SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

224 CHAPTER 5. SEMICONDUCTOR JUNCTIONS

EFm

eVbi

eφB

x

z

y

n

n(E-EC)

EC

EFs

Figure 5.5: Schottky Barrier in equilibrium

Also shown is the electron distribution:

n(E−EC)=2f(E−EC)·N(E−EC) (5.3.6)

similartothecaseofap−njunction, the factor of 2 in accounting for electron spin. Thermionic
emission assumes that all electrons in the semiconductor with kinetic energy in the+zdirection
greater thaneVbi(Ez >eVbi)andkz > 0 , are capable of surmounting the barrier and con-
tributing to current flow from the semiconductor to the metal,Js→m. Note that the total kinetic
energyE−EC=Ex+Ey+Ez. At thermal equilibrium the current from the metal to the
semiconductor,Jm→s, will be equal in magnitude and opposite in sign toJs→m, making the net
current zero. To calculateJs→mone needs to sum the current carried by every allowed electron:

Js→m=e

∑

n(E−Ec)·vz (5.3.7)

forEz>eVbiandvz> 0. The methodology employed is to calculate the number of electrons
at energyEin a volume ofk-space(dk)^3 , multiply the number with the electron velocity in the
direction along the barrier, and sum or integrate over energy. Assuming a crystal of lengthL,
periodic boundary conditions yield allowedkvalues given by

k=2πN (5.3.8)

whereNis an integer and the separation between allowedk’s isΔk=2π/L. The number of
electrons in a volume elementdkx,dky,dkzis therefore

dN=2f(E−EC)

dkxdkydkz

Δk^3

(5.3.9)

Assuming(E−EC)EFand writingE−EF=E−EC+EC−EFgives

dN=2 exp

(

−((E−EC)+(EC−EF))

kBT

)

dkxdkydkz

Δk^3

(5.3.10)