Transcript ภาพรวมของโครงการ

ENE 311
Lecture 9
Junction Breakdown
• When a huge reverse voltage is applied to a p-n
junction, the junction breaks down and
conducts a very large current.
• Although, the breakdown process is not
naturally destructive, the maximum current
must be limited by an external circuit to avoid
excessive junction heating.
• There are two mechanisms dealing with the
breakdown: tunneling effect and avalanche
multiplication.
Tunneling Effect
• If a very high electric
field is applied to a p-n
junction in the reverse
direction, a valence
electron can make a
transition from the
valence band to the
conduction band by
penetrating through the
energy bandgap called
tunneling.
• The typical field for Si
and GaAs is about 106
V/cm or higher.
Tunneling Effect
• To achieve such a high field, the doping
concentration for both p- and n-regions must be
very high such as more than 5 x 1017 cm-3.
• The breakdown voltage for Si and GaAs
junctions about 4Eg/e is the result of the
tunneling effect. With the breakdown voltage is
more than 6Eg/e, the breakdown mechanism is
the result of avalanche multiplication.
• As the voltage is in between 4Eg/e and 6Eg/e, the
breakdown is due to a mix of both tunneling
effect and avalanche multiplication.
Avalanche Multiplication
• Let consider a p+-n one-sided abrupt junction
with a doping concentration of ND  1017 cm-3 or
less is under reverse bias.
• As an electron in the depletion region gains
kinectic energy from a high electric field, the
electron gains enough energy and acceleration
in order to break the lattice bonds creating an
electron-hole pair, when it collides with an
atom.
Avalanche Multiplication
• A new electron also
receives a high kinetic
energy from the electric
field to create another
electron-hole pair. This
continue the process
creating other electronhole pairs and is called
avalanche multiplication.
Avalanche Multiplication
• Assume that In0 is incident current to the
depletion region at x = 0.
• If the avalanche multiplication occurs, the
electron current In will increase with distance
through the depletion region to reach a value of
Mn In0 at W, where Mn is the multiplication
factor.
I n (W )
Mn 
I n0
(1)
Avalanche Multiplication
• The breakdown voltage VB
for one-sided abrupt
junctions can be found by
EcW  s Ec2
1
VB 

 NB 
2
2e
• The breakdown voltage
for linearly graded
junctions is expressed as
2 EcW 4 Ec3 / 2  2 s 
VB 



3
3  ea 
1/ 2
Avalanche Multiplication
The critical field Ec can be calculated
for Si and GaAs by using the plot in the
above figure.
Ex. Calculate the breakdown voltage for a Si onesided p+-n abrupt junction with ND = 5 x 1016 cm-3
Ex. Calculate the breakdown voltage for a Si onesided p+-n abrupt junction with ND = 5 x 1016 cm-3
Soln From the figure, at the given NB, Ec is about
5.7 x 105 V/cm.
VB 

 s Ec2
2e
 NB 
1
11.9  8.85 10
 21.4 V
14
  5.7 10
2 1.6  10

5 2
19
5 10 
16 1
Avalanche Multiplication
Avalanche Multiplication
• Assume the depletion layer reaches the n-n+
interface prior to breakdown.
• By increase the reverse bias further, the device
will break down.
• This is called the punch-through.
Avalanche Multiplication
• The critical field Ec is the same as the previous case,
but the breakdown voltage VB for this punch-through
diode is
VB  W 
W 

 2 

VB  Wm 
Wm 
(4)
• Punch-through occurs when the doping concentration
NB is considerably low as in a p+--n+ or p+--n+ diode,
where  stands for a lightly doped p-type and 
stands for a lightly doped n-type.
Avalanche Multiplication
Breakdown voltage for p+-π-n+ and p+-v-n+
junctions.
W is the thickness of the lightly doped region
Avalanche Multiplication
• Ex. For a GaAs p+-n one-sided abrupt junction
with ND = 8 x 1014 cm-3, calculate the depletion
width at breakdown. If the n-type region of this
structure is reduced to 20 μm, calculate the
breakdown voltage if r for GaAs is 12.4.
Avalanche Multiplication
Soln For the figure, we can find VB is about 500 V
(VB >> Vbi)
2 s Vbi  V 
W
eN B
2  8.85 1014 12.4   500 

