chapter 11 bjt static characteristics
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Semiconductor Device Physics. Chapter 11 BJT Static Characteristics. Deviations from the Ideal. Chapter 11. BJT Static Characteristics. Common base. Deviations , due to model limitations. Common emitter. Chapter 11. BJT Static Characteristics. W. I E. I C. P+. N. P. . +. - PowerPoint PPT PresentationTRANSCRIPT
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Lecture 10
Semiconductor Device Physics
Dr.-Ing. Erwin SitompulPresident University
http://zitompul.wordpress.com
2 0 1 3
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Chapter 11BJT Static Characteristics
Semiconductor Device Physics
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Deviations, due to model limitations
Deviations from the IdealChapter 11 BJT Static Characteristics
Common base
Common emitter
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dc 2
E B
B E E B
B E Edc
E B
1
12
D N W WD N L LD N L
D N W
Wx
D pB(x)
EBB0 ( 1)qV kTp e (VCB=0)
0
W
P+ N P
+ VEB
IE IC
Increasing –VCB
C dc B CE0I β I I
If –VCB increases→ W decreases→ bdc increases→ IC increases
Base-Width ModulationChapter 11 BJT Static Characteristics
Common-Emitter ConfigurationActive Mode Operation
Recalling two formulas,
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WB = xnEB + xnCB
Punch-ThroughChapter 11 BJT Static Characteristics
Punch-Through: E-B and C-B depletion regions in the base touch each other, so that W = 0.
As –VCB increases beyond the punch-through point, the E-B potential hill decreases and therefore increases the carrier injections and IC.
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Punch-through Avalanche
Increasing reverse bias of C-B junction
EC EB CB=V V V
Breakdown MechanismsChapter 11 BJT Static Characteristics
In the common-emitter configuration, for high output voltage VCE, the output current IC will increase rapidly due to the two mechanisms: punch-through and avalanche.
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pnp BJT
dc CB0C B
dc dc
α
1 α 1 α
M II I
M M
M : multiplication factor
Avalanche MultiplicationChapter 11 BJT Static Characteristics
Holes [0] are injected into the base [1], then collected by the C-B junction.
Some holes in the C-B depletion region have enough energy do impact ionization [2].
The generated electrons are swept into the base [3], then injected into the emitter [4].
Each injected electron results in the injection of IEp/IEn holes from the emitter into the base [5].
For each pair created in the C-B depletion region by impact ionization, (IEp/IEn) + 1 > bdc additional holes flow into the collector.
This means that carrier multiplication in C-B depletion region is internally amplified.
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Geometrical EffectsChapter 11 BJT Static Characteristics
Emitter area is not equal to collector area.Current does not flow in one
direction only.Series resistance.
Voltage drop occurs not only across the junction.
Current crowding.Due to lateral flow, current is
larger around emitter periphery than the collector periphery.
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diff
kT q
xE
xdiff : exponential decay constant
Graded BaseChapter 11 BJT Static Characteristics
Dopants are injected through diffusion.More or less falling exponential
distribution with distance into beneath of the semiconductor.
The doping within the base is not constant as assumed in ideal analysis.A function of position, having
maximum at E-B junction and minimum at C-B junction.
Creating a built-in electric field.The electric field enhances the
transport of minority carrier across the quasineutral width of the base. Increase of IE and IC.
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Due to recombination in emitter depletion region
Due to high level injection in base, base series resistance,
and current crowding
Gummel Plot
Figures of MeritChapter 11 BJT Static Characteristics
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E1 E2E1 E2 d p d p
qD qDdx dx
D D
Continuity of hole current in emitter
(1 polysilicon; 2 Si) Shallower slope less JP
higher g, b
E1 E2 E2 E2 E2
E1 E1
d p D d p d p
dx D dx dx
D D D
Polysilicon Emitter BJTChapter 11 BJT Static Characteristics
bdc is larger for a poly-Si emitter BJT as compared with an all-crystalline emitter BJT.
This is due to reduced dpE(x)/dx at the edge of the emitter depletion region.
Lower mp
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Summary on BJT PerformanceChapter 11 BJT Static Characteristics
High gain (bdc >> 1)One-sided emitter junction, so that emitter efficiency g 1Emitter doped much more heavily than base (NE >> NB).
Narrow base, so base transport factor aT 1.Quasi-neutral base width << minority-carrier diffusion length
(W << LB). IC determined only by IB (IC function of VCE or VCB)
One-sided collector junction, so that quasineutral base width W does not change drastically with changes in VCE or VCB.
Base doped more heavily than collector (NB > NC), W = WB – xnEB – xnCB for pnp BJT.
