Prof. Iqbal A. Khan, EED, UQU Course Code: 802311 Course Name: Electronic Devices By Prof. Iqbal Ahmad Khan Department of Electrical Engineering Faculty

Prof. Iqbal A. Khan, EED, UQU

Course Code: 802311Course Name: Electronic Devices

By

Prof. Iqbal Ahmad KhanDepartment of Electrical Engineering

Faculty of Engineering & Islamic ArchitectureUmm Al Qura University, Makka Al Mukarrama

Kingdom of Saudi Arabia

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Course Number: 802311 Units: (Lec., Lab., Tot.): (3, 1 , 4)

Course Name: ELECTRONIC DEVICES

Prerequisite: 802301 & 403102 Contact Hours: 6

Course Topics:1. Semiconductor Theory 2. PN Junction Diode 3. Other Diodes and Devices4. Transistors 5. DC and AC Analysis of Transistors6. Field-Effect Transistors

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Text Book: Thomas l. Floyd, “ELECTRONIC DEVICES”, Nineth Edition, Pearson Education International, 2012.

References : 1.R. Boylestad and L. Nashelsky, “Electronic Devices and Circuit Theory”, 10th Edition, Pearson Education International, 2010.2.A. S. Sedra and K. C. Smith, “Microelectronics Circuits”, Oxford University Press, 5th Edition, 2008.

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Classification of MaterialsElectrically Materials can be classified into three categories: 1. Insulators, 2. Conductors, 3. Semiconductors.1. Insulators:

The materials in which all electrons are tightly bounded to atoms are Insulators.

Examples: Glass, Ceramics, Plastic, Rubber.2. Conductors:

The materials in which the outermost atomic electrons are free to move around are Conductors.

Conductors typically have ~1 “free electron” per atom.Examples: Gold, Silver, Copper, Aluminum.

3. Semiconductors.The materials in which electrons are not tightly bound and can be easily “promoted” to a free state are Semiconductors.

Examples: Silicon, Germanium, Gallium Arsenide.

4


Conductors, Semiconductors and Insulators

Conductors Semiconductors Insulators

Gold Silicon Glass

Silver Germanium Plastic

Copper Gallium Arsenide Ceramics

Aluminum Rubber

5


The diagram above shows the structure and lattice of a 'normal' pure crystal of Silicon

Semi-conductors:The most commonly used semiconductor material is Silicon. It has four valence electrons in its outer most shell which it shares with its adjacent atoms in forming covalent bonds. The structure of the bond between two silicon atoms is such that each atom shares one electron with its neighbour making the bond very stable.

Si

A Silicon AtomAtomic Number = 14

A Silicon AtomShowing 4 Electrons in

outermost orbit

Si

Si

Si Si

Si

Co-valent Bond

Silicon Crystal Lattice

6

Prof. Iqbal A. Khan, EED, UQU 7

I II III IV V VI VII ZERO

H He

Li Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Zn Ga Ge As Se Br Kr

Rb Cd In Sn Sb Te I Xe

Donors and Acceptors in the Periodic Table:

AcceptorsImpurity

DonorsImpurity


The diagram above shows the structure and lattice of the donor impurity atom Antimony.

N-type Semiconductor

In order for our silicon crystal to conduct electricity, we need to introduce an impurity atom such as Arsenic, Antimony or Phosphorus into the Si crystalline structure. These atoms have five outer electrons in their outermost co-valent bond to share with other atoms and are commonly called "Pentavalent" impurities.

The resulting semiconductor material has an excess of current-carrying electrons, each with a negative charge, and is therefore referred to as "N-type" material with the electrons called "Majority Carriers"

Si

Si

Sb Si

Si

Co-valent Bond

N-Type Semiconductor

Free Electron

A Donar Antimony Atom with 5

Electrons in outermost orbit

Sb

Impurity Atom(Donar)

8


P-Type Semiconductor

If a "Trivalent" (3-electron) impurity is introduced into the Si crystal structure, such as Aluminum, Boron, Gallium or Indium, only three valence electrons are available in the outermost covalent bond meaning that the fourth bond cannot be formed. The vacancy of an electron in the bond is known as a hole. Therefore, a complete connection is not possible, giving the semiconductor material an abundance of positively charged carriers known as "holes" in the structure of the Si crystal.

The diagram above shows the structure and lattice of the acceptor impurity atom Boron.

Addition of Boron causes conduction to consist mainly of positive charge carriers results in a "P-type" material and the positive holes are called "Majority Carriers" while the free electrons are called "Minority Carriers".

