SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

226 CHAPTER 5. SEMICONDUCTOR JUNCTIONS

where

A∗=

4 πem∗k^2 B

2 π^3

= 120Acm−^2 K−^2 ×

m∗

m 0

(5.3.21)

is the Richardson constant andφB=Vbi+(EC−EF), the barrier seen by electrons in the
metal of the Schottky barrier height. We have calculatedJs→matV=0. The analysis can be
easily extended to a forward bias ofVF, the only change being replacing the barrier,Vbiby the
new barrierVbi−VF. This changesIzto

Iz=

kBT



exp

(

−

eVbi

kBT

)

·exp

(

eVF

kBT

)

(5.3.22)

or

Js→m(V=VF)=Js→m(V=0)·exp

(

eVF

kBT

)

(5.3.23)

Since the current flow from the metal to the semiconductor is unchanged:

J(V=VF)=Js→m(V=VF)−Jm→s(V=VF) (5.3.24)

=A∗T^2 exp

(

−qφB

kBT

)[

exp

(

eVF

kBT

)

− 1

]

(5.3.25)

Example 5.2InaW-n-type Si Schottky barrier the semiconductor has a doping of 1016

cm−^3 andanareaof 10 −^3 cm^2.

(a) Calculate the 300 K diode current at a forward bias of 0.3 V.

(b) Consider an Sip+−njunction diode with the same area with doping of

Na=10^19 cm−^3 andNd=10^16 cm−^3 ,andτp=τn=10−^6 s. At what forward bias will

thep-ndiode have the same current as the Schottky diode?Dp=10. 5 cm^2 /s.

From table 5.2 the Schottky barrier ofWon Si is 0.67 V. Using an effective Richardson

constant of 110 A cm−^2 K−^1 , we get for the reverse saturation current

Is =(10−^3 cm^2 )×(110 A cm−^2 K−^2 )×(300K)^2 exp

(

− 0 .67(eV)

0 .026(eV)

)

=6. 37 × 10 −^8 A

For a forward bias of 0.3 V, the current becomes (neglecting 1 in comparison to

exp (0.3/0.026))

I =6. 37 × 10 −^8 Aexp(0. 3 / 0 .026)

=6. 53 × 10 −^3 A

In the case of thep-ndiode, we need to know the appropriate diffusion coefficients and

lengths. The diffusion coefficient is 10.5 cm^2 /s, and using a value ofτp=10−^6 sweget

Lp=3. 24 × 10 −^3 cm. Using the results for the abruptp+−njunction, we get for the