Chapter 3: Bipolar Junction Transistor

Chapter 3:Chapter 3:BIPOLAR JUNCTION

TRANSISTOR

Learning Outcomes

At the end of this chapter, the students shouldbe able to:

● Understand the basic transistor construction,operation & configuration

● Discuss transistor parameters andcharacteristicscharacteristics

● Develop working knowledge of a transistorthrough use of specification sheet

● In the nineteenth century, scientists were rarelyinventors: Samuel F.B. Morse, AlexanderGraham Bell, Thomas Alva Edison

● In the twentieth century, scientists invaded the● In the twentieth century, scientists invaded thedomain of invention: John Fleming invented thevacuum diode tube and Lee De Forest inventedthe triode tube

● The transistor can be viewed, as can the laser,as an invention of physicists.

– Source: Bunch and Hellemans, The Timetables ofTechnology, Simon and Schuster, 1993

William B. Shockley (1910-1989)

● Known as the “Father of the Transistor”

joined Bell Labs in 1936 in the vacuum tube● joined Bell Labs in 1936 in the vacuum tubedepartment (solid state physicist)

● Moved to the semiconductor laboratory:

– “It has today occurred to me that an amplifier usingsemiconductors rather than vacuum tubes is inprinciple possible.”principle possible.”

William B. Shockley

Walter Houser Brattain

● Experimental physicist who also worked onvacuum tubesvacuum tubes

● Joined Shockley and Bardeen insemiconductor research.

Walter Houser Brattain

John Bardeen (1908-1991)

● Physicist, Naval Ordnance Laboratory 1941-1945

● Research Physicist, Bell Telephone Laboratories● Research Physicist, Bell Telephone Laboratories1945-1951 (theorist)

● Professor of Electrical Engineering,

– University of Illinois, 1951-1978

● Nobel Prize in Physics: 1956 and 1972

transistor (1956) and superconductivity (1972)● transistor (1956) and superconductivity (1972)

– “I knew the transistor was important, but I neverforesaw the revolution in electronics it would bring.”

John Bardeen

Nobel Prize in 1956

● Shockley, Brattain and Bardeen startworking with p- and n- type germanium andsilicon semiconductors in 1946silicon semiconductors in 1946

● Bardeen and Brattain put together the firsttransistor in December 1947:

– a point-contact transistor consisting of a singlegermanium crystal with a p- and an n- zone.Two wires made contact with the crystal nearTwo wires made contact with the crystal nearthe junction between the two zones like the“whiskers” of a crystal-radio set.

Point-contact-transistor

● Shockley immediately set out to define theeffects that they had observed, i.e., to explainthe physics of transistors

A few months later, Shockley devised the● A few months later, Shockley devised thejunction transistor, a true solid-state devicewhich did not need the “whiskers” of the point-contact transistor.

● AT&T licensed the transistor very cheaply toother manufacturers and waived patent rightsother manufacturers and waived patent rightsfor the use of transistors in hearing aids, in thespirit of its founder, Alexander Graham Bell

Shockley’s sandwitch transistor

Manufacturing transistors on a chip

● Shockley Semiconductor Laboratories,

Palo Alto, CA (1954)

– the beginnings of “Silicon Valley”

● Fairchild Semiconductors founded in MountainView, CA (1957) by eight Shockley employeesincluding Gordon Moore and Robert Noyce

● Bell Labs had made several improvements inthe manufacturing of crystals of silicon andthe manufacturing of crystals of silicon andgermanium with the impurities needed to createsemiconductors

Meanwhile….

● Jack Kilby worked for Texas Instruments

Conceived of a manufacturing method that● Conceived of a manufacturing method thatallowed the miniaturization of electroniccircuits on semiconductor chips, calledintegrated circuits or ICs.

● Kilby had reduced the transistor to the size ofa match heada match head

● Texas Instruments sold these for $450.

Introduction

• The basic of electronic system nowdays is semiconductordevice. The famous and commonly use of this device isdevice. The famous and commonly use of this device is

BJTs (Bipolar Junction Transistors).

• It can be use as amplifier and logic switches.

