242 CHAPTER 5. SEMICONDUCTOR JUNCTIONS
Problem 5.13 In some narrow-bandgap semiconductors, it is difficult to obtain a good
Schottky barrier (with low reverse current) due to the very small barrier height. Consider
ann-type InGaAs sample. Describe, on physical bases, how the “effective” Schottky
barrier height can be increased by incorporating a thinp-type doped region near the
surface region.
Problem 5.14 In the text, when we discussed the current flow in a Schottky barrier, we
assumed that the current was due to thermionic emission only. This is based on classical
physics where it is assumed that only particles with energy greater than a barrier can pass
through. Consider aW-n-type GaAs Schottky barrier in which the Schottky barrier
triangular potential is described by a field of 105 V/cm. The Schottky barrier height is 0.8
V. Calculate the tunneling probability through the triangular barrier as a function of
electron energy fromE=0. 4 eV toE=0. 8 eV. The tunneling current increases the
Schottky reverse current above the value obtained by thermionic current considerations.
Problem 5.15 Consider an Al-n-type Si Schottky diode. The semiconductor is doped at
1016 cm^3. Also consider ap-ndiode made from Si with the following parameters (the
diode is ideal):
Nd = Na=10^18 cm−^3
Dn =25cm^2 /s
Dp =8cm^2 /s
τn = τp=10ns
Calculate the turn-on voltages for the Schottky andp-ndiode. Assume that the current
density has to reach 10^5 A/cm^2 for the diode to be turned on.
Problem 5.16An important problem in very high-speed transistors (to be discussed in
chapter 8) based on the InAlAs/InGaAs system is the reliability of the Schottky barrier.
Consider a Schottky barrier formed on an InAlAs dopedn-type at 10^16 cm−^3. Calculate
the saturation current density if the Schottky barrier height is (i) 0.7 V; (ii) 0.6 V at 300 K.
The mass of the electrons in InAlAs is 0.08 mo. The Richardson constant has a value
R∗= 120
(
m∗
mo
)
=9. 6 Acm−^2 K−^2
The saturation current density then becomes
Js(φb=0.7V) = R∗T^2 exp
[
−
(
eφb
kBT
)]
=1. 8 × 10 −^6 A/cm^2
Js(φb=0.6V) = 8. 2 × 10 −^5 A/cm^2
Thus the current density varies by a very large value depending upon the Schottky barrier
value. The Schottky barrier height depends upon the metal-semiconductor interface
