Download - Differential Amplifier with Active Loadsese319/Lecture_Notes/Lec_16_DiffAmpActLoad_09.pdfESE319 Introduction to Microelectronics 2008 Kenneth R. Laker updated 02Nov09 KRL 1 Differential

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker updated 02Nov09 KRL

Differential Amplifier with Active Loads

● Active load basics● PNP BJT current mirror● Small signal model● Design example and simulation● Comparison of CMRR with resistive load design



Differential Amp – Active Loads Basics 1

Rc1 Rc2

Rb1 Rb2

Rref

Vee

VccIref

Vcg1Vcg2

Rref1

Rref2

Iref1

Iref2

-Vee

Vcc

Q1

Q3 Q4

Q5 Q6 Q7

Vc1

Q2

Vc2

Vi1 Vi2

RC1⇒ ro6

RC2⇒ r o7

PROBLEM: Op. Pt. VC1

, VC2

very sensitive to mismatch I

ref1 ≠ I

ref2.

+

+ +

+




IC2

, IC7

VBE2

VCE2

VCE2

VCE2

VBE2

slope = -1/RC2 slope = -1/r

o7

IC2

VCC

VCC




PROBLEM: Op. Pt. VC1

, VC2

very sensitive to mismatch I

ref1 ≠ I

ref2.

SOLUTION: all currents referenced to I

ref1. Op. Pt. sensitivity eliminated.

COST: output single-ended only. GOOD NEWS: CMRR is much improved over resistive-load differential amp single-ended CMRR.

Vc1 Vc2 Vc2Vc1



Quick Review - PNP BJT

I

IE

IC

IB

Note reference currentdirections!

Voltage bias equations:

I E=V CC−V BV EB

RE

I≈I E

10

I E=V CC−V B−0.7

RE

I E=V R1−0.7

RE=

I R1−0.7RE

0.7 V

VB

R1

R2

RE

RC



Quick Review - PNP BJT

iE

iC

iB

Usual Large signal equations:

iC=I S ev EB

V T

iB=iC

iE=iCiB=1

iC

iC= iE

R1

R2

RE

RC



Quick Review - PNP – Small Signal Model

ib

ie

ic

ib

ie

ic



PNP Mirror

I

I=V CC−0.7

RREF

Q1 = Q2: gm1

= gm2

= gmp

;Small signal model:

i x=v eb1

r pg mp veb1= 1

r pg mpveb1

i x= gmp

pg mpv eb1= p1

p g mp veb1=veb1

rep

I

Q1 Q2

veb

ix

B1 C1

E1

B2

E2

C2

Current through rπ and gmveb1:

re1

E1 E2

B1 C1 B2 C2

V CC=V EB1 I RREFr p=

p

gmb

rep=r p

p1

recall

rπ1 ro1gm1veb1 gm2veb1rπ2

ro2

ropgmpveb1rop rπ2p

rep

RREF

vEB2=v EB1

re1

= re2

= rep

; rπ1

= rπ2

= rπp

; β

1 = β

2 = β

p



Simplified Mid-band DM Small Signal Model

NOTE:1. Due to imbalance created by active load current mirror, only single-ended output is available from common collector of Q2 and Q4.2. Q3, Q4 emitter symmetry creates virtual ground at amplifier emitter connection.

Q1 Q2

Q3 Q4

vi-dm/2

vi-dm/2

VEE

VCC

I ieie

Q3 = Q4

vc4-dm

vc4-dm is single-ended output.

vi-dm/2 vi-dm/2

B3 C3

E3 E4

B4C4B1=C1

E1

B2 C2

E2

virtual groundv

e = 0, i = 0

i

vdm/2 vdm/2

vc4-dm

ie ie

Q1 = Q2

gm2veb1

gm3vi-dm/2

gm4vi-dm/2

rπ3

re1||ro1 rπ2 ro2

ro4 rπ4

ro3

vbe2 = veb1

ro

ve



Simplified DM Small Signal Model cont.

B3

E3

B4

E4

C3

E1 E2

B1=C1=B2 C4C2

vi-dm/2 vi-dm/2

virtual ground

re1∥ro1∥r 2∥ro3≈re1

vdm/2 vdm/2

combining parallel resistances to ground

Matched NPN: Q3 = Q4g m3=gm4=g mn

ro3=r o4=ron

Matched PNP Q1 = Q2gm1=gm2=gmp

ro1=r o2=r op

r1=r 2=r p

r3=r4=rn

re3=r e4=r en

re1=r e2=rep

gmpveb1

gmnvi-dm/2

gmnvi-dm/2

From previous slide

vc4-dm

rep

rπn rπnron

rop

re1∥ro1∥r 2∥ro3≈re1vc4-dm



Simplified DM Small Signal Model - cont.

veb1≈gmnvi−dm

2rep

Matched NPN & PNP transistors have beenassumed throughout.

ic2=g mp veb1=g mpgmnv i−dm

2r ep

where: rep= p1 p

1gmp

≈ 1gmp

ic2≈gmnvi−dm

2

vi-dm

/2 vi-dm

/2vdm/2 vdm/2

B1C2

C4

ic2

E1 E2

B3 B4

E3 E4 hence:

virtual ground

ic4

Determine ic2

and ic4

:

ic4=gmnvi−dm

2by inspection:

