Differential Amplifier with Active Loadsese319/Lecture_Notes/Lec_16_DiffAmpActLoad_09.pdfESE319...
Transcript of Differential Amplifier with Active Loadsese319/Lecture_Notes/Lec_16_DiffAmpActLoad_09.pdfESE319...
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
ESE319 Introduction to Microelectronics
22008 Kenneth R. Laker updated 02Nov09 KRL
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.
+
+ +
+
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker updated 02Nov09 KRL
Differential Amp – Active Loads Basics 2
IC2
, IC7
VBE2
VCE2
VCE2
VCE2
VBE2
slope = -1/RC2 slope = -1/r
o7
IC2
VCC
VCC
ESE319 Introduction to Microelectronics
42008 Kenneth R. Laker updated 02Nov09 KRL
Differential Amp – Active Loads Basics 3
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
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker updated 02Nov09 KRL
Quick Review - PNP – Small Signal Model
ib
ie
ic
ib
ie
ic
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
112008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker updated 02Nov09 KRL
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
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q2
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
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker updated 02Nov09 KRL
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
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q2
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
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
172008 Kenneth R. Laker updated 02Nov09 KRL
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
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker updated 02Nov09 KRL
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.