Sedra/Smith
Microelectronic Circuits 6/E
Chapter 4-A
Bipolar Junction Transistors (BJTs)
S. C. Lin, EE National Chin-Yi University of Science and Technology
1
【Outline】
4.1 Device Structure and Physical Operation
4.2 Current－Voltage Characteristics
4.3 BJT Circuits at DC
4.4 Applying the BJT in Amplifier Design
4.5 Small-Signal Operation and Models
4.6 Basic BJT Amplifier Configurations
4.7 Biasing in BJT Amplifier Circuits
4.8 Discrete-Circuit BJT Amplifiers
4.9 Transistor Breakdown and Temperature Effects
S. C. Lin, EE National Chin-Yi University of Technology
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4.1 Device structure and physical operation
Emitter and collector regions having identical physical dimensions
(C > E > B) and doping concentrations (E > C > B)
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BJT modes operation
Mode
Cutoff
Active
Reverse active
Saturation
EBJ
Reverse
Forward
Reverse
Forward
CBJ
Reverse
Reverse
Forward
Forward
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Figure 5.3 Current flow in an npn transistor biased to operate in
the active mode. (Reverse current components due to drift of
thermally generated minority carriers are not shown.)
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n p (0) = n p 0eVBE / VT = 103 e0.7 / 0.025 highly
n p (W ) = n p 0eVCB / VT = 103 e −1/ 0.025 ≅ 0
Figure 5.4 Profiles of minority-carrier concentrations in the base and
in the emitter of an npn transistor operating in the active mode: vBE > 0
and vCB ≥ 0.
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According to the law of the junction (sec. p.61, eq.1.57)
the concentration n p (0) will be propotional to evBE / VT
n p (0) = n p 0 evBE / VT
(4.1)
where n p 0 is the thermal- equilibrium value of the
minority-carrier (electron) concentration in the base region
VT is the thermal-voltage (VT ≈ 25mV)
This electron diffusion current (ref.1.44) I n as follows :
I n = AE qDn
dn p ( x)
dx
⎛ n p (0) ⎞
= AE qDn ⎜ −
⎟
W
⎝
⎠
n p (0)
np
(4.2)
where AE is the cross-sectional area of B-E junction,
W
q is the magnitude of the base,
Dn is the electron diffusivity in the base,
W is the effective width of the base,
I n flows from right to left(in the negtive direction of x)
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The Collector Current
iC = − I n = AE qDn
where I s = AE qDn
n p (0)
W
np0
W
= AE qDn
n p 0evBE / VT
W
,subtituting n p 0
= I s evBE / VT
ni 2
=
NA
ni 2
⇒ I s = AE qDn
N AW
(4.4)
The I s is inversely propotional to the base width W and is directly
propotional to the area of the EBJ.
Typically I s is in the range of 10−12 A to 10−18A
(depending on the size of the device)
The I s is propotional to ni 2 , it is a strong function of temperature,
approximately doubling for every 5oC rise in temperature.
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The Base Current
iB1 =
Dn ⋅ AE qD p ni 2 ⋅ N AW
N AW ⋅ N D Lp ⋅ Dn
evBE / VT
D p N AW vBE / VT
Dn AE qni 2 D p N AW vBE / VT
=
⋅
e
= Is
e
N AW
N D Lp Dn
N D Lp Dn
Is
Lp is the hole diffusion length in the emitter
iB2 =
Qn
τb
, τ b is minority-carrier lifetime,
Qn is replenished by electron injection from the emitter
1
⇒ Qn = AE q ⋅ n p (0)W .
