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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


2

Chapter 3:

Semiconductors


3Application of pn Junction: Diodes


4Application of pn Junction: Solar Cells


5Application of pn Junction: LEDs


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


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.


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


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


10

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


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

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.



12Si Lattice Structure

3D View of the Si Lattice

TEM (Transmission

Electron Microscopy)

Image of Si Lattice


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…



14

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.


15

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.



16

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



17

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.



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)



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.


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...

numbernumberacceptorholes

atomsin-type

(eq3.6) ( ) ( )p A

p

p N



21

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…




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


23


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


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



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...



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)


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!


27

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


28

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


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


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


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


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.


dx

dpqDJ pp


33

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


Diffusion Current

npD

nnpp

III

AJIAJI

;

Current

through

Area A


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


35

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


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.


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


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


393.4.1 Physical Structure

pn junction (diode) structure



metal contact for connection


40Creating a pn junction


41SEM (Scanning Electron Microscopy) Images: pn junction

Zener Diode

LED


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


43


filled with free electrons


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


44

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.



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



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


47

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.



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



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.

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.



50

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



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)



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.


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


54

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


55

Note that the magnitude of drift current (IS) is

unaffected by level of diffusion and / or V0. It

will be, however, affected by temperature.


drift current (IS)

The Drift Current IS and Equilibrium


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 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


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


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


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


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


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


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


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:


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

---

-+++

+


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


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


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


64pn junction: current vs voltage


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.


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



67


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



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



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



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)


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


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


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

Reverse Bias: Breakdown


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.


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


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


77Important Equations

Table 3.1 on p. 159


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


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

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