⇒ ic2≈ic4

gmp

veb1

Matched NPN: Q3 = Q4Matched PNP: Q1 = Q2

rep

gmn vi−dm /2

gmn vi−dm /2

vc4−dm

rop

ronrn

rn

From previous slide:re1∥r o1∥r o3∥r2≈re1=r ep

1

1

2

3

2

3



Simplified DM Small Signal Model - cont.

ic2≈gmn

v i−dm

2=ic4

vc4−dm=g mn Ro vi−dm

Av−dm4=vc4−dm

v i−dm=gmn rop∥ron

Ro=rop∥ron≠ ro∥ro=ro

2

Av−cm4=vc4−cm

v i−cm=−

rop

p r oalso


From previous slide:

Note that the resistors ro2 = rop(PNP)& ro4 = ron (NPN) are in parallel, andare driven by two nearly equal parallelcurrent sources.

vc4−dm=ic2ic4Ro=2 g mnv i−dm

2Rogmn vi−dm /2

gmn vi−dm /2

repvc4−dm



vi−cm=r n

1ie2 r o ie≈2 r o ie

veb1≈rep∥r pie=rep∥rbv i−cm

2 ro

vi-cm

B3 C3

E3 E4

B4C4B1=C1 = B2

E1 = E2

B2 C2

i = 2ie ro

ve

vcm vcm

vc4-cm

ie ie

re1∥ro1∥r2≈r ep∥r p

n ib n ib

ic4

ic4≈ie

ro3=r o4=ron

r3=r4=rn

re3=r e4=r en

Matched NPN: Q3 = Q4 Matched PNP: Q1 = Q2

ro1=r o2=r op

r1=r2=r pre1=r e2=rep

3=4=ngm1=gm2=gmp

gm3=gm4=gmn

1=2=p

gmpveb1

rep∥rb= p

g mp

p

gmp 1 p

p

gmp

p

gmp 1 p

=1

gmp

p2

1p∗

1 p

p1 p p

vo−cm=r opgmp veb1−ie=rop

2 r ogmprep∥rb−1v i−cm

rep∥rb p

gmp 1p∗

1 p

2p

veb1

ie gmp veb1−ic4

Simplified Mid-band CM Small Signal Model

vi-cm

re1∥ro1∥r2≈r ep∥r p

rn rnron ron

rop

ie≈vi−cm

2 ro



Simplified CM Small Signal Model - cont.

Av−cm4=vc4−cm

vi−cm=

rop

2 r ogmp

1gmp

p

2 p−1

.=r op

2 r o

p

2 p−1=

rop

2 r o

−22p

=> Av−cm4=−rop

2 pro≈−

r op

p r o

rep∥rb=1

gmp

p

2p


vo−cm=rop

2 rog mp rep∥rb−1v i−cm

From previous slide

re1∥ro1∥r2≈r ep∥r p gmpveb1

rn rnronron

rop

vi-cm vi-cm

vc4-cm

ro

n ib n ib

ieie

ic4

ve

ii



DM Diff Amp 2N3906 PNP Active Loads

vi-dm

vc2-dm

Av−dm2=vo−dm

v i−dm=gm ro2∥ro8

Av-dm2(dB) = 55.5 dB @ 1 kHz = 52.5 dB @ 1.43 MHz

Design: set R3 for IREF = 10 mA

R3=23.3V10 mA

=2.33 k

IREF

Matched NPN: Q1 = Q2 = Q3 = Q4

Matched PNP: Q7 = Q8



CM Diff Amp 2N3906 PNP Active Loads

Av−cm2=vc2−cm

v i−cm

vi-cm

vc2-cm

Av-cm2(dB) = -79.9 dB @ 1 kHz = -76.9 dB @ 0.54 MHzAv-dm2(dB) = 55.5 dB @ 1 kHz

CMRR=20 log10 ∣Av−dm2∣∣Av−cm2∣

.=Av−dm2 dB −Av−cm2dB

= 135.4 dB @ 1 k HzMatched NPN: Q1 = Q2

= Q3 = Q4Matched PNP: Q7 = Q8



CMRR Comparison

vc2

vi-dm/2

CMRR = 82.8 dB @ 1 kHz

v-dm/2 v-dm/2

vc2

CMRR = 135 dB @ 1 kHz

vi-dm/2vi-cm

vi-dm/2 vi-dm/2vi-cm

2.33k Ohm

I ref 10 mA

Av-cm2(dB) = -79.9 dB @ 1 kHzAv-dm2(dB) = 55.5 dB @ 1 kHz

Av−cm2 dB=20log10 0.043=−27.3dB @ 1 kHz

56k Ohm56k Ohm

Av-dm2(dB) = 55.5 dB @ 1 kHz



Summary

Active load advantages:1. Minimizes number of passive elements needed.2. Can produce very high gain in one stage.3. Much larger single-ended CMRR then single-ended

CMRR for resistive load differential amplifier.3. Inherent differential-to-single-ended conversion.

Active load disadvantages:1. No differential output available.

Download - Differential Amplifier with Active Loadsese319/Lecture_Notes/Lec_16_DiffAmpActLoad_09.pdfESE319 Introduction to Microelectronics 2008 Kenneth R. Laker updated 02Nov09 KRL 1 Differential

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