Bipolar Junction Transistor - I.K. International ... Two major types of ‘three terminal’...

40 Semiconductor Devices

CHAPTER 2

Bipolar Junction Transistor

2.1 TRANSISTORS VS. VACUUM TUBES

Transistor was invented by John Bardeen, William Shockley and Walter Brattin ofBell Laboratory in year 1948. Invention of transistor revolutionized electronicindustry. Transistors have replaced bulky vacuum tubes. Main advantages oftransistor over the vacuum tubes are given below.

(1) Filament (or heater) is not needed. Delays due to heating are very less.Heating power required is also very less or almost negligible.

(2) These are light in weight and small in size.(3) Transistors require very low operating voltages.(4) Transistors have very low power consumption. Thus, they increase circuit

efficiency.(5) There is no effect of ‘ageing’ on transistors. They have very long life.(6) Transistors are shockproof.

Two major types of ‘three terminal’ semiconductor devices are:(1) Bipolar junction transistors (BJTs) and(2) Field effect transistors (FETs)Bipolar junction transistors (BJTs) consist of two p-n junctions, which are

constructed in a special way. These junctions are connected in series and/or backto back. Electric current flows due to both charge carriers viz. electrons andholes. That is why, these are called ‘bipolar’.

BJTs are very often referred to as “transistors”. BJTs are widely used indiscrete circuits and also in IC design (both analog and digital).

2.2 WHAT IS A TRANSISTOR?

It is basically a(1) Three-terminal device.


value of ‘α’ near to unity and a very large value of ‘β’. Cross-sectional view ofBJT as shown in Figure 2.2 shows that collector base junction has larger area thanthat of emitter-base junction.

Unlike other transistors, BJT is not a symmetrical device. Interchangingcollector and emitter makes transistor to come out of forward active mode andstart operating in reverse mode. Transistor’s internal structure is optimized toforward mode operation. Interchanging collector and emitter simply reverses thevalues of α and β as a result their values become much smaller than those inforward operation. ‘α’ of reverse mode is less than 0.5. Lack of symmetry isattributed to doping ratios of emitter and collector. Emitter is “heavily doped”.Collector is “lightly doped”. Thus, it allows application of a large reverse biasvoltage before collector-base junction breakdown occurs. Collector base junctionis reverse biased normally. It is due to the reason that emitter is heavily doped,which increases ‘emitter injection efficiency’. ‘Emitter injection efficiency’ is theratio of carriers injected by emitter to the carriers injected by base. For ‘highcurrent gain’, carriers which are injected into emitter-base junction should comefrom emitter.

2.2.3 Transportation Model of n-p-n TransistorThe transportation model of a n-p-n transistor (symbol shown in Figure 2.3 (b)) isshown in Figure 2.3(a), below.

Figure 2.3: (a) n-p-n transportation model (b) Symbol n-p-n transistor

p ‘Base’

n ‘Collector’

n ‘Emitter’

iB

iC

iE

iB

Collector (C)

Emitter (E)

Base (B)

Arrow directionis always from “p to n”.

p n

(a) (b)

–+

+–

vBC

vBE

iC

iE

Chief points are given below.(1) Narrow base region is coupling between two ‘back-to-back’ p-n junctions.(2) Emitter injects electrons into base region of transistor. All electrons travel

across narrow base. These electrons are removed by collector.

Bipolar Junction Transistor 43

(3) Base-emitter-voltage ‘VBE’ and base-collector-voltage ‘VBC’ determinecurrent in transistor. If VBE and VBC are positive, they forward bias theirp-n junctions.

(4) Terminal currents are collector current (iC), base current (iB) and emittercurrent (iE).

2.2.4 Transportation Model of p-n-p TransistorThe transportation model of a p-n-p transistor (symbol shown in Figure 2.4 (b)) isshown in Figure 2.4(a), below.

Figure 2.4: (a) p-n-p transportation model (b) Symbol of p-n-p transistor

n ‘Base’

p ‘Collector’

p ‘Emitter’

iB

iE

iC

iB

Collector (C)

Emitter (E)

Base (B)

(a) (b)

–+

+–

vEB

vCB

iE

iC

C

E

Chief points are given below.(1) If voltages ‘VEB’ and ‘VCB’ are positive, then they will forward bias their

respective p-n-junctions.(2) Collector current and base current exit transistor terminals. Emitter current

enters the transistors.

