Course Outline - Weber State Universityfaculty.weber.edu/snaik/EE3110/01Chapter 3.pdf ·...
Transcript of Course Outline - Weber State Universityfaculty.weber.edu/snaik/EE3110/01Chapter 3.pdf ·...
EE 3110 Microelectronics I Suketu Naik
1Course Outline
1. Chapter 1: Signals and Amplifiers
2. Chapter 3: Semiconductors
3. Chapter 4: Diodes
4. Chapter 5: MOS Field Effect Transistors (MOSFET)
5. Chapter 6: Bipolar Junction Transistors (BJT)
6. Chapter 2 (optional): Operational Amplifiers
EE 3110 Microelectronics I Suketu Naik
2
Chapter 3:
Semiconductors
EE 3110 Microelectronics I Suketu Naik
3Application of pn Junction: Diodes
EE 3110 Microelectronics I Suketu Naik
4Application of pn Junction: Solar Cells
EE 3110 Microelectronics I Suketu Naik
5Application of pn Junction: LEDs
EE 3110 Microelectronics I Suketu Naik
6Objectives [1/2]
The basic properties of semiconductors and, in
particular, Silicon (Si) – the material used to
make most modern electronic circuits
How doping a pure silicon crystal dramatically
changes its electrical conductivity – the
fundamental idea in underlying the use of
semiconductors in the implementation of
electronic devices
EE 3110 Microelectronics I Suketu Naik
7Objectives [2/2]
The two mechanisms by which current flows in
semiconductors – drift and diffusion charge
carriers.
The structure and operation of the pn junction –
a basic semiconductor structure that
implements the diode and plays a dominant role
in semiconductors.
EE 3110 Microelectronics I Suketu Naik
83.1 Intrinsic Semiconductors
Semiconductor – a material whose conductivity lies
between that of conductors (copper) and insulators
(glass).
Single-element – such as Germanium (Ge) and
Silicon (Si).
Compound – such as Gallium-Arsenide (GaAs).
Single-element crystal Compound crystal
EE 3110 Microelectronics I Suketu Naik
93.1. Intrinsic Semiconductors
What does a Semiconductor look like?
Where is it used?
Si wafer
SiO2 under SEM (Scanning
Electron Microscope)Raw Silicon
Procesed Wafer
and
Electronics
Components
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Valence electron – is an electron that participates in the formation of chemical bonds.
Lies in the outermost electron shell of an element
The number of valence electrons that an atom has determines the kinds of chemical bonds that it can form.
Covalent bond – is a form of chemical bond in which two atoms share a pair of electrons
By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability
3.1. Intrinsic Semiconductors
valence
electron
covalent
bond
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Why use Si?
Cheap and abundant
Thermally stable
SiO2 is strong dielectric
Si atom
Has four valence electrons (Carbon group or group IV)
Requires four more to complete outermost shell and form tetrahedral symmetry which is more stable
Each pair of shared forms a covalent bond
Diamond cubic structure repeats and forms a lattice structure
Figure 3.1 Two-dimensional representation of the
silicon crystal. The circles represent the inner core
of silicon atoms, with +4 indicating its positive
charge of +4q, which is neutralized by the charge of
the four valence electrons. Observe how the
covalent bonds are formed by sharing of the valence
electrons. At 0K, all bonds are intact and no free
electrons are available for current conduction.
3.1 Intrinsic Semiconductors
EE 3110 Microelectronics I Suketu Naik
12Si Lattice Structure
3D View of the Si Lattice
TEM (Transmission
Electron Microscopy)
Image of Si Lattice
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13
Silicon at low temp
all covalent bonds – are intact
no electrons – are available for conduction
conductivity – is zero
Silicon at room temp
some covalent bonds – break, freeing an electron and creating hole, due to thermal energy
some electrons – will wander from their parent atoms, becoming available for conduction
conductivity – is greater than zero
The process of freeing electrons, creating holes, and filling
them facilitates current flow…
3.1 Intrinsic Semiconductors
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3.1: Intrinsic
Semiconductors
silicon at low temps:
all covalent bonds are
intact
no electrons are available
for conduction
conductivity is zero
silicon at room temp:
sufficient thermal energy
exists to break some
covalent bonds, freeing an
electron and creating hole
a free electron may wander
from its parent atom
a hole will attract
neighboring electrons
the process of freeing electrons, creating holes, and filling them facilitates current flow
Figure 3.2: At room temperature, some of the covalent bonds are
broken by thermal generation. Each broken bond gives rise to a
free electron and a hole, both of which become available for
current conduction.
