ECE 305 Homework: Week 12: Solutions
ECE 305 Homework: Week 12: Solutions
ECE 305 Homework: Week 12: Solutions
1a)
Sketch
the
electrostatic
potential
vs.
position
inside
the
semiconductor.
Solution:
Note:
We
assume
that
there
is
no
charge
in
the
oxide
and
no
charge
at
the
oxide-‐Si
interface.
1c)
Do
equilibrium
conditions
apply
inside
the
semiconductor?
Explain
Solution:
YES.
The
Fermi
level
is
constant,
but
even
if
the
Fermi
level
in
the
metal
does
not
align
with
the
Fermi
level
in
the
semiconductor
(as
is
the
case
when
a
gate
voltage
is
applied)
the
oxide
insures
that
no
current
flows,
so
the
metal
and
semiconductor
are
two
separate
systems
in
equilibrium
with
possibly
different
Fermi
levels.
(Note:
We
assume
that
light
is
not
shining
on
the
semiconductor.)
1d)
Roughly
sketch
the
hole
concentration
vs.
position
inside
the
semiconductor.
Solution:
1g)
What
is
the
surface
potential?
Solution:
φS = φ ( x = 0 ) − φ ( x → ∞ ) = 0.51 V
1h)
What
is
the
gate
voltage?
Solution:
The
Fermi
level
in
the
metal
aligns
with
the
Fermi
level
in
the
semiconductor,
so
the
gate
voltage
must
be
zero.
(Note:
The
fact
that
there
is
a
volt
drop
across
the
oxide
and
the
semiconductor
with
VG
=
0
indicates
that
there
is
a
workfunction
difference
between
the
metal
and
the
semiconductor.)
1i)
What
is
the
voltage
drop
across
the
oxide?
Solution:
The
electric
field
at
the
surface
of
the
semiconductor
is
given
by
eqn.
(16.27)
in
SDF
as:
2qN AφS
ES =
KSε0
We
find
the
electric
field
in
the
oxide
from:
K OE ox = K SE S
so
KS 2qN AφS
E ox =
KO KSε0
K Oε 0
CEOT =
(The
capacitance
of
a
layer
of
SiO2
of
thickness,
EOT.
EOT
By
definition,
the
above
two
capacitances
must
be
equal:
K HfO ε 0 K ε
CHfO = 2
= CEOT = O 0
2
xHfO EOT
2
KO 3.9
EOT = xHfO = 4× = 0.6 nm
EOT = 0.6 nm
2
K HfO 25
2
Note
that
0.6
nm
of
SiO2
would
be
too
leaky
because
of
quantum
mechanical
tunneling,
but
4
nm
of
HfO2
gives
the
same
capacitance
with
less
leakage
current
due
to
quantum
mechanical
tunneling.
3) Assume
an
MOS
capacitor
on
a
p-‐type
Si
substrate
with
the
following
parameters:
N A = 2.7 × 1018 cm -3
for
the
bulk
doping
Oxide
thickness:
xo = 1.1 nm
K O = 3.9
QF = 0
(no
charge
at
the
oxide-‐Si
interface)
T = 300 K
VG = 1 V
Also
assume
that
the
structure
is
ideal
with
no
metal-‐semiconductor
workfunction
difference.
Determine
the
following
quantities
by
analytical
calculations.
You
should
use
the
depletion
or
delta-‐depletion
approximation
for
these
calculations.
3a)
The
flatband
voltage,
VFB .
Solution:
VFB = 0
because
there
is
no
workfunction
difference
and
no
charge
at
the
interface.
ES =
KO
KS
E ox =
3.9
11.8
( )
2.4 × 106 = 7.9 × 105
Solution:
ION
ROUT gm
RON
DIBL
S
IOFF
On
current
in
mA/µm :
≈ 800 µA/µm
The
on-‐current
is
the
current
for
the
maximum
gate
and
drain
voltages.
Off
current
in
µ A/µm :
≈ 6 µA/µm
The
off
current
is
the
current
for
zero
gate
voltage
when
the
drain
voltage
is
at
its
maximum
value.
The
source
and
drain
series
resistors
increase
the
linear
region
resistance,
and
the
source
series
resistance
lowers
the
saturation
current.