Transistor Best Concepts

7/29/2019 Transistor Best Concepts

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Lecture 8:

BIPOLAR JUNCTION

TRANSISTORS

Semester II2010/2011

Code:EEE2213


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BJT STRUCTUREBasic structure of the bipolar junction transistor (BJT) determines its

operating characteristics.

Constructed with 3 doped semiconductor regions called emitter, base,

and collector, which separated by two pn junctions.

2 types of BJT;

(1) npn: Two n regions separated by a p region

(2) pnp: Two p regions separated by an n region.

BIPOLAR:

refers to the useofboth holes &

electrons as

current carriers

in the transistor

structure.


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Base-emitter junction: the pn junction joining the base region & the

emitter region.

Base-collector junction: the pn junction joining the base region & the

collector region.

A wire lead connects to each of the 3 regions. These leads labeled as;

E: emitter

B: base

C: collector


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BASE REGION: lightly doped, & very thin

EMITTER REGION: heavily doped

COLLECTOR REGION: moderately doped


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Standard BJT Symbols


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BASIC BJT OPERATIONFor a BJT to operate properly as an amplifier, the two pn junctions

must be correctly biased with external dc voltages.

Figure: shows a bias arrangement for npn BJTs for operation as an

amplifier.

In both cases, BE junction is forward-biased & the BC junction isreverse-biased. called forward-reverse bias.

Look at this one circuit as two separate

circuits, the base-emitter(left side) circuit and

the collector-emitter(right side) circuit. Note

that the emitter leg serves as a conductor for

both circuits. The amount of current flow in

the base-emitter circuit controls the amount of

current that flows in the collector circuit.

Small changes in base-emitter current

yields a large change in collector-current.


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The heavily doped n-type emitter region has a very high density ofconduction-band (free) electrons.

These free electrons easily diffuse through the forward-based BE

junction into the lightly doped & very thin p-type base region

(indicated by wide arrow).

The base has a low density of holes, which are the majority carriers

(represented by the white circles).

A small percentage of the total number of free electrons injected into

the base region recombine with holes & move as valence electrons

through the base region & into the emitter region as hole current

(indicated by red arrows).


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BJT operation showing electron flow.


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When the electrons that have recombined with holes as valence

electrons leave the crystalline structure of the base, they become free

electrons in the metallic base lead & produce the external base

current.

Most of the free electrons that have entered the base do not recombine

with holes because the base is very thin.

As the free electrons move toward the reverse-biased BC junction,

they are swept across into the collector region by the attraction of the

positive collector supply voltage.

The free electrons move through the collector region, into the external

circuit, & then return into the emitter region along with the base

current.

The emitter current is slightly greater than the collector current

because of the small base current that splits off from the total current

injected into the base region from the emitter.


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Transistor CurrentsThe directions of the currents in both npn and pnp transistors and their

schematic symbol are shown in Figure (a) and (b). Arrow on the emitter

of the transistor symbols points in the direction of conventional

current. These diagrams show that the emitter current (IE) is the sum of

the collector current (IC) and thebase current (IB), expressed as follows:

IE = IC + IB


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BJT CHARACTERISTICS &

PARAMETERSFigure shows the proper biasarrangement fornpn

transistor foractive

operation as an amplifier.

Notice that the base-emitter

(BE) junction is forward-biased by VBB and the base-

collector (BC) junction is

reverse-biased by VCC. The dc

current gain of a transistor is

the ratio of the dc collector

current (IC) to the dc base

current (IB), and called dc beta

(DC).DC = IC/IB

The ratio of the dc collector current (IC)

to the dc emitter current (IE) is the dc

alpha. DC = IC/IE


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Ex 4-1Determine DC and IE for a transistor where IB= 50 A and IC = 3.65 mA.


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Ex 4-1Determine DC and IE for a transistor where IB= 50 A and IC = 3.65 mA.

7350

65.3

A

mA

I

I

B

C

DC

IE = IC + IB= 3.65 mA + 50 A = 3.70 mA

986.070.3

65.3

mA

mA

I

I

E

C

DC


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The collector current is

determined bymultiplying the base

current by beta.

Thus, IC= DC * IB

Analysis of this transistor circuit to predict the dc voltages and currentsrequires use ofOhms law, Kirchhoffs voltage law and the betafor thetransistor;

Application of these laws begins with the base circuit to determine the

amount of base current. Using Kichhoffs voltage law, subtract the VBE

=0.7 V, and the remaining voltage is dropped across RB .