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Intrinsic semiconductor – is one which is not doped
One example is pure silicon.
Generation – is the process of free electrons and holes being created.
generation rate – is speed with which this occurs.
Recombination – is the process of free electrons and holes
disappearing.
recombination rate – is speed with which this occurs
Thermal generation – effects a equal concentration of free electrons
and holes: electrons move randomly throughout the material.
In thermal equilibrium, generation and recombination rates are
equal.
1) Generation can be effected by thermal energy (heat)
2) Both generation and recombination rates are functions of
temperature.
3.1 Intrinsic Semiconductors
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ni = number of free electrons and holes in a unit volume for intrinsic semiconductor
B = parameter which is 7.3E15 cm-3K-3/2 for silicon
T = temperature (K)
Eg = bandgap energy which is 1.12eV for silicon
(energy between top of valence band and conduction band, see the figure above)
k = Boltzman constant (8.62E-5 eV/K)
/ 23 / 2
equal to and
(eq3.1) gE kT
p n
in BT e
3.1 Intrinsic Semiconductors
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Q: Why can thermal generation not be used to effect
meaningful current conduction?
A: Silicon crystal structure described previously is not
sufficiently conductive at room temperature.
Additionally, a dependence on temperature is not
desirable.
Q: How can this “problem” be fixed?
A: doping
Doping – is the intentional introduction of impurities
into an extremely pure (intrinsic) semiconductor for
changing carrier concentrations.
3.1 Intrinsic Semiconductors
EE 3110 Microelectronics I Suketu Naik
183.2 Doped Semiconductors
p-type semiconductor
doped with trivalent
impurity atom
(e.g. Boron)
n-type semiconductor
doped with pentavalent impurity atom (e.g. Phosphorus)
EE 3110 Microelectronics I Suketu Naik
193.2 Doped Semiconductors
p-type semiconductor Silicon is doped with element
having a valence of 3.
To increase the concentration of holes (p).
One example is boron, which is an acceptor.
n-type semiconductor Silicon is doped with element
having a valence of 5.
To increase the concentration of free electrons (n).
One example is phosphorus, which is a donor.
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p-type doped semiconductor
Concentration of acceptor atoms is NA
If NA is much greater than ni …
Then the concentration of holes in the p-type is
defined as below.
they will be equal...
numbernumberacceptorholes
atomsin-type
(eq3.6) ( ) ( )p A
p
p N
3.2 Doped Semiconductors
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n-type doped semiconductor
Concentration of donor atoms is ND
If ND is much greater than ni …
Then the concentration of electrons in the n-type is
defined as below.they will be equal...
number numberfree donor
e-trons atomsin -type
(eq ( ) ( )3.4) n D
n
n N
Important: now the number of free electrons (aka.
conductivity) is dependent on doping concentration, not
temperature…
3.2 Doped Semiconductors
EE 3110 Microelectronics I Suketu Naik
223.2 Doped Semiconductors
Free electrons in p-type semiconductor
numbernumber numberof freeof holes of free
electronsin -type electronsand holes
: combine this with equationon
in -typein thermal
e
previous slide
qu
2
il
2
.
(eq3.7)
pp
p p i
ip
A
p n n
nn
n
action
Holes in n-type
semiconductor
number number numberof holes of free of freein n-type electrons electrons
in n-type and holes
: combine this with equationon previous
in
sli
thermalequ
d
2
i
2
e
l.