2
subtituting n p (0) = n p 0 e
vBE / VT
and n p 0
ni 2
AE qWni 2 vBE /VT
e
=
, gives Qn =
2N A
NA
1 AE qWni 2 vBE / VT 1 Dn AE qW 2 ni 2 vBE / VT 1 Dn AE qni 2 W 2 vBE / VT
⇒ iB2 = ⋅
e
= ⋅
e
= ⋅
⋅
e
2 N Aτ b
2 Dn N Aτ bW
2
N AW
Dnτ b
Is
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⎛ Dp N A W 1 W 2 ⎞ v /V
⋅
⋅ + ⋅
e BE T
iB = iB1 + iB2 = I s ⎜
⎟
⎜ D N L 2 Dτ ⎟
D
p
n b ⎠
⎝ n
iC ⎛ is ⎞ vBE / VT
iB = = ⎜ ⎟ e
(4.6)
β ⎝β ⎠
1
where β =
⎛ Dp N A W 1 W 2 ⎞
⋅
⋅ + ⋅
⎜⎜
⎟⎟
⎝ Dn N D L p 2 Dnτ b ⎠
The β is called the common- emitter current gain,
For moden npn transistor, β is in the range 50 to 200,
but it can be as high as1000 for special devices
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The Emitter current
iE = iB + iC
=
1+ β
β
iC =
1+ β
β
is evBE / VT
(4.9)
β
α
, β=
∵α =
1+ β
1−α
∴ iE =
is
α
evBE / VT
(4.12)
The α is common-base current gain,
that is less than but very close to unity.
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11
iC
iC
I s evEB / VT
iB
vBE
−
DE
( I SE / α )
α F iE
iB
iE
vBE
−
iE
DE
( I SE / α )
iE
(a)
(b)
Figure 4.5 Large-signal equivalent-circuit models of the npn BJT
operating in the forward active mode.
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12
iC
iB
vBE DB
− ( I SB = I S / β )
I s evBE / VT
(c)
iE
iC
iB
iB
DB
( I SB = I S / β )
vBE
−
iE
β iB
(d)
Figure 4.5 Large-signal equivalent-circuit models of the npn BJT
operating in the forward active mode.
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iC
DC
I SC = ( I S / α R )
α RiC
iE
Figure 5.7 Model for the npn transistor when operated in the reverse
active mode (i.e., with the CBJ forward biased and the EBJ reverse
biased).
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Ebers-Moll (EM) Model
iDC
iDE
iE
iC
α RiDC
iE = iDE − α RiDC ,
iB
α F iDE
iC = −iDC + α F iDE ,
iB = iE − iC
Figure The Ebers-Moll (EM) model of the npn transistor.
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The diode D C represents the collector-base junction,
the scale current is I SC
The diode D E represents the emitter-base junction,
the scale current is I SE
I SC >> I SE
α R is in the range of 0.01 to 0.5
β R is in the range of 0.01 to 0.1
α F I SE = α R I SC = I S
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iDE = I SE ( e
(
vBE / VT
iDC = I SC e
vBC / VT
− 1) =
)
−1 =
IS
vBE / VT
e
− 1)
(
(a)
IS
(e
(b)
αF
αR
vBC / VT
)
−1
The transistor terminal current equtions:
iE = iDE − α R iDC
(c)
iC = −iDC + α F iDE
(d)
iB = iE − iC = (1 − α F )iDE + (1 − α R )iDC
(e)
Subtituting (a) and (b) into (c),(d)and(e), we have
S. C. Lin, EE National Chin-Yi University of Technology
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iE =
IS
αF
(
(
iC = I S e
iB =
IS
βF
(
(
)
)
evBE / VT − 1 − I S evBC / VT − 1
vBE / VT
e
)
−1 −
vBE / VT
)
IS
αR
−1 +
αF
where β F =
,
1−αF
(
IS
βR
)
evBC / VT − 1
(
)
evBC / VT − 1
αR
βR =
1−αR
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Application of the EM model
(A) Opoerating in the forward avtive mode
iE =
IS
αF
(
(e
