bipolar junction transistor (bjt) - introductionsiva/2015/ee101/bjt.pdf · bipolar junction...

Bipolar Junction Transistor (BJT) - Introduction

It was found in 1948 at the Bell Telephone Laboratories.

It is a three terminal device and has three semiconductorregions.

It can be used in signal amplification and digital logic circuitsas well.

Its current conduction process is due to both holes andelectrons. That is why the name bipolar.

S. Sivasubramani EE101 - BJT 1/ 60

NPN Transistor

E C

B

n type p type n type

Emitter region Base region Collector region

EB Junction CB Junction

B

C

E

Figure: Symbol


PNP Transistor

E C

B

p type n type p type

Emitter region Base region Collector region

EB Junction CB Junction

B

C

E

Figure: Symbol


BJT - Modes of Operation

Mode EB Junction CB Junction

Active Forward Reverse

Saturation Forward Forward

Cutoff Reverse Reverse

Active mode - Amplification

Saturation and Cutoff - Digital logic circuits


Active Mode - NPN Transistor

E IE C

IC

B

IB

N P N

Emitter Base Collector

VBE VCB

iC = IS exp(vBEVt

)

iB =iCβ

where β is common-emitter current gain.Its typical value is in the range of 50 to 200.


Active Mode

By KCLiE = iC + iB

iE = iCβ + 1

β

iC = αiE

where α is a constant.

α =β

β + 1

iE =ISαexp(

vBEVt

)

If β = 100, then α ≈ 0.99. α is called as common-base currentgain.


i − v Characteristics - CE

VBE

IB

VCE

IC

IE

vBE (V)

iB

0.5 0.7


i − v Characteristics - CE

vCE

iC

vBE = · · ·

vBE = · · ·

vBE = · · ·

vBE = · · ·

Active Region

Saturation Region

Cutoff Region

As long as vCE > vCE sat , BJT is in active region. vCE sat = 0.2 V.If vCE falls below vCE sat , BJT will enter into saturation region.


BJT Model - Active Mode

BiB

+

−

vBE

iE

E

CiC

βiB

This model is nonlinear. (large signal)As E is common to both input and output, it is called as commonemitter configuration.


i − v Characteristics - Practical BJT

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(A) BJT_Characteristics (active)

V_V2

0V 5V 10VIC(Q1)

0A

0.5A

1.0A


Early Voltage

As Practical BJTs have finite slope in the iC Vs vCE characteristicswhile acting in active mode.This is accounted for by the following relation.

iC = IS exp(vBEVt

)(1 +vCEVA

)

where VA is the Early voltage whose value is in the range of 10 Vto 100 V.The nonzero slope indicates the output resistance is finite.


Large Signal Model Including Early Voltage

BiB

+

−

vBE

iE

E

CiC

βiB ro

It is the large signal model of common emitter NPN transistor inactive region. As there is no restriction in the signal range, it iscalled as a large signal model.


Summary - Active Mode

iC = IS exp(vBEVt

)

iB =iCβ

=ISβ

exp(vBEVt

)

iE =iCα

=ISα

exp(vBEVt

)

iC = βiB ; iC = αiE

α =β

β + 1

β =α

α− 1


BJT - Various Modes of Operation

IB

IC

+

−VBE

+

−VBC +

−

VCE

VBE = VBC + VCE

VBC = VBE − VCE

1 Active mode : VBE = 0.7V , VCE > 0.2V .

2 Saturation mode : VBE = 0.7V , VCE ≤ 0.2V .

3 Cutoff mode: VBE < 0.5V , VBC < 0.5V .


BJT Models - Active Mode

B

E

CiC

βiB

IB > 0

+

−

VBE =0.7 V

+

−

VCE >0.2 V

EBJ : Forward biasedCBJ: Reverse biased


BJT Models - Saturation Mode

B

E

CiC

IB > 0

+

−

VBE =0.7 V

+

−

VCE =0.2 V

EBJ : Forward biasedCBJ: Forward biased


BJT Models - Cutoff Mode

B

E

CIB = 0 IC = 0

+

−

VBE< 0.5 V

EBJ : Reverse biasedCBJ: Reverse biased


BJT - DC Analysis

Whenever we want to analyze BJT circuits only with DC voltages,use the following steps.