(eq3.5)
n n i
in
D
p n n
np
n
action
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p-type semiconductor
np will have the same
dependence on
temperature as ni2
the concentration of
holes (pn) will be much
larger than electrons
holes are the majority
charge carriers
free electrons are the
minority charge
carriers
n-type semiconductor
pn will have the same dependence on temperature as ni
2
the concentration of free electrons (nn) will be much larger than holes
electrons are the majority charge carriers
holes are the minority charge carriers
3.2 Doped Semiconductors
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p-type or n-type semiconductor
is electrically neutral by itself (as standalone unit)
charge of majority carriers (holes in p-type and electrons in n-type) is neutralized by the bound charges associated with impurity atoms
A bound charge (polarization charge)
is charge of opposite polarity to free electron (proton)
neutralizes the electrical charge of these majority carriers
However if you put p-type and n-type together, electron flow happens...
3.2 Doped Semiconductors
EE 3110 Microelectronics I Suketu Naik
253.3 Current Flow in Semiconductors
Summary
Holes (absence of electrons, p) and free electrons (n):
p-type semiconductor: holes are majority carriers ( pp ),
free electrons (np) are minority carriers
n-type semiconductor: free electrons are majority
carriers (nn), holes are minority carriers ( pn )
Two distinct mechanisms for current flow (movement
of charge carriers)
Drift Current (IS)
Diffusion Current (ID)
EE 3110 Microelectronics I Suketu Naik
263.3.1 Drift Current Q: What happens when an electrical field (E) is applied
to a semiconductor crystal?
A: Holes are accelerated in the direction of E, free electrons are repelled.
Q: How is the velocity of these holes defined?
p pp p
p pp p
hole mobility electron mobility
electric field electric fie
P P
P Pld
(eq3.8) (eq3.9)
p n
p drift p n drif n
E E
tv E v E
.E (V/ cm), μp (cm2/Vs) = 400 for doped Si,.μn (cm2/Vs) = 1110 for
doped Si
Electrons
move faster
than holes!
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Assume that, for the single-crystal silicon bar on previous
slide, the concentration of holes is defined as p and
electrons as n.
Q: What is the current components attributed to the flow
of holes (Ip) and electrons (In)?
3.3.1 Drift Current (IS)
p
p
p
p
current flow attributed to holes cross-sectional area of silicon
magnitude of the electron charge concentration of holes
drift velocity of holes
(eq3.10)
p
p drift
IA
p p dr f
qp
v
i tI Aqpv
IpIn
IS = Ip+In
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Conductivity (s.) –relates current density (J=Is/A) and electrical field (E)
Resistivity (r.) – Inverse of conductivity
Example 3.3 - FYI: how to calculate resistivity of a substrate
Ohm's Law1
( )
1
(eq3.14)
(eq3.16)
(eq3.15)
(eq3.1
( )
/
1
(
7)
)
(
)
)
(
1
p n
p n
p
p n
n
p n
q p n
q p n
J E
q
q p
p n
J E
q p n
n
s
s
r
r
3.3.1 Drift Current (IS)
1) Resistivity of the intrinsic silicon is reduced
significantly when it is doped (see example 3.3)
2) Also, doping reduces carrier mobility
EE 3110 Microelectronics I Suketu Naik
29Doping and Mobility
1) For low doping concentrations, the mobility is almost
constant
2) At higher doping concentrations, the mobility decreases due
to ionized impurity scattering with the ionized doping atoms
EE 3110 Microelectronics I Suketu Naik
30Mobility
Holes have less mobility than free electrons
Why?
Free electrons are loosely tied to the nucleus and are closer
to the conduction band (higher orbits, see slide 19)
Holes are absence of electrons in the covalent bond
between Si atoms and B
Holes are locked or subjected to the stronger atomic force
pulled by the nucleus than the electrons residing in the
higher shells or farther shells
So, holes have a lower mobility
EE 3110 Microelectronics I Suketu Naik
313.3.2 Diffusion Current (ID)
Example of Diffusion Process
Inject holes – By some
unspecified process, one
injects holes in to the left
side of a silicon bar.
Concentration profile arises
– Because of this continuous
hole injection, a
concentration profile arises.
Diffusion occurs – Because
of this concentration
gradient, holes will flow
from left to right.
inject holes
concentration profile arises
diffusion occurs
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Carrier diffusion – is the flow of charge carriers from area of high concentration to low concentration.