2.3 HETEROJUNCTION BIPOLAR TRANSISTOR (HBT)

HBT is an improvement over BJT. HBTs can handle very high frequency signalstypically up to several hundred gigahertz (GHz). HBTs are very commonly used inultra fast circuits like R.F. systems. H.B.T has different semiconductors for transistorelements. Emitter of HBT is made up of a “Large band gap” material than its basematerial. Two common HBTs are silicon-germanium and aluminum-gallium-arsenide. Other semiconductors can also be conveniently used for HBT. HBTs aredeveloped by epitaxy techniques like “MOCVD” and “MBE”.


2.4 TRANSISTOR CONFIGURATIONS

Three transistor configurations have been discussed below.

2.4.1 Common Emitter (CE)In this CE configuration, input is applied between base and emitter. Output is takenfrom collector and emitter. Emitter of transistor is “Common” to both input andoutput circuits and that is why it is called “Common-emitter”.

InputE

B

C

Output

Figure 2.5: CE configuration

2.4.2 Common Base (CB)In CB configuration, input is applied between emitter and base. Output is takenfrom collector and base. Base of transistor is common to both input and outputand that is why is called ‘Common-base’.

Figure 2.6: CB configuration

Input OutputB

E C

2.4.3 Common Collector (CC)In CC configuration, inputs are applied between base and collector. Output istaken between the emitter and collector. Collector of transistor is common to bothinput and output and that is why it is called ‘Common-collector’.


2.5 TRANSISTOR “ALPHA (ααααα)” AND “BETA (βββββ)”

Number of electrons which cross the transistor base and reach transistor collectoris a measure of BJT efficiency. Heavy doping of emitter region and light doping ofbase region causes more and more electrons injection from emitter into base. Lessholes are injected from base into emitter. “βF” or “hfe” represent common emittercurrent gain. ‘β’ is called “Base current amplification factor”. ‘β’ is ratio of changein D.C. collector current to the change in D.C. base current in forward activeregion. ‘β’ is around 100. However, it can vary between 20 to 500.

Another parameter is common base current gain. It is denoted by ‘αF’, whichis called “current amplification factor”. ‘αF’ is ratio of change in collector currentto the change in emitter current for a constant collector base voltage (VCB). ‘α’and ‘β’ are thus given as

α = C

E

II

∆∆ and β =

C

B

II

∆∆

Also, β = ( )1α− α and α = ( )

β1 β+

Current amplification factor ‘α’ is always less than unity. As, base current decreases,‘α’ approaches unity. ‘α’ ranges from 0.9 to 0.99.

2.6 COMPARISON OF THREE CONFIGURATIONS

Table 2.1 below gives a comparison between three transistor configurations, asdiscussed earlier.

Figure 2.7: CC configuration

InputE

B

Output

C


2.7 CURRENT COMPONENTS

Figure 2.8 below, shows three currents viz. iE, iB, and iC, and their flow inside thetransistor and in peripheral circuit.

Reverse current components are due to drift of thermally generated minoritycarriers. These are not shown in Figure 2.8.

Table 2.1: Comparison table

Sr.No. Parameter CB CE CC

(1) Input dynamic resistance Low (20 Ω – 100 Ω) Moderate Very high(750 Ω–1000 Ω) (750K Ω)

(2) Output dynamic Very high Very high Lowresistance (450 K – 1 MΩ) (10 K Ω–1 KΩ) (50 Ω)

(3) Current gain Less than unity High (about100) —(about 0.98)

(4) Leakage current Very small Very large(1 µA – 5 µA) (20 µA–500 µA) —

(5) Application For high freqn For Audio freqn Impedancematching

(6) Voltage gain about 150 about 500 less than 1

Figure 2.8: Current components in a n-p-n transistor in active mode

Injectedelectrons

Diffusingelectrons

Collectedelectrons

iC

iE

iB

n p n

Injected holes (i )B1

Forward bias Reverse bias

iB

RecombinedElectrons (i )B2

C

iCiE

E

BE

– + – +CB

– +– +

iE iE iC iCB

CBBE

2.8 MODES OF OPERATION

There are five (05) distinct modes of transistor operations. Figure 2.9 showsinput-output characteristics of transistor.

Figure 2.10 depicts various transistor operation modes.


Figure 2.9: Input-output characteristics

Figure 2.10: Transistor operation modes

Table 2.2: Operation condition and modes

Condition JE JC Region ofEmitter junction Collector junction operation

F R Forward biased Reverse biased ActiveF F Forward biased Forward biased SaturationR R Reverse biased Reverse biased CutoffR F Reverse biased Forward biased Inverted

Forwardactive

Saturation

IB

Cut off

Reverseactive

Saturation

VCE

IB

IC

Cut off

Reverse Forward

Forwardactive

Cutofforoff

Saturationoron

Reverseactive

Revers

eF

orw

ard

Emitter junction

Colle

cto

rju

nction


2.8.1 CutoffIf both junctions ‘JE’ and ‘JC’ are reverse biased, transistor is doing nothing. Avery little current flows. It corresponds to a logical “OFF’, or an open switch.

