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 <10 3 S/cm, according to doping - conductors: s >10 3 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 . 10 22 atoms/cm 3  GaAs is similar, but with two different species of atoms

Where do bands come from?  Atomic Si outer shell is 3s 2 p 2  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 states For a < a o (it’s the states real case) states states • for T =0 - the valence band (VB) is completely full states - the conduction band (CB) is completely empty -> no conduction Inter-atom distance

Where do bands come from? • for T >0 - some electrons jump from BV to BC - VB is almost completely full, i.e. states there are some holes states - CB is almost states states completely empty - conduction is given by both holes in BV and electrons in BC states Inter-atom distance

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 r n/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  r n/p (E) depends on - number of available states per unity of volume and energy, N n/p (E) - probability of occupation of each state, f n/p (E)

Density of states  For electrons, it’s proportional to (E-E C ) 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 as due to different m* (see later)

Fermi function  Probability of occupation of an allowed state at anergy E for electrons: Fermi Dirac function with k B Boltzmann constant, 1.38*10 -23 J/K E F Fermi level, the energy for which f(E)=1/2 (for any T )  For holes, it’s the complement to one: f p (E)=1-f n (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>>E F ) f(E)~ exp (-(E-E F )/(k B T)) this is feasible, and we get with N C/V ~(Tm n/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=n i =p i

n i  depends on the material, and increases with T  at T= 300K , n i ~1.5 . 10 10 cm -3

In an intrinsic material  Let us call E Fi its Fermi level  From n i =p i , i.e. we get (with N C ~N V )

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 other 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 n  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

Doping and E F  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 2 , then in  In particular, for an intrinsic semiconductor, n i p i =n i thermal equilibrium (but also for doped material) 2 np=n i  Shockley equations:

Doping  Let N D and N A be the concentrations of donor and acceptors (typically, 10 14 ~10 19 cm -3 ) (remember that Si atoms concentration is 5 . 10 22 cm -3 ) - and N D + be the concentrations of ionized donor and  Let N A - =N A and N D + =N D acceptors; normally, N A  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, at if N D >>N A : if N A >>N D : 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=t R • we get • and the solution is

Tendence to neutrality [Nichols] • With “normal” values for e and s we get t R ~ 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 grad s . This does not apply in high injection regime (see later).