iC = I S e
iB =
IS
βF
vBE / VT
vBE / VT
(e
⎛ v /V
⎞ IS v /V
⎛
1 ⎞
BC
T
BE
T
e
−1 − IS ⎜ e
− 1⎟ ≅
+ I S ⎜1 −
⎟
⎜
⎟ α
α
⎝
F
F ⎠
⎝ 0
⎠
)
⎞
⎛ 1
⎞
I S ⎛ vBC / VT
vBE / VT
−1 −
− 1⎟ ≅ I S e
+ IS ⎜
− 1⎟
⎜⎜ e
⎟
αR ⎝ 0
⎝ αR ⎠
⎠
vBE / VT
)
⎞ IS v /V
⎛ 1
I S ⎛ vBC / VT
1 ⎞
BE
T
－I S ⎜
e
−1 −
− 1⎟ ≅
−
⎜⎜ e
⎟
⎟ β
βR ⎝ 0
β
β
⎝ F
F
R ⎠
⎠
)
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4.1.4 Operation in the Saturation Mode
VCC
β forced I B
RC
IB
VBC
VCE
VBE
sat
evBE / VT >> 1
evBC / VT >> 1
iC = β forced I B
βR
αR =
1 + βR
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The EM expression for iC
iC = I S ( e
− 1) −
vBE / VT
≈ I S evBE / VT
iC = I S e
vBE / VT
−
X
β forced iB = X −
1
αR
1+ βR
βR
1+ βR
βR
(
)
I S evBC / VT − 1
≈ I S evBC / VT
I S evBC / VT
Y
Y ⇒ iB =
X
β forced
The EM expression for iB ⇒ iB =
−
X
βF
1+ βR
β R β forced
－
Y
Y
βR
S. C. Lin, EE National Chin-Yi University of Technology
(a)
(b)
21
(a ) = (b)
X
1 + βR
X
Y
⇒
−
－
Y=
β forced β R β forced
βF βR
β F − β forced
1 + β R + β forced
X =
Y
β F β forced
β R β forced
1 + β R + β forced
X
=
Y
βR
β F − β forced
βF
=
(1 + β R + β forced ) β F
( β F − β forced ) β R
X I S evBE / VT
( vBE − vBC ) / VT
vCE / VT
∵ =
=
=
e
e
Y I S evBC / VT
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⇒ VCE ( sat ) = VT
1+ β + β
β
(
)
ln
(β − β ) β
R
F
VCE ( sat ) = VT ln
forced
forced
1
F
⋅
R
βF βR
1
βF βR
1 + ⎡⎣(1 + β forced ) / β R ⎤⎦
1 − ( β forced / β F )
The VCE ( sat ) for the case β F = 50, and β R = 0.1
β forced
vCE ( sat )
50
48
45
40
30
20
10
0
∞
235
211
191
166
147
123
60
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Figure 4.8 The iC –vCB characteristic of an npn transistor fed with a
constant emitter current IE. The transistor enters the saturation mode of
operation for vCB < –0.4 V, and the collector current diminishes.
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Figure 4.10 Current flow in a pnp transistor biased to operate in the
active mode.
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25
iE
vBE
iB
−
( IS /αF )
iE
DE
I s evEB / VT
vBE
iB
−
DB
( I SB = I S / β )
I s evBE / VT
iC
iC
Figure 4.11 Large-signal model for the pnp transistor operating in the
active mode.
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26
4.2 Current－Voltage Characteristics
4.2.1 Circuit symbols and Conventions
Figure 4.13 Voltage polarities and current flow in transistors biased in
the active mode.
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The constant n
The constant n , its value is between 1 and 2.
For modern BJT the constant n is close to unity except in special cases:
XAt high currents, the iC - vBE relationship exhibits a value for
n that is close to 2 .
YAt low currents, the iB - vBE relationship shows a value for
n approximately 2 .
The Collector-Base Reverse Current
( I CBO )
X The current I CBO is the reverse current flowing from collector
to base with the emitter open-circuited.
Y The current I CBO depends strongly on temperature,
approximately doubling for 100C rise.