Figure 2.11: ‘Cutoff’ in transistor

oA

E

+10 V

VBE

oVVBE

RL

2.8.2 ActiveIt is of two types viz. forward active and reverse active. Both these modes havebeen described below.

2.8.2.1 Forward ActiveIn this, emitter-base junction is forward biased. Base-collector junction is reversebiased. BJTs possess very large “common-emitter current gain (β)” in this forwardactive mode. In this case, collector-emitter current is proportional to base current.

Figure 2.12: Forward active mode

IC

E

+10 V

IB

RL

VBE

+

–


VCC – VCE = IC RC

or IC =1 CC

CEC C

VVR R

− +

Comparing above with “y = (mx + c)” equation, which is of straight line Here,

slope is 1

CR −

and intercept ‘C’ is CC

C

VR

with IC axis. This straight line in the

case of transistors is known as “D.C. load line”. It is shown in Figure 2.14 below.

Figure 2.14: D.C. load line

O D VCC

B

Q

AVCC

RC

V (Volts)CE

C

I (mA)C

2.9.1 Importance of D.C. Load LineCurrent ‘IC’ and voltage ‘VCE’ of transistor are represented by D.C. load line.When IC is maximum (= VCC/RC), then VCE = 0. If IC = 0, then VCE is maximum.Also, at IC = 0, VCE is equal to VCC. From DC load line it can be seen that forcollector current ‘OC’, collector-emitter voltage ‘VCE’ is ‘OD’. Thus, load line isa far more convenient and direct solution.

Also, resistance ‘RC’ is called ‘load’ or simply ‘load resistance’. Therefore,D.C. load line is also called as “load line”.

2.9.2 Operating PointZero signal values of ‘IC’ and ‘VCE’ are called ‘operating point’.

In a circuit, operating conditions of transistor are described in terms of ‘VCE’and ‘IC’. ‘VCE’ and ‘IC’ fix up operating points of transistor. Operating point isalso decided by number of other factors like VCC, RC, RB, VBE and VBB. Firstly,values of VCC and RC are determined, which insure ‘operating-point’ of transistorlying on its ‘D.C.-load line’. Exact operating point lying on D.C. load line is decidedby value of base current ‘IB’. Base current ‘IB’ is decided by value of ‘VBE’ (oftransistor), RB and VBB. By applying KVL to base circuit of transistor, we get thefollowing.


Collector leakage current is greatly affected by temperature changes. A rise of10°C doubles the collector leakage current. Collector leakage current may be ashigh as 0.2 mA in the case of low powered Germanium transistors.

Self-destruction of an untablized transistor is called “thermal runaway”. Toavoid thermal runaway, ‘IC’ should be kept constant.

2.10.2 Biasing MethodsThree biasing methods are described below.

2.10.2.1 Fixed biasingOperating point in fixed biasing is obtained by following three steps

Step 1: Find base current IB

IB =( )CC BE

B

V VR−

As, VBE << VCC, therefore

IB ≈ (VCC/RB)

Figure 2.16: Fixed-biasing

+

–VBE

RB RC

+ VCC

Step 2: Find collector current

‘IC’. C BI I≅ β provided CC

CC

VIR

≤

Step 3: Find collector emitter voltage ‘VCE’.

VCE = (VCC – IC RC)

2.10.2.1.1 AdvantagesAdvantages of fixed biasing are given below.


(1) Only one resistance ‘RB’ is needed. It is very simple circuit.(2) Biasing conditions are very easily achieved, Calculations are simple.(3) There is no loading of source by biasing circuit. Because, there is no

resistor across base-emitter junction.

2.10.2.1.2 DisadvantagesDisadvantages of fixed-biasing are given below.

(1) It provides poor stabilization.(2) Stability factor is very high.

2.10.2.2 Biasing with feedback resistorIt is shown in Figure 2.17 below.

Figure 2.17: Biasing with feedback resistor

IB

RB

VCC

IB

(I + ICB )

+

RC

IE

Here, resistor ‘RB’ connects collector with base (input). Due to feedback,base current is dependent on collector voltage. This dependence nullifies changesin base current. This method is also called as “voltage-feedback” bias circuit.

2.10.2.2.1 DisadvantagesIts disadvantages are given below.

(1) Stability factor is high.(2) It provides negative feedback. Thus, gain of amplifier is reduced.

