Chemistry 140a Lecture #5 Jan,

Fermi-Level Equilibration When placing two surfaces in contact, they will equilibrate; just like the water level in a canal lock. The E F of the semi-conductor will always lower to the E F of the metal or the solution. This can be understood by looking at the density of states for each material/soln. Initial E F Semi-ConductorMetal/Soln. Eq. E F Initial E F Eq. E F

Fermi-Level Equilibration Charge comes from the easiest thing to ionize, the dopant atoms. This leads to a large region of (+) charges within the semi-conductor. In the metal all of the charge goes to the surface. (Gauss’s Law) The more charge transferred the more band bending. E x EFEF EFEF E x EFEF E CB E VB E CB V bi E x EFEF EFEF E x EFEF E CB E VB E CB V bi

Depletion Approximation All donors are fully ionized to a certain distance, W, from the interface. W=W(N D,V bi ) X W NDND W V bi W

Final Picture E x EFEF EFEF E x EFEF E CB E VB E CB V bi E Vac  sc mm EgEg

Useful Equations Poisson’s Eqn: E(x) E(x) = Electric Field (V/cm)  (x) = Electric Potential (V)  (x) = Electric Potential Energy (J)

Electric Potential (V) Integrate Poisson’s Eqn. B.C.’s Result: X W Q=qN D W  x) qN D -qN D W 2 (2K  0 )  (x) x quadratic  V bi =

Depletion Width Rearranging for W: As expected, W increases w/ V bi and decreases w/ N D If one accounts for the free carrier distribution’s tail around x=W

Typical Values V bi max (V)N D (cm -3 ) W (  m) Q (C/cm 2 ) *10 11

Electric Potential Energy E(x) = -q  (x)  (0) = -V bi qV bi = (E F,SC -E F,M )  B = V bi + V n –Barrier height –Independent of doping –V bi and V n are doping dependent  (x) x E x EFEF E CB E VB V bi BB e-e- h+h+ Net = Eq. VnVn

Electric Field (V/cm) E(x) W E max =-qN D W/(K    x E  x)

I-V Curve No Band Bending Low Band Bending High Band Bending I V

Review N-typeP-type E x EFEF E CB E VB V bi E Vac  sc mm EgEg E x EFEF E CB E VB V bi E Vac  sc mm EgEg __ -

Solution Contact 10^17 atoms in 1mL of 1mM solution D.O.S. argument holds Difference in exchange current across the interface A- Li Angstroms *Significantly less than typical W ~ 10nm

Semiconductor Contacting Phase No longer 1-Sided Abrupt Jxn. as the semi-conductor doesn’t have infinite capacity to accept charge Assume N D (n-type)=N A (p-type), then W n =W p e- h+ n-type p-type Diode directionalized current

Degenerate Doping Dope p-type degenerately N A >>N D --> 1-sided Abrupt Jxn. WnWn WpWp P-N Homojunction N-type B B B P + -type

Heterojunctions 2 different semiconductors grown w/ the same cyrstal structure (difficult) –Ge/GaAsa o ~5.65 angstroms NormalStaggered Broken

LASERs 3 Pieces --> 2 Heterjunctions –p - (Al,Ga)As | GaAs | n - (Al, Ga) As e-e- h+h+ h Traps electrons and holes

Fermi-Level Pinning  E F,M 1 Ideal Case (only works for very ionic semiconductors like TiO 2 and SnO 2) Never works for Si Fermi-Level Pinning Sze p. 278 Slope  A-B 1 CdS TiO 2 SnO 2 Si GaAs

What’s Missing? Fermi-Level pinning hurts –Hinders our ability to fine tune V bi V bi/Ni ~V bi/Pt ~V bi/Au Why does this happen? E x EFEF E CB E VB E Vac BB E x EFEF E CB E VB E Vac BB vs. *Solution contact for GaAs sees Fermi-level pinning, while the barrier height correlates well with the electro-chemical potential for solution contact to Si

Devious Experimenter Given a Si sample with a magic type of metal on the surface X Thus the Fermi-level will alwaysequilibrate to the Fermi-level of X Thin interface --> e - ’s tunnel through it and no additional potential drop is observed E x E F,X E CB E VB E x E F,X E CB E VB E F,X -E F,M

What is X? Any source or sink for charge at the interface –Dangling bonds –Surface states –etc.

Questions –Abrupt 1-sided junction (What is it?) –Sign of Electric P.E. and Electric Potential (Are they correct? I put them as they were in the notes, but this doesn’t seem to agree with the algebra to me)