• BJT consists of three terminal:collector : Ccollector : Cbase : Bemitter : E

BJT consists of threeterminal:terminal:

collector : C base : Bemitter : E

Two types of BJT :pnp andTwo types of BJT :pnp andnpn

• NPN / PNP Block Diagrams

Bipolar Transistor

Emitter

Collector

Base Collector

Transistor Construction (Size)

There are two types of transistors: (a) pnp and (b) npn-type.

Transistor ConstructionTransistor Construction

• 3 layer semiconductor device consisting:

• 2 n- and 1 p-type layers of material npn• 2 n- and 1 p-type layers of material npntransistor

• 2 p- and 1 n-type layers of material pnptransistor

• The term bipolar reflects the fact that holes andelectrons participate in the injection process intothe oppositely polarized material

• A single pn junction has two different types of bias:

• forward bias

• reverse bias

Position of the terminals and symbolof BJT.

• Base is located at the middle• Base is located at the middle

and more thin from the levelof collector and emitter

• The emitter and collectorterminals are made of thesame type of semiconductormaterial, while the base of theother type of materialother type of material

Transistor currents

-The arrow is always drawn

on the emitter

-The arrow always point-The arrow always point

toward the n-type

-The arrow indicates the

direction of the emitter

current:

pnp:E B

npn: B EIC=the collector current

npn: B EIC=the collector currentIB= the base currentIE= the emitter current

• By imaging the analogy of diode, transistor can be

construct like two diodes that connected together.

• It can be conclude that the work of transistor is base on

work of diode.work of diode.

Transistor OperationTransistor Operation

• The basic operation will be described using the pnptransistor. The operation of the pnp transistor isexactly the same if the roles played by the electronand hole are interchanged.

• One p-n junction of a transistor is reverse-biased,• One p-n junction of a transistor is reverse-biased,whereas the other is forward-biased.

Forward-biased junctionof a pnp transistor

Reverse-biased junctionof a pnp transistor

• Both biasing potentials have been applied to a pnptransistor and resulting majority and minority carrierflows indicated.

• Majority carriers (+) will diffuse across the forward-biased p-n junction into the n-type material.

• A very small number of carriers (+) will through n-typematerial to the base terminal. Resulting IB is typically inorder of microamperes.

• The large number of majority carriers will diffuseacross the reverse-biased junction into the p-typematerial connected to the collector terminal.

• Majority carriers can cross the reverse-biasedjunction because the injected majority carriers willappear as minority carriers in the n-type material.

• Applying KCL to the transistor :• Applying KCL to the transistor :

IE = IC + IB

• The comprises of two components – the majority

and minority carriers

IC = ICmajority + ICOminority

• ICO – IC current with emitter terminal open and is

called leakage current.called leakage current.

CommonCommon--Base ConfigurationBase Configuration

• Common-base terminology is derived from the fact that

- base is common to both input and output of the- base is common to both input and output of the

configuration.

- base is usually the terminal closest to or at

ground potential.

• All current directions will refer to conventional (hole)

flow and the arrows in all electronic symbols have a

direction defined by this convention.

• Note that the applied biasing (voltage sources) are such

as to establish current in the direction indicated for each

branch.

Common-base configuration (CB)• Fig below shows the common-base configuration for pnpand npn transistor.• CB is derived from the fact that the :- base is common to both i/p and o/p of the configuration.- base is usually the terminal closest to or at ground potential- base is usually the terminal closest to or at ground potential

Fig. 4.3: Symbols used with the CB configuration

Analysis of Common-base configuration for pnpStep 1:• B-E junction must be forward bias

E C Bp pn

Analysis of Common-base configuration for pnpStep 2:• B-C junction must be reverse bias• ICBO=ICO=0 A is a reverse saturation current and normallyknown as leakage currentknown as leakage current

Analysis of Common-base configuration for pnp

• Current base, IB (A) is small compare to currentemitter, IE (mA) and current collector,IC (mA).

• The relationship among these current can be analysewith KCL : IE = IB + IC

• Current collector is produce from the total sum ofcurrent emitter and leakage current.

• Current emitter that flow through collector known as

DC IE . The value is big compare to leakage current.

IC = IC(majority) + IC(minority)

IC = IE + ICBO

The analysis can be understand by expression below:

IC = IE + ICBO

IC = IE (ignore ICBO due to small value)

= Ideally = 1, but in reality it is between 0.9 and

0.998.

thus,E

is a common base current gain factor that shows theefficiency

● To describe the behavior of common-base amplifiers

requires two set of characteristics:

- Input or driving point characteristics.

