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ESE319 Introduction to Microelectronics
12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 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 (based on P. V. Lopresti 2006) updated 18Nov08 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
Vcg1
Q2
Vcg2
Vi1 Vi2
RC1⇒ r o6
RC2⇒ r o7
PROBLEM: Op. Pt. VCG1
, VCG2
very sensitive to mismatch I
ref1 ≠ I
ref2.
+
+ +
+
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 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 (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Differential Amp – Active Loads Basics 3
PROBLEM: Op. Pt. VCG1
, VCG2
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.
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Quick Review - PNP BJT
I
IE
IC
IB
Note reference currentdirections!
Voltage bias equations:
I E=VCC−V BG0.7
RE
I≈I C
10
I E=VCC−V BG−0.7
RE
I E=VR1−0.7
RE=
I∗R1−0.7RE
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Quick Review - PNP BJT
I
IE
IC
IB
Usual Large signal equations:
iC=I S evEB
VT
iB=iC
iE=iCiB=1
iC
iC= iE
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Quick Review - PNP – Small Signal Model
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
PNP Mirror
I
I=VCC−0.7
RREF
Current sourcemodel Q1 = Q2:
Small signal model:
i x=veb1
r
gm veb1= 1r
gmveb1
i x= g m
g mveb1= 1
g mveb1=veb1
r e
I
Q1 Q2
veb
ix
B1 C1
E1
B2
E2
C2
Current through rπ and gmveb1:
re1
E1 E2
B1 C1 B2 C2
VCC=VEB1I RREFr=
g m
r e=r
1
recall
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified Midband 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. Symmetry creates virtual ground at amplifier emitter connection.
Q1 Q2
Q3 Q4
vdm/2
vdm/2
VEE
VCC
I ieie
Q3 = Q4
vo-dm
vo-dm is single-ended output.
vdm/2 vdm/2
B3 C3
E3 E4
B4C4B1=C1
E1
B2 C2
E2
virtual groundv
eg = 0 , i = 0
iro
veg
vdm/2 vdm/2
vo-dm
ie ie
Q1 = Q2
gm2 veb1
gm3 vdm/2 gm4 vdm/2
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified DM Small Signal Model cont.
B3
B3
E3
B4
E4
C3
E1 E2
B1=C1=B2 C4C2
E3 E4
B4B1=C1
E1 E2
B2 C4C2
C3
vdm/2 vdm/2
vdm/2 vdm/2
From previous slide
virtual ground
re1∥ro1∥ro3∥rpi2≈re1
vdm/2 vdm/2
vdm/2 vdm/2
vo-dm
virtual groundcombining parallel resistances to ground
Matched NPN: Q3 = Q4g m3=gm4=g mn
r o3=r o4=r on
Matched PNP Q1 = Q2gm1=gm2=gmp
r o1=r o2=r op
r1=r 2=r p
r3=r 4=rn
r e3=r e4=r en
r e1=r e2=r ep
gm2 veb1
gm3 vdm/2 gm4 vdm/2
ESE319 Introduction to Microelectronics
112008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified DM Small Signal Model - cont.
vdm/2
B3 B4
E3E3 E4
C4B1
E1 E2
C2
vdm/2 vdm/2
vo-dm
virtual ground
From previous slide with re1∥ro1∥ro3∥rpi2≈re1
vdm/2
gm3 vdm/2 gm4 vdm/2
gm2 veb1
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q4
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified DM Small Signal Model - cont.
veb1≈gmnvdm
2r ep
Matched transistors has been assumedthroughout.
ic2=g mpveb1=g mpgmnvdm
2r ep
where: r ep=1
1gmp
≈1
gmp
ic2≈gmnvdm
2
vdm/2 vdm/2vdm/2 vdm/2
B1C2
C4
ic2E1 E2
B3 B4
E3 E4
vo-dm
hence:virtual ground
ic4
Determine ic2
and ic4
:
ic4=gmnvdm
2by inspection:
⇒ic2≈ic4
gm3 vdm/2 gm4 vdm/2
gm2 veb1
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q4
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified DM Small Signal Model - cont.Note that the Early resistors ro2 = rop(PNP) & ro4 = ron (NPN) are in parallel, and are driven by two nearly equalparallelcurrent sources.
ic2≈gmnvdm
2=ic4
vo−dm=ic2ic4Ro=2gmnvdm
2Ro
vo−dm=2 gmn Rovdm
2
Gdm=vo−dm
vdm=gmn r op∥r on
vdm/2 vdm/2
vdm/2 vdm/2
B1 C2C4
Ro=r op∥r on≠ r o∥r o=r o
2vo-dm
ic2 = ic
ic4 = ic
ic
ic
From previous slide:
virtual ground
E1
Gcm=vo−cm
vcm=−
r op
p r oalso
gm3 vdm/2 gm4 vdm/2
gm2 veb1
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q4
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified Midband CM Small Signal Model
vcm=rn
1ie2 r o ie≈2 r o ie⇒ ie≈
vcm
2 r o
veb1≈r ep∥r pie=r ep∥rbvcm
2 r o
vcm vcm
B3 C3
E3 E4
B4C4B1=C1 = B2
E1 = E2
B2 C2
iro
veg
vcm vcm
vo-cm
ie ie
re1∥ro1∥rpi2≈re1∥rpi2
n ib n ib
ic4
ic4≈ie
r o3=r o4=r on
r3=r 4=rn
r e3=r e4=r en
Matched NPN: Q3 = Q4 Matched PNP: Q1 = Q2
r o1=r o2=r op
r1=r2=r pr e1=r e2=r ep
3=4=n
gm1=gm2=gmpgm3=gm4=gmn
1=2=p
gm2 veb1
r ep∥rb= p
g mp
p
gmp1 p
p
gmp
p
gmp1 p
=1
gmp
p2
1p∗
1 p
p1 p p
vo−cm=r op g mp veb1−ie=r op
2 r og mp r ep∥rb−1vcm
rb∥r ep p
gmp1p∗
1 p
2p
veb1ie
re1∥rpi2
gm2veb1−ic4
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
Simplified CM Small Signal Model - cont.
Gcm=vo−cm
vcm=
r op
2 r ogmp
1g mp
p
2 p−1
.=r op
2 r o
p
2 p−1=
r op
2 r o
−22p
=> Gcm=−r op
2pr o≈−
r op
p r o
rb∥r ep=1
gmp
p
2p
Matched NPN: Q3 = Q4Matched PNP: Q1 = Q4
vo−cm=r op
2 r ogmp r b∥r ep−1vcm
From previous slide
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
DM Diff Amp 2N3906 PNP Active Loads
vin-dm
vo-dm
Adm=vo−dm
vin−dm=gmr o2∥r o8
Adm = 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
172008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
CM Diff Amp 2N3906 PNP Active Loads
Acm=vo−cm
vin−cm
vin-cm
vo-cm
Acm = -79.9 dB @ 1 kHz = -76.9 dB @ 0.54 MHz
Adm = 55.5 dB @ 1 kHz
CMRR=20 log10 ∣Adm∣∣Acm∣
Adm−dB−Acm−dB== 135.4 dB @ 1 k Hz
Matched NPN: Q1 = Q2 = Q3 = Q4
Matched PNP: Q7 = Q8
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
CMRR Comparison
vo
v-dm/2 v-dm/2
CMRR = 66 dB @ 1 kHz
v-dm/2 v-dm/2
vo
CMRR = 135 dB @ 1 kHz
ESE319 Introduction to Microelectronics
192008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 18Nov08 KRL
SummaryActive 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 advantages:1. No differential output available.
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