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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

(2)

ESE319 Introduction to Microelectronics

+

-

A Vo = A (V1 - V2)

Most “Famous” Differential Amplifier

VCC

VEE

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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 1V 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

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ESE319 Introduction to Microelectronics

Replace V

1

, V

2

with DM & CM Sources V

d

, V

c

I1 + I2

V c=V 1V 2

2 V d=V 1−V 2

Definition Solve for V1 and V2

V 1=V cV d/ 2

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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

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ESE319 Introduction to Microelectronics Z1 Z3 V 1=V cV d/ 2 V 2=V cV 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

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ESE319 Introduction to Microelectronics

Common and Differential Mode Currents

Solving for I1 and I2:

I1=Ic/2Id I 2=Ic/2I d Z1 Z3 Z2 I1 = Ic/2 + Id Vc Vd/2 - Vd/2 Ic I2 = Ic/2 - Id Definition: I d= Ic=I1I2 I1−I 2 2 + + + ±. ±. ±.

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ESE319 Introduction to Microelectronics

CM-DM Circuit Description

Z1 Z3 Z2 Ic/2 Vc Vd/2 - Vd/2 Ic/2 Ic 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 = 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/2I d

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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)

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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=I1I2= 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=I1I 2=IdI d=0 Id= I1−I2 2= Vd 2 −V x 2 Z1−Vd 2 −Vx 2 Z2 = Vd 2 Z

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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∣∣Z2Z3 = 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

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ESE319 Introduction to Microelectronics

Summary

<=> Definitions: V c=V1V 2 2 Vd=V1−V 2 Ic=I1I2 I d= I1−I 2 2 V 1=V cV d/ 2 V 2=V cV d/2 I1=Ic/2Id I 2=Ic/2I d Ic I1 I2

All V's & I's have

differential & com-mon mode com-ponents Z1 Z2 Z3 I2 Id

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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=Z1Z2=2 Z

The common mode input impedance

of a balanced circuit is: Zi n−cm=Z1∥Z2Z3=

Z

2 Z3

IId

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ESE319 Introduction to Microelectronics

BJT Differential Amplifier – DC Bias View

IC=Icm/2 Icm/2 IB IB I B= Icm 21 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/ 2Id

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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=vi1vi2 2 vid=vi1−vi2 ro

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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

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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

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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=1re Z3=1ro 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=ie1d1iic/2 & ie2=−ie2d1iic/2

ib1=iic/2iid

iic /2

iid iid

ib2=iic/2iid

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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=21reibd Z3=1ro Z1=Z2=Z =1re vid 2 =ied reied revid 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 =21 re Zin−dm= vid ibd Z Z Z3 vid /2 vic Vx -vid /2

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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 1reRC re vodm=RCibd−−RC ibd

vodm=vo2dvo1d

ibd=vid 2 1 1re Avdm2=−Avdm1=vo2d vid =  RC 21reRC 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=ibie2−dm=−ie−dm=ie2=ie vid/2 RB ro RC RC βibd βibd re re R B vo2d vo1d ic icd icd ibd i bdib2−dm=−ib−dm=ib2=ib ic1−dm=ic−dm=ic1=icic2−dm=−ic−dm=ic2=ic ic−dm=ib−dm vid /2 vx ibd=vid 2 1 1re

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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[re2ro]ibc Z1=Z2=1re Z3=1ro Balanced circuit Zin−cm=vic icm= vic 2 ibc=1 re 2 ro≈1ro 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∥ZZ3= Z 2 Z3 ibc icc icc vic=1 reibc21roibc Note: iec ibc=iic/2 ibc=iic/2 Z Z Z3 vid /2 vic Vx -vid /2

(22)

ESE319 Introduction to Microelectronics

Superposition Small-signal Analysis

Common Mode (v

id

= 0)

vic + -ibc iec ibc icc vocm=−RCibc−−RCibc=0 ibc= vic 1re2ro≈ vic 21ro

CM Differential voltage gain:

vocm=vo2cvo1c

Balanced circuit

Avcm=vvocm

ic

=0

CM Single-ended voltage gains: Avcm2=Avcm1=vo1c vic =− RC 21ro≈− RC 2ro vo2c=vo1c=−RCibc= −RC  21ro 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 1iic ibc=iic/2 vic=1re2roibc

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ESE319 Introduction to Microelectronics

Summary

Differential mode: Zi n−dm=21re Avdm=vodm vidRC re

Differential DM Voltage gain:

Common mode: Zi n−cm=1

re

2 ro

≈1ro

Differential 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. Balanced

(24)

ESE319 Introduction to Microelectronics

vi

Practical Signal Excitation to Test CM

vo2 vo1

(25)

ESE319 Introduction to Microelectronics

Practical Signal Excitation to Test DM

vo1 vo2

References

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