chapter 10 bjt fundamentals
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Chapter 10 BJT Fundamentals. Chapter 10. BJT Fundamentals. Bipolar Junction Transistors (BJTs). - PowerPoint PPT PresentationTRANSCRIPT
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President University Erwin Sitompul SDP 9/1
Dr.-Ing. Erwin SitompulPresident University
Lecture 9Semiconductor Device Physics
http://zitompul.wordpress.com
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Chapter 10BJT Fundamentals
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Bipolar Junction Transistors (BJTs)Chapter 10 BJT Fundamentals
Over the past decades, the higher layout density and low-power advantage of CMOS (Complementary Metal–Oxide–Semiconductor) has eroded away the BJT’s dominance in integrated-circuit products.Higher circuit density better system performance
BJTs are still preferred in some digital-circuit and analog-circuit applications because of their high speed and superior gainFaster circuit speed (+)Larger power dissipation (–)
• Transistor: current flowing between two terminals is controlled by a third terminal
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EB E B
CB C B
EC E C
EB CB
V V VV V VV V V
V V
BE B E
BC B C
CE C E
CB EB
V V VV V VV V V
V V
IntroductionChapter 10 BJT Fundamentals
There are two types of BJT: pnp and npn.
The convention used in the textbook does not follow IEEE convention, where currents flowing into a terminal is defined as positive.
We will follow the normal convention: . . . . . .
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Common-Emitter I–V Characteristics
Cdc
B
100II
Most popular configuration
Saturation ModeIC < IB
Active Mode
In active mode, dc is the common
emitter dc current gain
Circuit ConfigurationsChapter 10 BJT Fundamentals
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Mode E-B Junction C-B JunctionSaturation forward bias forward bias
Active/Forward forward bias reverse biasInverted reverse bias forward bias
Cutoff reverse bias reverse bias
Modes of OperationChapter 10 BJT Fundamentals
Common-Emitter Output Characteristics
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AE DB ACN N N
CB EBW W
B nEB nCBW W x x
W : quasineutral base width
BJT ElectrostaticsChapter 10 BJT Fundamentals
Under equilibrium and normal operating conditions, the BJT may be viewed electrostatically as two independent pn junctions.
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BJT ElectrostaticsChapter 10 BJT Fundamentals
Electrostatic potential, V(x)
Electric field, E(x)
Charge density, ρ(x)
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pnp BJT, active mode
BJT DesignChapter 10 BJT Fundamentals
Important features of a good transistor: Injected minority carriers do not recombine in the neutral
base region short base, W << Lp for pnp transistorEmitter current is comprised almost entirely of carriers
injected into the base rather than carriers injected into the emitter the emitter must be doped heavier than the base
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EMITTER BASE COLLECTOR
p-type n-type p-type3
1
4
2
CB0i
Base Current (Active Bias)Chapter 10 BJT Fundamentals
The base current consists of majority carriers (electrons) supplied for:
1. Recombination of injected minority carriers in the base2. Injection of carriers into the emitter3. Reverse saturation current in collector junction4. Recombination in the base-emitter depletion region
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Decrease relative to to increase transport factor
Decrease relative to and to increase efficiency
CpT
Ep
II
dc T Common base dc current gain:
Ep Ep
E Ep En
I II I I
IEn
IEp
ICn
ICp
Negligible compared to holes injected
from emitter
15
21 2
BJT Performance Parameters (pnp)Chapter 10 BJT Fundamentals
Emitter Efficiency Base Transport Factor
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C dc C B CB0α ( )I I I I
C dc E CB0αI I I
Common emitter dc current gain:dc C
dcdc B1
II
ICB0 :collector current when IE = 0
23
dc CB0C B
dc dc
α1 α 1 α
II I
CB0I
Collector Current (Active Bias)Chapter 10 BJT Fundamentals
The collector current is comprised of:Holes injected from emitter, which do not recombine in the
base Reverse saturation current of collector junction
C dc B CE0I β I I
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Chapter 11BJT Static Characteristics
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E AE
E N
E n
E N
E0 p02i E
N ND D
L Ln n
n N
B DB
B P
B p
B P
B0 n02i B
N ND D
L Lp p
n N
C AC
C N
C n
C N
C0 p02i C
N ND D
L Ln n
n N
Minority carrier
constants
Notation (pnp BJT)Chapter 11 BJT Static Characteristics
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2E E
E 2E
0 d n nDdx
EB
E
E E0
( ) 0( 0) ( 1)qV kT
n xn x n e
