Differential Amplifier Common & Differential Modes
Transcript of Differential Amplifier Common & Differential Modes
ESE319 Introduction to Microelectronics
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Differential AmplifierCommon & Differential Modes
● Common & Differential Modes● BJT Differential Amplifier● Diff. Amp Voltage Gain and Input Impedance● Small Signal Analysis – Differential Mode● Small Signal Analysis – Common Mode
ESE319 Introduction to Microelectronics
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+-
Vo = A (V
1 - V
2)A
Most “Famous” Differential AmplifierV
CC
VEE
V1 & V
2 are single-ended input voltages defined w.r.t. ground.
ESE319 Introduction to Microelectronics
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Common and Differential ModesConsider the two-input conceptual circuit shown below.
Z1
Z3
Z2
I1
I2
V2
V1
Define (convention) the voltages:
V d=V 1−V 2
Vc = “common mode voltage” Vd = “differential mode voltage”
+
+
±a
±a
V c=V 1V 2
2
1. Let V1 = V2 = V => Vc = V & Vd = 0
2. Let V1 = -V2 = V => Vc = 0 & Vd = 2V
Vc “common” to V
1 & V
2; V
d is split between V
1 & V
2 s.t. “difference” V
1 - V
2 = V
d
I1 + I2
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Replace V1, V2 with DM & CM Sources Vd , V
c
I1 + I2
V c=V 1V 2
2V d=V 1−V 2Definition Solve for V
1 and V
2
V 1=V cV d /2V 2=V c−V d /2
ESE319 Introduction to Microelectronics
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+-
Vo = A (V
1 - V
2)
A
Our Most “Famous” Differential AmplifierV
CC
VEE
V1 & V
2 are single-ended input voltages defined w.r.t. ground.
+-
Vo = A (V
1 - V
2)
A
VCC
VEE
Vo V
o
V o=Av−dmV dAv−cm V c CMRR=∣Av−dm∣∣Av−cm∣
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Z1
Z3
V 1=V cV d /2
V 2=V c−V d /2
Vx
<=>R
C1R
C2
RE1
RB1
RB2
RE2
Vd/2
-Vd/2V
d/2
-Vd/2
Vc
Vc
Typical Op-Amp Input Stage
vo1 vo2
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Common and Differential Mode Currents
Solving for I1 and I
2:
I 1= I c /2 I d
I 2= I c /2−I d
Z1
Z3
Z2
I1 = I
c/2 + I
d
Vc
Vd/2
- Vd/2
Ic
I2 = I
c/2 - I
d
Definition: I c= I 1 I 2I d=I 1−I 22
+
++
±.
±.
±.
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CM-DM Circuit Description
Z1
Z3
Z2
Ic/2
Vc
Vd/2
- Vd/2 Ic Ic/2
Id
1. Voltage sources defined as common and differential mode quantities.
2. Differential mode currents take “blue” path.
3. Common mode current takes “red” path.
4. When Z1 = Z
2 = Z, the differential
circuit is said to be balanced.
OBSERVATIONSi. No DM I
d flows through Z
3.
ii. No CM Ic flows around Z
1, Z
2 loop
+
++
±.
±.
±.
I1
I2
I 1= I c /2 I d I 2= I c /2−I d
ESE319 Introduction to Microelectronics
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Z1
Z3
Vx
<=>R
C1R
C2
RE1
RB1
RB2
RE2
Vd/2 -V
d/2
Typical Diff Amp Op-Amp Input Stage
Z2
Z 1=Z 2=Zbalanced circuit => Z1 = Z2 => Q1 = Q2, RC1 = RC2,RE1 = RE2 and RB1 = RB2
Balanced Circuit Differential Mode (VC = 0)
vo −vo
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Analyze Circuits by SuperpositionBalanced Circuit Differential Mode (V
C = 0)
Z1
Z3
Z2
I1 = Id
Vd/2
- Vd/2
I2 = -Id
Id
Z 1=Z 2=Z
V x
I c=0⇒V x=0
+
+
±.
±.
