ELE 2110A Electronic Circuits
Week 4: Bipolar Junction Transistor
ELE2110A © 2008
Lecture 04 - 1
Topics to cover …
z
Physical Operation and I-V Characteristics
z
BJT Circuit Models
z
DC Analysis
z
BJT Biasing
z
Reading Assignment:
Chap 5.1-5.3, 5.5-5.9, 5.11, 10.1-10.3 of Jaeger & Blalock, or
Chap 5.1-5.5 of Sedra & Smith
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Lecture 04 - 2
Bipolar Junction Transistor (BJT)
z
npn transistor:
z
3 terminals:
– emitter, base, and collector
z
2 pn junctions:
– emitter-base junction (EBJ)
– collector-base junction (CBJ)
Emitter: heavily doped
z Base: very thin
z
Vertical npn
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Lecture 04 - 3
Regions of Operation
z
Depends on the biasing across each of the junctions,
different regions of operation are obtained:
Base-Collector Junction
Base-Emitter
Junction
Reverse Bias
Forward Bias
Forward Bias
Forward-active region Saturation region
(Active region)
(Closed switch)
(Good amplifier)
Reverse Bias
Cutoff region
(open switch)
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Reverse-active region
(Poor amplifier, rarely used)
Lecture 04 - 4
BJT in Active Region
E-field
Biasing:
E-B: Forward
C-B: Reverse
1
2
3
4
Operation:
1.
2.
3.
4.
Forward bias of EBJ causes electrons to diffuse from emitter into
base.
As base region is very thin, the majority of these electrons diffuse to
the edge of the depletion region of CBJ, and then are swept to the
collector by the electric field of the reverse-biased CBJ.
A small fraction of these electrons recombine with the holes in base
region.
Holes are injected from base to emitter region. (4) << (1).
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Lecture 04 - 5
Minority Concentration Profiles
E-field
z
Current dominated by electrons from emitter to base (by design) b/c
of the forward bias and minority carrier concentration gradient through
the base
– some recombination causes bowing of electron concentration (in the
base)
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Lecture 04 - 6
Diffusion Current Through the Base
iC
z
Diffusion of electrons through the base is set by concentration profile
at the EBJ
v /V
n p (0) = n p 0 e
z
T
Diffusion current of electrons through the base is (assuming an ideal
straight line case):
I n = AE qD n
z
BE
The collector current:
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dn p ( x )
i C = −I n
dx
= AE qD n ( −
n p (0)
W
)
(- sign: current into collector)
Lecture 04 - 7
Collector Current
z
The collector current:
iC = I S e
vBE / VT
z
z
z
where
IS =
qAE Dn n p 0
W
qAE Dn ni2
=
N AW
Note that iC is independent of collector voltage.
The current at the collector is controlled by the voltage
across the other two terminals – a voltage controlled
current source. This control is the basic transistor action.
Saturation current IS is
– inversely proportional to W and directly proportional to AE
z
Want short base and large emitter area for high currents
– dependent on temperature due to ni2 term
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Lecture 04 - 8
Base Current
1
2
3
4
Base current consists of two components: iB1 and iB2:
z iB1 , due to forward bias of EBJ, is an exponential function of vBE.
z iB2 , due to recombination, is directly proportional to the numbers of
electrons injected from the emitter, which in turn is an exponential
function of vBE.
v BE / VT
B
i ∝e
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Lecture 04 - 9
The Beta (β)
z
Can relate iB and iC by the following equation
iB =
z
iC
β
=
IS
β
e v BE / V T
β=iC/iB
–
–
–
–
β is constant for a particular transistor
On the order of 100-200 in modern devices (but can be higher)
called common-emitter current gain.
also denoted as βF.
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Lecture 04 - 10
Emitter Current
1
2
3
4
Treat transistor as a super-node, by KCL we obtain
iE = iB + iC =
∴
iC = α iE
1
β
iC + iC =
where
1+ β
β
α=
iC
β
1+ β
or iC =
or
β
1+ β
β=
iE
α
1−α
α is called common-base current gain. α < 1 but very close to 1.
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Lecture 04 - 11
Circuit Symbols and Conventions
pnp transistor structure:
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Lecture 04 - 12
I-V Characteristics
z
Collector current vs. vCE shows the BJT looks like a
current source (ideally)
– Plot only shows values where BCJ is reverse biased and so BJT
in active region
z
However, real BJTs have non-ideal effects
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Lecture 04 - 13
Early Effect
iC = I S e
vBE /VT
z
Early Effect:
–
–
–
–
z
vCE
(1+ )
VA
iC
Current in active region depends (slightly) on vCE
Account for Early effect with additional term in collector current equation
VA is a parameter for the BJT (50 to 100) and called the Early voltage
Nonzero slope means the output resistance is NOT infinite.
At low value of vCE, the CBJ becomes forward-biased and the
transistor enters the saturation region.