It requires non-uniform distribution of carriers.
Diffusion current – is the current flow that results from diffusion.
Current flow due to mobile charge diffusion is proportional to the carrier concentration gradient.
The proportionality constant is the diffusion constant.
3.3.2 Diffusion Current (ID)
dx
dpqDJ pp
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Diffusion Current Density p
pp
2
p
pp
J current flow density attributed to holes
magnitude of the electron charge
diffusion constant of (12cm /s for silicon h )
(
J
Joles
(( )
eq3.19)
p
p
p
J
q
D
x
p
d xJ qD
dx
p
hole diffusion current density :p
pp
pp
) hole concentration at point
/ gradient of hole concentration
current flow density attributed t
J
J
o
(eq3 .2 ) ( )
0
n
n
x
d dx
n
J
d xJ qD
dx
p
electron diffusion current den ty : n
si
pp
pp
pp
2
pp
free electrons
diffusion constant of electrons
( ) free electron concentration at point
/ gradient of free electron concentra
(
tion
35cm /s for silicon
J
)
J
J
J
nD
x x
d dx
n
n
3.3.2 Diffusion Current (ID)
Diffusion Current
npD
nnpp
III
AJIAJI
;
Current
through
Area A
EE 3110 Microelectronics I Suketu Naik
343.3.3 Relationship Between D and .?
Q: What is the relationship between diffusion constant (D) and
mobility ()?
A: thermal voltage (VT)
Q: What is this value?
A: at T = 300K, VT = 25.9mV
the relationship between diffusion constantand mobility is defined by thermal voltage
(eq3.21) pn
T
n p
DDV
q
kTD
Where, VT = kT
EE 3110 Microelectronics I Suketu Naik
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Drift current density (Jdrift)
effected by – an electric field (E).
Diffusion current density (Jdiff)
effected by – concentration gradient in free electrons and holes.pp
pp
cross-sectional area of silicon, magnitude of the electron charge,
concentration of holes, concentration of free elect
J
r Jons,
( )
A q
p
drift p drift n drift
n
p nJ J J q p n E
drift current density :
pp
2
hole mobility, electron mobility, electric field
diffusion constant of holes (12 m s
J
c /
( ) (
)
p n
p
diff p diff n diff p n
E
D
d x d xJ J J qD qD
dx dx
diffusion currep
nt densityn
:
pp
pp
2 for silicon), diffusion constant of electrons (35cm /s for silicon),
( ) hole concentration at point , ( ) free electron concentration at point ,
/ gradient of hole concent
J
J
ration,
nD
x x x x
d dx
p n
p pp / gradient of free electron concentrat nJiod dxn
Summary
Drift current IS = Jdrift A ; Due to electric field
Diffusion current ID = Jdiff A ; Due to concentration gradient
EE 3110 Microelectronics I Suketu Naik
36Example 3.2: Doped Semiconductor
Consider an n-type silicon for which the dopant
concentration is ND = 1017/cm3. Find the electron and
hole concentrations at T = 300K.
EE 3110 Microelectronics I Suketu Naik
37Example 3.3: Resistivity of Intrinsic and Doped Semiconductor
Q(a): Find the resistivity of intrinsic silicon using following
values:
μn = 1350cm2/Vs, μp = 480cm2/Vs, ni = 1.5E10/cm3.
Q(b): Find the resistivity of p-type silicon with NA = 1016/cm2
and using the following values:
μn = 1110cm2/Vs, μp = 400cm2/Vs, ni = 1.5E10/cm3
EE 3110 Microelectronics I Suketu Naik
38Exercise 3.4: Find drift current
n-type silicon:
length = 2 μm, Vd = 1 V, ND=1016/cm3, μn=1350 cm2/Vs
Find the drift current in the silicon across cross sectional
area A=0.25μm2
EE 3110 Microelectronics I Suketu Naik
393.4.1 Physical Structure
pn junction (diode) structure
p-type semiconductor
n-type semiconductor
metal contact for connection
EE 3110 Microelectronics I Suketu Naik
40Creating a pn junction
EE 3110 Microelectronics I Suketu Naik
41SEM (Scanning Electron Microscopy) Images: pn junction
Zener Diode
LED
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42
VD = 0 VD > 0VD < 0
In order to understand the operation of pn junction (diode), it is
necessary to study its behavior in three operation regions:
equilibrium, reverse bias, and forward bias.
pn junction: modes of operation
+
-VD
p
n
EE 3110 Microelectronics I Suketu Naik
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n-type semiconductor
filled with free electrons
p-type semiconductor
filled with holes p-n junction
Step #1: The p-type and n-type semiconductors are joined at the
junction.
