L9,10,11 Transistor [EngineeringDuniya.com]

7/28/2019 L9,10,11 Transistor [EngineeringDuniya.com]

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Department of Electronics and Communication Engineering,Manipal Institute of Technology, Manipal, INDIA

Subject Code : ECE

101/102

BASIC ELECTRONICS

COURSE MATERIALFor

1st & 2nd Semester B.E.

(Revised Credit System)

DEPARTMENT OF

ELECTRONICS & COMMUNICATION ENGINEERING


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Bipolar Junction Transistor

Solid state transistor was invented by a team of scientists at

Bell laboratories during 1947-48

Advantages of solid state transistor over vacuum devices:

Smaller size, light weightNo heating elements required

Lower power consumption and operating voltages

Low price


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Different transistor packages


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Introduction

Bipolar Junction Transistor (BJT) is a three layer, 2 junctionsemiconductor device

It is a sandwich of one type of semiconductor material

between two layers of another type

Two kinds of BJT sandwiches are: NPN and PNP


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Introduction

The three layers of BJT are called Emitter, Base and Collector

Base is narrower compared to the other two layers

Base is lightly doped, Emitter is heavily doped and Collector is

moderately doped NPNEmitter and Collector are made of N-type

semiconductors; Base is P-type

PNPEmitter and Collector are P-type, Base is N-type

BJT has two junctionsEmitter-Base (EB) Junction andCollector-Base (CB) Junction


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Introduction

Note: Arrow direction from P to N (like diode) which

indicates the direction of the flow of conventional current

The device is called bipolar junction transistor because

current is due to the motion of two types of charge carriers-free electrons & holes

Transistor symbols:


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Transistor Operation Operation of NPN transistor and PNP is similar with roles of

free electrons and holes interchanged Depending upon the bias condition (forward or reverse) of

each of the two junctions, different regions of operation for

the BJT are obtained

Active region- transistor operating as an amplifier and saturation &

cutoff region- Switching applications, e.g. in logic circuits

Region


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Transistor Operation


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Note the current directions in NPN and PNP transistors

For both varieties:

Collector current has two components:

One due to injected charge carriers from emitter

Another due to thermally generated minority carriers

Therefore,

C

E

B

IC

IE

IB

NPN

C

E

B

IC

IE

IB

PNP

BCEIII

CBOEdcCIII


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Department of Electronics and Communication EngineeringManipal Institute of Technology, Manipal, INDIA

Transistor Configurations

Common Base configuration

Base is common between input and output

Input voltage: VEB Input current: IE

Output voltage: VCB Output current: IC


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As the currents constituting the collector current are in the

same direction, we can writeICas,

SinceICBO value is very low, we can neglect it compared toIE

andIC . Therefore,

wheredc is the fraction of charge carriers emitted from

emitter, that enter into the collector region

This parameterdc is called common base DC current gain

Value ofdc varies from 0.90 to 0.998 and is defined for themajority carriers

Therefore,

epartment of Electronics and Communication Engineering,Manipal Institute of Technology, Manipal, INDIA

CBOEdcCIII


EdcCII

E

C

dc

I

I


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CB Input characteristics

A plot ofIEversus VEB

for various values ofVCB

It is similar to forwardbiased diode

characteristics

As VCB is increased,IE

increases only slightly


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Base Width modulation

As the reverse bias voltage VCB is increased, the depletionregion width at the C-B junction increases. Part of this

depletion region lies in the base layer. So, effective base width

decreases. Hence number of electron-hole combination at the

base decreases. So base current reduces and collector current

increases. Since IEIC,emitter current also shoots up early.

This phenomenon is also called as the Early effect


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CB Output characteristics A plot ofICversus VCB for various values ofIE Three regions are identified: Active, Cutoff, Saturation


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In case of Active region

E-B junction forward biased C-B junction reverse biased ICincreases withIE For givenIE,ICis almost constant; increases onlyslightly with increase in V

CB

. This is due to base-widthmodulation

In the cut off region, E-B and C-B junctions of the transistor

are reverse biased

Collector current is 0A

In the saturation region, the E-B and C-B junctions of the

transistor are forward biased


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Common Emitter configuration

Emitter is common between input and output

Input voltage: VBE Input current: I

B

Output voltage: VCE

Output current: IC


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CE Input characteristics

Plot ofIB versus VBEfor various values ofVCE. As VCEis increased,

IBdecreases slightly. This is due to base-width modulation


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Transistor Configurations CE Output characteristics

