Slide 1
18 - 322 Fall 2003 Lecture 21
Bipolar Junction Transistor (BJT)
• NPN Cross-section and Masks• BJT Notation• Hand Analysis Models• NPN Modes of Operation• Ebers - Moll Model• BJT Inverter• TTL
Chapter 2.4
Slide 2
NPN Transistor Cross-section
Slide 3
NPN Transistor Layout
p
n
p
p
n
Slide 4
BipolarJunctionTransistor (BJT) Notation
Collector
Base
Emitter
nn
n
pB
C C
n-p-n transistor
C
B
ICn
IBB
p
IE
n
E EE
Slide 5
NPN Regions of Operation
V
V BE
BC
Forward Active
Reverse Active Saturation
Cut-off
Slide 6
NPN Forward Active PolarizationVBE Forward-biased &
VBC Reverse-biased
E C
Electron Flow(Current in opposite direction)B
Base current:IB = IC / βF βF = IC / IB
βF - current gain (~100!)
Slide 7
Common-Emitter Characteristics
B
VBE
VCE
C
E
C I [mA]
3
2
1
0
5
4
7
6
10
9
8
12
11
14
13I = 120µAB
VCE [V]5.04.03.02.01.0
I = 100µAB
I = 80µAB
I = 60µAB
I = 40µAB
I = 20µAB
Slide 8
Saturation Region C I [mA]
3
2
1
0
5
4
7
6
10
9
8
12
11
14
13I = 120µAB
VCE [V]5.04.03.02.01.0
I = 100µAB
I = 80µAB
I = 60µAB
I = 40µAB
I = 20µAB
• VBE Forward-biased• VBC Reverse-biased
• No Current Gain relationship.
Slide 9
Hand Analysis Model
IB IC
VCEVBE
IC (active) = βF IBIB = IS (eVbe/Vt -1)
VBE(on) = 0.7 V VCE(sat) = 0.1 V
Slide 10
Hand Analysis Example
IC
IE
RC
5V • RC = 300ΩIB =
IC =
VCE =
Operation Mode:
• RC = 1k ΩIB =
IC =
VCE =
Operation Mode:
βF=100
RB=20kΩ
IB+-
2.7V
Slide 11
BJT Parasitics
• Junction Capacitances:– Base-Emmitter– Base-Collector
• Excess Base Charge:– QR when– VBC forward-biased– Must be removed to
switch modes
CCSQR Cbc S
QF Cbe
Slide 12
NPN Transistor Doping Levels
Depletion layers
BE C
nn p
1020
1017cm
-3
2 103
N A10
15cm
-3
2
2 105
cm-3
N DN D
Slide 13
np Junctions Revisted
• Forward-biased:– Dominant current: diffusion of majority
carriers• Reverse-biased:
– Drift of minority carriers
Slide 14
NPN Forward Active PolarizationVBE > 0 & VBC < 0
VBE VBCB
nn p
+ -
Concentration of minority
carriers
forward-biased
reverse-biased
E C+-
Slide 15
NPN Forward Active PolarizationVBE > 0 & VBC < 0
VBE VBCB
nn p
+ -
Current determined by concentration
of minority carriers
Current determined by concentration
of majority carriers
forward-biased
reverse-biased
E C+-
Slide 16
NPN Forward Active PolarizationConcentration proportional to
VBE > 0 & VBC < 0
VBE VBCB
nn p
forward-biased
+ -
VBEe VT
E C+-
Slide 17
NPN Forward Active PolarizationVBE > 0 & VBC < 0
VBE VBCB
nn p
reverse-biased
+ -
Concentration proportional to
VBCVTe
E C+-
Slide 18
NPN Forward Active PolarizationVBE > 0 & VBC < 0
BE C
nn pCollector current
determined by the slope of the concentration
of minority carriers
(electrons) in the base.Collector current:
IC = IS [exp(VBE/VT) -1]
IS - saturation current
Slide 19
NPN Transistor Reverse ActiveVBE < 0 & VBC > 0
βF > βR
n p
reverse-biased
forward-biased n
BE C
Slide 20
NPN Transistor SaturationVBE > 0 & VBC > 0
forward-biased
forward-biased
n p n
BE C
Slide 21
NPN_Cut-off
BE C
nn p
VBE < 0 & VBC < 0
Slide 22
Ebers-Moll Modelnn
B
C
E
C
B
E
IDC
IDC
DEI
DEI
αF
αR
Slide 23
Ebers - Moll Model
Equations:IDE = IES [exp(VBE/VT) -1]IDC = ICS [exp(VBC/VT) -1]
Typical values:αF = .99 IES = 10-15 AαR = .66 ICS = 10-15 A
Slide 24
BJT Inverter & Fan-Out Analysis
• BJT Inverter
• Voltage Transfer Characteristics
• Logic Level Description
• Fan-Out Analysis
Slide 25
BJT InverterCCV
Base diffusion
= 5 VoutV CCV
out
R
V
C1 kΩ
inV = 0 V inV = 5 V
inV RB
10 k Ω100 Ω/ 10 = 1kΩ
Slide 26
NPN BJT Parameters
VBE(on) = 0.7 VVBE(sat) = 0.8 VVCE(sat) = 0.1 V
Forward active mode:• current gain βF = 70
Slide 27
VOH and VIL
V = +5 V
BE C
nn p
outV = 5 V
= 5 VCCV
RC1 k Ω
Q0
EBV = 0 V
CB
BI = 0CI = 0
inV = 0 V10 k Ω
I C
R Bcut-off
I B
EBV = 0 - 0.7 V Cut-off
Slide 28
VOH and VILCut-offVout [V]
5VOH BP1
4
3
2
1
Vin [V]0
0 1 2 3 4 5VIL0.7 V
Slide 29
VOL
BE C
nn p
outV = 0.1V
= 5 VCCV
RC1 k Ω
Q0
Saturation V = 0.1 V
EBV = 0.7 V
out
BI = (5-0.7)/10kΩ = 0.43 mACI = I βF = 30 mAB
outV = 5 - I R = -25 V ?C C
CI < I βF = 30 mAB βF < 70
NOT Forward Active !