3.4.2 Operation with Open Circuit Terminals
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positive bound
charges
negative bound
charges
Step #1A: Bound charges are attracted (within the material) by
free electrons and holes in the p-type and n-type semiconductors,
respectively. They remain weakly “bound” to these majority
carriers; however, they do not recombine.
3.4.2 Operation with Open Circuit Terminals
EE 3110 Microelectronics I Suketu Naik
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Step #2: Diffusion begins.Those free electrons and holes
which are closest to the junction will recombine and,
essentially, eliminate one another.
Diffusion:
1) Concentration of holes is higher in p-type than in n-type (thermally
generated holes): holes travel from p-type to n-type
2) Concentration of electrons is higher in n-type than in p-type
(thermally generated electrons): electrons travel from n-type to p-type
3.4.2 Operation with Open Circuit Terminals
EE 3110 Microelectronics I Suketu Naik
46Movement of Holes and Electrons
Holes are absence of electrons in covalent bond which travels
across the lattice
Donor atoms have extra
electrons, which are loosely
bound to the donor nuclei,
and travel the other way
EE 3110 Microelectronics I Suketu Naik
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The depletion region is filled with “uncovered” bound charges – who
have lost the majority carriers to which they were originally linked.
Step #3: The depletion region begins to form – as
diffusion occurs and free electrons recombine with holes.
3.4.2 Operation with Open Circuit Terminals
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Q: Why does diffusion occur even when bound charges neutralize the electrical attraction of majority carriers to one another?
A: Diffusion current, as shown in (3.19) and (3.20), is effected by a gradient in concentration of majority carriers – not an electrical attraction of these particles to one another.
In other words the bound charges can not effectively neutralize the majority carriers while the pn junction seeks equilibrium
3.4.2 Operation with Open Circuit Terminals
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Step #4: The “uncovered” bound charges effect a voltage
differential across the depletion region. The magnitude of this
barrier voltage (V0) differential grows, as diffusion continues.
volt
age
po
ten
tial
location (x)
barrier voltage (Vo)
No voltage differential exists across regions of the pn-junction
outside of the depletion region because of the neutralizing effect
of positive and negative bound charges.
3.4.2 Operation with Open Circuit Terminals
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pn-junction built-in voltage (V0) –is the equilibrium value of barrier voltage.
Vo ~ 0.7 V for Si, Vo ~ 0.3 V for Ge
This voltage is applied across depletion region, not terminals of pnjunction.
Power cannot be drawn from V0
It can not be measured
0 barrier voltage thermal voltage
acceptor doping concentration donor doping concentration
concentration of free electrons... ...in intrinsic
2
sem
0
ic
(eq3.22)
T
A
D
i
VV
NN
n
A DT
i
N NV V
n
ln
onductor
3.4.2 Operation with Open Circuit Terminals
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Step #5: The barrier voltage (V0) is an electric field whose
polarity opposes the direction of diffusion current (ID). As the
magnitude of V0 increases, the magnitude of ID decreases.
diffusion
current (ID)
drift
current (IS)
3.4.2 Operation with Open Circuit Terminals
EE 3110 Microelectronics I Suketu Naik
52The Drift Current IS
In addition to majority-carrier diffusion current (ID), a component of current due to minority carriers, i.e. drift current (IS) exists.
Specifically, some of the thermally generated holes in the n-type material and thermally generated electrons in p-type material move toward and reach the edge of the depletion region.
There, they experience the electric field (V0) in the depletion region and are swept across it.
