Lecture #15
OUTLINE
The Bipolar Junction Transistor
– Fundamentals
– Ideal Transistor Analysis
Reading: Chapter 10, 11.1
Spring 2003
EE130 Lecture 15, Slide 1
Bipolar Junction Transistors (BJTs)
• Over the past 3 decades, the higher layout density and
low-power advantage of CMOS technology has eroded
away the BJT’s dominance in integrated-circuit products.
(higher circuit density Æ better system performance)
• BJTs are still preferred in some digital-circuit and
analog-circuit applications because of their high speed
and superior gain.
9 faster circuit speed
8 larger power dissipation
Æ limits integration level to ~104 circuits/chip
Spring 2003
EE130 Lecture 15, Slide 2
1
Introduction
• The BJT is a 3-terminal device
– 2 types: PNP and NPN
VEB = VE – VB
VCB = VC – VB
VEC = VE – VC
= VEB - VCB
VBE = VB – VE
VBC = VB – VC
VCE = VC – VE
= VCB - VEB
• The convention used in the textbook does not follow IEEE
convention (currents defined as positive flowing into a terminal)
Spring 2003
EE130 Lecture 15, Slide 3
• We will follow the convention used in the textbook
Charge Transport in a BJT
• Consider a reverse-biased pn junction:
– Reverse saturation current depends on rate of
minority-carrier generation near the junction
⇒ can increase reverse current by increasing rate of
minority-carrier generation:
¾Optical excitation of carriers
¾Electrical injection of minority carriers into the
neighborhood of the junction
Spring 2003
EE130 Lecture 15, Slide 4
2
PNP BJT Operation (Qualitative)
“Active Bias”: VEB > 0 (forward bias), VCB < 0 (reverse bias)
ICn
“Collector”
“Emitter”
“Base”
ICp
Spring 2003
EE130 Lecture 15, Slide 5
β dc ≅
IC
IB
BJT Design
• Important features of a good transistor:
– Injected minority carriers do not recombine in the
neutral base region
– Emitter current is comprised almost entirely of carriers
injected into the base (rather than carriers injected into
the emitter
Spring 2003
EE130 Lecture 15, Slide 6
3
Base Current Components
The base current consists of majority carriers supplied for
1.
2.
3.
Recombination of injected minority carriers in the base
Injection of carriers into the emitter
Reverse saturation current in collector junction
•
4.
Reduces | IB |
Recombination in the base-emitter depletion region
Spring 2003
EE130 Lecture 15, Slide 7
Circuit Configurations
Spring 2003
EE130 Lecture 15, Slide 8
4
Modes of Operation
Common-emitter output characteristics
(IC vs. VCE)
Spring 2003
EE130 Lecture 15, Slide 9
BJT Electrostatics
• Under normal operating conditions, the BJT may be
viewed electrostatically as two independent pn junctions
Spring 2003
EE130 Lecture 15, Slide 10
5
BJT Performance Parameters (PNP)
• Emitter Efficiency:
• Base Transport Factor:
I Ep
I Ep + I En
γ=
αT =
– Decrease (5) relative to (1+2)
to increase efficiency
– Decrease (1) relative to (2)
to increase transport factor
• Common-Base d.c. Current Gain:
Spring 2003
I Cp
I Ep
EE130 Lecture 15, Slide 11
α dc ≡ γα T
Collector Current (PNP)
•
The collector current is comprised of
•
Holes injected from emitter,
which do not recombine in the base ← (2)
•
Reverse saturation current of collector junction ← (3)
I C = α dc I E + I CB 0
where ICB0 is the collector current
which flows when IE = 0
I C = α dc (I C + I B ) + I CB 0
IC =
α dc
I
I B + CB 0
1 − α dc
1 − α dc
= βI B + I CE 0
Spring 2003
• Common-Emitter d.c.