I c= I 1 I 2=
V d
2−V x
Z1
−V d
2−V x
Z 2=−2V x
Z
balanced circuit =>
+ -VZ1
+- VZ2
Z i n−dm=V d
I d=2 Z
NOTE: If , then => Vx ≠ 0.Z 2=Z 1Z Ic ≠ 0
Ic = 0
I c=I 1I 2=I d−I d=0
I d=I 1− I 22
=
V d
2−V x
2Z 1−
−V d
2−V x
2 Z 2=
V d
2Z
ESE319 Introduction to Microelectronics
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Analyze Circuits by SuperpositionBalanced Circuit Common Mode (V
d = 0)
Z1
Z3
Z2
Ic/2
Vc
IcIc/2
Id = 0
Z 1=Z 2=Z
I c=V c
Z 1∣∣Z 2Z 3=
V c
Z2Z 3
+ ±.
balanced circuit =>
I1
I2
Z i n−cm=V c
I c=
Z2Z 3
I 1= I c /2I 2=I c /2
=> I d=0
ESE319 Introduction to Microelectronics
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Summary
<=>
Definitions:
V c=V 1V 2
2V d=V 1−V 2
I c= I 1 I 2 I d=I 1−I 22
V 1=V cV d /2V 2=V c−V d /2I 1= I c /2 I d
I 2= I c /2−I d
I c
I 1
I 2
All V's & I's have differential & com-mon mode com-ponents
Z 1
Z 2
Z 3
I 2
Id
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Summary cont.In a balanced differential circuit (Z1 = Z2 = Z): 1. Differential mode voltages result in differential mode currents.2. Common mode voltages result in common mode currents.
The differential mode input impedance of a balanced circuit is: Z i n−dm=Z 1Z 2=2ZThe common mode input impedance of a balanced circuit is: Z i n−cm=Z 1∥Z 2Z 3=
Z2Z 3
Id Id = 0
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BJT Differential Amplifier – DC Bias View
I C= I cm /2 I cm /2
I BI B
I B=I cm
21
RC1=RC2=RC
RB1=RB2=RB
Collector bias path inherentlycommon mode.
I C=I cm
2≈
I cm
2
Choose Icm and RC for approximate½ VCC drop across RC.
balanced circuit
Icm
Q1=Q2
IE
IE
I E=I cm
2
RB1 RB2
RC2RC1
I 1=I c /2 I d
I 2= I c/2−I d
Recall00
for dc bias
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vid/2
-vid/2
vic
BJT Differential Amplifier – AC Signal View
RB1 RB2
RC1 RC2
vo1
vo2+-v
od
vod=vo2−vo1
vi1
vi2
v ic=v i1v i2
2v id=v i1−v i2
ro
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vid /2
rin-dm
+-
Z in−dm=v id
ib1d
Avdm=vodm
v idvo1d
Single-ended DM outputs: vo1d, vo2d
Differential DM output: vodm
=vo1d
- vo2d
vid /2
RC1
Avdm1 ,2=vo1 ,2d
vid
(diff Adm
)
(s-e Adm
)+
-
ro
NOTE: ib1d=ib1
; ib2d
=-ib2
ib1d i
b2d
RC2
RB1RB2
vodm vo2d
BJT Differential Amplifier – AC Diff Mode View
ro
ie1d
ie2d
DM input impedance
vidm=vid
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ro
vic
ib1c
rin-cm
+-vocm
Avcm=vocm
vicm
Single-ended CM outputs: vo1c
, vo2c
Differential CM output: vocm
= vo2c
- vo1c
vo1c vo2c
RB2Avcm1 ,2=
vo1 ,2c
vic
(diff Acm
)
(s-e Acm
)
NOTE: ib1c
=ib1
; ib2c
=ib2
BJT Differential Amplifier – AC Cmn Mode View
RB1
RC1RC2
ICM
ib2c
Z in−cm=v ic
ib1c
CM input impedance
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BJT Differential Amplifier – Small-signal View
ICM current source output impedance
ib1 ib2ie1
ic1ic2
ie2
vic
vid /2 vid /2
NOTE: 1. ro for amplifier Q1 & Q2 is ignored.2. vid /2 and vic are ac small signals
Z 1=Z 2=1re
Z 3=1r o
RB RB
RCRC
βib1re
ro
re
βib2V
x
Z1
Z2
Z3
vid
/2
vic
Vx
-vid
/2
re1=re2=re
RC1=RC2=RC
RB1=RB2=RB
ie1=ie1d1i ic /2 ie2=−ie2d1 i ic /2&
ib1=i ic /2iid
iic
/2
iid iid
ib2=i ic /2−i id
iic
/2
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Superposition Small-signal AnalysisDifferential Mode (vic = 0)
DM Path (ignoring both RB):
Balanced circuit
+-vo-dm
ie
icib
ied
vid=2 re ied=21r e ibd
Z 3=1roZ 1=Z 2=Z=1r e
vid
2=ied reied r e−
v id
2
ib1ib
vid /2 vid /2
ied
ie
RB
ro
βib-dm β ib-dm
RCRC
RBre re
⇒ ie2=−ie1=ie , ib2=−ib1=ib , ic2=−ic1=ic
Z i n−dm=2 ZRecall:
ibdibdic
icdicd
ie1−dm=ie−dm=ie1=i e
ib1−dm=ib−dm=i b1=i b
ie2−dm=ie−dm=−i e2=−ie
ib2−dm=ib−dm=−i b2=−ib
ic1−dm=ic−dm=ic1=ic ic2−dm=ic−dm=−ic2=−i c
vx
Z i n−dm=vid
ibd=21 re
Z in−dm=vid
ibd
Z Z
Z3
vid
/2
vic
Vx
-vid
/2
ESE319 Introduction to Microelectronics
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Superposition Small-signal AnalysisDifferential Mode (vic = 0) cont.
