Download I n

1
Course 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
Application of pn Junction: Diodes
EE 3110 Microelectronics I
3
Suketu Naik
Application of pn Junction: Solar Cells
EE 3110 Microelectronics I
4
Suketu Naik
Application of pn Junction: LEDs
EE 3110 Microelectronics I
5
Suketu Naik
6
Objectives [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
7
Objectives [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
8
3.1 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
9
3.1. Intrinsic Semiconductors
 What does a Semiconductor look like?
 Where is it used?
SiO2 under SEM (Scanning
Raw Silicon
Si wafer
Electron Microscope)
Procesed Wafer
and
Electronics
Components
EE 3110 Microelectronics I
Suketu Naik
10
3.1. Intrinsic Semiconductors
 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
EE 3110 Microelectronics I
valence
electron
covalent
bond
Suketu Naik
3.1 Intrinsic Semiconductors
11
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
Figure 3.1 Two-dimensional representation of the
 Each pair of shared forms a
silicon crystal. The circles represent the inner core
of silicon atoms, with +4 indicating its positive
covalent bond
charge of +4q, which is neutralized by the charge of
 Diamond cubic structure repeats
the four valence electrons. Observe how the
covalent bonds are formed by sharing of the valence
and forms a lattice structure
electrons. At 0K, all bonds are intact and no free
electrons are available for current conduction.
EE 3110 Microelectronics I
Suketu Naik
12
Si Lattice Structure
TEM (Transmission
Electron Microscopy)
Image of Si Lattice
3D View of the Si Lattice
EE 3110 Microelectronics I
Suketu Naik
13
3.1 Intrinsic Semiconductors
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…
EE 3110 Microelectronics I
Suketu Naik
14
Figure 3.2: At room temperature, some of the covalent bonds are
3.1: Intrinsic
broken by thermal generation. Each broken bond gives rise to a
free Semiconductors
electron and a hole, both of which become available for
current conduction.
 silicon at low temps:
 silicon at room temp:
 all covalent bonds are
 sufficient thermal energy
intact
exists to break some
covalent bonds, freeing an
 no electrons are available
electron and creating hole
for conduction
the process of freeing electrons,
creating
holes,