Current Gain:
β dc = 1−αα
dc
EE130 Lecture 15, Slide 12
dc
6
Notation (PNP BJT)
NE = NAE
DE = DN
τE = τn
LE = LN
nE0 = np0 = ni2/NE
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NB = NDB
DB = DP
τ B = τp
LB = LP
pB0 = pn0 = ni2/NB
NC = NAC
DC = DN
τ C = τn
LC = L N
nC0 = np0 = ni2/NC
EE130 Lecture 15, Slide 13
Ideal Transistor Analysis
• Solve the minority-carrier diffusion equation in each quasi-neutral
region to obtain excess minority-carrier profiles
– different set of boundary conditions for each region
• Evaluate minority-carrier diffusion currents at edges of depletion
regions
∆nE
I En = −qADE ddx
"
I Cn = qADC d∆dxn'C
x "= 0
x '= 0
I Ep = −qADB d∆dxpB
I Cp = −qADB d∆dxpB
x =0
x =W
• Add hole & electron components together Æ terminal currents
Spring 2003
EE130 Lecture 15, Slide 14
7
Emitter Region Formulation
• Diffusion equation:
0 = DE
d 2 ∆n E
dx "2
− ∆τnEE
• Boundary Conditions:
∆nE ( x" → ∞) = 0
∆nE ( x" = 0) = nE 0 (e qVEB / kT − 1)
Spring 2003
EE130 Lecture 15, Slide 15
Base Region Formulation
• Diffusion equation:
0 = DB
d 2 ∆p B
dx 2
− ∆τpBB
• Boundary Conditions:
∆pB (0) = pB 0 (e qVEB / kT − 1)
∆pB (W ) = pB 0 (e qVCB / kT − 1)
Spring 2003
EE130 Lecture 15, Slide 16
8
Collector Region Formulation
• Diffusion equation:
0 = DC
d 2 ∆nC
dx '2
− ∆τnCC
• Boundary Conditions:
∆nC ( x' → ∞) = 0
∆nC ( x' = 0) = nC 0 (e qVCB / kT − 1)
Spring 2003
EE130 Lecture 15, Slide 17
Current Formulation
∆nE
I En = −qADE ddx
"
I Ep = −qADB d∆dxpB
I Cp = −qADB d∆dxpB
I Cn = qADC d∆dxn'C
Spring 2003
x "= 0
x =0
x =W
x '= 0
EE130 Lecture 15, Slide 18
9
Emitter Region Solution
• The solution of 0 = DE
d 2 ∆n E
dx "2
− ∆τnEE is:
∆nE ( x" ) = A1e − x"/ LE + A2 e x"/ LE
• From the boundary conditions: ∆nE ( x" → ∞) = 0
∆nE ( x" = 0) = nE 0 (e qVEB / kT − 1)
qV / kT
− x "/ LE
we have: ∆nE ( x" ) = nE 0 (e EB − 1)e
and:
Spring 2003
I En = qA DLEE nE 0 (e qVEB / kT − 1)
EE130 Lecture 15, Slide 19
Collector Region Solution
• The solution of 0 = DC
d 2 ∆nC
dx '
2
− ∆τnCC is:
∆nC ( x' ) = A1e − x '/ LC + A2 e x '/ LC
• From the boundary conditions: ∆nC ( x' → ∞) = 0
∆nC ( x' = 0) = nC 0 (e qVCB / kT − 1)
qV / kT
− x '/ L
• we have: ∆nC ( x' ) = nC 0 (e CB − 1)e C
and:
Spring 2003
I Cn = −qA DLCC nC 0 (e qVCB / kT − 1)
EE130 Lecture 15, Slide 20
10
Base Region Solution
• The solution of
0 = DB
d 2 ∆n B
dx 2