Electrons moved by drift from p to n and holes moved by drift from n to p: add together to form combined drift current IS.
EE 3110 Microelectronics I Suketu Naik
53
Drift current (IS) – is due to the movement of these minority carriers.
electrons from p-side to n-side
holes from n-side to p-side
determined by number of minority carriers that make it to the depletion region
Because these holes in n-type and electrons in p-type are produced by thermal energy, IS is heavily dependent on temperature
Any depletion-layer voltage, regardless of how small, will cause the transition across junction.
Therefore IS is independent of V0.
The Drift Current IS
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Step #6: Equilibrium is reached, and diffusion ceases, once the
magnitudes of diffusion and drift currents equal one another –
resulting in no net flow.
diffusion current (ID)
drift current (IS)
Once equilibrium is achieved, no net current flow exists (Inet = ID – IS)
within the pn-junction while under open-circuit condition.
p-type n-typedepletion region
The Drift Current IS and Equilibrium
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Note that the magnitude of drift current (IS) is
unaffected by level of diffusion and / or V0. It
will be, however, affected by temperature.
diffusion current (ID)
drift current (IS)
The Drift Current IS and Equilibrium
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The depletion region is not
symmetrical
Typically NA > ND
More holes can travel from
p-type to n-type than
electrons can travel from n-
type to p-type
The width of depletion
layer differs on two sides
The depletion region will
extend deeper in to the “less
doped” material, a
requirement to uncover the
same amount of charge.
Depletion Region
EE 3110 Microelectronics I Suketu Naik
57With of the Depletion Region
ppp
pp
0
p width of depletion region
electrical permiability of silicon (11.7 1.04 12 )
magnitude of electron charge
concentration of acceptor ato
P
P/
P
m
0
s
(eq3.22 1
6) 1
S
A
W
q
F cm
Sn p
A D
N
W x x Vq N N
E
pp
pp
pp0
concentration of donor atoms
barrier / junction built-in volta Pge
P
P
(eq3.27)
(eq3.28)
D
An
A D
Dp
N
V
A D
Nx W
N N
Nx W
N N
EE 3110 Microelectronics I Suketu Naik
58Summary
The pn junction is composed of two silicon-based
semiconductors, one doped to be p-type and the other n-
type
Majority carriers: holes are present on p-side, free electrons
are present on n-side
Bound charges: charge of majority carriers are neutralized
electrically by bound charges
Diffusion current ID: majority carriers close to the junction
will diffuse across, resulting in their elimination
(concentration gradient)
Depletion region: carriers disappear and release bound
charges and uncovered bound charges create a voltage
differential V0
EE 3110 Microelectronics I Suketu Naik
59Summary
Depletion-layer voltage: as diffusion continues, the depletion layer voltage (V0) grows, making diffusion more difficult and eventually bringing it to halt
Minority carriers
Are generated thermally (due to heat)
Free electrons are present on p-side, holes are present on n-side
Drift current IS
The depletion-layer voltage (V0) facilitates the flow of minority carriers to opposite side
Open circuit equilibrium ID = IS
Drift current IS = Jdrift A ; Due to minority charge carriers
generated by heat and electric field in the depletion region
Diffusion current ID = Jdiff A ; Due to majority charge carriers
generated by doping and nonuniform concentration profile
EE 3110 Microelectronics I Suketu Naik
60pn junction: modes of operation
(a) Open-circuit:voltage drop across depletion region = V0 , ID = IS
(b) Reverse bias:voltage drop across depletion region = V0 +VR, ID < IS
(c) Forward bias:voltage drop across depletion region = V0 -VF, ID > IS
EE 3110 Microelectronics I Suketu Naik
61Reverse-Bias Case
Observe that increased barriervoltage will be accompanied by…
(1) Increase in stored uncovered charge on both sides of junction
(2) wider depletion region
pp
p0p
0
0
width of depletion region
electrical permiability of silicon (11.7 1.04 12 )
magn
replace with
itude of electron ch
/
0
arge
P
P
(eq3.31)2 1 1
( )
S
R
F cm
Sn p R
VV V
W
q
A D
W x x V Vq N N
action:
Epp
pp
pp
p0 p
pp
concentration of acceptor atoms
concentration of donor atoms
barrier / junction built-in voltage
externally applied reverse-bias volta
P
P
P
g Pe
P
(eq3.3 22)
A
D
R
N
N
V
J
V
Q A
0
pp
0
magnitude of charge store
0
d on either side of depletion re
replace with
gi Pon
( )
J
R
VV V
A DS R
A
Q
D
N Nq V V
N N
action:
EE 3110 Microelectronics I Suketu Naik
62Reverse-Bias Case
ID reduces to nearly zero, Why?
