Chapter 5 Transistors BJT Part I

8/12/2019 Chapter 5 Transistors BJT Part I

1/20

Bipolar Junction Transistors

The bipolar junction transistor (BJT) is constructedwith three doped semiconductor regions called emitter,base, and collector. These three regions are separated bytwopnjunctions: The base-emitter junction and thebase-collector junction.


2/20


3/20

Transport Model for the npnTransistor

Base-emitter voltage vBEand base-collector voltage vBCdetermine currents intransistor and are said to be positive when they forward-bias their respective

pnjunctions.

The terminal currents are collector current(iC), base current (iB) and emitter

current (iE).


4/20

npn Transistor | Forward Characteristics

Collector (Forward transport) current

is

ISis the BJT saturation current

Base current iB is given by

bF is the forward common-emitter current gain

Emitter current iE is

aF is the forward common-base current gain

In this forward-active region of operation

195.0

11exp

F

F

FF

T

BE

F

SBCE

V

vIiii

a

b

ba

a

E

CF

B

CF

i

i

i

i ab

VT= kT/q =0.025 V at room temperature


5/20

Emitter current iE is

1expT

BCSRE

V

vIii

1exp

T

BC

R

S

R

RB V

vIii

bb

Collector current iC is given by

bRis the reverse common-emitter current

gain

aRis the reverse common-base current gain

Base current iB is given by

Base currents in forward and reverse modes

are different due to asymmetric doping levels

in emitter and collector regions.

95.001

R

ab

ba

R

RR

npn Transistor | Reverse Characteristics


6/20

First term in both emitter and collector current expressions gives current transported

completely across base region.

npn Transistor | Complete Transport ModelEquations for Any Bias

1exp1exp

1expexpexp

1expexpexp

T

BC

R

S

T

BE

F

SB

T

BE

F

S

T

BC

T

BE

SE

T

BC

R

S

T

BC

T

BESC

V

vI

V

vIi

V

vI

V

v

V

v

Ii

VvI

Vv

VvIi

bb

b

b


7/20

Transport Model Calculation| Example

Problem: Find terminal voltages and currents. Given data: VBB = 0.75 V, VCC= 5.0 V, IS = 10

-16 A,

bF= 50,bR = 1

Assumptions: Room temperature operation, VT= 25.0 mV.

Analysis: VBE= 0.75 V,

VBC= VBB -VCC= 0.75 V- 5.00V = - 4.25 V

AI

mAI

mAI

B

E

C

4.211

025.0

25.4exp

1

101

025.0

75.0exp

50

10

09.11025.0

75.0exp

50

10

025.0

25.4exp

025.0

75.0exp10

07.11

025.0

25.4exp

1

10

025.0

25.4exp

025.0

75.0exp10

1616

1616

1616


8/20

i-v Characteristics of Bipolar Transistors

1. Common-

Emitter Output

Characteristics


9/20

i-v Characteristics of Bipolar TransistorsCommon-Emitter Output Characteristics

For iB = 0, the transistor is cutoff. If iB > 0, iCalso increases.

For vCE> vBE, npn transistor is in forward-active region, iC = bFiB is nearlyindependent of vCE.For vCE< vBE, transistor is in saturation.

For vCE< 0, roles of collector and emitter reverse.

Three regions are recognized


10/20

Operation Regions of BipolarTransistors

Base-Emitter Junction Base-Collector Junction

Reverse Bias Forward Bias

Forward Bias

Forward-Active

Region(Good Amplifier)

Saturation Region

(Closed Switch)

Reverse BiasCutoff Region

(Open Switch)

Reverse-Active

Region

(Poor Amplifier)


11/20

i-v Characteristics of Bipolar Transistors

2. Transfer Characteristics

Defines relation between collector

current and base-emitter voltage of

transistor.

Almost identical to transfer

characteristic of apnjunction diode

Setting vBC = 0 in the collector-current

expression yields

Collector current expression has the same form as that of the diode equation

1exp

T

BESC

VvIi


12/20

Simplified Cutoff Region Model

In the cutoff region, both junctions

are reverse-biased; the transistor is

said to be in off state

vBE< 0, vBC< 0

If we assume that

where 4kT/q = 0.1 V, then the

transport model terminal current

equations simplify to

q

kTv

q

kTv BCBE 4and4

F

S

F

SB

F

SE

R

SC

IIi

IiIi

bb

bb

and


13/20

Simplified Cutoff Region ModelExample

Problem: Estimate terminal currents using the transport model

Given data: IS = 10-16 A, aF= 0.95, aR = 0.25, VBE= 0 V, VBC= -5 V

Assumptions: Simplified transport model assumptions

Analysis: From given voltages, we know that transistor is in cutoff.

