ESE319 Introduction to Microelectronics
Differential Amplifier
Common & Differential Modes
●
Common & Differential Modes
●BJT Differential Amplifier
●
Diff. Amp Voltage Gain and Input Impedance
●Small Signal Analysis – Differential Mode
ESE319 Introduction to Microelectronics
+
-
A Vo = A (V1 - V2)Most “Famous” Differential Amplifier
VCC
VEE
ESE319 Introduction to Microelectronics
Common and Differential Modes
Consider the two-input conceptual circuit shown below.
Z1 Z3 Z2 I1 I2 V2 V1
Define (convention) the voltages:
V d=V1−V 2 Vc = “common mode voltage”
Vd = “differential mode voltage”
+ + ±a ±a Vc=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 V1 & V2; Vd is split between V1 & V2 s.t. “difference” V1 - V2 = Vd
ESE319 Introduction to Microelectronics
Replace V
1, V
2with DM & CM Sources V
d, V
cI1 + I2
V c=V 1V 2
2 V d=V 1−V 2
Definition Solve for V1 and V2
V 1=V cV d/ 2
ESE319 Introduction to Microelectronics
+
-Vo = A (V1 - V2)
A
Our Most “Famous” Differential Amplifier
VCC
VEE
V1 & V2 are single-ended input voltages defined w.r.t. ground.
+
-Vo = A (V1 - V2) A VCC VEE Vo Vo V o=Av−dmV dAv−cmV c CMRR=∣Av−dm∣ ∣Av−cm∣ESE319 Introduction to Microelectronics Z1 Z3 V 1=V cV d/ 2 V 2=V c−V d/2 Vx <=> RC1 RC2 RE1 RB1 RB2 RE2 Vd/2 -Vd/2 Vd/2 -Vd/2 Vc Vc
Typical Op-Amp Input Stage
vo1 v
ESE319 Introduction to Microelectronics
Common and Differential Mode Currents
Solving for I1 and I2:I1=Ic/2Id I 2=Ic/2−I d Z1 Z3 Z2 I1 = Ic/2 + Id Vc Vd/2 - Vd/2 Ic I2 = Ic/2 - Id Definition: I d= Ic=I1I2 I1−I 2 2 + + + ±. ±. ±.
ESE319 Introduction to Microelectronics
CM-DM Circuit Description
Z1 Z3 Z2 Ic/2 Vc Vd/2 - Vd/2 Ic/2 Ic Id1. 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 = Z2 = Z, the differential
circuit is said to be balanced. OBSERVATIONS
i. No DM Id flows through Z3.
ii. No CM Ic flows around Z1, Z2 loop
+ + + ±. ±. ±. I1 I2 I1=Ic/2Id I 2=Ic/2−I d
ESE319 Introduction to Microelectronics Z1 Z3 Vx <=> RC1 RC2 RE1 RB1 RB2 RE2 Vd/2 -V d/2
Typical Diff Amp Op-Amp Input Stage
Z2
Z1=Z2=Z
balanced circuit => Z1 = Z2 => Q1 = Q2, RC1 = RC2, RE1 = RE2 and RB1 = RB2
Balanced Circuit Differential Mode (V
C= 0)
ESE319 Introduction to Microelectronics
Analyze Circuits by Superposition
Balanced Circuit Differential Mode (V
C= 0)
Z1 Z3 Z2 I1 = Id Vd/2 - Vd/2 I2 = -Id Id Z1=Z2=Z V x Ic=0⇒V x=0 + + ±. ±. Ic=I1I2= Vd 2 −V x Z1 −Vd 2 −V x Z2 = −2V x Z balanced circuit => + VZ1 -+ - VZ2 Zi n−dm=Vd Id =2 Z NOTE: If , then => Z2=Z1 Z I Vx ≠ 0. c ≠ 0 Ic = 0 Ic=I1I 2=Id−I d=0 Id= I1−I2 2= Vd 2 −V x 2 Z1 − −Vd 2 −Vx 2 Z2 = Vd 2 Z
ESE319 Introduction to Microelectronics
Analyze Circuits by Superposition
Balanced Circuit Common Mode (V
d= 0)
Z1 Z3 Z2 Ic/2 Vc Ic Ic/2 Id = 0 Z1=Z2=Z Ic= Vc Z1∣∣Z2Z3 = V c Z 2 Z3 + ±. balanced circuit => I1 I2 Zi n−cm=V c Ic = Z 2 Z3 I1=Ic/2 I2=Ic/2 => Id=0