I CBO 2 = I CBO1 ⋅ 2(T2 −T1 ) /10
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Example 4.2 The transistor in the circuit of Fig (a) has β = 100 and
exhibits vBE of 0.7V at iC = 1mA. Design the circuit so that a current
of 2mA flows through the collector and a voltage of +5V appear at
the collector.
Sol :
10V
= 5kΩ ▲
2mA
⎛ I2 ⎞
= 0.7 + VT ln ⎜ ⎟ = 0.717V
⎝ I1 ⎠
RC =
VBE
VE = −0.717V
β = 100 ⇒ α = 0.99
I E = I C / α = 2mA / 0.99 = 2.02mA
VE − (−15V )
RE =
= 7.07kΩ ▲
IE
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4.4.2 Graphical Representation of Transistor Characteristic
iC = I S evBE / VT
VT ∝ T ∴ T ↑ VT ↑ iC ↓
Figure (a) The iC–vBE characteristic for an npn transistor.
(b) Effect of temperature on the iC–vBE characteristic. At a constant
emitter current (broken line), vBE changes by –2 mV/°C.
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30
4.2.3 Dependence of ic on the Collector Voltage–The Early effect
iC (mA)
α ×5
5mA
α ×4
4mA
α ×3
3mA
α ×2
2mA
α ×1
iE = 1mA
vCB (V)
iC = I S evBE / VT
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WB
Vo − VEB
Vo
VEB
VCB
WB '
W
Early effect has three consequences:
(1) α increases with increasing VCB
(2) I C increases with increasing reverse collector voltage.
(3) punch through
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∆iC
iC '
vCE
iC′ + ∆iC
∆iC
=
VA + vCE
vCE
iC′ vCE + ∆iC vCE = ∆iCVA + ∆iC vCE ⇒ ∆iC =
vCE
iC′
VA
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The collector current ( operation in the active mode)
neglected the Early effect: iC ' = I S evBE / VT
vCE
including the Early effect: iC = iC + ∆iC =iC + iC
VA
'
= IS e
'
vBE / VT
'
⎛
vCE ⎞
⎜1+
⎟
V
⎝
A ⎠
The nonzero slopeof the iC − vCE straight line indicates
that the output resistance looking into the collector
is not infinite. Rather, it is finite and defined by
⎡ ∂i
ro ≡ ⎢ C
⎢⎣ ∂vCE
−1
⎤
VA + VCE VA
⎥ =
= '
IC
IC
vBE = cons tan t ⎥
⎦
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34
iC
iB
vBE
−
I s evBE / VT
DB
( IS / β )
ro
(a)
iE
iC
iB
iB
DB
( IS / β )
vBE
−
iE
β iB
ro
(b)
Figure 4.18 Large-signal equivalent-circuit models of an npn BJT operating
in the active mode in the common-emitter configuration. Those in Fig.5.5(c)
and (d), P.13, with the resistance ro connected between the C and E terminals
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35
4.2.4 An Alternative Form of The Common-Emitter Characteristics
Common-Emitter Current Gain
β dc ≡
β ac ≡
I CQ
I BQ
(hFE ),
∆iC
(h fe )
∆iB
Figure 4.19(a)(b) Common-emitter characteristics. Note that the
horizontal scale is expanded around the origin to show the saturation
region in some detail.
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36
The saturation voltage VCEsat and saturation resistance RCEsat
I Csat < β F I B
β forced ≡
I Csat
IB
β forced < β F
RCEsat
∂VCE
≡
∂iC
iB = I B
iC = I Csat
Figure 4.19(c) An expanded view of the common-emitter
characteristics in the saturation region.