- Output or collector characteristics

● The output characteristics has 3 basic regions:● The output characteristics has 3 basic regions:

- Active region –defined by the biasing arrangements

- Cutoff region – region where the collector current is 0A

- Saturation region- region of the characteristics to the left of

- VCB = 0V

VCB=-10VVCB=-1V

IE(mA)

IC(mA)

IE=4mA

IE=5mA

IE=6mA

Saturationregion

Activeregion

VCB(V)IE=0mA

IE=1mA

IE=2mA

IE=3mA

IE=4mA

VBE(V)

0.2 0.4 0.6 0.81.0

Input characteristics for acommon-base pnp transistor

Cutoffregion

VCB(V)IE=0mA

-5 -10-15-20

Output characteristics for acommon-base pnp transistor

Energy band of a NPN transistor

Try to do analysisTry to do analysisCommon-base configuration for

npn !!!

Transistor as an amplifier

Common-emitter configuration (CE)

• It is called common-emitter configuration since :

- emitter is common or reference to both i/p and o/pterminals.terminals.

- emitter is usually the terminal closest to or at groundpotential.

• Almost amplifier design is using connection of CEdue to the high gain for current and voltage.

• Two set of characteristics are necessary to describe thebehavior for CE ;input (base terminal) and output (collectorterminal) parameters.

IB VCC

Proper Biasing common-emitter configuration in active region

C ICIC

(a)npntransistorconfiguration

BCE III

IB VCC

(b)pnptransistorconfiguration

Fig4.7:Common-emitter configuration

Common Emitter Characteristics

Treating the transistor as a current node:Treating the transistor as a current node:

● Also:

coEC IIαI

BCE III

● Hence:

which after some rearrangement givesCOBCC I)IΙαI

ICOC BII

● Define a common emitter current-transfer ratio

Such that:

● Since reverse saturation current is negligiblethe second term on the right hand side of thisthe second term on the right hand side of thisequation can usually be neglected (eventhough (1- α) is small)

● Thus

● We note, in passing that, if β can be regardedas a constant for a given transistor thenas a constant for a given transistor then

bc ● For a practical (non-ideal) transistor this is only

true at a particular bias (operating) point.

● A small change in α causes a much biggerchange in ß which means that ß can varychange in ß which means that ß can varysignificantly, even from transistor to transistorof the same type.

● We must try and allow for these variations incircuit design.

● We can therefore draw an input characteristic(plotting base current IB against base-emitter(plotting base current IB against base-emittervoltage VBE) and

● an output characteristic (plotting collectorcurrent Ic against collector-emitter voltage VCE)

● We will be using these characteristic curvesextensively to understand:extensively to understand:

● How the transistor operates as a linearamplifier for a.c. signals.

● The need to superimpose the a.c. signals ond.c. bias levels.

The relationship between the transistor and the● The relationship between the transistor and thecircuit in which it is placed.

● Once these basics are understood we willunderstand:understand:

● Why we can replace the transistor by asmall signal (a.c.) equivalent circuit.

● How to derive a simple a.c. equivalentcircuit from the characteristic curves.circuit from the characteristic curves.

● Some of the limitations of our simpleequivalent circuit.

IDEAL CE INPUT (Base)Characteristics

IDEAL CE INPUT Characteristics

● The plot is essentially that of a forward biaseddiode.diode.

● We can thus assume VBE 0.6 V whendesigning our d.c. bias circuits.

● We can also assume everything we know aboutincremental diode resistance when deriving oura.c. equivalent circuit.a.c. equivalent circuit.

● In the ‘non-ideal’ case IB will vary slightly withVCE. This need not concern us.

IDEAL CE OUTPUT (Collector)Characteristics

IDEAL CE OUTPUT(Collector) Characteristics

Avoid thisAvoid thissaturationregionwhere wetry toforwardforwardbias bothjunctions

IDEAL CE OUTPUT

Avoid this cut-off region where we try to reversebias both junctions (IC approximately 0)

IDEAL CE OUTPUT (Collector)Characteristics

● The plots are all parallel to the VCE axis (i.e. ICdoes not depend on VCE)does not depend on VCE)

● The curves strictly obey IC = βIB● In particular IC = 0 when IB = 0.