2.10.2.3 Voltage divider biasing circuitIt is shown in Figure 2.18 below.

Here, two resistors ‘R1’ and ‘R2’ are put across supply voltage ‘VCC’. Theseresistors provide biasing. Emitter resistance ‘RE’ provides stabilization. Voltagedivider circuit is formed by ‘R1’ and ‘R2’. Voltage drop across ‘R2’, forwardbiases base-emitter junction. As a result, base current and collector current flow inzero signal conditions.


Terminal ‘3’ : EmitterIin : Base current (ib)io : Collector current (ic)Vin : Base to emitter voltage (VBE)Vo : Collector to emitter voltage (VCE)

h-parameters are defined ashin = hie; Input impedance of transistor (It corresponds to emitter

resistance ‘re’.)hrx = hre; Dependence of transistor’s ‘IB – VBE’ curve on ‘VCE’. It is

very small and can be neglected.hfx = hfe; Current gain of transistor. This parameter ‘hfe’ is specified as ‘hfe’.

It is same as D.C. current gain ‘βDC’, which is given in datasheets.hox = hoe; Output impedance of transistor. It is basically an admittance

and needs to be inverted so as to convert it into an impedance.‘h’-parameters with subscripts signify A.C. conditions or analyses. DC conditionsare specified in upper case. For CE topology, h-parameter analysis model iscommonly used. h-parameter model simplifies circuit analysis. ‘hoe’ and ‘hre’parameters are set to infinity and zero, respectively. h-parameter model is wellsuited for low frequency and small signal analysis. For high frequency analyses,inter electrode capacitors must be considered.

2.11.2 High Frequency Model; Ebers-Moll ModelHigh frequency model was developed by Ebers-Moll in 1954. It is also termed as“Coupled-diode model”. It is an ideal model for a BJT. This model fits well forforward active mode, reverse active mode, saturation and cut-off modes ofoperation.

Ebers Moll model has two diodes and two current sources. It is shown inFigure 2.20, below.

Two diodes in this model are base-emitter and base collector diodes. Currentsources signify minority carriers flowing through base region. Current sourcesdepend on current flowing through each diode. Parameters of this model are:

I = I eF ES

VBE

VT – 1 I = I eR CS

VBE

VT – 1

IRR IFF

IE IC

IB

VBE VBC

– + – +

Figure 2.20: Ebers-Moll model


Here, WB: Physical base width (thickness of doped base region) and

WBeff : Effective base width

If emitter-base junction is forward biased, then ‘WD’ does not change. Ifcollector-base junction is reverse biased, then ‘WD’ changes with collector voltagechange. It is so because collector is less heavily doped with respect to base. Theprinciple involved is called ‘charge-neutrality’. This principle governs these twowidths. Emitter-base junction remains unchanged as emitter-base voltage remainssame.

Two consequences of ‘base-modulation’ affecting current are mentioned below.(1) There is very little chance of recombination within the “smaller” base

region.(2) Charge gradient is increased across the base. Therefore, current of minority

carriers, which are injected in emitter junction, increases.These two factors increase collector or “output” current of transistor as

collector voltage is increased. This increased current is shown in Figure 2.23,below.

Figure 2.22: ‘Early-effect’

WB eff

WB

E B C

p-type n-type p-type

Figure 2.23: Early voltage in output characteristic plot of a BJT

IC

V = Early voltageA

– VAVCE


Tangents to these characteristics especially at large voltages are extrapolatedbackward which intercept with voltage axis at a single voltage called as “Earlyvoltage”. Early voltage is denoted as ‘VA’.

2.13 PHOTOTRANSISTOR

A phototransistor is normally connected in common-emitter configuration. Base iskept open and radiation is concentrated on a region near collector junction. Thistransistor works in “active region”. In forward bias, minority carriers are thermallygenerated. Electrons cross from base to collector and holes and from collector tobase. They constitute reverse saturation collector current ‘ICO’. As, base is openIB = 0.

Thus, IC = (β + 1) ICO

If light is turned ‘ON’, additional minority carriers are photo generated. Theseadditional minority carriers contribute to reverse saturation current, similar tothermally generated minority charges. Let ‘Il’ be the component of reverse saturationcurrent due to light, then total collection current ‘IC’ is given as

IC = (β + 1) (ICO + Il)

Current due to radiation is multiplied by (β + 1), a very large factor, due to transistoraction.

Figure 2.24: A phototransistor

+ –

n p n

IC

C

JC JE

E

Radiations

VCE

IMPORTANT RELATIONSHIPS

Important relationships are given below.