Recall that ID is the result of diffusion of Holes
from p type to n type and diffusion of Electrons
from n type to p type
Holes (absence of an electron in the covalent
bond in Si atoms in order to create covalent
bond between B and Si) have to overcome higher potential barrier
In other words, the electrons being supplied by the external supply
now enter p region and fill up the holes in Si atoms which uncovers
more bound charges (Boron, negative): this makes it harder for the
holes to move across the depletion region
Similarly, holes supplied by external supply enters the n region and
combine with free electrons here which uncovers more bound charges
(Phospohorus, positive): this makes it harder for free electrons to
move
---
-+++
+
EE 3110 Microelectronics I Suketu Naik
63
Observe that decreased barrier voltage will be accompanied by
(1) Decrease in stored uncovered charge on both sides of junction
(2) Smaller depletion region
0
0
pp
pp
pp
0
width of depletion region
electrical permiability of silicon (11.7 1.04 12 )
magnitude of electron charge
con
replac
P
P
P
/
e with
0
2 1 1( )
A
F
S F c
V
W
q
m
Sn p F
A D
N
V V
W x x V Vq N N
action:
E
pp
pp
pp0
pp
centration of acceptor atoms
concentration of donor atoms
barrier / junction built-in voltage
externally applied forward-bias voltage
P
P
P
0
P
2 (
D
F
A DJ S F
A D
N
V
V
N NQ A q V V
N N
0
pp
0
magnitude of charge stored on either side of
rep
dep
lace wit
letion region
P
h
)
J
FV V
Q
V
action:
Forward-Bias Case
EE 3110 Microelectronics I Suketu Naik
64pn junction: current vs voltage
EE 3110 Microelectronics I Suketu Naik
653.5.2. The Current-Voltage Relationship of the Junction
Step #1: Initially, a small forward-bias voltage (VF) is
applied. Due to its polarity, it pushes majority (holes in p-
region and electrons in n-region) carriers toward the
junction and reduces width of the depletion zone.
VF Note that, in
this figure, the
smaller circles
represent
minority
carriers and
not bound
charges –
which are not
considered
here.
EE 3110 Microelectronics I Suketu Naik
66
Step #2: As the magnitude of VF increases, the depletion zone
becomes thin enough such that the barrier voltage (V0 – VF) cannot
stop diffusion current VF
3.5.2. The Current-Voltage Relationship of the Junction
EE 3110 Microelectronics I Suketu Naik
67
diffusion current (ID)
drift current (IS)
Step #3: Majority carriers (free electrons in n-region and holes in
p-region) cross the junction and become minority charge carriers
at boundary of the depletion region.
VF
3.5.2. The Current-Voltage Relationship of the Junction
EE 3110 Microelectronics I Suketu Naik
68
Step #4: The concentration of minority charge carriers increases
on either side of the junction. They diffuse and recombine with
majority charge carriers.
min
ori
ty c
arri
er
con
cen
trat
ion
location (x)
VF
3.5.2. The Current-Voltage Relationship of the Junction
EE 3110 Microelectronics I Suketu Naik
69
Step #5: Diffusion current is maintained – in spite low
diffusion lengths (e.g. microns) and recombination – by
constant flow of both free electrons and holes towards the
junction by the external supply
VF
flow of holes flow of electrons
flow of diffusion current (ID)
recombination
3.5.2. The Current-Voltage Relationship of the Junction
EE 3110 Microelectronics I Suketu Naik
70Diffusion Current
2 / /( 1)(eq3. ( 140) )T T
S
p V V V Vni
n
I
S
p D A
D DI Aqn e I e
L N L N
Saturation current (IS) (drift current):
maximum reverse current which will
flow through pn-junction.