+- vodm
ie ie
icib
iedied
Balanced circuit
DM Differential voltage gain:
Avdm=vodm
vid=
RC 1r e
≈RC
r e
vodm=RCibd −−RC ibd
vodm=vo2d−vo1d
ibd=v id
21
1r e
Avdm2=−Avdm1=vo2d
v id=
RC
21re≈
RC
2 re
vodm=2RC
1 re
v id
2
DM Single-ended voltage gains:
ib
ie1−dm=ie−dm=ie1=i e
ib1−dm=ib−dm=i b1=i b
−ie2−dm=−ie−dm=i e2=ie
vid /2
RB
ro
RC RC
βibd β ibdre re RB
vo2d vo1d
icicd
icdibd ibd
−ib2−dm=−ib−dm=i b2=ib
ic1−dm=ic−dm=ic1=ic −ic2−dm=−ic−dm=ic2=ic
ic−dm= i b−dm
vid /2vx
ibd=v id
21
1r e
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Superposition Small-signal AnalysisCommon Mode (vid = 0)
CM path: both re are in parallel
vic
+-ibc
iec
icm /2=ie−cm=i e1=i e
.=1[r e2ro]ibc
Z 1=Z 2=1re Z 3=1r o
Balanced circuit
Z in−cm=vic
icm=
vic
2 ibc=1
re
2ro≈1r o
ib1−cm=i b−cm=ib1=ib
RB
βibcβibcre
re
ro
RB
RC RC
i ic /2=ibc
Recall:ic1−cm=ic−cm=ic1=ic
icm /2=ie−cm=i e2=ie
ib2−cm=ib−cm=i b2=ib
ic2−cm=ic−cm=ic2=ic
vo-cm
Z i n−cm=Z∥ZZ 3=Z2Z 3
ibc
icc i
cc
vic=1 re ibc21 ro ibc
Note:
iec
ibc=i ic /2 ibc=i ic /2
Z Z
Z3
vid
/2
vic
Vx
-vid
/2
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Superposition Small-signal AnalysisCommon Mode (vid = 0)
vic
+-ibc
iec
ibc
icc vocm=−RC ibc−−RC ibc=0
ibc=v ic
1 re2 ro ≈
v ic
2 1 ro
CM Differential voltage gain:
vocm=vo2c−vo1cBalanced circuit
Avcm=vocm
vic=0
CM Single-ended voltage gains:
Avcm2=Avcm1=vo1c
v ic=−
RC 2 1 ro
≈−RC
2 ro
vo2c=vo1c=−RC ibc=−RC 2 1r o
vic
ro
RC RC
RB RBre re
βibcβibc
icm /2=ie−cm=i e1=i e icm /2=ie−cm=i e2=ieib1−cm=i b−cm=ib1=ib ib2−cm=ib−cm=i b2=ib
ic1−cm=ic−cm=ic1=ic ic2−cm=ic−cm=ic2=ic
ic−dm= i b−dm
vocm
ic−dm= i b−dm
icc
iec
ibc=i ic /2 ibc=i ic /21i ic
vic=1 r e2 ro ibc
ESE319 Introduction to Microelectronics
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Summary Differential mode:
Z i n−dm=21r e
Avdm=vodm
vid≈
RC
re
Differential DM Voltage gain:
Common mode:
Z i n−cm=1 r e
2 ro≈1ro
Differential CM Voltage gain:
Avcm=vocm
v ic=0
Single-Ended DM Voltage gain:
Avdm1=−Avdm2=vo1d
v id≈−
RC
2 reAvcm1=Avcm2=
vo1c
v ic≈−
RC
2 ro
Single-Ended CM Voltage gain:
CMRR=Avdm1 ,2
Avcm1 , 2≈20 log10 ro
re Single-ended-output (balanced)Differential-output
CMRR=Avdm
Acdm=∞ i.f.f. Balanced
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v i
Practical Signal Excitation to Test CM
vo2vo1
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Practical Signal Excitation to Test DM
vo1 vo2
vo1 vo2