1.6 1019  8 1014
 2.93 103  29.3  m
Avalanche Multiplication
Soln When the n-type region is reduced to 20 μm,
the punch-through will occur first.
VB  W 
W 

 2 

VB  Wm 
Wm 
20 
 20 
VB  500  
2


  449 V
29.3 
 29.3 
Heterojunction
A heterojunction is
defined as a junction
formed by two
semiconductors with
different energy bandgaps
Eg, different dielectric
permittivities s, different
work function es, and
different electron
affinities eχ.
Heterojunction
• The difference energy
between two conduction band
edges and between two
valence band edges are
represented by EC and EV,
respectively, as
EC  e  2  1 
EV  Eg1  e1   Eg 2  e 2   Eg  EC
where Eg is the difference
energy bandgap of two
semiconductors.
Heterojunction
• Generally, heterojunction has to be formed
between semiconductors with closely matched
lattice constants.
• For example, the AlxGa1-xAs material is the most
important material for heterojunction.
• When x = 0, the bandgap of GaAs is 1.42 eV with
a lattice constant of 5.6533 Å at 300 K.
• When x = 1, the bandgap of AlAs is 2.17 eV with
a lattice constant of 5.6605 Å.
Heterojunction
• We clearly see that the lattice constant is
almost constant as x increased. The total builtin potential Vbi can be expressed by
Vbi  Vb1  Vb 2
 2 N 2 Vbi  V 
Vb1 
1 N1   2 N 2
1 N1 Vbi  V 
Vb 2 
1 N1   2 N 2
(7)
where N1 and N2 are the doping concentrations in
semiconductor 1 and 2, respectively.
Heterojunction
• The depletion widths x1 and x2 can be found by
x1 
x2 
21 2 N 2 Vbi  V 
eN1 1 N1   2 N 2 
21 2 N1 Vbi  V 
eN 2 1 N1   2 N 2 
(8)
Heterojunction
Ex. Consider an ideal abrupt heterojunction with a
built-in potential of 1.6 V. The impurity
concentrations in semiconductor 1 and 2 are 1 x
1016 donors/cm3 and 3 x 1019 acceptors/cm3, and
the dielectric constants are 12 and 13,
respectively. Find the electrostatic potential
and depletion width in each material at thermal
equilibrium.
Heterojunction
Soln
Vb1 
13   3  1019   1.6
12  1 10
 1.6 V
  13  3 10 
12  1 10   1.6
V 
 4.9  10 V
12  1 10   13   3  10 
2  12  13   8.85  10   3  10  1.6
x 
 4.608  10
1.6  10  1 10  12  10  13  3  10 
2  12  13   8.85  10   1 10  1.6
x 
 1.536  10
1.6  10  3  10  12  10  13  3  10 
16
19
16
4
b2
16
19
14
19
5
1
19
16
16
14
2
19
19
19
cm
16
16
8
19
Note: Most of the built-in potential is in the
semiconductor with a lower doping concentration and
also its depletion width is much wider.
cm
Metal-Semiconductor Junctions
• The MS junction is more
likely known as the
Schottky-barrier diode.
• Let’s consider metal band
and semiconductor band
diagram before the
contact.
Metal-Semiconductor Junctions
• When the metal and
semiconductor are joined,
electrons from the
semiconductor cross over
to the metal until the
Fermi level is aligned
(Thermal equilibrium
condition).
• This leaves ionized donors
as fixed positive charges
that produce an internal
electric field as the case
of one-sided p-n junction.
Metal-Semiconductor Junctions
• At equilibrium, equal number of electrons across the
interface in opposite directions.
• Hence, no net transport of charge, electron current Ie
equals to zero. The built-in voltage Vbi = m - s.
• The barrier for electrons to flow from the metal to
semiconductor is given by eb = e(m - χs) or it is called the
barrier height of MS contact.
Metal-Semiconductor Junctions
• When a voltage is applied, the barrier height
remains fixed but the built-in voltage changes as
increasing when reverse biased and decreasing
when forward biased.
Metal-Semiconductor Junctions
• Reverse bias
• Few electrons move across the interface from metal to
semiconductor due to a barrier, but it is harder for
electrons in the semiconductor to move to the metal.
• Hence, net electron transport is caused by electrons
moving from metal to semiconductor. Electron current
flows from right to left which is a small value.
Metal-Semiconductor Junctions
• Forward bias
• Few electrons move across the interface from metal to
semiconductor, but many electrons move across the
interface from semiconductor to metal due to the
reduced barrier.
• Therefore, net transport of charge flows from
semiconductor to metal and electron current flows from
left to right.
Metal-Semiconductor Junctions
• Under forward bias, the electrons emitted to
the metal have greater energy than that of the
metal electrons by about e(m - χs).
• These electrons are called hot-carrier since
their equivalent temperature is higher than that
of electrons in the metal.
• Therefore, sometimes, Schottky-diode is called
“hot-carrier diode”.
Metal-Semiconductor Junctions
• This leads to the thermionic emission with
thermionic current density under forward bias
as
J F  A**T 2e
 em   s VF  / kT 
where J s  A**T 2e
 J s eeVF / kT
 em   s  / kT 
 saturation current
m*
A A.
 effective Richardson's constant
m0
**
*
Metal-Semiconductor Junctions
• This behavior is referred to a rectification and
can be described by an ideal diode equation of
J  J s e
eV / kT
 1
(10)
where V positive for forward bias and negative for
reverse bias.
Metal-Semiconductor Junctions
• The space-charge region width of Schottky diode is
identical to that of a one-sided p-n junction.
• Therefore, under reverse bias, they can contain the
charges in their depletion region and this is called
Schottky diode capacitance.
Metal-Semiconductor Junctions
Ex. A Schottky junction is formed between Au and
n-type semiconductor of ND = 1016 cm-3. Area of
junction = 10-3 cm2 and me*= 0.92 m0. Work
function of gold is 4.77 eV and eχs = 4.05 eV. Find
current at VF = 0.3 volts.
Metal-Semiconductor Junctions
Soln
*
m
A**  A* e  120  0.92  110 A/  cm 2 .K 
m0
J s  A**T 2e
 em   s  / kT
J  J s  eeV / kT  1