a
free
electron
may
wander
 conductivity is zero
and filling them facilitates
current
flow
from
its parent
atom
 a hole will attract
neighboring electrons
EE 3110 Microelectronics I
Suketu Naik
15
3.1 Intrinsic Semiconductors
 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.
EE 3110 Microelectronics I
Suketu Naik
3.1 Intrinsic Semiconductors
(eq3.1) ni  BT
16
3/ 2  Eg / 2 kT
e
equal to p and n
 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)
EE 3110 Microelectronics I
Suketu Naik
17
3.1 Intrinsic Semiconductors
 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.
EE 3110 Microelectronics I
Suketu Naik
3.2 Doped Semiconductors
 p-type semiconductor
 doped with trivalent
impurity atom
(e.g. Boron)
18
 n-type semiconductor
 doped with pentavalent
impurity atom
(e.g. Phosphorus)
EE 3110 Microelectronics I
Suketu Naik
3.2 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.
19
 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.
EE 3110 Microelectronics I
Suketu Naik
3.2 Doped Semiconductors
20
 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...
(eq3.6) (pp )  (NA )
number
holes
in
p -type
number
acceptor
atoms
EE 3110 Microelectronics I
Suketu Naik
21
3.2 Doped Semiconductors
 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...
(eq3.4) (nn )  (ND )
number
free
e-trons
in n -type
number
donor
atoms
Important: now the number of free electrons (aka.
conductivity) is dependent on doping concentration, not
temperature…
EE 3110 Microelectronics I
Suketu Naik
22
3.2 Doped Semiconductors
 Free electrons in p-type
semiconductor
action: combine this with equation
on previous slide
pp  np  n
2
i
number
of holes
in p -type
number
of free
electrons
in p -type
number
of free
electrons
and holes
in thermal
equil.
2
i
n
(eq3.7) np 
nA
 Holes in n-type
semiconductor
action: combine this with equation
on previous slide
pn  nn  n
2
i
number
of holes
in n-type
number
of free
electrons
in n-type
number
of free
electrons
and holes
in thermal
equil.
2
i
n
(eq3.5) pn 
nD
EE 3110 Microelectronics I
Suketu Naik
23
3.2 Doped Semiconductors
 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 ni2
 the concentration of free
electrons (nn) will be
much larger than holes
 electrons are the
majority charge
carriers
 holes are the minority
charge carriers
EE 3110 Microelectronics I
Suketu Naik
3.2 Doped Semiconductors
24
 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...
EE 3110 Microelectronics I
Suketu Naik
3.3 Current Flow in Semiconductors
25
 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
26
3.3.1 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 hole mobilityPpp
n electron mobilityPpp
(eq3.8) vpdrift  pE
(eq3.9) vndrift   nE
E electric fieldPpp
E electric fieldPpp
Electrons
move faster
than holes!
.E (V/ cm), μp (cm2/Vs) = 400 for doped Si,.μn (cm2/Vs) = 1110 for
doped Si
EE 3110 Microelectronics I
Suketu Naik
27
3.3.1 Drift Current (IS)
 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)?
Ip  current flow attributed to holes
A cross-sectional area of siliconp
q magnitude of the electron chargep
p concentration of holesp
vpdrift  drift velocity of holesp
(eq3.10) Ip  Aqpvpdrift
Ip
In
IS = Ip+In
EE 3110 Microelectronics I
Suketu Naik
28
3.3.1 Drift Current (IS)
 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 Law
1
(eq3.14) J  s E
q(pp  nn )
(eq3.16) s  q(pp  nn )
q(p
1
(eq3.15) J  E / r
q(p p  nn )
1
(eq3.17) r 
q(pp  nn )
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
Doping and Mobility
29
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
30
Mobility
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
31
3.3.2 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
EE 3110 Microelectronics I
diffusion occurs
concentration
profile arises
Suketu Naik
3.3.2 Diffusion Current (ID)
32
 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.
dp
J p   qD p
dx
EE 3110 Microelectronics I
Suketu Naik
3.3.2 Diffusion Current (ID)
 Diffusion Current Density
Jp  current flow density attributed to holesJpp
q  magnitude of the electron chargeJpp
Dp  diffusion constant of holes (12cm2 /s for silicon)Jpp
p( x ) hole concentration at point xJpp
dp / dx  gradient of hole concentrationJpp
33
Current
through
Area A
dp(x)
(eq3.19) hole diffusion current density : Jp  qDp
dx
dn(x)
(eq3.20) electron diffusion current density : Jn  qDn
dx
Jn  current flow density attributed to free electronsJpp
Dn  diffusion constant of electrons (35cm2 /s for silicon)Jpp
n( x ) free electron concentration at point xJpp
dn / dx  gradient of free electron concentrationJpp
 Diffusion Current I p  J p  A; I n  J n  A
ID  I p  In
EE 3110 Microelectronics I
Suketu Naik
3.3.3 Relationship Between D and .?
34
 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 constant
and mobility is defined by thermal voltage
(eq3.21)
Dn
n