1 Assume that the transistor is operating in active mode.

2 Determine IC , IB , VCE and VBE using the active mode model.

3 Check for consistency of results with active-mode operationsuch as VCE > VCE sat .

4 If it is satisfied, analysis is over.

5 If not, assume saturation mode and repeat the analysis likeactive mode.

This analysis is mainly used to identify the operating point.


Check yourself

Which mode is transistor operating? β=100 and VBE = 0.7 V.

4 V

4.7 k Ω

10 V

3.3 kΩ

By KVL in Base Emitter loop,

4 = VBE + 3.3IE

IE = 1 mA

IB =IE

β + 1=

1

101= 9.9µA

IC = IE − IB = 0.99 mA

By KVL in CE loop,

VCE = 10−4.7× IC −3.3× IE = 2.047 V

Since VCE > VCE sat , it is operating inactive mode.


Amplifier - An Introduction

io

RL

+

−

vO(t)

iI

vI (t)

vI (t)vO(t)



vO(t) = AvI (t)

As vO and vI are linearly related, it is called as a linear amplifier.

vI

vO

← Slope= A



io

RL

+

−

vO(t)

iI

vI (t)

An amplifier takes power from battery to amplify the signal givenat its input. This is how it amplifies the input signal.


BJT- Amplifier

+−vBE

RC

iC

VCC

+

−

vCE

vBE is the input voltage.RC is the load resistance.The output vCE is taken betweenthe collector and ground.

vCE = VCC − iCRC

vCE = VCC − IS exp(vBEVt

)RC


BJT - Voltage Transfer Characteristics (VTC)

vBE

vCE

0.5 V

0.2 V

Cutoff SaturationActive


BJT - Voltage Transfer Characteristics

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(A) BJT_Characteristics (active)

V_V1

0V 0.5V 1.0V 1.5VV(Q1:c)

0V

2.5V

5.0V


BJT - Biasing

VBE

RC

IC

VCC

+

−

VCE vBE

vCE

VCE Q ← Quiescent

VBE

point

VCE = VCC − IS exp(VBE

Vt)RC


BJT- Amplifier

VBE

vbe +

−

vBE

RC

iC

VCC

+

−

vCE


BJT-Amplifier

vBE

vCE

Y

0.5 V

Z0.2 V

Cutoff SaturationActive

VCE Q

VBE


Small Signal Voltage Gain

If the input signal vbe is very small, the output signal vce will be

vce = Avvbe

where Av is the voltage gain that is determined by evaluating theslope of VTC at Q point.

Av =dvcedvbe

∣∣∣VBE

Av =d(VCC − IS exp(

vBEVt

)RC )

dvbe

∣∣∣VBE

=−IS exp(

VBE

Vt)RC

Vt

Av =−ICRC

Vt


Graphical Analysis - To determine Bias point

+−vBE

RC

iC

VCC

+

−

vCE

vCE

iC

vBE = · · ·

vBE = · · ·

vBE = · · ·

vBE = · · ·

VCC

RC

VCC

←i C

=

V CC−v C

E

R C

Z

Q

Y

VCE

IC


How to choose a bias point

There are mainly two factors that determine the location of Q orbias or Operating point.

1 Gain

2 Swing limit at the output

It is clear from graphical analysis, there are three operating points(Z, Q & Y).

If we choose Z, gain will be more but distorted negativeportion of signal.

If we choose Y, reduction in gain and distortion in positiveportion of signal.

If Q is chosen, gain as well as swing in the output signal areachieved.

Choose an operating point that should give gain andmaximum swing


Check yourself

Find the operating point. Assume VBE = 0.7 V.

+

−

vi

500 Ω

VCC = 5V

+

−

vo

215 kΩ

VCE (V)

IC (mA)

IB = 50µA

IB = 40µA

IB = 30µA

IB = 20µA

IB = 10µA2

4

6

8

10

1 2 3 4 5


Solution

5 V

215 kΩ

500 Ω

5 V

By KVL in BE loop,

5 = IB215× 103 + VBE

IB =5− 0.7

215× 103= 20µA

KVL in CE loop, (Load line)

5 = IC500 + VCE

IC =5− VCE

500


Solution - contd...