Co-valent Bond

Si

Si

B Si

Si

P-Type Semiconductor

Hole

An Acceptor Boron Atom with 3 Electrons

in outermost orbit

B

Impurity Atom(Acceptor)

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In solid-state physics, the electron mobility characterizes how quickly an electron can move through a metal or semiconductor, when pulled by an electric field. In semiconductors, there is an analogous quantity for holes, called hole mobility. The term carrier mobility refers in general to both electron and hole mobility in semiconductors.Electron and hole mobility are special cases of electrical mobility of charged particles in a fluid under an applied electric field.When an electric field E is applied across a piece of material, the electrons respond by moving with an average velocity called the drift velocity ( vd ). Then the electron mobility μ is defined as

Electron mobility is almost always specified in units of cm2/(V·s). This is different from the SI unit of mobility, m2/(V·s). They are related by 1m2/(V·s) = 104cm2/(V·s). The hole mobility is smaller than that of the electron. μN = 580 Cm2/ V.Sec and μP = 230 Cm2 / V.Sec

Evd

Electron and Hole Mobility


The PN-junction

When the P-type and N-type materials are joined (or fuse) together then a P-N Junction is formed.the resulting device that has been made is called a PN-junction Diode or Rectifier Diode.

Symbol of the PN-Junction Diode

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The Basic Diode Symbol and Static I-V Characteristics.

ID = IS(eVD /VT – 1)

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The diode V-I relationship is characterized by the following equation:

ID = IS(eVD/VT – 1)

• VD = Bias Voltage• ID = Current through Diode. ID is Negative for Reverse Bias and Positive for Forward Bias• IS = Reverse Saturation Current is the emission coefficient for the diode. For a silicon diode is

around 2 for low currents and goes down to about 1 at higher currents

• VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is:

k = 1.38 x 10-23 J/K, T = temperature in Kelvin, q = 1.6 x 10-19 C

Diode Charateristic Equation:

q

kTVT

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Forward Biased Diode

Forward Characteristics Curve for a Diode.

With forward biased the depletion layer is reduced and after the threshold level of voltage the majority charge carriers cross the depletion layer and the diode conducts or ON. The diode in forward mode after the threshold has low resistance.

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A Reverse Biased Diode

Reverse Characteristics Curve for a Diode.

With the reversed biased junction the depletion layer is increased and theMajority charge carriers can not cross the depletion region. This condition represents the high resistance and the diode is said to be OFF.

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Then we can say that an ideal small signal diode conducts current in one direction (forward-conducting) and blocks current in the other direction (reverse-blocking). Signal Diodes are used in a wide variety of applications such as a switch in rectifiers, limiters, snubbers or in wave-shaping circuits.

Forward and Reversed Biased Diode

Anode is positive with respect to Cathode

Cathode is positive with respect to Anode

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Half-wave Rectifier CircuitDuring each "positive" half cycle of the AC sinewave, the diode is Forward Biased and current flows through it. The voltage across the load is then Vout = Vs.During each "negative" half cycle of the AC sinewave, the diode is Reverse Biased and No current flows through it. Therefore, in the negative half cycle of the supply, The output voltage Vout = 0.

rmsmm

avdcOUT

rmsrmsmm

dc

mmmπmdc

π

πm

π

m

π

dcav

V.V.V

VVV

V..V..V.π

VV

π

V

π

VCosCosπ

π

VCosθ

π

VV

dθdSinVπ

dSinVπ

VV

4503180

41413180231803180

22

]1)1([2

)]0([2

][2

]0[2

1

2

1

0

2

0

2

0

The current on the DC side of the circuit flows in one direction only making the circuit Unidirectional and the value of the DC voltage VDC (i,.e., the average value) across the load resistor is calculated as follows.

π 2π00 π 2π

=VmSinθ

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Half-wave Rectifier with Smoothing Capacitor

Half-wave Rectifier with Smoothing Capacitor

When rectification is used to provide a direct voltage power supply from an alternating source, the amount of ripple can be reduced by using larger value capacitors as shown in the Figure.After the peak voltage the capacitor discharges through the load resistor at the slower rate and thus increases the average value or the DC value.

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Example-1:Calculate the current (IDC) flowing through a 100Ω resistor connected to

a 240v single phase half-wave rectifier as shown above, and also the power consumed by the load.