Dp
p
 VT
D
kT

 q
Where, VT = kT
EE 3110 Microelectronics I
Suketu Naik
35
Summary
Drift current density (Jdrift)
 effected by – an electric field (E).
Diffusion current density (Jdiff)
 effected by – concentration gradient in free electrons and holes.
A cross-sectional area of silicon, q  magnitude of the electron charge,Jpp
p concentration of holes, n concentration of free electrons,Jpp
p  hole mobility, n  electron mobility, E  electric fieldJpp
drift current density : Jdrift  Jpdrift  Jndrift  q(pp  nn )E
diffusion current density : Jdiff  Jpdiff  Jndiff
dp(x)
dn(x)
 qDp
 qDn
dx
dx
Dp  diffusion constant of holes (12cm2 / s for silicon), Dn  diffusion constant of electrons (35cm2 /s for silicon),Jpp
p( x ) hole concentration at point x , n ( x ) free electron concentration at point x ,Jpp
dp / dx  gradient of hole concentration, dn / dx gradient of free electron concentrationJpp
 Drift current IS = Jdrift A ; Due to electric field
 Diffusion current ID = Jdiff A ; Due to concentration gradient
EE 3110 Microelectronics I
Suketu Naik
36
Example 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
Example 3.3: Resistivity of Intrinsic and Doped Semiconductor
37
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
38
Exercise 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
3.4.1 Physical Structure
39
 pn junction (diode) structure
 p-type semiconductor
 n-type semiconductor
 metal contact for connection
EE 3110 Microelectronics I
Suketu Naik
Creating a pn junction
EE 3110 Microelectronics I
40
Suketu Naik
SEM (Scanning Electron Microscopy) Images: pn junction
41
Zener Diode
LED
EE 3110 Microelectronics I
Suketu Naik
42
pn junction: modes of operation
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.
VD = 0
VD
VD < 0
VD > 0
p
+
-
n
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
43
Step #1: The p-type and n-type semiconductors are joined at the
junction.
p-type semiconductor
filled with holes
p-n junction
EE 3110 Microelectronics I
n-type semiconductor
filled with free electrons
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
44
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.
negative bound
charges
positive bound
charges
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
45
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
EE 3110 Microelectronics I
Suketu Naik
46
Movement 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
3.4.2 Operation with Open Circuit Terminals
47
Step #3: The depletion region begins to form – as
diffusion occurs and free electrons recombine with holes.
The depletion region is filled with “uncovered” bound charges – who
have lost the majority carriers to which they were originally linked.
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
48
 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
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
49
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.
voltage potential
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.
barrier voltage
(Vo)
location (x)
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
 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
50
V0  barrier voltage
VT  thermal voltage
NA  acceptor doping concentration
ND  donor doping concentration
ni  concentration of free electrons...
...in intrinsic semiconductor
 This voltage is applied across
depletion region, not terminals of pn
junction.
 NA ND 
(eq3.22) V0  VT ln  2 
 Power cannot be drawn from V0
 ni 
 It can not be measured
EE 3110 Microelectronics I
Suketu Naik
3.4.2 Operation with Open Circuit Terminals
51
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)
EE 3110 Microelectronics I
Suketu Naik
52
The 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
The Drift Current IS
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.
EE 3110 Microelectronics I
Suketu Naik
54
The Drift Current IS and Equilibrium
Step #6: Equilibrium is reached, and diffusion ceases, once the
magnitudes of diffusion and drift currents equal one another –
resulting in no net flow.
Once equilibrium is achieved, no net current flow exists (Inet = ID – IS)
within the pn-junction while under open-circuit condition.
diffusion
current drift
(ID)
p-type
depletion
region
current
(IS)
n-type
EE 3110 Microelectronics I
Suketu Naik
The Drift Current IS and Equilibrium
diffusion
current drift
(ID)
55
current
(IS)
Note that the magnitude of drift current (IS) is
unaffected by level of diffusion and / or V0. It
will be, however, affected by temperature.
EE 3110 Microelectronics I
Suketu Naik
Depletion Region
56
 The depletion region is not
symmetrical
 Typically NA > ND
 More holes can travel from
p-type to n-type than
electrons can travel from ntype 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.
EE 3110 Microelectronics I
Suketu Naik
57
With of the Depletion Region
W  width of depletion regionPpp
 S  electrical permiability of silicon (11.7 0 1.04 E12 F / cm)Ppp
q  magnitude of electron chargePpp
NA  concentration of acceptor atomsPpp
ND  concentration of donor atomsPpp
V0  barrier / junction built-in voltagePpp
2 S  1
1 
(eq3.26) W  xn  x p 


V0
q  NA ND 
NA
(eq3.27) xn  W
NA  ND
(eq3.28) x p  W
EE 3110 Microelectronics I
ND
NA  ND
Suketu Naik
Summary
58
 The pn junction is composed of two silicon-based
semiconductors, one doped to be p-type and the other ntype
 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
Summary
59
 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
pn junction: modes of operation
60
(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
61
Reverse-Bias Case
 Observe that
increased barrier
voltage will be
accompanied by…
W  width of depletion regionPpp
 S  electrical permiability of silicon (11.7 0 1.04 E12 F / cm)Ppp
q  magnitude of electron chargePpp
NA  concentration of acceptor atomsPpp
ND  concentration of donor atomsPpp
V0  barrier / junction built-in voltagePpp
VR  externally applied reverse-bias voltagePpp
2 S  1
1 
 (1) Increase in
(eq3.31) W  xn  x p 