Proportional to cross-section of
junction (A).
Typical value is 10-18A.
Is depends on ni2 which depends
strongly on temperature T
(recall that ni=BT3/2e-Eg/2kT)
EE 3110 Microelectronics I Suketu Naik
71
Reverse bias case
the externally applied voltage VR adds to (aka. reinforces) the barrier voltage V0 (increases barrier)
this reduces rate of diffusion, reducing ID
if VR > 1V, ID will fall to 0A
the drift current IS is unaffected, but dependent on temperature
result is that pn junction will conduct small drift current IS
Forward bias case
the externally applied voltage
VF subtracts from the barrier
voltage V0 (decreases barrier)
this increases rate of diffusion,
increasing ID
the drift current IS is
unaffected, but dependent on
temperature
result is that pn junction will
conduct significant current
ID - IS
Minimal current flows in
reverse-bias caseSignificant current flows in
forward-bias case
Summary
EE 3110 Microelectronics I Suketu Naik
72Reverse Bias: Breakdown
As V decreases to VZ, dramatic
increase in reverse current
occurs: this is known as junction
breakdown
Breakdown is not destructive:
pn junction can still be operated
Can operate with a max value
(set by resistor)
Why does breakdown occur?(1) Zener effect: when VZ < 5 V
(2) Avalanche effect: when Vz > 7 V (3) Either effect when 5 V< Vz < 7 V
EE 3110 Microelectronics I Suketu Naik
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Why does breakdown occur?
1) Zener effect: when VZ < 5 V As electric field increases, covalent bonds begin to
break: new hole-electron pairs are created
Electrons are swept into n side and holes into p
side
At V=VZ very large number of carriers are
generated and large reverse current appears
We can control over the value of reverse current
Voltage is capped at V=VZ
2) Avalanche effect: when Vz > 7 V Ionizing collision: under electric field minority
charge carriers (electrons in p side and holes in n
side) collide with atoms and break covalent bonds
Resulting carriers have high energy to cause more
carriers to be liberated in further ionizing collision
Process keeps repeating as avalanche
We can control over the value of reverse current
Voltage is capped at V=VZ
Reverse Bias: Breakdown
EE 3110 Microelectronics I Suketu Naik
74Junction Capacitance
A reverse-biased pn junction can be viewed as a capacitor
The depletion width (Wdep) hence the junction capacitance
(Cj) varies with VR.
EE 3110 Microelectronics I Suketu Naik
75Capacitance: Voltage Dependence
dep
sij
WC
si 10-12 F/cm is the permittivity of silicon.
0
0
0
0
1
2
1
VNN
NNqC
V
V
CC
DA
DAsij
R
j
j
EE 3110 Microelectronics I Suketu Naik
76Reverse Biased pn Junction: Application
LCf
res
1
2
1
A very important application of a reverse-biased pn junction is
in a voltage controlled oscillator (VCO)
By changing VR, we can change C, which changes the
oscillation frequency
EE 3110 Microelectronics I Suketu Naik
77Important Equations
Table 3.1 on p. 159
EE 3110 Microelectronics I Suketu Naik
78Example 3.6: Current in pn-Junction
Consider a forward-biased pn junction conducting a
current of I = 0.1mA with following parameters:
NA = 1018/cm3, ND = 1016/cm3, A = 10-4cm2, ni =
1.5E10/cm3, Lp = 5um, Ln = 10um, Dp (n-region) =
10cm2/s, Dn (p-region) = 18cm2/s
Q(a): Calculate IS .
Q(b): Calculate the forward bias voltage (V).
Q(c): Component of current I due to hole injection and
electron injection across the junction
EE 3110 Microelectronics I Suketu Naik
79Exercise 3.11 : Change in Current due to Change in Carriers
Forward-biased pn junction:V = 0.605 V with same
parameters as Example 3.6
ND = 0.5 x 1016/cm3
Q(a): Calculate IS .
Q(b): Calculate current I