Solution:

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Full-wave Rectifier Circuit-Full Wave RectifierIn a full-wave rectifier circuit two diodes are now used, together with a transformer whose secondary winding is split equally into two and has a common center tapped connection, (C). Now each diode conducts in turn when its Anode terminal is positive with respect to the center point C as shown in Figure. The output voltage(Vdc) can be analysed as follows.

+ve HalfCycle

-ve HalfCycle

rmsmm

avdcOUT

rmsrmsmdc

mmmdc

mπmdc

π

π

πm

m

π

dcav

V.V.V

VVV

V..V..V.V

V

π

V

π

VV

CosCosCosCosππ

VCosCosθ

π

VV

dθSindSinπ

VdSinV

πVV

9063602

41416360263606360

24

2)]1(11)1([

2

]}2[)]0{[2

}][]{[2

][22

1

20

2

0

2

0

π0 2ππ0

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The Diode Bridge Rectifier

The Positive Half-cycleThe Negative Half-cycle

Another type of circuit that produces full-wave rectification is that of the Bridge Rectifier. This type of single phase rectifier uses 4 individual diodes connected in a "bridged" configuration to produce the desired output but does not require a special center tapped transformer, thereby reducing its size and cost. The single secondary winding is connected to one side of the diode bridge network and the load to the other side as shown below.

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Full-wave Rectifier with Smoothing Capacitor

The full-wave bridge rectifier however, gives us a greater mean DC value (0.636Vmax) with less superimposed ripple while the output wveform is twice that of the frequency of the input supply frequency. We can therefore increase its average DC output level even higher by connecting a suitable smoothing capacitor across the output of the bridge circuit.

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Zener Diode I-V CharcateristicsZener Diodes are used in the "REVERSE" bias mode. We can see that the zener diode has a region in its reverse bias characteristics of almost a constant voltage regardless of the current flowing through the diode. This voltage across the diode (it's Zener Voltage, Vz) remains nearly constant even with large changes in current through the diode caused by variations in the supply voltage or load. This ability to control itself can be used to great effect to regulate or stabilise a voltage source against supply or load variations. The diode will continue to regulate until the diode current falls below the minimum Iz value in the reverse breakdown region.

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The Zener RegulatorZener Diodes can be used to produce a stabilized voltage output by passing a small current through it from a voltage source via a suitable current limiting resistor, (RS). The DC output voltage from the half or full-wave rectifiers contains ripple. By connecting a simple zener stabilizer circuit as shown below across the output of the rectifier a more stable dc output voltage can be produced.

Zener Diode Stabiliser

=(VS - VZ ) / IZ

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Example -1.

A 5.0v stabilized power supply is required from a 12V d.c. input source. The maximum power rating of the Zener diode is 2W. Using the circuit above calculate:

a) The maximum current flowing in the Zener Diode.

b) The value of the series resistor, RS

c) The load current IL if a load resistor of 1kΩ is connected across the Zener diode.

d) The total supply current IS

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BZX55 Zener Diode Power Rating 500mW

2.4V 2.7V 3.0V 3.3V 3.6V 3.9V 4.3V 4.7V

5.1V 5.6V 6.2V 6.8V 7.5V 8.2V 9.1V 10V

11V 12V 13V 15V 16V 18V 20V 22V

24V 27V 30V 33V 36V 39V 43V 47V

BZX85 Zener Diode Power Rating 1.3W

3.3V 3.6V 3.9V 4.3V 4.7V 5.1V 5.6 6.2V

6.8V 7.5V 8.2V 9.1V 10V 11V 12V 13V

15V 16V 18V 20V 22V 24V 27V 30V

33V 36V 39V 43V 47V 51V 56V 62V

Zener Diodes with different voltages and power ratings

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Vrms = 1.11 Vav

rms and average relationship:

11.122

,2

,2

av

rms

mav

mrms

V

V

VV

VV

Form Factor =

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Diode ClippersThe diode in a series clipperseries clipper “clips” any voltage that does not forward bias it:•A reverse-biasing polarity•A forward-biasing polarity less than 0.7 V (for a silicon diode)

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Parallel ClippersThe diode in a parallel clipperparallel clipper circuit “clips” any voltage that forward bias it.

DC biasing can be added in series with the diode to change the clipping level.

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Summary of Clipper CircuitsSummary of Clipper Circuits

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ClampersClampersA diode and capacitor can be combined to “clamp” an AC signal to a specific DC level.

31


Biased Clamper CircuitsBiased Clamper Circuits

The input signal can be any type of waveform such as sine, square, and triangle waves.

The DC source lets you adjust the DC clamping level.