βF = 70 in Forward Active
inV = 5 V10 k Ω
I C
RB
I B
Slide 30
VOL
Saturation
Cut-offVout [V]
5
0
VOH
VOL
VIL
BP1
0.7 V
0.1 V
4
3
2
1
Vin [V]
0 1 2 3 4 5
Slide 31
VIH
Edge of saturation:
EBsV = 0.8 V
βF = 70
outV = 0.1V
outV = 0.1V Q0
= 5 VCCV
RC1 k Ω
BI = ((5-0.1)/1kΩ)/70 = 70 µA
in
V = V + I R = 0.8 + 0.7B B EBs
V = 1.5 V
inV = ? 10 k Ω
I C
RB
I B
Slide 32
VIH
Saturation
Cut-offVout [V]
5
0
VOH
VOL
VIL VIH
BP1
BP2
0.7 V
0.1 V
1.5 V
4
3
2
1
Vin [V]
0 1 2 3 4 5
Slide 33
Transition Region
Q0
= 5 VCCV
RC1 k Ω
inV = 0.7 - 1.5VoutV = 5.0 - 0.1 V
10 k Ω
RB
Forward Active !
B
CI = I βFB
I = (V - 0.7)/R in B
outV = V - I R C C CC
Slide 34
Transition Region:Load Line Analysis
CI [mA]
I = 120µAB
VCE [V]5.01.0
I = 100µAB
I = 80µAB
I = 60µAB
I = 40µAB
I = 20µAB
14
B
CI = I βFB
I = (V - 0.7)/R in B
outV = V - I R C C CC
13
12
11
10
9
8
BI = (1.5 - 0.7)/ 10kΩ = 80 µΑ7
6
V = 0.7 - 1.5Vin5
4
3
2
1
02.0 3.0 4.0
Slide 35
Voltage Transfer Characteristic
Saturation
Forward Active
Cut-offVout [V]
5
0
VOH
VOL
VIL VIH
BP1
BP2
4
3
2
1
Vin [V]
0 1 2 3 4 5
Slide 36
Voltage Transfer Characteristic
Vout [V]
Vin [V]
5
4
3
2
1
0
0 1 2 3 4 5
VOH
VOL
VIL VIH
BP1
BP2
• Transition Width:TW= VIH - VIL = .8 V
• High Noise Margin:NMH = VOH - VIH = 3.5 V
• Low Noise Margin:NML = VIL -VOL = .6 V
• Logic Swing:LS = VOH - VOL = 4.9 V
Slide 37
Inverter Fan-Out
= 5 V
outV R B
10 k Ω
CCV
inV Q1
RC1 k Ω
Q010 k Ω
RC1 k Ω
R B
QN
RB
10 k Ω
RC1 k Ω
inV = 0 R B
n
Slide 38
Inverter Fan-Out
outV = 4.6 V
inV = 0.1 V
= 5 VCCV
R B
10 k Ω
RC1 k Ω
R B
10 k Ω
RC1 k Ω
Q0 Q1cut-offsaturation
• LOAD =1 VOH = 4.6 V
R B
BR
C +10
V = ( V CC - 0.8 ) out + 0.8 = R
( 5.0 - 0.8 ) + 0.8 = 4.61 V 10 +1
Slide 39
Equivalent Circuit
outV
= 5 VCCV
R /NB10 k Ω
RC1 k Ω
Q0
VOH = ?
R BinV V BE(sat)
R / N B
R /N B
RC +
V = ( V CC - 0.8 )
10 /10 10 /10 1 + + 0.8 = 2.9 V
out + 0.8 =
( 5.0 - 0.8 )
Slide 40
Maximum Number of Load Gates: VOH = VIH
R / N B
R /N B
RC +
V = ( V CC - 0.8 ) out
outV
inV
= 5 VCCV
R /NB
R B
10 k Ω
RC1 k Ω
Q0 V BE(sat)
VOH = ?
VIH = ?
N = ?
+ 0.8
10 /N 10 /N 1 +( 5.0 - 0.8 ) + 0.8 = 1.5 V
Slide 41
Transistor-Transistor Logic (TTL)
• Disadvantages of TTL gates:• Large # of components, area --> VLSI not
feasible• Large power consumption• Saturated transistors in either high or low state
• Large propagation times
• Advantage:• Can drive large capacitive loads since large
output currents available
Slide 42
TTL Inverter= 5 VCCV
Q 2
R4130 Ω
Q3
Q1
D1
R14 k Ω
R 31.6 k Ω
Q 4
A
R21 k Ω
INPUT STAGE PHASE-SPLITTERor
LEVEL-SHIFT
OUTPUT STAGE
Slide 43
TTL NAND Gate= 5 VCCV
Q 2
R14 k Ω
R 31.6 k Ω
R4130 Ω
Q3
D1
R21 k Ω
Q 4
INPUT STAGE OUTPUT STAGEPHASE-SPLITTERor
LEVEL-SHIFT
Q1AA
Q1B
B
Slide 44
Multi-Emitter TransistorB
pn
E1 E2 B C
E1
E2
C
E1
E2
B
C