VCE (V)

IC (mA)

IB = 50µA

IB = 40µA

IB = 30µA

IB = 20µA

IB = 10µA2

4

6

8

10

1 2 3 4 5

QICQ

VCEQ


Transistor Biasing

Biasing means establishing a constant current in the collectorof BJT.

The Q point or bias or Operating point should not besensitive to temperature variations and variations in β.

The location of DC bias point should allow for maximumoutput signal swing.

Transistor is operated in active region by biasing. To calculate ICand VCE corresponding to DC bias, DC analysis should be carriedout.

(Instead of using two different batteries, one DC battery VCC canbe used for both EB and CB junctions.)


Two Basic Schemes - Not to be used

IB

RB2

RB1

VCC

RC

IC

VCC

VCE

+

−VBE

(a)

IB

RB

VCC

RC

IC

VCC

VCE

+

−VBE

(b)

Figure: Bad SchemesS. Sivasubramani EE101 - BJT 36/ 60

Bad Schemes

1 First scheme (Figure a)

It applies constant VBE through voltage divider action.For a small change in VBE due to temperature variation, IBchanges drastically because of exponential relationship betweenthem.It will change IC as IC = βIB .Hence, It is not a good approach.

2 Second scheme (Figure b)

Figure b applies constant IB .If there is a change in β, IC will change.Hence, It is also not a good approach

As these two schemes are sensitive to variation in temperature andvariation in β, they should not be used for biasing a transistor.


Classical Biasing Scheme

IB

R2

R1

VCC

RC

IC

VCC

VCE

RE

+

−VBE

In order to make biasingscheme insensitive totemperature and βvariations, RE is included inemitter circuit.


Classical Bias Scheme - Thevenin Equivalent

IB

VBB = VCCR2

R1 + R2

RC

IC

VCC

VCE

RE

+

−VBE

RB = R1||R2


Classical Bias Scheme

The current IE can be determined by writing KVL in base-emitterloop.

VBB = IBRB + VBE + IERE

Substituting IB =IE

β + 1

VBB = IERB

β + 1+ VBE + IERE

IE =VBB − VBE

RE + RB/(β + 1)

As IC = αIE and α =β

β + 1

IC =β(VBB − VBE )

RE (β + 1) + RB


Contd...

If we choose RE (β + 1) >> RB ,

IC ≈β(VBB − VBE )

RE (β + 1)

Normally β lies in the range of 10 to 100.

β + 1 ≈ β

IC can be written as

IC ≈(VBB − VBE )

RE

Now IC is independent of β. As VBB >> VBE , change in VBE willalso not affect IC much. In this bias scheme, Q point is insensitiveto variation in β and variation in temperature.


Transistor - Small Signal Analysis

To perform small signal analysis, small signal equivalent circuitshould be drawn.

1 Calculate the operating point using DC analysis Click here to refer .

2 Draw the small signal equivalent circuit at this operatingpoint.

3 As small signal circuit is linear, calculating the voltage,current would be easy.

Small signal analysis is also called as ac analysis.


Circuit with small signal

Let us consider this circuit to draw small signal equivalent andanalyze.

VBE

vbe

+

−

vBE

RC

iC

+

−

vCE

VCC


Circuit with small signal - DC Analysis

To find the operating point by DC analysis, ac signal should beremoved.

IB

VBE

RC

IC

IE

+

−

VCE

VCC

Calculate IC , VCE as follows.

IC = IS exp(VBE

Vt)

VCE = VCC − ICRC

Mostly IC is determined as

IC = βIB

where IB is found using KVL inbase emitter loop.


Small signal model - BJT

(refer lecture notes on two port network)Let us consider BJT as a two port network with emitter as acommon terminal.