ESE319 Introduction to Microelectronics
Summary
<=> Definitions: V c=V1V 2 2 Vd=V1−V 2 Ic=I1I2 I d= I1−I 2 2 V 1=V cV d/ 2 V 2=V c−V d/2 I1=Ic/2Id I 2=Ic/2−I d Ic I1 I2All V's & I's have
differential & com-mon mode com-ponents Z1 Z2 Z3 I2 Id
ESE319 Introduction to Microelectronics
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: Zi n−dm=Z1Z2=2 Z
The common mode input impedance
of a balanced circuit is: Zi n−cm=Z1∥Z2Z3=
Z
2 Z3
IId
ESE319 Introduction to Microelectronics
BJT Differential Amplifier – DC Bias View
IC=Icm/2 Icm/2 IB IB I B= Icm 21 RC1=RC2=RC RB1=RB2=RB
Collector bias path inherently
common mode.
IC= Icm 2 ≈
Icm
2
Choose Icm and RC for approximate
½ VCC drop across RC. balanced circuit Icm Q1=Q2 IE I E IE= Icm 2 RB1 RB2 RC2 RC1 I1=Ic/2Id I2=Ic/ 2−Id
ESE319 Introduction to Microelectronics
vid/2 -vid/2
vic
BJT Differential Amplifier – AC Signal View
RB1 RB2 RC1 RC2 vo1 -vod + vo2 vod=vo2−vo1 vi1 vi2 vic=vi1vi2 2 vid=vi1−vi2 ro
ESE319 Introduction to Microelectronics vid /2 rin-dm + -Zin−dm= vid ib1d Avdm= vodm vid vo1d
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 RB1 RB2 vodm vo2d
BJT Differential Amplifier – AC Diff Mode View
ro
ie1d ie2d
DM input impedance
ESE319 Introduction to Microelectronics ro vic ib1c rin-cm + -vocm Avcm= vocm vicm
Single-ended CM outputs: vo1c, vo2c
Differential CM output: vocm = vo2c - vo1c
vo1c vo2c RB2 Avcm1 , 2= vo1 ,2c vic (diff Acm) (s-e Acm) NOTE: ib1c=ib1; ib2c=ib2
BJT Differential Amplifier – AC Cmn Mode View
RB1 RC1 RC2 ICM ib2c Zin−cm= vic ib1c CM input impedance
ESE319 Introduction to Microelectronics
BJT Differential Amplifier – Small-signal View
ICM current source output impedance ib1 ie1 ib2 ic1 ic2 ie2 vic vid /2 v id /2 NOTE:
1. ro for amplifier Q1 & Q2 is ignored. 2. vid /2 and vic are ac small signals
Z1=Z2=1re Z3=1ro RB R B RC RC βib1 re ro re βib2 Vx Z1 Z2 Z3 vid /2 vic Vx -vid /2 re1=re2=re RC1=RC2=RC RB1=RB2=RB
ie1=ie1d1iic/2 & ie2=−ie2d1iic/2
ib1=iic/2iid
iic /2
iid iid
ib2=iic/2−iid
ESE319 Introduction to Microelectronics
Superposition Small-signal Analysis
Differential Mode (v
ic= 0)
DM Path (ignoring both RB):
Balanced circuit + -vo-dm ie ic ib ied vid=2 reied=21reibd Z3=1ro Z1=Z2=Z =1re vid 2 =ied reied re− vid 2 ib1i b vid /2 vid /2 ied ie RB ro βib-dm βib-dm RC RC RB re re
⇒ie2=−ie1=ie,ib2=−ib1=ib,ic2=−ic1=ic
Zi n−dm=2 Z Recall: ibd icd icd ic ibd ie1−dm=ie−dm=ie1=ie ib1−dm=ib−dm=ib1=ib ie2−dm=ie−dm=−ie2=−ie ib2−dm=ib−dm=−ib2=−ib
ic1−dm=ic−dm=ic1=ic ic2−dm=ic−dm=−ic2=−ic
vx Zi n−dm=vid ibd =21 re Zin−dm= vid ibd Z Z Z3 vid /2 vic Vx -vid /2
ESE319 Introduction to Microelectronics
Superposition Small-signal Analysis
Differential Mode (v
ic= 0) cont.