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Lecture 04 - 14
What causes the Early effect?
z
When VCB increases:
– depletion region of CBJ widens
– so the effective base width decreases (base-width modulation)
z
Shorter effective base width → higher dn/dx
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Lecture 04 - 15
BJT Breakdown
z
z
z
z
If reverse voltage across either of the two pn junctions in
the transistor is too large, corresponding diode will break
down.
Emitter is the most heavily doped region and collector is
the most lightly doped region.
Due to doping differences, base-emitter diode has
relatively low breakdown voltage (3 to 10 V). Collectorbase diode can be designed to break down at much
larger voltages.
Transistors must be selected in accordance with
possible reverse voltages in circuit.
ELE2110A © 2008
Lecture 04 - 16
Topics to cover …
z
Physical Operation and I-V Characteristics
z
BJT Circuit Models
z
DC Analysis
z
BJT Biasing
ELE2110A © 2008
Lecture 04 - 17
Equivalent Circuits for Active Mode
iC = β iB
iC = I S ev
BE / VT
z
Reverse saturation of DB = Is/β, because:
iB =
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iC
β
=
IS
β
e v BE / V T
Lecture 04 - 18
Equivalent Circuits for Active Mode
iC = α iE
iC = I S ev
BE / VT
z
Reverse saturation of DE = Is/α, because:
iE =
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iC
α
=
IS
α
ev
BE
/ VT
Lecture 04 - 19
Simplified Model
Simplified model:
z
Base-emitter diode is replaced by a constant voltage drop (VBE = 0.7
V) since it is biased in forward-active region.
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Lecture 04 - 20
Simplified Circuit Model for Saturation Mode
EBJ Forward-biased; CBJ Forward–biased.
0.5V
VCEsat = 0.2V
0.7V
z
Forward voltage drop VBC is small b/c collector doping level is low
(by design)
z
No expressions for terminal currents other than iC + iB = iE.
z
The ratio of IC to IB is called the forced current gain,
βˆ = I Csat / I Bsat
which is much smaller than β Æ next slide.
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Lecture 04 - 21
Forced Beta
Assume BJT works in active mode:
z Increase IB (or VI) causes
– IC to increase proportionally (IC = βIB)
– VC to drop
z
VC will drop to a point where BCJ
becomes forward-biased, then:
–
–
–
–
VCE clamps to VBE - 0.5V, or 0.2V
Further increases of IB will not increase IC
IC is said “saturated”
IC/IB becomes smaller than β and is called
“forced β”.
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Lecture 04 - 22
Simplified Circuit Model for Cutoff Mode
Cutoff Region:
EBJ Reverse-biased
CBJ Reverse-biased
i B = 0,
i E = 0,
iC = 0 :
No expression for terminal voltages.
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Lecture 04 - 23
Models for pnp Transistors
PNP model in active mode
VEB ≅ 0.7 V
PNP model in saturation mode
VEB ≅ 0.7 V
VEC ≅ 0.2 V
ELE2110A © 2008
Lecture 04 - 24
Topics to cover …
z
Physical Operation and I-V Characteristics
z
BJT Circuit Models
z
DC Analysis
z
BJT Biasing
ELE2110A © 2008
Lecture 04 - 25
Example 1
•
•
•
•
Problem: Determine the Q-point.
Given data: β = 100
Assumptions: VBE = 0.7 V
Analysis:
V = 4 − V ≅ 4 − 0.7 = 3.3 V
Assume active
mode of operation,
E
IE =
BE
VE
3.3
=
= 1 mA
RE 3.3
For active mode operation, IC = αI E
β
100
≅ 0.99
β + 1 101
∴ IC = 0.99 × 1 = 0.99 mA
Qα =
=
VC = 10 − IC RC = 10 − 0.99 × 4.7 ≅ +5.3 V
VBC= –1.3 V. The CBJ is reversebiased and the transistor is indeed in
the active mode as assumed.
IB =
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IE
1
=
≅ 0.01 mA
β + 1 101
Lecture 04 - 26
Example 2
• Problem: VB changes to 6V. Find Q-point.
• Analysis:
Assume active mode operation, we have
VE = 6 − VBE ≅ 6 − 0.7 = 5.3 V
IE =
6v
VE
5.3
=
= 1.6 mA
RE 3.3
For active mode operation,
IC = αI E = 0.99 × 1.6 ≅ 1.6 mA
VC = 10 − IC RC = 10 − 1.6 × 4.7 ≅ +2.48 V
Since VBC= 6 – 2.48 = 3.52 V, the CBJ
is forward-biased and the original
assumption of active mode operation
is INCORRECT. A Second guess must
be tried.
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Lecture 04 - 27
Example 2 (Cont.)
Assume the transistor is in saturation region.
Use the saturation model, we have
VE = 6 − VBE ≅ 6 − 0.7 = 5.3 V
IE =
VE
5.3
=
= 1.6 mA
RE 3.3
VC = VE + VCEsat ≅ +5.3 + 0.2 = 5.5 V
10 − 5.5
= 0.96 mA
4.7
I B = I E − IC = 1.6 − 0.96 = 0.64 mA
IC =
Thus the transistor is operating at a forced β of
β forced =
IC 0.96
=
= 1. 5
I B 0.64
Since βforced << normal β of 100, the
transistor is indeed saturated.