Emitter RegionChapter 11 BJT Static Characteristics
Diffusion equation:
Boundary conditions:
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2B B
B 2B
0 d p pDdx
EB
CB
B B0
B B0
(0) ( 1)( ) ( 1)
qV kT
qV kTp p ep W p e
Base RegionChapter 11 BJT Static Characteristics
Diffusion equation:
Boundary conditions:
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2C C
C 2C
0 d n nDdx
CB
C
C C0
( ' ) 0( ' 0) ( 1)qV kT
n xn x n e
Collector RegionChapter 11 BJT Static Characteristics
Diffusion equation:
Boundary conditions:
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EEn E
0x
d nI qADdx
B
Ep B0x
d pI qADdx
BCp B
x W
d pI qADdx
Cn C
0
C
x
d nI qADdx
E
B
C
( ), ( ),( )
n xp xn x
ICIB
IEE Ep En
C Cp Cn
B E C
I I II I II I I
Ideal Transistor AnalysisChapter 11 BJT Static Characteristics
Solve the minority-carrier diffusion equation in each quasi-neutral region to obtain excess minority-carrier profilesEach region has different set of boundary conditions
Evaluate minority-carrier diffusion currents at edges of depletion regions
Add hole and electron components together terminal currents is obtained
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2E E
E 2E
0 d n nDdx
E EE 1 2( ) x L x Ln x Ae A e
EB
E
E E0
( ) 0( 0) ( 1)qV kTn x
n x n e
EB EE E0( ) ( 1)qV kT x Ln x n e e
EEn E
0x
d nI qADdx
Emitter Region SolutionChapter 11 BJT Static Characteristics
Diffusion equation:
General solution:
Boundary conditions:
Solution
EBEE0
E
( 1)qV kTDqA n eL
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C CC 1 2( ) x L x Ln x Ae A e
CB CC C0( ) ( 1)qV kT x Ln x n e e
2C C
C 2C
0 d n nDdx
CB
C
C C0
( ) 0( 0) ( 1)qV kTn x
n x n e
CCn C
0x
d nI qAD
dx
Collector Region SolutionChapter 11 BJT Static Characteristics
Diffusion equation:
General solution:
Boundary conditions:
Solution
CBCC0
C
( 1)qV kTDqA n eL
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B BB 1 2( ) x L x Lp x Ae A e
2B B
B 2B
0 d n pDdx
EB
CB
B B0
B B0
(0) ( 1)( ) ( 1)
qV kT
qV kTp p ep W p e
B BEB
B B
B BCB
B B
( ) ( )
B B0
B0
( ) ( 1)
( 1)
W x L W x LqV kT
W L W L
x L x LqV kT
W L W L
e ep x p ee e
e ep ee e
Base Region SolutionChapter 11 BJT Static Characteristics
Diffusion equation:
General solution:
Boundary conditions:
Solution
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sinh( )2
e e
EB
CB
BB B0
B
BB0
B
sinh ( )( ) ( 1)
sinh( )sinh( ) ( 1)sinh( )
qV kT
qV kT
W x Lp x p e
W Lx Lp eW L
as
B BEB
B B
B BCB
B B
( ) ( )
B B0
B0
( ) ( 1)
( 1)
W x L W x LqV kT
W L W L
x L x LqV kT
W L W L
e ep x p ee e
e ep ee e
Base Region SolutionChapter 11 BJT Static Characteristics
Since
We can write
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CBEB
BEp B
0
B BB0
B B B
cosh( ) 1( 1) ( 1)sinh( ) sinh( )
x
qV kTqV kT
d pI qADdx
D W LqA p e eL W L W L
CBEB
BCp B
B BB0
B B B
cosh( )1 ( 1) ( 1)sinh( ) sinh( )
x W
qV kTqV kT
d pI qADdx
D W LqA p e eL W L W L
sinh( ) cosh( )2 2
d d e e e ed d
Base Region SolutionChapter 11 BJT Static Characteristics
Since
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EB
CB
E B BE E0 B0
E B B
BB0
B B
cosh( ) ( 1)sinh( )
1 ( 1)sinh( )
qV kT
qV kT
D D W LI qA n p eL L W L
D p eL W L
EB
CB
BC B0
B B
C B BC0 B0
C B B
1 ( 1)sinh
cosh( ) ( 1)sinh( )
qV kT
qV kT
DI qA p eL W L
D D W Ln p eL L W L
E En Ep ,I I I C Cn CpI I I
B E CI I I
Terminal CurrentsChapter 11 BJT Static Characteristics
Since
Then
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EB
CB
/B B0
/B0
( ) ( 1) 1
( 1)
qV kT
qV kT
xp x p eWxp eW
Due to VEB
Due to VCB
0 2
0
limsinh( )
lim cosh( ) 12
B B B0 B( ) (0) ( ) (0) xp x p p W pW
Simplified RelationshipsChapter 11 BJT Static Characteristics
To achieve high current gain, a typical BJT will be constructed so that W << LB.
Using the limit value
We will have
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E B
B E E
1
1 D N WD N L
dc 2
E B
B E E B
1
12
D N W WD N L L
T 2
B
1
112WL
dc 2
E B
B E E B
1 ,112
D N W WD N L L
Performance ParametersChapter 11 BJT Static Characteristics
For specific condition of “Active Mode”: emitter junction is forward biased and
collector junction is reverse biasedW << LB, nE0/pB0 NB/NE
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Homework 7
Deadline: 21.06.2012, at 08:00.
1.(10.17)Consider a silicon pnp bipolar transistor at T = 300 K with uniform dopings of NE = 5×1018 cm–3, NB = 1017 cm–3, and NC = 5×1015 cm–3 . Let DB = 10 cm2/s, xB = 0.7 μm, and assume xB << LB. The transistor is operating in saturation with JP = 165 A/cm2 and VEB = 0.75 V. Determine:(a) VCB, (b) VEC(sat), (c) the number/cm2 of excess minority carrier holes in the base, and (d) the number/cm2 of excess minority carrier electrons in the long collector, take LC = 35 μm.
2.(10.14)Problem 10.4, Pierret’s “Semiconductor Device Fundamentals”.
Chapter 11 BJT Static Characteristics