1. In a n-p-n transistor, minority carrier concentration in base (p-region) is givenas

( ) /BE T

o

V Vpo eη = η


Here,IB : Base currentα : Current amplification factor

and ICBO : Collector-base current with emitter open (Leakage current)

(c) Input characteristic is shown in Figure 2.25 below.(d) Input resistance ‘ri’ is

Figure 2.25: Input characteristic of CB configuration

0 10 20 30 40 50

0.5

1.0

1.5

2.0

2.5

3.0

V (V)EB

I (mA)E

at constant CBCB

i VC

Vr

I∆

=∆

(e) Output resistance ‘ro’ is

at constant EBE

o IE

VrI

∆=∆

(f) Output characteristic is shown in Figure 2.26 below.

Figure 2.26: Output characteristic of CB configuration

IC

(mA)

V (volts)CB

I = 1 mAE

I = 2 mAE

I = 3 mAE

I = 4 mAE

I = 5 mAE

I = 0 mAE

0

1

2

3

4

5

Collector-Basevoltage


5. Common emitter (CE) connection/configurationImportant terms in CE connection have been summarized below.

(a) Current amplification factor ‘β’ is

C

B

II

∆β =

∆

(b) Collector current ‘IC’ is

IC = β IB + ICEO ( )αβ

1 α

= −

Here, ICEO : Collector-emitter current with base open(c) Emitter current ‘IE’ is

IE = (β + 1) IB + ICEO

( ) ( )( )

αHere, β 11 α

β α

+ = − =

(d) Input characteristic is shown in Figure 2.27 below.

Figure 2.27: Input characteristics of CE configuration

I ( A)B

V (volts)BE0

1

2

3

4

0.7 1.4 2.1

V=

1V

CE

V=

10

V

CE

(e) Input resistance ‘ri’ is

at constant CEBE

i VB

VrI

∆=∆

(f) Output resistance ‘ro’ is


at constant B

CEo I

C

Vr

I∆

=∆

(g) Output characteristic is shown in Figure 2.28 below.

Figure 2.28: Output characteristics of CE configuration

IC

VCE

I = 5 AB

I = 10 AB

I = 15 AB

I = 20 AB

I = 25 AB

0

1 mA

2 mA

3 mA

4 mA

5 mA

6. Common collector (CC) connection/configurationImportant terms in CC connection have been summarized below.(a) Current amplification factor ‘ γ ’is

E

B

II

∆γ = ∆

or ( ) ( )1γ 1 β

1 α= +

−

(b) Collector current ‘IC’ is

IC = (β + 1) IB +(β + 1) ICBO

Here, ICBO : Collector-base current with emitter open

7. Avalanche multiplication factor ‘M’ is

( )1

1 nCB CBO

MV B V

=−


Stability factor ‘S’ is

( )at constt and

1

1B

CI

COB

C

ISId I

d I

ββ + ∂

= =∂

− β

In fixed bias, ‘IB’ is independent of IC.

Thus, 0B

C

d Id I

=

and ( )1S = β +

Here, ICO : Collector leakage current

11. In voltage divider bias method

IC = ( )2 BE

E

V VR−

Here, CC2 2

1 2

VV R

R R

= +

Thus, VCE = [VCC –IC (RC + RE)]

Here, stability factor ‘S’ is

S = 1

and IE = ( )2 BE

CE

V VI

R−

≈

12. In collector to base bias circuit

VCE = VCC – (IC + IB) RC

or IB = ( )( ) ( )

CE BE CC

C B B C

V V VR R R R

−≈

+ + β


As, IC = βIB

Thus, VCE = (VCC – ICRC)

Collector to base bias circuit is shown in Figure 2.29 below.

Figure 2.29: Collector to base bias

VCE

IB

RB– +

VBE

+–

RC

I + IBC

IC

Figure 2.30: Transistor amplifier

IB

RC

IC

+ VCC

Signal IE

Output

CC

X

Y

13. Transistor as an amplifierFigure 2.30 below, depicts an amplifier circuit using a transistor.


Output is available either across ‘RC’ or across terminal ‘X’ and ‘Y’ as shown inFigure 2.30. Output magnitude remains same in either case.

PerformanceIn common emitter (CE) amplifier

Input resistance ‘Ri’ is

Ri = BE

ieB

V hI

∆ =∆

Output resistance ‘Ro’ is

Ro = 1CE

C oe

VI h

∆= ∆

Effective collector load ‘RAC’ is

RAC = RC || RO

= ( )C O

C O

R RR R

×+

In a single stage amplifier, effective load is same as collector load ‘RC’. If amplifiersare cascaded, input resistance ‘Ri’ of next stage plays an important role. Effectivecollector load ‘Reff’ is parallel combination of RC, RO and Ri. It is obtained as

Reff = (RC || RO || Ri)

= ( )C i

C i

R RR R

×+

Figure 2.31: CE amplifier

VBE

IB

Rin


SOLVED PROBLEMS

Q1. In a p-n-p transistor, forward emitter current is 2 mA. Collector circuitis open. Find(i) Junction voltage ‘VC’ and ‘VE’(ii) VCE

(ICO = 2 µA ,oEI = 1.6 µA and ‘αN’ = 0.98, where ‘αN’ represents

normal mode of operation of transistor)Also, investigate whether transistor is in ‘saturation’ or ‘cutoff’ or‘active’ region.

