differential amplifier cont.ese319/lecture_notes/lec_13_diffampl-ii_13.pdfdifferential amplifier...

ESE319 Introduction to Microelectronics

1Kenneth R. Laker, update KRL 18Oct13

Differential Amplifier Cont.● Differential amplifier with emitter resistors● Single-ended output● Common mode rejection ratio (CMRR)



VCC

CM DC

Quick Review

VEE

DC Bias is ALWAYS COMMON MODE

VEE

= - VCC

RC1=RC2=RC=V CC−V C

I CM

2

VC

VCI

B IB

RB1=RB2=RB=.large



Quick Review

Iff balancedCMRRdif =

Avdm

Avcm=∞

Zin-dm Z

in-cm

CMRRs−e=Avdm1

Avcm1=

r o

r e



Differential Amplifier With Emitter Resistors

Balance assumed:

RC1=RC2=RC

RE1=RE2=RE

RB1=RB2=RB

Q1=Q2 Matched Pair

IB1

IE1 I

E2

IB2

IC1 I

C2

RB1

RC1 RC2

RB2RE1 RE2

I

What is the purpose for RE1

and RE2

?



Differential Mode Analysis (vic = 0)

ied

ib

ib

ic

ie

ie

vid /2

vo-dm

+-

vid /2

RC RC

RE RE

re

ro

β ibdβibdre

ied

ibd ibdic

icd icd

Balanced circuit

Z i n−dm=21reRE

Avdm=vodm

v id≈

RC

REr e

Avdm1=vo1d

vid=

−RC

2 1RE≈

−RC

2REre

Avdm2=vo2d

v id=

RC

21RE≈

RC

2REre

Differential DM Voltage gain:

Single-Ended DM Voltage gains:

NOTE:1. ro for Q1 and Q2 are ignored.2. Both RB's are ignored.



Common Mode Analysis (vid = 0)

ibc

icc

icc

ie1

ie2

vic

vocm

+-

iic/2

RCRC

RE RE

ro

βibcβibc rere

i ic /2=ibc

Balanced circuit

Z i n−cm=1 reRE

2r o≈1 ro

Differential CM Voltage gain:

Avcm=vocm

vic=0

Single-Ended CM Voltage gains:

Avcm1=vo1c

v ic=−

RC

1reRE2 r 0≈−

RC

2 ro

Avcm2=vo2c

vic=Avcm1

NOTE:1. ro for Q1 and Q2 are ignored.2. Both RB's are ignored.

ibc

iic/2



Common Mode Rejection RatioCommon mode rejection ratio is the magnitude ratio of thedifferential mode gain over the common mode gain expressed in dB:

CMRR=20 log10∣Avdm1 ,2∣∣Avcm1,2∣

CMRR≈20 log10 RC

2 REr e∗2 ro

RC CMRRdb≈20 log10 ro

REr e

Avcm2=Avcm1≈−RC

2 ro

Avdm2=−Avdm1≈RC

2 REr e

Single-Ended:

Differential:

Avdm≈RC

REre

Single-Ended:

Differential:Avcm=0

Single-Ended:

Differential: CMRRdb=20 log10∣Avdm∣∣Avcm∣=∞

(iff balanced) (iff balanced)



ib1

ib2

ic1

ic2

ie1

ie2

vic

vocm

+-

Let RC2

> RC1

, s.t. RC1

= RC & RC2=RCRC

Avcm1=vo1c

vic≈−

RC

reRE2 r0≈−

RC

2 ro

Avcm2=vo2c

vic≈−

RCRC

2r0

.=−RC

2 r o1

RC

RC

Avcm1≈−RC

2 ro≠ Avcm2

Avcm=Avcm2−Avcm1=−RC

2 ro

RC

RC≠ 0

ie2cie1c

RC1 RC2

RE RE

ro

βib2re

βib1re

Hence!

What if the Diff Amp isn't Perfectly Balanced?

where ΔRC << RC

iic/2 iic/2

Differential Output CM Gain



What if the Diff Amp isn't Perfectly Balanced?

ib1

ib2

ic1

ic2

ie1

ie2

vic

vocm

+-

ie2cie1c

RC1 RC2

RE RE

ro

βib2re

βib1re

iic/2 iic/2

Let RC1

= RC and RC2=RCRC

Avcm unbal =Avcm2−Avcm1=−RC

2 r o

RC

RC

From previous slide

Avdm unbal =Avdm2−Avdm1≈RC

RC

2RE

≈RC

RE

Avdm1≈−RC

2REreAvdm2≈

−RCRC 2REre

CMRRdb=20 log10 ∣Avdm∣∣Avcmunbal ∣=20 log10 2 ro

REre

1RC /RC

≠∞

Avdm unbal ≈Avdm bal =AvdmDifferential Output



What if the Diff Amp isn't Perfectly Balanced?Single-ended outputs

Avdm2 bal =−Avdm1bal ≈RC

2 REr e

Avdm2 unbal ≠−Avdm2bal ≈RCRC

2REre≈

RC

2 REreAvdm1unbal =−Avdm1bal

Avcm2 bal =Avcm1 bal ≈−RC

2 ro

Avcm2 unbal ≠ Avcm2 bal ≈−RC RC

2 ro≈−

RC

2 roAvcm1unbal =−Avcm1bal

CMRRdbunbal ≈CMRRdbbal =20 log10 r o

REre



Differential mode (balanced) Z in−dm=vid

ibd=12r eRE

Common mode (balanced) Z in−cm=v ic

2 ibc=1

reRE

2r0

Avdm=vodm

vid≈

RC

REr eAvdm2=

vo2d

v id=−Avdm1≈

RC

2REr e

Avcm1=vo1c

vic=Avcm2≈−

RC

2 r o

CMRRdb=Avdm1 ,2

Avcm1 ,2≈20 log10 ro

reRE

Avcmunbal =vocm

vic≈

RC

2 ro

RC

RC

Single-ended-output (balanced)

CMRRdb=20 log102ro

rERE

1RC /RC

Differential-output

if Unbalanced due to RC1

≠ RC2

CMRR=∞ i.f.f. Balanced

Avcm=vocm

vic=0 i.f.f. Balanced

Summary

Avdm unbal ≈Avdm bal =Avdm

Unbalanced



Differential Amplifier Design● Design with ideal emitter current source bias.● Differential and common mode gain results● Add finite output resistance r

o to ideal current source.