Thus, VRB= VBB - VBE.

Determining the current for the base with this information is a matter ofapplying of Ohms law. VRB/RB = IB


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What we ultimately

determine by use ofKirchhoffs voltage law

for series circuits is that,

in the base circuit, VBB is

distributed across the

base-emitter junction

and RB in the base

circuit. In the collector

circuit we determine that

VCC is distributedproportionally across

RC and the

transistor(VCE).


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BJT Circuit AnalysisThere are three key dc voltages and three key dc currents to be

considered. Note that these measurements are important fortroubleshooting.

IB: dc base current

IE: dc emitter current

IC: dc collector current

VBE: dc voltage acrossbase-emitter junction

VCB: dc voltage acrosscollector-base junction

VCE: dc voltage fromcollector to emitter


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When the base-emitter junction is forward-biased,

VBE 0.7 V

VRB = IBRB: by Ohms law

IBRB = VBBVBE : substituting for VRB

IB = (VBBVBE) / RB: solving for IB

VCE = VCCVRc: voltage at the collector with respect to the

grounded emitter

VRc = ICRC

VCE = VCCICRC: voltage at the

collector with

respect to the emitter

The voltage across the reverse-biased

collector-base junction

VCB = VCEVBE where IC = DCIB


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Ex 4-2Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. Thetransistor has a DC = 150.


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Ex 4-2Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. Thetransistor has a DC = 150.

When the base-emitter junction is forward-biased,VBE 0.7 V

IB = (VBBVBE) / RB

= (5 V0.7 V) / 10 k = 430 A

IC

= DC

IB

= (150)(430 A)

= 64.5 mA

IE = IC + IB

= 64.5 mA + 430 A

= 64.9 mAVCE = VCCICRC

= 10 V(64.5 mA)(100 )

= 3.55 V

VCB

= VCE

VBE

= 3.55 V0.7 V

= 2.85 V

Since the collector is at a higher

voltage than the base, the collector-

base junction is reverse-biased.


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Gives a graphicalillustration of the

relationship of collector

current and VCE with

specified amounts of

base current. Withgreater increases of VCC ,

VCE continues to increase

until it reaches

breakdown, but the

current remains about thesame in the linear region

from 0.7V to the

breakdown voltage.

Collector Characteristic Curves


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Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 A to 25 A in 5

A increment. Assume DC = 100 and that VCE does not exceed breakdown.


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Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 A to 25 A in 5

A increment. Assume DC = 100 and that VCE does not exceed breakdown.

IC= DC IB

IB IC5 A 0.5 mA10 A 1.0 mA

15 A 1.5 mA

20 A 2.0 mA

25 A 2.5 mA


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CutoffWith no IB , the transistor is in the cutoffregion and just as the

name implies there is practically no current flow in thecollectorpart of the circuit. With the transistor in a cutoff state,

the full VCC can be measured across the collector and

emitter(VCE).

Cutoff: Collector leakage current (ICEO) is extremely small and is usuallyneglected. Base-emitter and base-collector junctions are reverse-biased.


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SaturationOnce VCE reaches its maximum value, the transistor is said to be in

saturation.

Saturation: As IB increases due to increasing VBB, IC also increases and VCE

decreases due to the increased voltage drop across RC. When the transistor reaches

saturation, IC can increase no further regardless of further increase in IB. Base-

emitter and base-collector junctions are forward-biased.


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DC Load Line

The dc load line graphically illustratesIC(sat) and cutoff for a transistor.

DC load line on a family of collector characteristic curves illustrating the

cutoff and saturation conditions.

Active

region of

the

transistors

operation.


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Ex 4-4Determine whether or not the transistors in Figure is insaturation. Assume VCE(sat) = 0.2 V.


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mAk

VV

R

VVI

C

satCECC

satC

8.90.1

2.010

)(

)(

Ex 4-4Determine whether or not the transistors in Figure is insaturation. Assume VCE(sat) = 0.2 V.

First, determine IC(sat)

mAmAII

mAk

V

k

VV

R

VVI

BDCC

B

BEBB

B

5.11)23.0)(50(

23.010

3.2

10

7.03

Now, see if IB is large enough to produce IC(sat).

Thus, IC greater than

IC(sat). Therefore, the

transistor is saturated.