32

Summary of Clamper CircuitsSummary of Clamper Circuits

3333Prof. Iqbal A. Khan, EED, UQU 33


Voltage Doubler:• Positive Half-Cycle

o D1 conductso D2 is switched offo Capacitor C1 charges to Vm

• Negative Half-Cycleo D1 is switched offo D2 conductso Capacitor C2 charges to 2Vm

Vout = VC2 = 2Vm

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Voltage Tripler and Quadrupler

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Light-Emitting Diodes:

Light-emitting diodes are designed with a very large bandgap so movement of carriers across their depletion region emits photons of light energy. Lower bandgap LEDs (Light-Emitting Diodes) emit infrared radiation, while LEDs with higher bandgap energy emit visible light. Many stop lights are now starting to use LEDs because they are extremely bright and last longer than regular bulbs for a relatively low cost.

Schematic Symbol for a Light-Emitting Diode

A KThe arrows in the LED representation indicate emitted light.


Photodiodes:Photodiodes are sensitive to received light. They are constructed so their PN junction can be exposed to the outside through a clear window or lens.In Photoconductive mode the saturation current increases in proportion to the intensity of the received light. This type of diode is used in CD players.In Photovoltaic mode, when the PN junction is exposed to a certain wavelength of light, the diode generates voltage and can be used as an energy source. This type of diode is used in the production of solar power.

Schematic Symbols Schematic Symbols for Photodiodesfor Photodiodes

A K

A K



LIGHT EMITTING DIODE-LED: LED are semiconductor p-n junctions that under forward bias conditions can emit radiation by electroluminescence in the UV, visible or infrared regions of the electromagnetic spectrum. The qaunta of light energy released is approximately proportional to the band gap of the semiconductor.

P-n junction

Electrical Contacts

Schematic Symbols Schematic Symbols for LEDfor LED

A K

Recombination produces light!!

Junction is biased to produce even more e-h and to inject electrons from n to p for recombination to happen


The BJT The BJT –– Bipolar Junction Bipolar Junction TransistorTransistor

The Two Types of BJT Transistors:The Two Types of BJT Transistors:

npnnpn pnppnp

nn pp nnEE

BB

CC pp nn ppEE

BB

CC

Cross SectionCross Section Cross SectionCross Section

Schematic Schematic SymbolSymbol

Schematic Schematic SymbolSymbol

• Collector doping is usually ~ 10Collector doping is usually ~ 1066

• Base doping is slightly higher ~ 10Base doping is slightly higher ~ 1077 –– 101088

• Emitter doping is much higher ~ 10Emitter doping is much higher ~ 101515

CC

BB

EE

CC

BB

EE

39


BJT Structure

• In this process, all steps are performed from the surface of the wafer

40


BJT Relationships - EquationsBJT Relationships - Equations

IIBB

--

++ ++

CCEE

IIEE IICC

BB

VVBEBE VVBCBC

--

++-- VVCECE

IIBB

IIEE IICC

--

++

VVEBEB VVCBCB

BB

CCEE++

--

++ --VVECEC

npnnpn

IIEE = I = IBB + I + ICC

VVCECE = -V = -VBCBC + V + VBEBE

pnppnp

IIEE = I = IBB + I + ICC

VVECEC = V = VEBEB - V - VCBCB

Note: The equations seen above are for Note: The equations seen above are for the transistor, not the circuit.the transistor, not the circuit.

41


Transistor Configurations:

Input = VInput = VBEBE & & IIBB Output = V Output = VCECE & & IICC

Input = VInput = VEBEB & I& IEE Output = Output = VVCBCB & I & ICC

Input = VInput = VBC BC & & IIBB Output = V Output = VECEC & I& IEE

42


DC DC and DC and DC = Common-base current gain = Common-base current gain

= Common-emitter current gain= Common-emitter current gain

The relationships between the two parameters are:The relationships between the two parameters are:

Note: Note: and and are sometimes referred to as are sometimes referred to as dcdc and and dcdc because the relationships being dealt with in the BJT are because the relationships being dealt with in the BJT are DC.DC.

