EE 3110 Microelectronics I Suketu Naik 1 Course Outline 1. Chapter 1: Signals and Amplifiers 2....

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 waferSiO2 under SEM (Scanning Electron Microscope)Raw Silicon

Procesed Wafer and

ElectronicsComponents


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

covalentbond


11

Why use Si? Cheap and abundant Thermally stable SiO2 is strong dielectricSi 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 zeroSilicon 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)


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.


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

3.2 Doped Semiconductors


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

p-type semiconductor np will have the same

dependence on temperature as ni

2

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



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 mobilityelectric field electric fie

P PP 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 () – relates current density (J=Is/A) and electrical field (E)

Resistivity () – 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

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 ) (

JJoles

(( )

eq3.19)

p

p

p

Jq

Dx

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

JJ

o

(eq3 .2 ) ( )

0

n

n

xd 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 siliconJ

)J

J

J

nDx x

d dx

nn


Diffusion CurrentnpD

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) pnT

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

Jr Jons,

( )

A qp

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

JJ

ration,

nDx x x x

d dx

p np p

p / 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 p-type semiconductor n-type semiconductor 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

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


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.

volta

ge p

oten

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 pn junction. 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)

TAD

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.


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

PP/

Pm

0

s

(eq3.22 1

6) 1

S

A

W

qF 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

PP

(eq3.27)

(eq3.28)

D

An

A D

Dp

NV

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 barrier voltage will be accompanied by…

(1) Increase in stored uncovered charge on both sides of junction

(2) wider depletion region

pp

p0p

00

width of depletion region electrical permiability of silicon (11.7 1.04 12 )

magn

replace with

itude of electron ch/

0

arge

PP

(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:

E

pp

pp

pp

p0 p

pp

concentration of acceptor atoms concentration of donor atoms

barrier / junction built-in voltage externally applied reverse-bias volta

P

PPg Pe

P

(eq3.3 22)

A

D

R

NN

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 CaseID 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

00

pp

pp

pp

0

width of depletion region electrical permiability of silicon (11.7 1.04 12 )

magnitude of electron charge con

replac

PP

P/

e with

0

2 1 1( )

A

F

S F c

V

W

qm

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

PP

P

0

P

2 (

D

F

A DJ S F

A D

NV

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.

VFNote 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

ority

car

rier

conc

entr

ation

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

Sp 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 ni

2 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 - ISMinimal 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 W

C

si 10-12 F/cm is the permittivity of silicon.

00

0

0

1

2

1

VNN

NNqC

VV

CC

DA

DAsij

R

jj


76Reverse Biased pn Junction: Application

LCfres

121

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

EE 3110 Microelectronics I Suketu Naik 1 Course Outline 1. Chapter 1: Signals and Amplifiers 2....

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