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Lecture 04 - 28
Example 3
• Problem: VB changes to 0V. Find Q-point.
• Analysis:
Assume the transistor operates in cutoff mode:
i B = 0, i E = 0,
vC = VCC
iC = 0
0v
VBE=0 V: BEJ reverse-biased
VBC=-10V: BCJ reverse-biased
Assumption is correct.
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Lecture 04 - 29
Summary: BJT DC Analysis
1.
Assume the BJT is biased in the active mode in which case VBE =
VBE(on), IB>0 and IC=βIB.
2.
Analyze the “linear” circuit with the active-mode equivalent circuit.
3.
Evaluate the resulting state of the BJT. If reverse-biasing of BCJ
(VCE>VCE(sat)) and forward-biasing of BEJ (IB>0) are true, then the
initial assumption is correct. However,
–
–
4.
If IB < 0, then the BJT is probably cutoff,
If VCE<VCE(sat), it is likely in saturation.
If the initial assumption is proven incorrect, then a new assumption
must be made and the new “linear” circuit must be analyzed. Step
3 must then be repeated.
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Lecture 04 - 30
Example 4
•
•
•
•
Problem: Determine the Q-point
Given data: β = 100
Assumptions: VBE = 0.7 V
EBJ must be forward-biased.
Analysis:
VE = VEB ≅ 0.7 V
V + − VE 10 − 0.7
IE =
=
= 4.65 mA
RE
2
Assume active-mode operation, we have
β
IC = αI E and α =
= 0.99
β +1
IC = 0.99 × 4.65 = 4.6 mA
VC = V − + IC RC = −10 + 4.6 × 1 = −5.4 V
Thus VBC= 5.4 V. The CBJ is reversebiased and the transistor is indeed in the
active mode as assumed.
I
4.65
≅ 0.05 mA
IB = E =
β + 1 101
ELE2110A © 2008
Lecture 04 - 31
Topics to cover …
z
Physical Operation and I-V Characteristics
z
BJT Circuit Models
z
DC Analysis
z
BJT Biasing
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Lecture 04 - 32
Example of Analog Electronic System:
FM Stereo Receiver
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Lecture 04 - 33
Amplification: Introduction
A periodic signal can be represented as the sum of many
individual sine waves. We consider only one component with
amplitude VS =1 mV and frequency ωS with 0 phase (signal is
used as the reference):
vs =Vs sinωst
Amplifier output is sinusoidal with same frequency but different
amplitude VO and phase θ:
vo =Vo (sinωst +θ )
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Lecture 04 - 34
Transfer characteristic of
the ideal amplifier
Voltage gain (A v ) ≡
vO
vI
iO
Current gain (A i ) ≡
iI
load power ( PL )
Power gain (A p ) ≡
input power ( PI )
Historically, electronic engineers
are used to express gain with a
logarithmic measure as:
Voltage gain in decibels = 20 log | A v | dB
Current gain in decibels = 20 log | A i | dB
Power gain in decibels = 10 log A p dB
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Lecture 04 - 35
Amplifier Saturation
z
The amplifier remains linear over only a limited range of input and
output.
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Lecture 04 - 36
The concept of Biasing
z
z
The transfer characteristic of a practical
amplifiers may not have a linear segment
around the origin.
To obtain linear amplification, one can bias
the circuit to operate at a point near the
middle of the linear segment. The point Q is
known as the quiescent point, the dc bias
point, or simply the operating point.
v I (t ) = VI + vi (t )
vO (t ) = VO + vo (t ) with vo (t ) = Av vi (t )
where
z
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Av =
dvO
dv I
at Q
The input signal must be kept sufficiently
small.
Lecture 04 - 37
Transfer Characteristic of
BJT Amplifier: Example 5
• Problem: Determine the dc voltage transfer
characteristic of the circuit for 0 < vI <5 V
• Analysis:
For v I ≤ 0.7 V, Q is cut off and v O = 5 V.
For v I > 0.7 V, Q turns on and is in the active mode, so that
iB =
v I − VBE v I − 0.7
=
RB
100kΩ
The output voltage is
v O = V + − i C RC = V + − β i B RC
⎡ v − 0.7 ⎤
v O = 5 − (100)⎢ I
⎥ 4kΩ
100
k
Ω
⎣
⎦
This equation is valid for v I ≥ 0.7 V and v O ≥ v CE (sat ) = 0.2V .
or
The input voltage for v O = 0.2 V is found to be v I = 1.9 V.
Now, for v I > 1.9 V, the transistor is in saturation.
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Lecture 04 - 38
Conceptual Bias Circuit for BJT
z
In order to create a linear amplifier, we must keep the transistor in the
active mode, establish a Q-point near the center of the active region,
and couple the time-varying signal to the base.
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Lecture 04 - 39