A plot ofICversus VCEfor various values ofIB

Three regions identified: Active, Cut-off, Saturation

VCESat


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Active region

Linear region in the output characteristics

E-B junction forward biased

C-B junction reverse biased

ICincreases withIB For givenIB, ICincreases slightly with increase in VCE; this

is due to base-width modulation

Saturation region

ICdecreases to zero at VCE=0

Cut off region

IB= 0, henceIC =ICEO


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Transistor Operation Parameterdc is common emitter DC current gain

Therefore, collector current expression is:

We have,

Substituting forIE, we get

CBOEdcCIII

CBOBCdcC IIII CBOBdcCdc III )1(

)1()1(dc

CBO

B

dc

dc

C

III

CEOBdcC III

where and

Values ofdc and dc vary from transistor to transistor

)1(dc

dc

dc

CBOdc

dc

CBO

CEOI

II 1

)1(

=

= =

B

C

dc

I

I

BdcCII


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Problems1. A BJT has alpha value as 0.998 andICBO of 1A. If emitter

current is 5mA, calculate the collector and base currents.

2. An npn transistor has collector current 4mA and base current

10 A. Calculate the alpha and beta value of the transistor

neglecting the reverse saturation currentICBO.

3. In a transistor, 99% of the carriers injected into the base cross

over to the collector region. If collector current is 4mA and

ICBO is 6 A, CalculateIEandIB.

4. A Ge transistor with = 100 hasICBO = 5A. If the transistor

is connected in common-emitter operation, find the collector

current for base current (a) 0 A (b) 40 A.

5. A Ge transistor has collector current of 51 mA when the base

current is 0.4 mA. If= 125, then what is the value ofICEO?


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Transistor Biasing

What is meant by biasing the transistor?

Applying external dc voltages to ensure that transistor

operates in the desired region

Which is the desired region?

For amplifier application, transistor should operate in active

region

For switch application, it should operate in cut-off and

saturation region

What is meant by quiescent point (Q-point)?

The point we get by plotting the dc values ofIC, IB and VCE

on the transistor characteristics


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Transistor BiasingTransistor characteristics showing Q-point:

Q-point is in the middle of active region.


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Transistor Biasing Types of biasing:

Fixed bias or base resistor bias Self bias or voltage divider bias

Fixed bias:

The value ofIBis fixed by choosing

proper value forRB Equations to be considered are:

B

BECC

B

R

VVI

CCCCCERIVV

Fixed bias circuit


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Transistor Biasing

Load Line

We have:

This is an equation of straight line with points VCC/RCand VCC

lying on y-axis and x-axis respectively

This line is called Load line because it depends on resistor

RCconsidered as Load and VCC

Intersection of load line with the transistor characteristic curve

is called Q-point or operating point for a particular value ofIB

,

hence giving a common solution

CCCCCERIVV


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Transistor Biasing If RB is varied, Q-point moves along the load line

If RB is held constant and RC is varied, then slope of load linevaries

If RB & RC are held constant and VCC is varied, then load line

shifts, maintaining same slope

With reference to the graphs, with everything else held constant If RB is increased, transistor goes towards cut-off and if RB is

decreased, transistor goes towards saturation

If RC is increased, transistor goes towards saturation and if

RC is decreased, transistor goes towards active region

If VCC is increased, transistor goes towards active region and

if VCC is decreased, transistor goes towards saturation



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Transistor BiasingVariation in load line with circuit parametersRB ,RC and VCC


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Transistor Biasing

Advantages of Fixed bias:

Simple circuit to analyze and design

Uses very few circuit components

Disadvantages of Fixed bias: Q-point is unstable i.e. if temperature increases, then

increases, henceICQ and VCEQvaries. Effectively Q-point

shifts

If the transistor is replaced with another transistor havingdifferentvalue, then also Q-point shifts


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Problems on Fixed bias

1. For a fixed bias circuit using Si transistor,RB = 500 k, RC= 2k,

VCC = 15 V, ICBO= 20 A and = 70. Find the collector currentICQ and VCEQ at Q-point.

2. A Si transistor is biased for a constant base current. If = 80,VCEQ = 8 V, RC= 3 k and VCC = 15 V, find ICQ and the value ofRB required.

Repeat the problem if the transistor is a germanium device.

3. For a fixed bias circuit, VCC = 12 V and RC= 4 k. The Getransistor used is characterized by = 50, ICEO = 0 A and

VCEsat = 0.2 V. Find the value of RB that just results in saturation4. A fixed bias circuit has VCC = 20 V, RC= 5 k andRB = 300 k.

The Si transistor has ICBO= 0 and = 50. Find ICQ and VCEQ.