Solution: (i) It is given that

IC = 0

IE = 2 mA

We know, ln 1o

EE T

E

IV VI

= −

In a p-n-p transistor, ‘oEI ’ is negative.

Therefore, oEI = – 1.6 µA

Now, VE = 0.026 ln 321

1 6 10− + ⋅ ×

or = 0.1853VFor forward bias case

o

n E

C

α Iln 1Ic TV V

= −

or = 0.026 ln 3

60.98 2 101

2 10

−

−

× ×− ×

or = 0.179 V(ii) ‘VCE’ is obtained as

VCE = (VC – VE)

or = 0.0063 V(iii) From above results, we find that transistor is in “saturation region”.


Q2. In a transistor ‘αdc’ is 0.99. Collector leakage current ‘ICo’ is 5 µA.Find base current and collector current, if ‘IE’ is 2 mA.

Solution: We know that

IC = (αdc IE + ICo)

Thus, ( )3 60.99 2 10 5 10CI − −= × × + ×

or IC = 1.985 × 10–3 ABase current is obtained as

IB = (IE – IC)

or IB = ( )3 32 10 1.985 10− −× − ×

or IB = 15 µAQ3. In a transistor, if emitter current changes by 4 mA, collector current

changes by 0.83 mA. What is short-circuit current gain of this transistor?Solution: Short circuit current gain ‘α’ is

α or hfb = C

E

ii

∆ ∆

Thus, α = 3

30.83 10

4 10

−

−

× ×

or α = 0.2075Q4. A common base circuit is shown in Figure 2.32, below. Find ‘VCB’ and

‘IC’. Transistor is made up of Silicon semiconductor material.

Figure 2.32: A common base circuit

IC

R = 3 kC

IE

V = 18 VCC

V = 10 VEE

R = 2 kE

IB


Solution: In the case of a Silicon transistor VBE = 0.7 V

Applying KVL in emitter loop yieldsVEE = (IE RE + VBE)

or IE = ( )EE BE

E

V VR−

Thus, IE = 310 0.72 10−

×

or IE = 4.65 mA

Here, 0 soB C EI I I≈ ≈

Thus, IC = 4.65 mAApplying KVL in collector loop yields

(VCC – VCB) = IC RCor VCB = VCC – IC RC

VCB = (18 – 4.65 × 10–3 × 3 × 103)or VCB (18 – 13.95)

VCB = 4.05 VQ5. What will be ‘α’ rating of transistor as shown in Figure 2.33 below?

Also, find value of IC.

Figure 2.33: A B.J.T.

I = 341 AB

IC

= 49

I = 15 mAE

Solution: We know that

α = ( )β

1 β+

In this case, α = ( )49

1 49+


or α = 0.98Further, IC = α IEor IC = 0.98 × 15× 10-3

or IC = 14.7 mA

Also, IC = β IB

or IC = 49× 341× 10-6Aor IC = 14.7 mA

Q6. In an application, maximum power dissipation ‘PD’ of a transistor is100 mW. Find maximum allowable collector current without causingany destruction to transistor. Consider VCF = 25 V.

Solution: We know thatPD = VCE × IC

In this case, 100 mW = 25 V× IC

or IC = 100mW

25V

or IC = 4 mAQ7. Potential divider method of biasing is used in a transistor circuit. Here,

R1 = 50 KΩ, R2 = 10 KΩ and RE = 1 KΩ. If VCC = 15 V, find(a) IC with VBE = 0.1 V, and(b) IC with VBE = 0.3 V

Solution: It is given thatR1 = 50 KΩ, R2 = 10 KΩ , RE = 1 KΩ and VCC = 15 V(a) If VBE = 0.1 V, voltage across R2 is

V2 = 2

1 2

RR R

+

VCC = ( )10 15

10 50×

+

= 2.5 V

Collector current ‘IC’ = ( )2 BE

E

V VR−

or = ( )2.5 0.1

1 K−Ω


Further, VCE = VCC – IC (RC + RE)

