bjt biasing & small signals
TRANSCRIPT
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Lecture 05 - 1ELE2110A © 2008
Week 5: BJT Biasing and Small Signal Model
ELE 2110A Electronic Circuits
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Lecture 05 - 2ELE2110A © 2008
Topics to cover …BJT Amplifier Biasing Circuits
Small Signal Operation and Equivalent Circuits
Reading Assignment: Chap 13.1 – 13.6 of Jaeger and Blalock , orChap 5.5 - 5.7 of Sedra & Smith
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Lecture 05 - 3ELE2110A © 2008
BJT as Amplifier: Example 1
V. and off cut is Q V,. For 570 =≤ OI vv
• Problem: Determine the dc voltage transfer characteristic of the circuit for 0 < vI <5 V• Analysis:
.saturation in is transistor the V,. for Now, V.. be to found is V . for voltage input The
919120
>
==
I
IO
vvv
..)sat( and V . for valid is equation This
kk
.)( or
is voltage output The
k.
that so mode, active the in is and on turns Q V,. For
Vvvv
vv
RiVRiVv
vR
Vvi
v
CEOI
IO
CBCCO
I
B
BEIB
I
2070
4100
701005
10070
70
=≥≥
Ω⎥⎦
⎤⎢⎣
⎡Ω
−−=
−=−=
Ω−
=−
=
>
++ β
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Lecture 05 - 4ELE2110A © 2008
Conceptual Bias Circuit for BJTKeep the transistor in the active mode;Establish a Q-point near the center of the active region;Couple the time-varying signal to the base.
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Lecture 05 - 5ELE2110A © 2008
Two Obvious Bias Circuits
Fix VBE Fix IB
Wide variations in IC and hence in VCE
“Bad” biasing schemes
IC=βIB , but β is a poorly controlled parameter
IC is an exponential function of VBE and thus VCC
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Lecture 05 - 6ELE2110A © 2008
Biasing Circuit in ERG2810 Experiment A3
Adjusting the variable resistor to give a desired amount of IB such that Vo is at about 8V (half the supply voltage).The transistor operates at the middle of active region.
BiasingIB
Vo
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Lecture 05 - 7ELE2110A © 2008
Classical Four-Resistor Bias Circuit
VB established by the voltage divider formed by R1 and R2.RE added reduce sensitivity to supply voltage, process, and temperature variations
to be discussed laterCoupling capacitor CC:– open circuit to DC, isolating the signal
source from the dc biasing current. – short circuit to AC signal (if the signal
frequency is large enough and CC is large enough).
The Q-point is usually specified by (IC, VCE) for npn transistor or (IC, VEC) for pnp transistor.
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Lecture 05 - 8ELE2110A © 2008
Example 2
21
1RR
RCCVEQV
+=
21
21RR
RREQR
+=
EIERBEVBIEQREQV ++=
µA68.261023.1V7.0-V4
)1(000,167.0000,124
=Ω×
=∴
+++=
BIBIBI β
KVL at loop 1:
Thevenin equivalent circuit at the base:
Assume BJT in active mode
Problem: find Q-point.
µA201== BICI β µA204)1( =+= BIEI β Q-point is (201 µA, 4.32 V)
V32.4=+−=
−−=
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
CIF
FRC
RCCV
EIERCICRCCVCEV
α
Given β = 200.
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Lecture 05 - 9ELE2110A © 2008
Example 2 (Cont’)All calculated currents > 0,
VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 VHence, base-collector junction is reverse-biased, assumption of active region operation is correct.
CICIERC
RCCVCEV 200,3812 −=+−=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
α
IB = 2.7 µA
intersection point Q-point.
Load-line for the circuit is:
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Lecture 05 - 10ELE2110A © 2008
Example 2 (Cont’)
21
1RR
RCCVEQV
+=
21
21RR
RREQR
+=
ERBI
EQRBEV
EQV
EI−−
=
VCC Temp. Process (β)
EIERBEVBIEQREQV ++=
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Lecture 05 - 11ELE2110A © 2008
Design Objectives: Q-point Insensitive to PVT Variations
1) IE linearly (not exponentially) related to VCC2) For IE to be less sensitive to IB (thus β):
Need small ReqLarge currents through R1 and R2 (I2 >> IB)
3) For IE to be less sensitive to VBE (due to temperature change):
VEQ >> VBE
)( BEVEQ
VBIEQR −<<
ER
BI
EQR
BEV
EQV
EI−−
=
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Lecture 05 - 12ELE2110A © 2008
Design Objectives: Low Power and Large Signal Swing
But …For low power consumption:
needs small I2contradict with constraint (2)set I2 = 10IB typically
For large output signal swing:VEQ cannot be too high(because VC∈ [VEQ, VCC])
VEQ =1/3 VCC typically
I2
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Lecture 05 - 13ELE2110A © 2008
Design GuidelinesChoose Thevenin equivalent base voltage
Select R1 to set I1 = 9IB
Select R2 to set I2 = 10IB
RE is determined by VEQ and desired IC
RC is determined by desired VCE
24CCV
EQVCCV≤≤
BIEQ
VR
91 =
BIEQ
VCC
VR
102
−=
CI
BEVEQ
V
ER−
≈
ERC
ICE
VCC
VCR −
−≈
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Lecture 05 - 14ELE2110A © 2008
Example 3• Problem: Design a 4-resistor bias circuit with given parameters.• Given data: IC =750 µA, VCE = 5 V, β = 100, VCC = 15 V, VBE = 0.7 V• Unknowns: VB, voltages across RE and RC; values for R1, R2, RC and RE.• Analysis: A common approach is to divide (VCC - VCE) equally between RE andRC. Thus, VE =5 V and VC =10 V.