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Maximum Transistor Ratings

A transistor has limitations on its operation. The product of VCEand IC cannot be maximum at the same time. If VCE is

maximum, IC can be calculated as

CE

D

CV

P

I

(max)

Ex 4-5A certain transistor is to be operated with VCE = 6 V. Ifits maximum power rating is 250 mW, what is the most collector

current that it can handle?

mAV

mW

V

PI

CE

D

C7.41

6

250(max)


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Ex 4-6The transistor in Figure has the following maximum ratings: PD(max) = 800mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC

can be adjusted without exceeding a rating. Which rating would be exceeded first?


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First, find IB so that you can determine IC.

The voltage drop across RC is.

PD = VCE(max)IC = (15V)(19.5mA) = 293 mW

VCE(max) will be exceeded first because the entire supply voltage, VCC will

be dropped across the transistor.

VRc = ICRC = (19.5 mA)(1.0 k) = 19.5 V

VRc

= VCC

VCE

when VCE

= VCE(max)

= 15 V

VCC(max) = VCE(max) + VRc = 15 V + 19.5 V = 34.5 V

mAAII

Ak

VV

R

VVI

BDCC

B

BEBB

B

5.19)195)(100(

19522

7.05

Ex 4-6The transistor in Figure has the following maximum ratings: PD(max) = 800mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC



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Derating PD(max)

P D(max) is usually specified at 25C.

At higher temperatures, P D(max) is less.

Datasheets often give derating factors for determining P D(max) at

any temperature above 25C.

Ex 4-7A certain transistor has a P

D(max)of 1 mW at 25C. The derating

factor is 5 mW/C. What is the P D(max) at a temperature of

70C?


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Transistor Datasheet

Refer Figure 4-20 (a partial datasheet for the 2N3904 npntransistor).

The maximum collector-emitter voltage (VCEO) is 40V.

The CEO subscript indicates that the voltage is measured from

collector to emitter with the base open. VCEO= VCE(MAX)

The maximum collector current is 200 mA.

* Other characteristics can be referred from the datasheet.


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A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC



35/65

A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC


PD(max) = 800 mW

VCE(max) = 15 V

IC(max) = 100 mA.

IB=195m A

IC= b

DCIB=19.5mA

VCC(max) = VCE(max) + VRc = 40 V + 19.5 V = 59.5 V

PD = VCE(max)IC = (40V)(19.5mA) = 780 mW

However at the max value of VCE, the power dissipation is

Power Dissipation exceeds the maximum of 645 mW specified on the

datasheet.


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THE BJT AS AN AMPLIFIERAmplification of a relatively

small ac voltage can be had byplacing the ac signal source in

the base circuit.

Recall that small changes in the

base current circuit causes large

changes in collector current

circuit.

The ac emitter current : Ie I

c = Vb/r

eThe ac collector voltage : Vc = IcRcSinceIc Ie, the ac collector voltage : Vc IeRcThe ratio of Vc to Vb is the ac voltage gain :Av = Vc/VbSubstituting IeRc for Vc and Iere for Vb : Av = Vc/Vb IcRc/Iere

The Ie terms cancel : Av Rc/re


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Ex 4-9Determine the voltage gain and the ac output

voltage in Figure if re = 50 .

The voltage gain : Av Rc/re = 1.0 k/50 = 20

The ac output voltage : AvVb = (20)(100 mV) = 2 V


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THE BJT AS A SWITCHA transistor when used as a switch is simply being biased so that it

is in cutoff (switched off) orsaturation (switched on). Rememberthat the VCE in cutoff is VCC and 0V in saturation.


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Conditions in Cutoff & Saturation

C

satCECC

satC

R

VVI

)(

)(

DC

satC

B

II

)(

(min)

A transistor is in the cutoff region when the base-emitter junction is notforward-biased. All of the current are zero, and VCE is equal to VCC

VCE(cutoff) = VCC

When the base-emitter junction is forward-biased and there is enough basecurrent to produce a maximum collector current, the transistor is saturated.

The formula for collector saturation current is

The minimum value of base current

needed to produce saturation is


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Ex 4-10(a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?(b) What minimum value of IB is required to saturate this transistor ifDC is

200? Neglect VCE(sat).

(c) Calculate the maximum value of RB when VIN = 5 V.

0


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Ex 4-10(a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?(b) What minimum value of IB is required to saturate this transistor ifDC is

200? Neglect VCE(sat).

(c) Calculate the maximum value of RB when VIN = 5 V.