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Transistor Biasing

Voltage divider bias or Self bias

ResistorREconnected betweenemitter and ground

Voltage-divider resistorsR1 &R2

replaceRB

Circuit can be analyzed in twomethods:

Exact method (using Thevenins

theorem)

Approximation method(neglecting base current)

Self bias circuit


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Transistor Biasing Exact method:

Input side of self-bias circuit (Fig. a) is transformed intoThevenins equivalent circuit (Fig. b) where,RTHis the

resistance looking into the terminalsA &B (Fig. c)

21

2

RR

RVV

CC

TH

21

2121 ||

RR

RRRRR

TH


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Transistor Biasing

Advantages of Self bias: The collector current and hence the Q-point is independent

of variation in temperature and replacement of transistor

Disadvantages of Self bias:

More circuit components are required

Analysis and design are complex



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Problems on Self bias1. For a self bias circuit, the transistor is a Si device, RE= 200 ,

R1 = 10R2= 10 k, RC= 2 k, = 100 and VCC = 15 V.Determine the values of ICQ and VCEQ.

2. Suppose if the transistor used in problem 1 failed and wasreplaced with a new transistor with = 75, is the new transistor

still biased for active region operation?

3. A self bias circuit uses Silicon transistor with RC= 3.3K ,

RE= 1K, R1= 39K , R2= 8.2K , = 120 and VCC = 18 V.

Determine the values of IB ,ICQ and VCEQ.


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Transistor Amplifier Amplifier

Circuit which increases the magnitude of the input signalapplied

BJT basically amplifies current: Collector current equalsbeta times Base current

By suitably designing the transistor circuit, we can getvoltage amplification and power amplification

For faithful amplification (with no distortion), BJT shouldoperate in Active region throughout the input cycle

This is achieved by proper use of biasing circuit

Biasing circuit fixes the operating point in the middle ofactive region


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Transistor Amplifier As vin varies, iin varies, thus base

current iB varies This variation in base current isamplified beta times to getvariation in collector current iC Output voltage voutis VCCiCRC Ifvin increases, there is

proportional decrease in vout

Similarly ifvin decreases, voutincreases proportionally

Thus output voltage of CEamplifier is 180o out of phase withinput voltage

CE amplifier circuit with fixed bias


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Transistor Amplifier

Gain of the amplifier is usually expressed in decibels

(AV)dB = 20 log10 |AV|

Usually a gain of 100 (i.e. 40 dB) can be obtained using single

transistor. For higher gain requirement, two or more amplifier

stages are to be cascaded

Overall gain is product of individual gains, but when expressed

in dB, overall gain is sum of individual gains (in dB)

VNVVVAAAA ........ 21

dBVNdBVdBVdBVAAAA )(.......)()()( 21


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RC Coupled Amplifier

CE amplifier employing self bias


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Transistor Amplifier Additional components used are CCand CE

CCis called coupling capacitorused to prevent dccomponent from entering or leaving amplifier stage

CE is called emitter bypass capacitorused to bypass theac emitter currentpreventing it from flowing throughRE

If ac emitter current is allowed to pass throughRE, then vBE

reduces and hence output voltage reduces

Frequency response of amplifier

It is important to know the behavior of the amplifier at

different frequencies Gain is NOT constant at all frequenciesdepends on

various factors

Frequency response is a plot of gain versus frequency


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Figure shows frequencyresponse plot

At lower and higher

frequencies, gain is less

Gain attains constantvalue at mid frequencies

Bandwidth of amplifier is

range of frequencies over

which gain is not less than3 dB of maximum gain


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Analysis of frequency response curve

At very low frequencies, reactance of coupling capacitors is

high, hence there is loss of signal voltage across capacitors,

resulting in reduced gain

Also at low frequencies, emitter bypass capacitor does not

fully bypass the ac emitter current, hence ac voltage drop

develops acrossRE, resulting in reduced gain

At very high frequencies, shunt capacitances due to wiring

and inter-layer junction capacitances will be prominent,

hence resulting in signal loss

At mid frequencies, gain is maximum and constant

i lifi


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Advantages

Cost is low

Offers high fidelity in audio frequency range (20 - 20KHz)

Circuit is quite compact

Disadvantages

Tends to be noisier with age

Gain is less


bl A lifi


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Problems on Amplifiers

1. A three-stage amplifier circuit has first stage gain of 45 dB,

second stage gain of 50 dB and third stage gain of5 dB.What is the overall gain? If input to the first stage is 0.1mV,

what is the output of final stage?

2. An amplifier has maximum gain of 200 and bandwidth of500 kHz. If lower cutoff frequency is 50 Hz, what is the

upper cutoff frequency and gain at this frequency?

3. The voltage amplifier has a voltage gain = 200 at the cut offfrequencies. Find the maximum voltage gain?


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End of Module 6

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