1. Muller & Kamins problem 2.20
a) The total sheet resistance is obtained by adding both conductances since both layers are
connected in parallel.
Rsq =
1
= 148 Ω/sq
q ( μ n ,1 N d ,1 x1 + μ n , 2 N d , 2 x 2 )
With Nd,1 = 1018 cm-3, μn,1 = 277 cm2/V-s and x1 = 10-4 cm and Nd,2 = 1017 cm-3, μn,2 = 721
cm2/V-s and x1 = 2 x10-4 cm
After processing the sheet resistance equals:
Rsq =
1
= 110 Ω/sq
qμ n , 3 N d , 3 x 3
With Nd,3 = 3 x 1017 cm-3, μn,3 = 474 cm2/V-s and x3 = 4 x 10-4 cm
b) The only way to increase the sheet resistance while adding dopants, is by adding acceptors.
Adding 6 x 1016 cm-3 acceptors yields 3.6 x 1017 cm-3 impurities with a corresponding mobility
μn = 438 cm2/V-s. The electron density reduces to 2.4 x 1017 cm-3.
This results in a sheet resistance of 148 Ω/sq. The corresponding dose equals Na x 4 x 10-4 = 2.4
x 1013 cm-2. The number of acceptors was obtained by iteration.
2. Derive equation 3.5.3 in the handout.
Starting from
qN d x d2 qN d x d d
+
φ i − Va =
2ε s
ε ox
One finds the depletion layer width:
x d = − A + A 2 + (φ i − Va ) B
where
A=d
εs
2ε s
and B =
ε ox
qN d
And the capacitance becomes
So that
C=
with
dQn
dx n
= qN d
= qN d
dVa
dVa
B
A 2 + (φ i − Va ) B
=
εs
LD
Vt
2(φ − Va )
*
i
qN d d 2
A2
= φi +
φ = φi +
2ε s
B
*
i
⎛ εs
⎜⎜
⎝ ε ox
2
⎞
⎟⎟ = φ i + Δφ
⎠
3. Derive equation 3.5.9 in the handout.
The current under forward bias equals:
I = I s exp
φi − φ n
Vt
= I s* exp
Va
nVt
Which combined with
⎡
φ n = φi − Va + 2Δφ ⎢1 − 1 +
⎣
φi − Va ⎤
⎥
Δφ ⎦
Can be rewritten as
⎡
φ − Va
Va − 2Δφ ⎢1 − 1 + i
Δφ
⎢⎣
I s exp
Vt
⎤
⎥
V
⎥⎦
= I s* exp a
nVt
is then given by:
⎡
φ i − Va ⎤ 1 I s* Va
Va − 2Δφ ⎢1 − 1 +
⎥ = ln +
n
Δφ ⎥⎦ Vt I s
⎣⎢
For Va close to 0 the second term from the left can be further expressed as:
⎡
Va
Δφ + φ i
− 2Δφ ⎢1 −
1−
Δφ
Δφ + φ i
⎢⎣
⎤
⎡
Δφ + φ i
⎥ = −2Δφ ⎢1 −
Δφ
⎥⎦
⎢⎣
⎤
Δφ
Va
⎥−
Δφ + φ i
⎥⎦
Using a Taylor series expansion for the square root.
Combining the terms containing Va one then finds the ideality factor, n.
1
n=
1−
And the saturation current Is* equals:
Δφ
Δφ + φ i
1
≅
1−
Δφ
φi
I s* = I s exp
[
2 Δφ 1 − 1 + φ i / Δφ
]
Vt
4. For a silicon Schottky diode with φB = 1 V and Nd = 1018 cm-3, calculate:
a) The built-in potential, depletion layer width and the maximum electric field in
thermal equilibrium.
b) The barrier lowering in thermal equilibrium.
c) The built-in potential as obtained from a Capacitance-Voltage measurement, if there
is a 3 nm interfacial SiO2 layer between the metal and the semiconductor.
d) The ideality factor, n, of the diode as obtained from a Current-Voltage measurement,
if there is a 3 nm interfacial SiO2 layer between the metal and the semiconductor.
The Schottky diode parameters are:
The built-in potential is obtained from:
φi = φ B −
E c − E F ,n
q
= 1 – 0.0616 = 0.938 V
The depletion layer width for Va = 0 V equals:
xd =
2ε s (φ i − V a )
= 3.52 x 10-6 cm
qN d
The electric field at the interface equals:
E ( x = 0) = −
qN d x d
εs
= - 53.4 kV/cm
The barrier lowering in thermal equilibrium equals:
Δφ B =
qE max
4πε s
= 80.3 mV
The built-in potential with the interfacial layer equals:
qN d 2
φ = φi + d
2ε s
*
i
2
⎛ εs ⎞
⎜⎜
⎟⎟ = φi + Δφ = 0.938 + 0.064 = 1.002 V
⎝ ε ox ⎠
The ideality factor is obtained from:
n=
1
= 1.34
1 − Δφ / φ i