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37
VCE sat = VCEoff + I Csat RCEsat
Typically, VCE sat ⇒ 0.1 V to 0.3V
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38
Example 4.3 For the circuit in Fig.4.21 has RB = 10kΩ and RC = 1kΩ,
it is required to determine the value of the voltage VBB that results the
transistor operating (a) in the active mode with VCE = 5V, (b) at the edge
of saturation, (c) deep in saturation with β forced = 10
For simplicity, assume that VBE = 0.7V. The transistor β = 50
VCC = 10V
Solution:
VBB
RB
IB
RC
IC
VBC
I B = I C / β = 5mA / 50 = 0.1mA
VCE
VBE
VCC − VCE (10 − 5 ) V
IC =
=
= 5mA
1kΩ
RC
VBB = I B RB + VBE
= 0.1×10 + 0.7 = 1.7V
▲
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39
(b) I C =
VCC − VCEsat
10 − 0.3) V
(
=
= 9.7mA
1kΩ
RC
I B = I C / β = 9.7mA / 50 = 0.194mA
VBB = I B RB + VBE = 0.194 × 10 + 0.7 = 2.64V
▲
(c) VCE = VCEsat ≈ 0.2V
IC =
IB =
VCC − VCEsat
RC
IC
β forced
10 − 0.2 ) V
(
=
= 9.8mA
1kΩ
9.8mA
=
= 0.98mA
10
the requred VBB can now be found as
VBB = I B RB + VBE = 0.98 × 10 + 0.7 = 10.5V ▲
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40
4.3 BJT circuits at DC
Example 4.4 Consider the circuit shown in below Fig. We wish to
analyze this circuit to determine all node voltages and branch
currents.We will assume that β is specified to be 100
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41
S. C. Lin, EE National Chin-Yi University of Technology
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Example 4.5 We wish to analyze this circuit below Fig. to determine
all node voltages and branch currents.We will assume that β is
specified to be at least 50.
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43
β forced
I C 0.96
=
=
= 1.5
I B 0.64
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Example 4.6 We wish to analyze this circuit below Fig., to determine
all node voltages and branch currents.
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Example 4.7 We desire to analyze this circuit below Fig., to determine
the voltages at all nodes and the currents though all branchs .Assume
β = 100
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46
Example 4.8 We want to analyze the circuit in below Fig., to determine
the voltages at all nodes and the currents in all branchs .Assume β = 100
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47
Example 4.9 We want to analyze the circuit in below Fig., to determine
the voltages at all nodes and the currents in all branchs .The minimum
value of β is specified to be 30.
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48
We want to analyze the circuit in below Fig., to
determine the voltages at all nodes and the currents though all branchs .
Assume β = 100
Example 4.10
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S. C. Lin, EE National Chin-Yi University of Technology
50
Example 4.12 We desire to evaluate the voltages at all nodes and the
currents though all branchs in the circuit of below Fig., Assume β = 100
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4.4 Applying the BJT in Amplifier Design
4.4.1 Obtaining a Voltage Amplifier
VCC
vCE
Cut Off
iC
vBE +
−
VCC
RC
X
Y
+
vo = vCE
−
Active
mode
Saturation
Edge of
Saturation
Z
≅ 0.3V
0
0.5V
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vBE
52
4.4.2 The Voltage Transfer Characteristic (VTC)
vCE
Cut Off
VCC
X
Y
Active
mode
Saturation
iC = I S evBE / VT
vCE = VCC − iC RC
Edge of
Saturation
= VCC − RC I S evBE / VT
Z
≅ 0.3V
0
0.5V
vBE
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53
4.4.3 Biasing the BJT to Obtain Linear Amplification
vCE
VCC
iC
VCC
RC
+
vBE +
−
vo = vCE
−
VCE
X
Y
Q
Z
VBE
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vBE
54
vCE
VCC
iC
RC
RB
vi +
−
VBB
iB
−
Figure 4.33(a)
IC = I S e
VCC X
+
vCE
VBE
Active
mode
Cut Off
Y
Saturation
Slop = A v
Q
≅ 0.3V
Z
0
vBE / VT
Time
0.5V VBE
vBE
vCE = VCC − RC I S evBE / VT
vBE (t ) = VBE + vbe (t )
Time
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4.4.4 The Small-Signal Voltage Gain
dVCE
Av ≡
dVBE
(4.29)
vBE =VBE
⎛ IC ⎞
dVCE
RC
vBE / VT
ISe
=−
⇒ Av = − ⎜ ⎟ RC
dVBE
VT
⎝ VT ⎠
VRC
I C RC
Av = −
, where VRC = VCC − VCE
=−
VT
VT
(4.30)
(4.32)
The observations on this expression for the voltage gain:
The gain is negative, which signifies that the amplifier is inverting
; that is, there is a 180o phase shift between the input and the output.