= ( )3 315 0.5 10 1 2 10−− × + ×

= 13.5 VCollector potential ‘VC’ = (VCC – IC RC)

or = ( )3 315 0.5 10 2 10−− × × ×

or = 14 VQ9. In the circuit shown below in Figure 2.35, operating point is chosen

such that IC = 5 mA and VCE = 1V. If RC = 1.1KΩ , VCC = 10 V and β= 50, find R1, R2 and RE. (assume VBE = 0.3 V and I1 = 5 IB)

Figure 2.35: Voltage divider circuit

RE

IE

IC

2.2 k

R1

R2

+ V = 10 VCC

I1

I1IB

VBE

VCE

I1 >> IB

Solution: It is given thatRC = 1.1 KΩ

VCC = 10 Vβ = 50

VBE = 0.3 Vand I1 = 5 IB‘IB’ is quite small in comparison with ‘I1’, therefore I1 flowing throughR1 also flows through R2 (approximately in same amount).

Base current ‘IB’ = CI

β

= 35 10

50

−×

= 0.1 mACurrent via ‘R1’ and ‘R2’ is obtained as below.


Given that I1 = 5 IBThus, I1 = 5 × 0.1 × 10–3

or = 0.5 mA

Further, I1 = ( )CC

1 2

VR R+

or ( )3

1 2

100.5 10R R

−× =+

or ( )1 2R R+ = 20 KΩ

Applying Kirchhoff’s voltage law at collector side, we getVCC = (IC RC + VCE + IE RE)

or VCC = (IC RC + VCE + IC RE)In this case,

10 = (0.5× 1.1 × 103 × 10–3 + 1 + 0.5 × 10–3× RE)or (10 – 5.5 – 1) = 0.5 × 10–3 RE

or RE 33.5

5 10−=

×or RE = 700 ΩVoltage across ‘R2’ is obtained as

V2 = (VBE + VE)or = (VBE + IE RE)or V2 = (0.3 + 5× 0.7)or V2 = 3.8 V

Resistance R2 = 2

1

VI

= 3.8

0.5 mA

= 7.6 KΩand R1 = (20 – 7.6) KΩ

= 12.4 KΩQ10. A Ge transistor (β = 50) is used in a self biasing circuit. In circuit, VCC

= 16V and RC = 1.5 KΩ. Quiescent point ‘Q’ is VCE = 8V and IC = 4mA. For stability factor ‘S’ = 12, find values of R1, R2 and RC.


Let ‘N’ be the ground point. ThenVBN = (VBE – IE RE)

or = (0.2 + 4.08 × 0.49)or = 2.2 Vand V = (VBN + IB RB)or = (2.2 + 0.08 × 7.05)or = 2.76 V

Also,1

CC

B

VVR R

=

= 1

16R

Thus, 116 BRR

V =

or = 16 7.05

2.76×

or = 41 KΩ

Further,( )

11

CC BNR

V VI

R−

=

or = 316 2.241 10−×

or = 0.337 mA

Also, ( )2 1R R BI I I= −

or = 0.257 mA

Thus, R2 = 2

BN

R

VI

or R2 =2.2

0.257

or R2 = 8.56 KΩNote that stability factor is improved. Power drained by biasing resistors‘R1’ and ‘R2’ from battery is increased.


Q11. h-parameters in case of a common-emitter transistor amplifier circuitare given as

hie = 100 Ω, hre = 2.5× 10–4, hfe = 50 and hoe = 25 µA.Find,(a) Maximum ‘RL’ for which Ri does not differ by more than 10% of itsvalue at RL = 0.(b) Maximum ‘RS’ for which Ro does not differ by more than 10% ofits value at RS = 0.

Solution: (a) We know that

Ri = ( )

1fe re

ie

oeL

h hh

hR

− +

Subscript ‘e’ denotes CE configuration. If RL = 0, then Ri = hie.

Thus, 0.9 Ri = 0.9 hie = 1fe re

ie

oeL

h hh

hR

− +

So0.1

1fe re

ie

oeL

h hh

hR

⋅ = +

or0.1

· 0.1 iefe re ie oe

L

hh h h h

R

= +

or ( )0.1 0.1iefe re ie oe

L

h h h h hR

= ⋅ −

or1

10 10ie ie oe

fe reL

h h hh hR

= ⋅ −

or

110

10

ie

Lie oe

fe re

hR

h hh h= ⋅ −


(b) Output voltage ‘vo’.(c) Value of ‘Re’ so as to act in other region (saturation / active).

Ans. (a) Saturation (b) 1.645 V (c) Re = 771.56 Ω (to work in active region)Q2. Consider a transistor circuit as shown in Figure 2.37 below. Find values of

‘Rc’ and ‘Re’. What should be ‘Rc’ and ‘Re’, if β is reduced to its 50%value?