For practical purposes, all three

currents are essentially zero.

AxII

AII

AxI

II

R

SB

SE

R

S

R

SC

16

16

16

103

10

1041

1

b

ab


14/20

Simplified Forward-Active RegionModel

In forward-active region, the emitter-base junction is forward-biased and thecollector-base junction is reverse-biased. vBE > 0, vBC< 0. if we assume

then the transport model terminal current equations simplify to

BJT is often considered a current-controlled device, though fundamental

forward-active behavior suggests a voltage- controlled current source.

1expexp

expexp

exp

BFB

T

BE

F

S

R

S

T

BE

F

SB

BFC

T

BE

F

S

F

S

T

BESE

EFC

R

S

T

BESC

IiV

vII

V

vIi

Ii

V

vII

V

vIi

IiI

V

vIi

bbbb

b

ab

ab

q

kTv

q

kTv BCBE 4and4


15/20

Simplified Forward-Active RegionModel | Example 1

Problem: Estimate transistor terminal currents and base-emitter voltage

Given data:IS=10-16 A,aF= 0.95, VBC= VB - VC = -5 V,IE= 100 A

Assumptions: Simplified transport model assumptions, room

temperature operation, VT = 25.0 mV

Analysis: Current source forward-biases base-emitter diode, VBE> 0,

VBC< 0, we know that transistor is in forward-active operation region.

VI

IVV

AAI

I

AAII

S

EFTBE

F

EB

F

FF

EFC

689.0ln

520

100

1

1995.01

95.0

1

9510095.0

a

b

a

ab

a


16/20

Problem: Estimate terminal currents, base-emitter and base-collectorvoltages for the transistor in the given circuit.

Given data:IS= 10-16 A,aF= 0.95, VC = +5 V,IB = 100 A

Assumptions: Simplified transport model assumptions, room

temperature operation, VT = 25.0 mV

Analysis: Current source causes base current to forward-bias base-emitter diode, VBE> 0, VBC


17/20

Simplified Circuit ModelForward-Active Region

Current in base-emitter diode is amplified by common-emitter current gain

bFand appears at collector; base and collector currents are exponentiallyrelated to base-emitter voltage.

Base-emitter diode is replaced by constant voltage drop model (VBE= 0.7 V)since it is forward-biased in forward-active region.

dc base and emitter voltages differ by 0.7-V diode voltage drop in forward-active region.


18/20

Simplified Forward-Active RegionModel | Example 3

Problem: Find transistor Q-point Given data: bF= 50, bR = 1

Assumptions: Forward-active region of operation, VBE = 0.7 V

Analysis:

correct.isregionactive-Forward

44.57.03.499.09

m990.08.1950

8.1951

01.1

1

01.18200

7.09

08200

VKmAV

VRIVV

AAII

AmAI

I

mAV

I

VIV

CE

BECCCCCE

BFC

F

EB

E

EEEBE

b

b


19/20

Simplified Circuit ModelSaturation Region

In the saturation region, both junctions are forward-biased, and thetransistor operates with a small voltage between collector and emitter.vCESAT is the saturation voltage for the npn BJT.

No simplified expressions exist for

terminal currents other than iC+ iB = iE.

Simplified Model

F

CB

BF

C

BR

C

R

TBCBECESAT

T

BC

R

S

T

BE

F

SB

T

BC

R

S

T

BESC

II

I

I

I

I

VVVV

V

VI

V

VII

V

VI

V

VII

b

b

b

a

bba

for

1

11

1ln

expexpexpexp


20/20

Transconductance Transconductance is defined by:

T

C

PtQT

BES

BEPtQBE

Cm

V

I

V

vI

dv

d

dv

dig

exp

Biasing for the BJT The goal of biasing is to establish known Q-point which in turn establishes

initial operating region of the transistor.

For a BJT, the Q-point is represented by (IC, VCE) for an npn transistor or (IC, VEC)

for apnp transistor.

Two practical biasing circuits used for a BJT are:

Four-Resistor Bias Network

Two-Resistor Bias Network

Documents

Chapter 5 Transistors BJT Part I