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Chapter 12BJT Dynamic Response Modeling
Semiconductor Device Physics
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Cut-off(OFF)
Idealized switching circuit
Saturation(ON)
Load line
Qualitative Transient ResponseChapter 12 BJT Dynamic Response Modeling
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cutoff
active saturation
saturation active
1
5
2 3
4
1
2 3 4
5
Qualitative Transient ResponseChapter 12 BJT Dynamic Response Modeling
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B BB
B
,dQ Q
idt
B B( , ) (0, ) 1x
p x t p tW
D D
B BB
B
0dQ Q
idt
B B B
0
( , ) (0, )2
W qAWQ qA p x t dx p t D D
Small
Interpreted as average lifetime of an excess
minority carrier
Charge Control RelationshipsChapter 12 BJT Dynamic Response Modeling
A pnp BJT biased in the active mode has excess minority-carrier charge QB stored in the quasineutral base.
While
In steady state,
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2
tB2
W
D
Interpreted as average time taken by minority carriers to diffuse across
the quasineutral base
BC B
( , )
x W
p x ti qAD
x
D
BC
t
Qi
(Active mode)
Base Transit Time tt Chapter 12 BJT Dynamic Response Modeling
B BB 2
B
(0, )( / 2 )
qAD Qp t
W W D D
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C Bdc
B t
I
I
B
Ct
Qi
BB
B
Q
i
The lifetime of a minority carrier before it recombines in the base is much longer than the time it requires to
cross the quasineutral base region
Relationship between tB and ttChapter 12 BJT Dynamic Response Modeling
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2
tB
,2
W
D B N n
kTD D
q
2
t
n2
W
kTq
From Figure 3.5, mn 801 cm2/(Vs)
ExampleChapter 12 BJT Dynamic Response Modeling
Given an npn BJT with W = 0.1 μm and NB = 1017cm-3. Find tt.
5 2
2
(10 cm)2.414 ps
2(25.86mV)(801cm /V s)
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B BBB
B
dQ QI
dt
BB BB B( ) tQ t I Ae
BB BB B( ) (1 )tQ t I e
BB BB Br
t tC
CCCC r
L
( )(1 ) for 0
( ) for
tQ t Ie t t
i tV
I t tR
SB BB
S
,V
i IR
tr : rise time, period of active mode
Turn-On TransientChapter 12 BJT Dynamic Response Modeling
During the turn-on transient:
The general solution is:
Initial condition: QB(0)=0, since transistor is in cutoff:
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CC tr B
BB B
1ln
1ItI
IBBtB > ICCtt
bdc IBB > ICC
bdc > ICC/IBB
bdc > bdc(saturation)
Turn-On TransientChapter 12 BJT Dynamic Response Modeling
In saturation mode:
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During the turn-off transient:
B BBB
B
dQ QI
dt
B/B BB B( ) tQ t I Ae
B/B BB B( ) (1 ) tQ t I e
B
CC sd/
BB BC Bsd
t t
for 0
(1 )( ) ( ) for
t
I t t
I ei t Q tt t
SB BB
S
,V
i IR
tsd : storage delay time
Turn-Off TransientChapter 12 BJT Dynamic Response Modeling
The general solution is:
Initial condition: QB(0)=IBBtB:
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CC tsd B
BB B
1ln It
I
Turn-Off TransientChapter 12 BJT Dynamic Response Modeling
The transient speed of a BJT depends on the amount of excess minority-carrier charge stored in the base and also the recombination lifetime tB.
By reducing tB, the carrier removal rate is increased, for example by adding recombination centers (Au atoms) in the base.
Tradeoff: bdc= tB/tt will decrease.
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Collector current (actual form)
Collector current (mathematical model)
Practical ConsiderationsChapter 12 BJT Dynamic Response Modeling
The foregoing analysis was highly simplified to avoid excessive amount of mathematics.
More realistic iC transient response is shown below.Added delay time td (due to charging of junction capacitance)
and fall time tf.
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Homework 8
Due: Monday, 02.12.2013.
1. (E11.36-B)The base of a silicon npn bipolar transistor is WB = 0.8 μm wide. The doping concentrations are NB = 2×1016 cm–3 and NC = 1015 cm–3. Determine the punch-through voltage.
Chapter 12 BJT Dynamic Response Modeling
3.Next slide.
2. (E11.37-B)The base impurity doping level is NB = 3×1016 cm–3 and the base width is WB = 0.7 μm. If the required minimum punch-through voltage is determined to be Vpt = 70 V, calculate the highest level of collector doping.
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Homework 8
Further data are given as follows: NE = 1019 cm–3, NB = 1017 cm–3, NC = 1015 cm–3, DE = 2 cm2/s, DB = 20 cm2/s, DC = 12 cm2/s, τE= 10–7 s, τB= τC= 10–6
3.(EIE130.AF00F)Consider a pnp silicon BJT of area A = 10–6 cm2 at 300 K. It operates in active region with VEB = 0.6 V and VCB = –1 V, so that WB = 0.6 μm. Assume that the emitter and collector regions are much longer than the respective minority carrier diffusion length.
Chapter 12 BJT Dynamic Response Modeling
a. What is the common-emitter d.c current gain of this transistor?
b. What are the excess minority-carrier concentrations at the edges of the depletion regions (at A, B, C, and D in the diagram above).
c. Calculate the collector current.