E−

+

vBE

B iB

−

+

vCE

C

iC

iC = IS exp(vBEVt

)

iB =iCβ


The small signal voltages and currents are related by linearequations.

ib = ybbvbe + ybcvce

ic = ycbvbe + yccvce

As iB and iC are only function of vBE , ybc and ycc will be zero.

ybb =∂iB∂vBE

∣∣∣VBE

=ISβVt

exp(VBE

Vt) =

ICβVt

ycb =∂iC∂vBE

∣∣∣VBE

=ICVt


Bib

rπ

EE

Cic

gmvbe

+

−

vce

+

−

vbe

where rπ =1

ybb=βVt

ICis the resistance between base and emitter

and gm = ycb =ICVt

is the transconductance .


Small signal model - circuit

VBE

vbe

+

−

vBE

RC

iC

+

−

vCE

VCC

Bib

rπ

EE

Cgmvbe

RC

+

−

vce

+

−

vbe


Transistor Small Signal Analysis - Steps to be followed

1 Eliminate the signal source and determine the DC operatingpoint in particular IC and VCE .

2 Calculate the value of small signal model parameters rπ andgm at the operating point.

3 Draw the small signal equivalent circuit by short circuiting theDC sources and open circuiting the DC current sources.

4 Analyze this circuit to determine voltage gain, input resistanceand output resistance.


Example - Small signal analysis

Calculate small signal gainvovi

. Assume β = 100.

100 kΩ

3 V

vi

3 kΩ

VO + vo

10 V


Example - Solution

DC Analysis:

100 kΩ

3 V

3 kΩ

10 V

Assume VBE =0.7 V.

IB =3− 0.7

100× 103

IB = 0.023 mA

IC = βIB = 2.3 mA

VCE = VCC − ICRC = 3.1 V


Solution - Contd

Small signal analysis (AC analysis):

vi

B

100 kΩ

rπ

+

−

vbe

EE

C

gmvbe 3 kΩ

+

−

vo

gm =ICVt

=2.3

25= 0.092f

rπ =β

gm=

100

0.092= 1086.96 Ω

vbe = virπ

100kΩ + rπ= 0.0108vi

vo = −gmvbe × 3000 = −2.96vivovi

= −2.96


AC Coupling

It is used to add input signal to the amplifier and outputsignal to the load without affecting bias point.

It is done with the help of large capacitance.

Large capacitance will become an open circuit in DC analysis.

Large capacitance will behave like a short circuit at signalfrequency in ac analysis.

In addition to two capacitances, one more capacitance is connectedacross RE . It is called as a bypass capacitor which will act as ashort circuit and ground emitter without RE in ac analysis.


CE Amplifier - Example

R2

R1RC

VCC

C2

RL

+

−

vo

RE CE

vi

RSC1


CE Amplifier - DC Analysis

To find the operating point, the circuit is redrawn like this.

R2

R1

RC

IC

VCC

C

REE

+

−

VCE


Small Signal Equivalent Circuit

B ib

rπ

+

−

vbe

E

C

ic

gmvbe RC ||RL

+

−

vcevi

RS

R2 R1


BJT - Digital Logic

1 Saturation : When EB junction and CB junction are forwardbiased, BJT will act as a closed switch between collector andemitter.

2 Cutoff : When EB junction and CB junction are reversebiased, BJT will act as an open circuit between collector andemitter.

The above two modes can be used to implement logic functions.


BJT - NOT Gate

RB

Vin

RC

VCC=5 V

Vout

When Vin < 0.7 V (Low), EBand CB junctions are reversebiased. IB = 0, IC = 0.

Vout = VCC = 5V (High)

When Vin = 5 V (High), EB andCB Junctions are forward biased.

Vout = VCEsat = 0.2V (Low)


Check yourself

RB

A

RC

VCC=5 V

C

RB

B

Positive Logic :0 V - Low ”0”5 V - High ”1”

A B C

0 0 10 1 11 0 11 1 0

NAND Gate


Check yourself

RB

A

RE

VCC=5 V

C

RB

B

Positive Logic :0 V - Low ”0”5 V - High ”1”

A B C

0 0 00 1 01 0 01 1 1

AND Gate


bipolar junction transistor (bjt) - introductionsiva/2015/ee101/bjt.pdf · bipolar junction...

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