+ - vodm ie ie ic ib ied ied Balanced circuit
DM Differential voltage gain:
Avdm=vodm vid = RC 1re ≈ RC re vodm=RCibd−−RC ibd
vodm=vo2d−vo1d
ibd=vid 2 1 1re Avdm2=−Avdm1=vo2d vid = RC 21re≈ RC 2 re vodm= 2 RC 1 re vid 2
DM Single-ended voltage gains:
ib ie1−dm=ie−dm=ie1=ie ib1−dm=ib−dm=ib1=ib −ie2−dm=−ie−dm=ie2=ie vid/2 RB ro RC RC βibd βibd re re R B vo2d vo1d ic icd icd ibd i bd −ib2−dm=−ib−dm=ib2=ib ic1−dm=ic−dm=ic1=ic −ic2−dm=−ic−dm=ic2=ic ic−dm=ib−dm vid /2 vx ibd=vid 2 1 1re
ESE319 Introduction to Microelectronics
Superposition Small-signal Analysis
Common Mode (v
id= 0)
CM path: both re are in parallel
vic + -ibc iec icm/2=ie−cm=ie1=ie .=1[re2ro]ibc Z1=Z2=1re Z3=1ro Balanced circuit Zin−cm=vic icm= vic 2 ibc=1 re 2 ro≈1ro ib1−cm=ib−cm=ib1=ib RB βibc βibc re re ro RB RC RC iic/2=ibc Recall: ic1−cm=ic−cm=ic1=ic icm/2=ie−cm=ie2=ie ib2−cm=ib−cm=ib2=ib ic2−cm=ic−cm=ic2=ic vo-cm Zi n−cm=Z∥ZZ3= Z 2 Z3 ibc icc icc vic=1 reibc21roibc Note: iec ibc=iic/2 ibc=iic/2 Z Z Z3 vid /2 vic Vx -vid /2
ESE319 Introduction to Microelectronics
Superposition Small-signal Analysis
Common Mode (v
id= 0)
vic + -ibc iec ibc icc vocm=−RC ibc−−RCibc=0 ibc= vic 1re2ro≈ vic 21roCM Differential voltage gain:
vocm=vo2c−vo1c
Balanced circuit
Avcm=vvocm
ic
=0
CM Single-ended voltage gains: Avcm2=Avcm1=vo1c vic =− RC 21ro≈− RC 2ro vo2c=vo1c=−RC ibc= −RC 21ro vic ro RC R C RB re re RB βibc βibc icm/2=ie−cm=ie1=ie icm/2=ie−cm=ie2=ie ib1−cm=ib−cm=ib1=ib ib2−cm=ib−cm=ib2=ib ic1−cm=ic−cm=ic1=ic ic2−cm=ic−cm=ic2=ic ic−dm=ib−dm vocm ic−dm=ib−dm icc iec ibc=iic/2 1iic ibc=iic/2 vic=1re2roibc
ESE319 Introduction to Microelectronics
Summary
Differential mode: Zi n−dm=21re Avdm=vodm vid ≈ RC reDifferential DM Voltage gain:
Common mode: Zi n−cm=1
re2 ro
≈1roDifferential CM Voltage gain: Avcm=vocm
vic =0
Single-Ended DM Voltage gain:
Avdm1=−Avdm2=vo1d vid ≈− RC 2 re Avcm1=Avcm2= vo1c vic ≈− 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. BalancedESE319 Introduction to Microelectronics
vi
Practical Signal Excitation to Test CM
vo2 vo1
ESE319 Introduction to Microelectronics
Practical Signal Excitation to Test DM
vo1 vo2