● Replace ideal current source & ro with current mirror.



Quick Amplifier Design

Neglecting IB: let I

B = 0

IC

IC

IE

IE

I C=I E=5mA⇒V Rc=I C

RC=5V

Avdm1=vo1d

vid≈

−RC

2REre=−4.76

+- vo

vid /2

vicSingle-ended voltage gains w.r.t. vi\d /2and vic (for Q2 side):

Avcm1=vo1c

vic≈−

RC

2 ro=0

ideal current source, i.e. ro = ∞

V Rc=5V ⇒V C=V CC−V Rc

=7V

vid /2

VC

I10 mA

By inspection DC bias (vi-dm = vi-cm = 0) for Q1 & Q2 is common mode:

RC RC

RE RERBRB

VC V BE=0.7V

I E=I2=5mA V C=7V

re=V T

I E≈5



Differential- Mode AC Gain Results

“Scope” output B at collectorof Q1, i.e. .

Input voltage 0.14 V

peak arg (0o) at f = 1 kHz.

Output voltage 1.33 V

peak arg (180o).

vB=vo1d

v A=vid

vB=vo1d

Measured gain:

Avdm1=v B

v A=

vo1d

vid=−0.664V0.141V

=−4.7≈−RC

2 REr e

vB=vo1dvA=v id

500 mV/Div

-664.2000 mV141.2799 mV



Common Mode AC Results

Input voltage 0.14 V


Since I is an ideal currentsource => .

“Scope” output B at collectorof Q1, i.e. .vB=vo1c=0V

v A=vic

vA=v icvB=vo1c

ro=∞ Avcm1=0

CMRRdb=20 log10∣Avdm1∣∣Avcm1∣

=20 log10∣4.70 ∣=∞

Avcm1=v B

v A=

vo1c

vic= −0V0.14V

=0



Common Mode Results - Add ro to Model

Avcm1=vo1c

vic≈−

RC

2 r o=

−1200

=−0.005

Avdm1=vo1d

vid≈

−RC

2REre=−4.76

vid /2 vid /2

vic insert finite ro

to model non-ideal current source

10 mAI

CMRRdb=20 log10∣Avdm1

Avcm1∣=59.6 dB

Theory

RC RC

RE RERB

RB

ro



Common Mode Results - Add Finite roro=100 k

Output voltage 0.007 V

peak arg (180o).

Avcm1≈−0.0071.4

=−0.005

CMRRdb≈20 log104.70.005

≈59.5dB

“Scope” output B at collectorof Q1, i.e. .

Input voltage 1.4 V


vB=vo1cm

v A=vic

vB=vo1cm

vB=vo1cvA=v ic

Avdm1=−4.7 (unchanged)



Simulation with 10 mA Current Mirror

I C3≈10mA

I REF=V CC−V BE4−V EE

Rref≈10mA

Matched 2N2222 BJTs(Q1, Q2, Q3 and Q4).

vi-cm

I C3≈10mA

I C1≈5mA I C2≈5mA

⇒ I C1= I C2≈5mA

I REF=10mA

V C1=7V

Rref =2.33k

RC RC

RE RERBRB

Rref

VBE4



Simulation with 10 mA Current Mirror cont.vA=v ic

vB=vo1c

Input voltage 1.4 V


v A=vic

Output voltage 60 mV

peak arg (180o).

vB=vo1c

Avcm1≈−RC

2 ro≈−0.06

1.4=−0.043

Avcm1≈−0.043

Infer ro with RC=1 k

ro=−RC

2 Avcm1=1k 0.086

=11.67 k

Infer VA with ro=11.67 k

ro=V A

I C⇒V A=ro I C=58.3V



Simulation with 10 mA CM - Bode Plots

10 mA current: Avcm1(dB) + 3 dB frequency

f B≈9.7MHz

Avcm1db f =1 kHz ≈20 log10RC

2 ro≈20 log10

0.061.4

=−27.3 dB

Avcm1db f =9.7MHz =−24.2dB

fB = - 3 dB A

vcm1 corner frequency



Simulate the 10 mA Design with 2 pF Parasitics

2 pF

2 pF

f B≈2.5MHz

-3dB common mode bandwidth with 2 pF base-emitter and base collector caps.

Avcm1dB f =1 kHz≈−27dB

Avcm1dB f =2.5MHz =−24.2dB

I REF=10mAI C1≈5mA

Parasitic caps reduces Av-cm(dB) break frequency f

B from 9.7 MHz to 2.5 MHz!

vi-cm

RC RC

RE RERB RB

Rref

RECALL: 2 pF is about the capacitance between 2 rows of Protoboard pins!



Observations

1). ro can be inferred from A

vcm.

2). Minimize parasitic capacitance around mirror transistor to in-crease common mode gain -3dB bandwidth.

3). Since no differential mode current flows through the mirror tran-sistor (Q3, i.e. r

o), it should have no effect on differential mode

performance.

differential amplifier cont.ese319/lecture_notes/lec_13_diffampl-ii_13.pdfdifferential amplifier...

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