− ∆τpBB is:
∆p B ( x) = A1e − x / LB + A2 e x / LB
qV / kT
• From the boundary conditions: ∆pB (0) = pB 0 (e EB − 1)
∆p B (W ) = p B 0 (e qVCB / kT − 1)
we have:
∆pB ( x ) = pB 0 ( e qVEB / kT − 1)
+ pB 0 ( e qVCB / kT − 1)
Spring 2003
(
(
e ( W − x ) / LB − e − ( W − x ) / LB
eW / LB − e −W / LB
e x / LB − e − x / L B
eW / LB − e −W / LB
EE130 Lecture 15, Slide 21
( )
• Now, we know sinh ξ =
• Therefore, we can write:
eξ − e − ξ
2
∆pB ( x) = pB 0 (e qVEB / kT − 1)
+ pB 0 (e qVCB / kT − 1)
as
∆pB ( x) = p B 0 (e
+ p B 0 (e
Spring 2003
)
)
qVCB / kT
(
(
e ( W − x ) / LB − e − ( W − x ) / LB
eW / LB − e −W / LB
e x / LB − e − x / LB
eW / LB − e −W / LB
qVEB / kT
)
)
sinh [(W − x ) LB ]
− 1)
sinh (W LB )
sinh [ x LB ]
− 1)
sinh (W LB )
EE130 Lecture 15, Slide 22
11
• We know cosh (ξ ) =
• Therefore, we have:
I Ep = qA DLBB pB 0
and:
I Cp = qA DLBB pB 0
eξ + e − ξ
2
[
(e qVEB / kT − 1) − sinh(W1 / LB ) e qVCB / kT − 1
[
(e qVEB / kT − 1) − sinh(W / LBB )) e qVCB / kT − 1
cosh(W / LB )
sinh(W / LB )
cosh(W / L
1
sinh(W / LB )
Spring 2003
(
)]
(
)]
(
(e
)]
− 1)]
EE130 Lecture 15, Slide 23
Terminal Currents
• We know:
I En = qA DLEE nE 0 (e qVEB / kT − 1)
[
p [
I Ep = qA DLBB pB 0
I Cp = qA DLBB
cosh(W / L B )
sinh(W / L B )
(e qVEB / kT − 1) − sinh(W1 / LB ) e qVCB / kT − 1
1
B 0 sinh(W / LB )
cosh(W / L
(e qVEB / kT − 1) − sinh(W / LBB ))
qVCB / kT
I Cn = −qA DLCC nC 0 (e qVCB / kT − 1)
• Therefore:
[(
= qA[(
)(e
DE
LE
nE 0 +
IC
DB
LB
pB 0 sinh(W1 / LB ) (e qVEB / kT − 1) −
Spring 2003
DB
LB
cosh(W / L B )
I E = qA
pB 0
sinh(W / L B )
)
qV EB / kT
(
DC
LC
− 1) −
(
nC 0 +
)]
)(
DB
LB
pB 0 sinh(W1 / LB ) e qVCB / kT − 1
DB
LB
pB 0
cosh(W / L B )
sinh(W / L B )
)(e
qVCB / kT
)]
−1
EE130 Lecture 15, Slide 24
12
Simplification
• In real BJTs, we make W << LB for high gain.
Then, since
sinh (ξ ) → ξ for ξ << 1
cosh (ξ ) → 1 + ξ2 for ξ << 1
2
we have:
∆pB ( x) ≅ pB 0 (e qVEB / kT − 1)(1 − Wx )
+ p B 0 (e qVCB / kT − 1)(Wx )
Spring 2003
EE130 Lecture 15, Slide 25
Performance Parameters (Active Mode)
1
Assumptions:
γ =
n
D N W
1+ n D N L
• emitter junction forward
iE 2
iB
αT =
α dc =
β dc =
2
E
B
B
E
E
1 + 12
( )
W 2
LB
• W << LB
1
1+
ni E 2 D N W
E
B
ni B 2 DB N E LE
1
ni E 2 D N W
E
B
ni B 2 DB N E LE
Spring 2003
biased, collector junction
reverse biased
1
+ 12
+ 12
( )
W 2
LB
( )
W 2
LB
EE130 Lecture 15, Slide 26
13