in the Boltzmann approximation (good if E>>EF) f(E)~ exp(-(E-EF)/(kBT)) this is feasible, and we get with NC/V~(Tmn/p*)3/2

Generation and recombination

 In a pure, or intrinsic, semiconductor at thermal

equilibrium, n and p are set by the equilibrium between

• generation of couples electrons-holes due to

 thermal generation  optical generation  generation by collision

with absorption of energy

• recombination of couples electrons-holes

with release of energy

 in any case, momentum is conserved (of course!)  In an intrinsic (i.e., not doped) material

n=p=ni=pi

ni

 depends on the material, and increases with T  at T=300K, ni~1.5 .1010 cm-3

In an intrinsic material

 Let us call EFi its Fermi level  From ni=pi, i.e.

we get (with NC~NV)

n doping

 By replacing Si atoms with donor atoms, with more than 4

electrons on the outer shell (e.g. V group: P, As, Sb): n doping

 The “fifth” electron is weakly bound to its atom, so at room

T it is free – in CB!

 The donor atom becomes a fixed positive ion  So we have a fixed ion and moving electron.

p doping

 By replacing Si atoms with acceptor atoms, with less than

4 electrons on the outer shell (e.g. III group: B, Al, Ga, In): p doping

 A covalence bound is missing... at room T, it’s easy for

• ther covalence electrons to move there

 We then have a missing covalent bound which is moving...

a hole in VB!

 The acceptor atom becomes a fixed negative ion  So we have a fixed ion and a moving hole.

Doping The same, seen on the band diagram:

 these are the distances (in

energy, meV) between the energy level of the extra/missing electron and the CB/VB

 these are shallow levels  Au, instead, has a deep

level (i.e. far from CB and VB): it will act as a trap (see later)

Dependence of n and p on T

 E.g. for n doping  Three regions

 freezing: only partial

ionization

 extrinsic: normal condition,

full ionization

 intrinsic: full ionization, but

thermal generation covers doping, so that the material is again intrinsic

If n>p, as here, electrons are majority carriers and holes are minority carriers; vice-versa if n<p

n

Doping and EF

 Doping moves the Fermi level: at T=300K,

 for n doping, from

we get

 similarly, for p doping

Only in thermal equilibrium

 we have, e.g. for n doping

Only in thermal equilibrium

 similarly, for p doping

Law of mass action and Shockley equations

 In a non-degenerate (i.e. when the Fermi level is in the

gap, see later) semiconductor (either intrinsic or not) it holds

 i.e. np depends only on the material and on T  In particular, for an intrinsic semiconductor, nipi=ni

2, then in

thermal equilibrium (but also for doped material) np=ni

2

 Shockley equations:

Doping

 Let ND and NA be the concentrations of donor and acceptors

(typically, 1014~1019 cm-3) (remember that Si atoms concentration is 5.1022cm-3)

 Let NA

• and ND

+ be the concentrations of ionized donor and

acceptors; normally, NA

• =NA and ND

+=ND

 With the exception of depletion layers (see later), the material is

always (almost) locally neutral (neutrality hypothesis), so that

 What normally matters is  At T=300K,

if ND>>NA : if NA>>ND :

at at

Tendence to neutrality [Nichols]

• With the exception of space-charge regions (see later), the

semiconductor is always almost electrically neutral

• This, because it has a strong tendency to neutrality
• Indeed, from the continuity equation and Gauss

we get

• and if we define the dielectric relaxation time e/s=tR
• we get
• and the solution is

Tendence to neutrality [Nichols]

• With “normal” values for e and s we get

tR ~ 10-10 – 10-15 s

• i.e. the material has a strong (fast) tendency to neutrality
• This does not apply in space-charge regions. Why?
• Actually
• and we had neglected the term grads. This does not apply in

high injection regime (see later).