● We shall work with the ideal characteristic andlater on base our a.c. equivalent circuit modelupon it.upon it.

ACTUAL CE OUTPUTCharacteristics

Characteristics of Common-Emitternpn transistor(Actual)

(a) - Collector characteristics = output characteristics.(b) - Base characteristics = input characteristics.

ACTUAL CE OUPUTCharacteristics

● Salient features are:

The finite slope of the plots (I depends on● The finite slope of the plots (IC depends onVCE)

● A limit on the power that can be dissipated.

● The curves are not equally spaced (i.e β varieswith base current, IB).

ACTUAL CE OUPUTCharacteristics

● You will get to measure these curves in thelab.lab.

Figure 4.3 Current flow for an npn BJT in the active region.Most of the current is due to electrons moving from the emitter

through the base to the collector. Base current consists of holescrossing from the base into the emitter and of holes that recombine

with electrons in the base.

Simulation of transistor as an amplifier

Example:

a) Using the o/p characteristics, determine IC at IB=30 Aa) Using the o/p characteristics, determine IC at IB=30 AVCE=10 V.b) Using the i/p characteristics, determine IC at VBE=0.7 Vand VCE =10 V.

Solution: (a)

IC(mA)

IB=60uA

IB=50uA

I =40uA4

Saturationregion

Active region

IB=0uA

IB=40uA

IB=30uA

IB=20uA

IB=10uA

Cutoff region

VCE(V)5 10 15VCE(sat)

CBOCEO II

Solution: (b)

IB(uA)

IC(mA)

IB=60uA

IB=50uA

405060

Saturationregion

Activeregion

IB=0uA

IB=40uA

IB=30uA

IB=20uA

IB=10uA

0.2 0.4 0.6 0.8 1.0Cutoffregion

CBOCEO II

Beta ()• The ratio of dc collector current (IC) to the dc base current

(IB) is dc beta (dc ) which is dc current gain where IC andIB are determined at a particular operating point, Q-point(quiescent point).

• It’s define by the following equation:

30 < dc < 300 2N3904dc

• On data sheet, dc=hFE with h is derived from ac hybridequivalent cct. FE are derived from forward-currentequivalent cct. FE are derived from forward-currentamplification and common-emitter configurationrespectivley.

• For ac conditions an ac beta has been defined as thechanges of collector current (IC) compared to thechanges of base current (IB) where IC and IB aredetermined at operating point.

• On data sheet, ac=hfe

• It can defined by the following equation:

VCE=constant

Example :

From o/p characteristics of CE configurationfind ac and dc with an operating point atIB=25 A and VCE =7.5V.

IC(mA)

IB=60uA

IB=50uA

IB=40uA

Solution:

Saturationregion

Active region

VCE(V)5 10 15V

IB=0uA

IB=40uA

IB=30uA

IB=20uA

IB=10uA

Q-point

Cutoff region5 10 15VCE(sat)

20V7.5VCE

Common-collector configuration (CC)

• Also called emitter-follower (EF).

• It is called common-emitter configuration since both thesignal source and the load share the collector terminal as asignal source and the load share the collector terminal as acommon connection point.

• The o/p voltage is obtained at emitter terminal.

• The input characteristic of CC configuration is similarwith CE configuration.with CE configuration.

• All the current relationship for CE configuration are truefor CC configuration.

IB VEE

VBBVEEVou t

(a) npn transistor configuration

IB VEEp

IB VEE Vou t

(b) pnp transistor configuration

Fig 4.8 : Common-collector configuration

VEE Vou t

IE(mA)

IB=60 uA

IB=50 uA5

Saturationregion

Active region

I =0 uA

IB=40 uA

IB=30 uA

IB=20 uA

IB=10 uA

Cutoff region

IB=0 uA

Fig 4.9: Outputcharacteristic in CC configurationfor npn transistor

Example :Calculate the emitter current for the common-collectorconfiguration below:

Limits of operation for transistor• Many BJT transistor used as an amplifier. Thus it isimportant to notice the limits of operations.• At least 3 maximum values is mentioned in data sheet.There are:

a) Maximum power dissipation at collector: PCmax or PD

b) Maximum collector-emitter voltage: VCEmax sometimesnamed as VBR(CEO) or VCEO.c) Maximum collector current: ICmax

• There are few rules that need to be followed for BJTtransistor used as an amplifier. The rules are:transistor used as an amplifier. The rules are:

i) transistor need to be operate in active region!ii) IC < ICmax

ii) PC < PCmax

Transistor limits of operation

Note: VCE is at maximum and IC is at minimum (ICmax=ICEO) in the cutoffregion.