 (V0  VR )
q  NA ND  action:
stored uncovered
replace V
charge on both
with V V
sides of junction
 (2) wider depletion
 NA ND 
(eq3.32) QJ  A 2 S q 
region
 (V0  VR )
0
 NA  ND 
0
R
action:
replace V0
with V0 VR
QJ  magnitude of charge stored on either side of depletion regionPpp
EE 3110 Microelectronics I
Suketu Naik
Reverse-Bias Case
62
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
Forward-Bias Case
 Observe that decreased
barrier voltage will be
accompanied by
W  width of depletion regionPpp
 S  electrical permiability of silicon (11.7 0 1.04 E12F / cm )Ppp
q  magnitude of electron chargePpp
NA  concentration of acceptor atomsPpp
ND  concentration of donor atoms Ppp
V0  barrier / junction built-in voltagePpp
VF  externally applied forward-bias voltagePpp
 (1) Decrease in stored
uncovered charge on both
sides of junction
 (2) Smaller depletion
region
2 S  1
1 
W  xn  x p 


 (V0  VF )
q  NA ND  action:
replace V0
with V0 VF
 NA ND
QJ  A 2 S q 
 NA  ND

 (V0  VF )
 action:
replace V0
with V0 VF
QJ  magnitude of charge stored on either side of depletion regionPpp
EE 3110 Microelectronics I
Suketu Naik
pn junction: current vs voltage
EE 3110 Microelectronics I
64
Suketu Naik
3.5.2. The Current-Voltage Relationship of the Junction 65
Step #1: Initially, a small forward-bias voltage (VF) is
applied. Due to its polarity, it pushes majority (holes in pregion and electrons in n-region) carriers toward the
junction and reduces width of the depletion zone.
Note that, in
VF
this figure, the
smaller circles
represent
minority
carriers and
not bound
charges –
which are not
considered
here.
EE 3110 Microelectronics I
Suketu Naik
3.5.2. The Current-Voltage Relationship of the Junction 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
EE 3110 Microelectronics I
Suketu Naik
3.5.2. The Current-Voltage Relationship of the Junction 67
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
diffusion
current (ID)
drift
current (IS)
EE 3110 Microelectronics I
Suketu Naik
3.5.2. The Current-Voltage Relationship of the Junction
68
Step #4: The concentration of minority charge carriers increases
on either side of the junction. They diffuse and recombine with
majority charge carriers.
minority carrier
concentration
VF
location (x)
EE 3110 Microelectronics I
Suketu Naik
3.5.2. The Current-Voltage Relationship of the Junction 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
recombination
VF
flow of diffusion current (ID)
flow of holes
flow of electrons
EE 3110 Microelectronics I
Suketu Naik
Diffusion Current
70

 Dp
Dn   V / VT
2
V / VT
(eq3.40) I   Aqni 

(
e

1)

I
(
e
 1)


S


L
N
L
N


p
D
n
A



IS
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
Summary
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 case
Significant current flows in
forward-bias case
EE 3110 Microelectronics I
Suketu Naik
Reverse Bias: Breakdown
72
 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
Reverse Bias: Breakdown
73
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
EE 3110 Microelectronics I
Suketu Naik
Junction Capacitance
74
 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
Capacitance: Voltage Dependence
Cj 
Cj 
C j0 
75
 si
Wdep
C j0
VR
1
V0
 si q N A N D
1
2 N A  N D V0
si  10-12 F/cm is the permittivity of silicon.
EE 3110 Microelectronics I
Suketu Naik
76
Reverse Biased pn Junction: Application
 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
f res
EE 3110 Microelectronics I
1

2
1
LC
Suketu Naik
Important Equations
77
 Table 3.1 on p. 159
EE 3110 Microelectronics I
Suketu Naik
78
Example 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
Exercise 3.11 : Change in Current due to Change in Carriers
79
 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
EE 3110 Microelectronics I
Suketu Naik