B

C

I

I

E

C

I

I

1)(1 BC

BC

CB

C

E

C

II

II

II

I

I

I

1)(1 EC

EC

CE

C

B

C

II

II

II

I

I

I

43


Example1: Example1: For a Common-Base NPN Circuit For a Common-Base NPN Circuit Configuration, Configuration, Given: IGiven: IBB = 50 = 50 A , I A , ICC = 1 mA, Find: I = 1 mA, Find: IEE , , , , and and . . Solution:Solution:

IIEE = I = IBB + I + ICC = 0.05 mA + 1 mA = 1.05 = 0.05 mA + 1 mA = 1.05 mAmA

= I= ICC / I / IBB = 1 mA / 0.05 mA = 20 = 1 mA / 0.05 mA = 20

= I= ICC / I / IEE = 1 mA / 1.05 mA = 0.95238 = 1 mA / 1.05 mA = 0.95238

could also be calculated using the could also be calculated using the value of value of with the formula from the with the formula from the previous slide.previous slide.

95238.0201

20

1

44


There are three regions of There are three regions of operation:operation:

1.1.Active Region: Active Region: The region The region where current curves are where current curves are practically flat. In this region practically flat. In this region the the JJEBEB is forward bias and the is forward bias and the JJCB CB is reversed bias. is reversed bias. In this In this region region transistor acts as an transistor acts as an amplifieramplifier. .

2. Saturation Region: 2. Saturation Region: In this In this region both the junctions region both the junctions JJEB EB and and JJCB CB are forward biasare forward bias, as a result , as a result the barrier potential of the the barrier potential of the junctions cancel each other out junctions cancel each other out causing a virtual short between causing a virtual short between collector and emitter terminals collector and emitter terminals i. e. the i. e. the transistor is ONtransistor is ON..3. Cutoff Region: 3. Cutoff Region: In this region both the junctions In this region both the junctions JJEB EB and J and JCB CB are are

reverse biasreverse bias, and thus the currents reduced to zero. In this , and thus the currents reduced to zero. In this region transistor region transistor

behaves like an open switch, i. e. the behaves like an open switch, i. e. the transistor is OFFtransistor is OFF..

IIB2B2

VVCECE

IICC

ActivActive e

RegioRegionn

Saturation Saturation Region Region (Transistor is (Transistor is ON) VON) VSAT SAT = 0.2V= 0.2V

Cutoff Cutoff Region IRegion IBB = =

0, I0, IC C = 0 = 0 ((Transistor is Transistor is

OFFOFF))

IIB4B4

IIB3B3

IIB1B1

Output Characteristics Output Characteristics of CE-Configurationof CE-Configuration

Output Characteristics of Common Emitter Output Characteristics of Common Emitter ConfigurationConfiguration

45


Self Bias Circuit for Active Region:

21

21

21

2

RR

RRR

VVRR

RV

BB

CCCCBB

+VBE

-IB

IC

IE

BE

BBCBE

BC

II

IIIII

II

)1(

46


Example: In the self bias circuit the R1 = 30K, R2 = 10K, RC = 4.3K,RE = 1.3K, VCC =12V and β = 100. Find IB , IC , IE and VB , VC , VE , VCE.

KRR

RRRBB 5.7

1030

1030

21

21

VVRR

RV CCBB 312

1030

10

21

2

Solution: The self bias circuit and its Thevenin’s equivalent is given as follows.

RC

VC

VE

VBVC

VB

VE

47


VBB = IERE + VBE + IBRBB and IE = (1+β)IB

Therefore,

3V = (1+100)IB x1.3 x1000 + 0.7V + IB x 7.5 x 1000

(101x1.3 + 7.5)x1000xIB = 3 – 0.7

138.8x1000xIB = 2.3

IB = 2.3 / (138.8x1000) A

IB = 2.3 / (138.8) mA

IB = 2.3x1000 / (138.8) µA

IB = 16.57 µA

Hence, IC = βIB =100x16.57 µA =1.657 mA

IE = (1+β)IB = (1+100)x16.5 µA = 1.673 mA48


VB = VBB – IB RB = 3V – 16.57x10-6x7.5x10+3 = 2.87V

VB = 2.87V

VE = IERE = 1.673 mA)(1.3K) = 2.17 V

VE = 2.17V

VBE =VB – VE = 2.87V – 2.17V = 0.7V

VC = 12 – ICRC = 12 - (1.657 mA)x(4.3K) = 4.87 V

VC = 4.87V

Thus

VC > VB ; The Collector Junction JC is reversed bias.

VB > VE ; The Emitter Junction JE is forward bias.

Therefore the transistor is in Active Region of operation.

VCE = VC – VE = 4.87 - 2.17 = 2.7V49

Documents

Prof. Iqbal A. Khan, EED, UQU Course Code: 802311 Course Name: Electronic Devices By Prof. Iqbal Ahmad Khan Department of Electrical Engineering Faculty