The gain is propotional to the collector bias current I C and to the load
resistance RC .
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56
Example 4.13 Consider an amplifier circuit using a BJT having
I s = 10−15 A, a collector resistance RC = 6.8kΩ, and a power supply
VCC = 10V.(a) Find VBE ,and I C , such that the BJT operate at VCE = 3.2V
(b)Find the voltage gain Av at this bias point. If vi = vbe = 5sin ωt (mV)
find vo .(c)Find the positive increment in vBE that drive the transistor to
the edge of saturation(vCEsat = 0.3V).(d)Find the negative increment in
vBE that drive the transistor to the within 1% of cut-off.
Solution:
VCC − VCE (10 − 3.2 ) V
=
= 1mA ▲
(a) I C =
6.8kΩ
RC
I C = I S evBE / VT ⇒ vBE = VT ln ( I C / I S ) = 690.8 mV ▲
VCC − VCE (10 − 3.2 ) V
=
= −272V/V ▲
(b) Av =
0.025V
VT
v = Vˆ = 272 × 0.005sin ω t = 1.36sin ω t (V ) ▲
o
ce
S. C. Lin, EE National Chin-Yi University of Technology
57
(c) For vCEsat = 0.3V
VCC − VCE (10 − 0.3) V
iC =
=
= 1.617mA
6.8kΩ
RC
To increase iC from 1mA to 1.617mA, vBE must be increase by
∆vBE
⎛ IC 2 ⎞
⎛ 1.617 ⎞
= VT ln ⎜
⎟ = 0.025ln ⎜
⎟ = 12mV ▲
⎝ 1 ⎠
⎝ I C1 ⎠
(d ) For vCE = 0.99VCC = 9.9V
VCC − vCE (10 − 9.9 ) V
=
= 0.0147mA
iC =
RC
6.8kΩ
To decrease iC from 1mA to 0.0147mA, vBE must be change by
∆vBE
⎛ 0.0147 ⎞
= 0.025ln ⎜
⎟ = −105.5mV ▲
⎝ 1 ⎠
S. C. Lin, EE National Chin-Yi University of Technology
58
4.4.5 Determining The VTC by Graphical Analysis
VCC
iC
RC
RB
vi +
−
VBB
iB
VBE
+
vCE
−
Figure 4.33(a)
Figure Graphical construction for the determination of the dc base
current in the circuit of Fig.4.33(a).
S. C. Lin, EE National Chin-Yi University of Technology
59
vCE = VCC − iC RC
VCC VCE
iC =
−
RC
RC
Figure 4.34 Graphical construction for determining the dc
collector current IC and the collector-to-emitter voltage VCE in the
circuit of Fig.4.33(a).
S. C. Lin, EE National Chin-Yi University of Technology
60
4.4.6 Locating the Bias Point Q
Figure 4.35 Effect of bias-point location on allowable signal swing: Load-line A results
in bias point QA with a corresponding VCE which is too close to VCC and thus limits the
positive swing of vCE. At the other extreme, load-line B results in an operating point too
close to the saturation region, thus limiting the negative swing of vCE.
S. C. Lin, EE National Chin-Yi University of Technology
61
S. C. Lin, EE National Chin-Yi University of Technology
62