kΩ67.6=−
=C
IC
VCC
VCR
V7.5
kΩ60.6
=+=
==
BEVEVBVEIEV
ER A5.6791
A75102µµ
====
BIIBII
A5.7 µβ == CIBI
kΩ124102
kΩ4.8491
=−
=
==
BIBV
CCV
R
BIBV
R
Now choose I2 = 10IB:
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Lecture 05 - 15ELE2110A © 2008
Example 4: Two-Resistor Bias Network • Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with
given parameters.• Given data: βF = 50, VCC = 9 V• Assumptions: Forward-active operation region, VEB = 0.7 V• Analysis:
V18.2V88.2)(10009
mA01.650
µA120000,69
V7.0V9
)51(1000000,189)(1000000,189
=
=+−=
==
=Ω
−=∴
++=∴
+++=
BCVBICIECV
BICIBI
BIBIEBVBICIBIEBV
Q-point is : (6.01 mA, 2.88 V)
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Lecture 05 - 16ELE2110A © 2008
Current Source Biasing
IE = I independent of the value of β and temperature
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Lecture 05 - 17ELE2110A © 2008
Current Source Implementation: Current Mirror
Use collector current of a transistor in active modeNeglect Early effect, IC2 is independent of VCE2 as long as VCE2 > VCEsat (i.e. BJT in active mode)For matched Q1 and Q2 , i.e., having identical IS, β, VA), we have
RBE
VBB
VREFI
−=
REFC II =2
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Lecture 05 - 18ELE2110A © 2008
Topics to cover …BJT Amplifier Biasing Circuits
Small Signal Operation and Equivalent Circuits
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Lecture 05 - 19ELE2110A © 2008
BJT as an Amplifier
If an ac+dc input signal the total vBE becomes
beBEBE vVv +=The collector current becomes
TbeTbeTBE
TbeBE
VvC
VvVVS
VvVSC
eIeeI
eIi///
/)(
==
= +
Superposition of DC with AC signal
ConceptualAmplifier circuit:
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Lecture 05 - 20ELE2110A © 2008
Small-signal TransconductanceFor small ac signal, i.e., vbe << VT :
The ac (or signal) component of the collector current is:
beT
Cc v
VIi =
gm is called the small signal transconductance. It represents the slope of iC-vBE curve at the Q-point.
321
AC
beT
C
DCC
TbeCVv
CC
vVII
VvIeIi Tbe
+=
+≅= )/1(/
T
C
be
cm V
Ivig =≡ We define:
CCbe IiBE
C
vbe
cm v
ivi
g=→
∂∂
=≡ 0
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Lecture 05 - 21ELE2110A © 2008
Signal Component of Base Current
Total base current: 43421
AC
beT
C
DC
CCB v
VIIii
βββ1
+==
bem
beT
Cb vgv
VIi
ββ==
1
rπ is the small-signal input resistance between base and emitter, looking into the base.
B
T
mb
be
IVr
givr ==≡ ππ
β or
Signal component of base current:
Define:
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Lecture 05 - 22ELE2110A © 2008
Signal Component of Emitter Current
The total emitter current iE: 43421
AC
beT
C
DC
CCE v
VIIii
ααα1
+==
beT
Ebe
T
Ce v
VIv
VIi ==
α1
re is the small-signal input resistance between base and emitter, looking into the emitter.
E
Te
me
bee I
Vrgi
vr ==≡ or α
It is easy to find out that eebe rriir )1()/( +== βπ
Define:
Signal component of emitter current:
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Lecture 05 - 23ELE2110A © 2008
Small Signal I-V Expressions
E
Te
e
be
IVr
iv
==B
T
b
be
IVr
iv
== πT
Cm
be
c
VIg
vi
==
(Hybrid-π small signal model of BJT)
Can be modeled by equivalent circuits:
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Lecture 05 - 24ELE2110A © 2008
Another model: T-model
T-model
E
Te
e
be
IVr
iv
==B
T
b
be
IVr
iv
== πT
Cm
be
c
VIg
vi
==
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Lecture 05 - 25ELE2110A © 2008
Early Effect: CI
CEV
Hybrid-π model including Early effect
C
A
C
CEA
vCE
Co
IV
IVV
vir
BE
≈+
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂∂
=
−1
fixed
Define
Include Early Effect in the model:
)1(/
A
CEVvSC V
veIi TBE +=
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Lecture 05 - 26ELE2110A © 2008
Graphic Analysis
Can be useful to understand the operation of BJT circuitsFirst, establish DC conditions by finding IB (or VBE)– Input load line:
Second, figure out the DC operating point for IC– Output load line:
BBBBBE iRVv −=
CECC
CCCCCCCCE v
RRViRiVv 1 −=⇒−=
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Lecture 05 - 27ELE2110A © 2008
Graphic Analysis (Cont.)
Apply a small signal input voltage and see ibSee how ib translates into VCECan get a feel for whether the BJT will stay in active region ofoperation– What happens if RC is larger or smaller?