Ans. (a) Rc = 255 Ω and Re = 127 Ω(b) Rc = 512.8 Ω and Re = 253.87 Ω

Figure 2.36: Transistor circuit

3 K

h = 100fe

500

7 K

RE–

+3 V

0

ICS

IB

– 10 V

IE


= 200

0

IC

IB

90 K

90 K

RC

IERe

+ 6 V

– 6 V

Q3. Consider transistor circuit as shown in Figure 2.38 below. Find ‘VBB’ (inputvoltage) required to just saturate the transistor.

Ans. VBB = 14.41 Volts


Q4. Consider a transistor circuit as shown in figure 2.39, below. Find outputvoltage ‘vo’, if transistor is in “saturation region”. Find region of operation.Also, find output voltage ‘vo’, if input voltage is modified to 1V.


= 100

VBB

500

1 K

+ 15 V

– 15 V

+

–

44 K


0

I1

2.2 K

100 K

+ 12 V

– 12 V

I2

15 K

V = 12 V

(initially)i

Ans. When transistor is in “saturation”, its output ‘vo’ is 0.208 V. Transistor goesinto “cutoff region” as input is modified to 1V. In this case, output ‘vo’ is11.999978 V ( )12 volt .≈

Q5. Find value of base current ‘IB’ needed to just saturate the transistor asshown in Figure 2.40, below.

Figure 2.40: Transistor

4 K

IE

– 12 V

IB


Ans. 56.96µAQ6. Consider a transistor circuit as shown in Figure 2.41 below. If transistor is

operating in “active region”, find(a) VE (b) VB (c) I2 (d) IC (e) β(f) I1 (g) V1 and (h) R1Here, the transistor is in “active region”.


IB

R1

20 K

100 = R e

3.3 K = R C

12 V

I1

R2

VE

I2

V1

I = 2 mAE

VB

Ans. (a) 0.2 V (b) 0.9 V (c) 0.045 mA (d) 11.96 mA(e) 49 (f) 0.085 mA (g) 5.25 V and (h) 51.19 KΩ

Q7. What is ‘vc’ (collector output) in transistor circuit as shown in figure 2.42,below?


+10 V

IB

vC

V = – 0.7 VE

10 K

–10 V

IC

Ans. 5.45 VQ8. Consider a transistor circuit as shown in Figure 2.43 below.

Consider Silicon transistor. Find


Ans.( )1

Hint : ' '1 B

C

SII

+ β = ∂ −β ∂

(RC + RE) >> RBQ11. Consider a transistor circuit as shown in Figure 2.46 below.


V = 16 VCC

IB

VO

R = 1.5 KC

R2

I1

I2

IE

VCC

R1

V = 8 V

I = 4 mA

S = 10

CE

CVB

RE

Find

(a) RE (b) VB (c) 1

2

RR

(d) If stability factor ‘S’ changes to 3, find the value of ‘R1’ , needed tomaintain the same ratio, as earlier.

Ans. (a) 490 Ω (b) 2.63 V (c) 1

2

RR = 4.19 and (d) R1 = 5.25 KΩ

MULTIPLE CHOICE QUESTIONS

Q1. In a transistor having normal bias, the emitter junction(a) has a high resistance.(b) has a low resistance.(c) is reverse biased.(d) emits carriers into base region.

Q2. Current crossing collector junction in a normally biased n-p-n transistor iscalled(a) diffusion current.


Q9. As temperature in transistor increases, the base emitter resistance(a) decreases.(b) increases.(c) remain same.(d) none of these

Q10. Phase difference between input and output voltages in a CE connectedtransistor is maximum.(a) True(b) False(c) Sometimes true(d) Conditionally true

Q11. BC-147 is a(a) Germanium (Ge) transistor.(b) Silicon (Si) transistor.(c) Carbon (C) transistor.(d) Gallium Arsenide (GaAs) transistor.

Q12. In voltage divider bias, current ‘I1’flowing through ‘R1’ and ‘R2’ for goodstabilization should be either equal to or greater than__________(a) 10 IB(b) 3 IB(c) 2 IB(d) 4 IB

Q13. Operating point stabilization in potential divider method is given by(a) RE(b) Rc(c) Vcc(d) none of these

Q14. Base resistor method is widely used in(a) amplifier circuits.(b) switching circuits.(c) rectifier circuits.(d) clipping and clamping circuits.

Q15. Best stabilization of operating point is provided in/by(a) Base resistor method.(b) Collector feedback bias.

Bipolar Junction Transistor - I.K. International ... Two major types of ‘three terminal’...

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Transcript of Bipolar Junction Transistor - I.K. International ... Two major types of ‘three terminal’...