 110  3002   e
  4.77  4.05  / 0.0259 

.  e0.3 / 0.0259  1
 0.897 A/(cm 2 )
I  A  J  103   0.897   0.897 mA
Metal-Semiconductor Junctions
Ex. Si-Schottky diode of 100 μm diameter has
(1/C2) v.s. VR slope of 3 x 1019 F-2V-1. Given r =
11.9 for Si. Find NB for this semiconductor.
Metal-Semiconductor Junctions
Soln
2 Vbi  VR 
1

;
2
Cj
e s N B
slope 
NB 
NB 
Cj 
C
[F/cm 2 ]
Area
2
[cm 4 F-2 V -1 ]
e s N B
2
slope  Area 2  e s
2
2


4 2 
 
19    100  10
19
12
3  10  
  1.6  10  8.85  10  11.9 

 

2
 



 6.414  1019 cm -3
Ohmic contact
• This contact is defined as a junction that will
not add a significant parasitic impedance to the
structure on which it is used and will not
sufficiently change the equilibrium-carrier
densities within the semiconductor to affect the
device characteristics.
• The I-V characteristic of ohmic contact is linear
for an ideal case.
Ohmic contact
• A specific contact resistance RC is given by
1
J

RC V
V 0
.cm2 
1
(11)
• A good ohmic contact should have a small
specific contact resistance about 10-6 Ω.cm2.
Ohmic contact
• When the semiconductor
is heavily doped with an
impurity density of 1019
cm-3 or higher, the
depletion layer of the
junction becomes very
thin so that carriers can
tunnel instead of going
over the potential
barrier.