AmAI

I

mAk

V

R

VI

DC

satC

B

C

CC

satC

50200

10

100.1

10

)(

(min)

)(

kA

V

I

VR

B

R

B

B 86

50

3.4

(min)

(max)

(a) When VIN = 0 VVCE = VCC = 10 V

(b) Since VCE(sat) is neglected,

(c) When the transistor is on, VBE 0.7 V.

VRB = VINVBE 5 V 0.7 V = 4.3 V

Calculate the maximum value of RB


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Transistor Construction

There are two types of transistors:

pnp npn

The terminals are labeled:

E - Emitter B - Base

C - Collector

pnp

npn

42


43/65

Transistor OperationWith the external sources, VEE and VCC, connected as shown:

The emitter-base junction is forward biased

The base-collector junction is reverse biased

43


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Currents in a Transistor

The collector current is comprised of two

currents:

BI

CI

EI

minorityCOI

majorityCI

CI

Emitter current is the sum of the collector and

base currents:

44


45/65

Common-Base Configuration

The base is common to both input (emitterbase) and

output (collectorbase) of the transistor.

45


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Common-Base Amplifier

Input Characteristics

This curve shows the relationship

between of input current (IE) to input

voltage (VBE) for three output voltage

(VCB) levels.

46


47/65

This graph demonstrates

the output current (IC) to

an output voltage (VCB) for

various levels of input

current (IE).

Common-Base Amplifier

Output Characteristics

47


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Operating Regions

ActiveOperating range of the

amplifier.

CutoffThe amplifier is basically

off. There is voltage, but little

current.

SaturationThe amplifier is full on.

There is current, but little voltage.

48


49/65

EI

CI

Silicon(forV0.7BEV

Approximations

Emitter and collector currents:

Base-emitter voltage:

49


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Ideally: = 1In reality: is between 0.9 and 0.998

Alpha ()Alpha () is the ratio of IC to IE:

EI

CI

dc

Alpha () in the AC mode:

EI

CI

ac

50


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Transistor Amplification

Voltage Gain:

V50k5ma10

mA10

10mA20

200mV

))((RLI

LV

iI

LI

EI

CI

iR

iV

iI

EI

Currents and Voltages:

51

250200mV

50V

iV

LV

vA


52/65

CommonEmitter Configuration

The emitter is common to both input

(base-emitter) and output (collector-

emitter).

The input is on the base and the

output is on the collector.

52


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Common-Emitter Characteristics

Collector Characteristics Base Characteristics

53


54/65

Common-Emitter Amplifier Currents

Ideal Currents

IE= IC+ IB IC= IE

Actual Currents

IC = IE + ICBO

When IB = 0 A the transistor is in cutoff, but there is some minoritycurrent flowing called ICEO.

A0

BI

CBO

CEO

II

1

where ICBO = minority collector current

54

ICBO is usually so small that it can be ignored, except inhighpower transistors and in high temperatureenvironments.


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Beta ()

In DC mode:

In AC mode:

represents the amplification factor of a transistor. ( issometimes referred to as hfe, a term used in transistor modelingcalculations)

B

C

I

I dc

constantac

CEV

B

C

II

55


56/65

Determining from a GraphBeta ()

108

A25

mA2.7 7.5VDC CE

100

A10

mA1

A)20A(30

mA)2.2mA(3.2

7.5V

AC

CE

56

B t ( )


57/65

Relationship between amplification factors and

1

1

Beta ()

Relationship Between Currents

BC II BE 1)I(I

57


58/65

CommonCollector Configuration

The input is on the

base and the output is

on the emitter.

58

C C C fi i


59/65

CommonCollector Configuration

The characteristics are

similar to those of the

common-emitterconfiguration, except the

vertical axis is IE.

59

O i Li i f E h C fi i


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VCE is at maximum and IC is at

minimum (ICmax= ICEO) in the cutoff

region.

IC is at maximum and VCE is at

minimum (VCE max = VCEsat = VCEO) inthe saturation region.

The transistor operates in the active

region between saturation and cutoff.

Operating Limits for Each Configuration

60

P Di i i


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Power Dissipation

Common-collector:

CCBCmax IVP

CCECmax IVP

ECECmax IVP

Common-base:

Common-emitter:

61

T i t S ifi ti Sh t


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Transistor Specification Sheet

62

T i t S ifi ti Sh t


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Transistor Specification Sheet

63

T i t T ti


64/65

Transistor Testing

Curve Tracer

Provides a graph of the characteristic curves.

DMM

Some DMMs measure DC or hFE. Ohmmeter

64


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Transistor Terminal Identification

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