IC is at maximum and VCE is at minimum (VCEsat ) in thesaturation region.The transistor operates in the active region between saturation and cutoff.

Example :

IC(mA)

IB=60uA

I =50uA

PD: maximum powerdissipation line

Refer to the fig.Step1:The maximum collectorpower dissipation,PD=ICmax x VCEmax (1)

= 18m x 20 = 360 mWStep 2:At any point on the characteristics the

Saturationregion

Activeregion

IB=50uA

IB=40uA

IB=30uA

IB=20uA

IB=10uA

At any point on the characteristics theproduct of and must be equal to 360mW.Ex. 1. If choose ICmax= 5 mA, subtituteinto the (1), we getVCEmaxICmax= 360 mWVCEmax(5 m)=360/5=7.2 V

Ex.2. If choose VCEmax=18 V, subtituteinto (1), we get3

Cutoffregion

IB=0uA

IB=10uA

20 VCEmax

into (1), we getVCEmaxICmax= 360 mW(10) ICmax=360m/18=20 mA

Derating PDmax

● PDmax is usually specified at 25°C.● PDmax is usually specified at 25°C.● The higher temperature goes, the less is PDmax

● Example;– A derating factor of 2mW/°C indicates the power

dissipation is reduced 2mW each degreecentigrade increase of temperature.

More Example:

Transistor 2N3904 used in the circuit withV = 20 V.This circuit used at temperatureVCE= 20 V.This circuit used at temperature1250C. Calculate the new maximum IC.Transistor 2N3904 have maximum powerdissipation is 625 mW. Derating factor is 5mW/0C.mW/ C.

Solution:

Step 1:

Temperature increase : 1250C – 250C = 1000C

Step 2:

Transistor 2N3904 used in the

circuit with VCE= 20 V.This

circuit used at temperature

1250C. Calculate the new

maximum IC. Transistor

Derate transistor : 5 mW/0C x 1000C = 500 mW

Step 3:

Maximum power dissipation at

1250C = 625 mW–500 mW=125 mW.

Step 4:

2N3904 have maximum power

dissipation is 625 mW. Derating

factor is 5 mW/0C.

Step 4:

Thus ICmax = PCmax / VCE=125m/20 = 6.25 mA.

Step 5:

Draw the new line of power dissipation at 1250C .

Example:

The parameters of transistor 2N3055 as follows:The parameters of transistor 2N3055 as follows:- Maximum power dissipation @ 250C=115 W- Derate factor=0.66 mW/0C.

This transistor used at temperature 780C.a) Find the new maximum value of power dissipation.b) Find the set of new maximum of IC if VCE=10V, 20 VC CE

and 40 V.

Solution:

Step 1:

Temperature increase : 780C – 250C = 530C

Step 2:

Derate transistor : 0.66mW/0C x 530C = 35 mW

Step 3:

Maximum power dissipation at 780C = 115W– 35W=80 mW.

Step 4:

ICmax = PCmax / VCE=80m/10 = 8 mA (point C)ICmax = PCmax / VCE=80m/10 = 8 mA (point C)

ICmax = PCmax / VCE=80m/20 = 4 mA. (point B)

ICmax = PCmax / VCE=80m/40 = 2 mA (point A)

Step 5:

Draw the new line of power dissipation at 780C .

IC(m A )

IB=60 uA

IB=50 uA

IB=40 uA

IB=30 uA

IB=20 uA

VC E(V )10 20 30

IB=0 uA

IB=20 uA

IB=10 uA

Transistor Specification Sheet

Transistor Testing

1. Curve TracerProvides a graph of the characteristic curves.

2. DMMSome DMM’s will measure DC or HFE.

1. Curve TracerProvides a graph of the characteristic curves.

2. DMMSome DMM’s will measure DC or HFE.

3. Ohmmeter

Some DMM’s will measure DC or HFE.3. Ohmmeter

Chapter 3: Bipolar Junction Transistor

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