SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

5.3. METAL SEMICONDUCTOR JUNCTION: SCHOTTKY BARRIER 225

The current density contributed by these electrons is

Jz=−evz

dN

L^3

(5.3.11)

ifkz> 0 andEz>eVbi. Note that all values ofExandEyare allowed as they represent motion
in thex−yplane which is not constrained by the barrier in the+zdirection. Note that

(Ex−EC)=

^2 k^2 x

2 m∗

(5.3.12)

with similar relationships for(Ey−EC)and(Ez−EC). Also employing the condition(Ez−Ec)>
eVbiyields a minimum value of

kmin=

√

eVbi

(

2 m∗

^2

)

(5.3.13)

Also,

vz=

kz

m∗

(5.3.14)

Therefore,

Jz=

−e

(2π)^3

∫+∞

−∞

dkx

∫+∞

−∞

dky

∫+∞

kmin

kz

m∗

dkz·

2 exp [−(Ex+Ey+Ez)/kBT]·exp [−(EC−EF)/kBT]exp

(

EC

kBT

)

=−

2 e

(2π)^3

∫

x

·

∫

y

·

∫

z

exp

(

−

EC−EF

kBT

)

(5.3.15)

where ∫

x

=

∫

y

=

∫∞

−∞

exp

(

^2 k^2 x

2 m∗kBT

)

dkx=

√

2 πm∗kBT



(5.3.16)

and ∫

z

=

∫∞

kmin

exp

(

−

^2 k^2 z

kBT

)

·

kz

m∗

·dkz (5.3.17)

=

kBT



exp

(

−^2 k^2 min/kBT

)

=

kBT



exp

(

−eVbi

kBT

)

(5.3.18)

Therefore,

Jz=

4 π

(2π)^3

·em∗k^2 BT^2 exp

(

−

(eVbi+(EC−EF))

kBT

)

(5.3.19)

or

Jz=A∗·T^2 exp

(

−eφB

kBT

)

=Js→m(V=0) (5.3.20)