microelectric circuit by meiling chen 1 lecture 11 operational amplifiers
TRANSCRIPT
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Microelectric Circuit by Meiling CHEN
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Lecture 11Lecture 11Operational AmplifiersOperational Amplifiers
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Microelectric Circuit by Meiling CHEN
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Topics Topics • Ideal op Amplifiers• Ideal OPA circuits analysis• Non-ideal op amplifiers• Non-ideal OPA circuit analysis
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Ideal operational amplifier
1. Infinite input impedance2. Zero output impedance3. Infinite bandwidth4. Infinite open-loop gain5. Zero common-mode gain (infinite CMRR)
Ideal OPA characters
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Microelectric Circuit by Meiling CHEN
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)( VVRuGV mo
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VCC
VEE
2
1
X1UA741
VCC
VEE4
3
X2UA741
R1
R2
stableVVV
VVAVVV
VVAV
VIRV
oo
oo
o
o
)(
)(
Negative feedback
Positive feedback
unstableVVVV
VVVV
VVAV
VIRV
satoo
satoo
o
o
)(
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CMRR
BW
A
R
R
o
in
0
Non-ideal cases
CMRR
BW
VVA
R
iiR
o
in
)(
0
)0(
Ideal OPA characters
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Consider finite open-loop gain
A
)2()(
)1(0
0
21
21
AvvvAv
R
vv
R
vv
vv
A
oi
1
2
12
12
/)/1(1
/
R
R
v
v
Aif
ARR
RR
v
v
i
o
i
o
A
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Input and output resistance
I
Oain v
vG
1
2
21
000
R
R
v
vG
R
v
R
v
I
Oain
OI
Example 2.11Ri
vR
i
iin
MR
MRif
100
1
2
1
An impractically large valueSo we may have the problem of input resistance.
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0I
VRout
V
I
)( VVAV
Viv
0oR
2R
1R
0
Output resistance
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Example 2.2
)1(
)2(000
)1(00
3
4
2
4
1
2
21
432
R
R
R
R
R
R
v
v
R
v
R
v
R
vv
R
v
R
v
i
o
XI
oXXX
Comparing with Example 2.1Design a amplifier with a gain –100 and an input resistance of 1M.
MRkR
MRMR
R
R
R
R
R
R
v
v
i
o
1,2.10
1,1
)1(
43
21
3
4
2
4
1
2
Example 2.2:
MR
MR
100
1
2
1
Example 2.1:
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Exercise 2.6
Vk
kv
k
v
k
vk
vv
k
v
oo
o
101
100
101
1
)2(0
)1(0101
1
mAii
mAi
mAk
i
mAk
i
Find
L
1
11
110
)10(0
101
10
21
0
2
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Weighted summer
nn
fffo
f
o
n
n
f
o
n
n
vR
Rv
R
Rv
R
Rv
R
v
R
v
R
v
R
v
R
v
R
v
R
v
R
v
22
11
2
2
1
1
2
2
1
1 )1(00000
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Exercise D2.8
xv
)()(
)(
)(
22
11
44
33
44
33
22
11
vR
Rv
R
R
R
Rv
R
Rv
R
Rv
vR
Rv
R
Rv
R
Rv
vR
Rv
R
Rv
aa
b
ccco
xb
ccco
aax
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Non-inverting amplifier
Io
oII
vR
Rv
R
vv
R
v
)1(
00
1
2
21
Ideal case:
Non-ideal case: A
)()1(
)2.().........(
)1....(00
21
21
2
21
RR
RR
R
vv
vvAv
R
vv
R
v
o
io
oii
A
v
v
RR
RR
i
o
1
2
1
2
11
1
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Voltage follower
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Exercise D2.9 Exercise D2.13
)32
)(32
32)(
9
1
1
1(9
)9
1
1
1(9)1(
)32
()3
1
2
1()2(
)2(032
)1(091
0
21
21
21
k
v
k
v
kk
kk
kkk
vkk
kv
k
v
k
v
kkv
k
vv
k
vvk
vv
k
v
vv
o
o
Vkkk
v
k
v
k
Vvv
o
o
109)9
1
1
1(
)1(09
1
1
01
1
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CMRR ( Common-Mode Rejection ratio )
cm
d
A
ACMRR log20
icmcmiddo vAvAv
Common-mode inputDifferent-mode input
Why use difference amplifier ?
)(21
vvv
vvv
icm
id
DC
DC
DC
DC
DCV
V
V V
V Vicmv
2idv
2idv
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Difference Amplifier
2343
11
2
212
21
12
43
2
21
1
1)
11()2(
)11
()()1(
)2(0
)1(0
i
ii
o
i
oi
vRRR
v
vR
Rv
RRR
R
v
R
vvRv
R
v
R
vv
R
vv
R
vv
vv
Method I:
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)()(
),()1(
12121
221
241321
22
1
2
43
42
11
21
iidiiooo
iio
io
vvAvvR
Rvvv
RRRRletvR
Rv
R
R
RR
Rv
vR
Rv
Different-mode inputMethod I: superposition
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Common-mode input
0,
]1[
][
1][
1
2413
1
2
4
3
43
4
1
2
43
3
43
421
2243
4
143
3
43
4
11
cm
cm
icmo
icmo
icmicmicm
ARRRR
R
R
R
R
RR
RA
vR
R
RR
R
RR
Rvii
RivRR
Rv
vRRR
Rv
RR
Rv
Ri
CMRR=infinite
0, 2413 cmARRRRif
CMRR=infinite
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Microelectric Circuit by Meiling CHEN
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1)(
111111
2
2
RR
iRRiRiv
diffin
id
Consider the problem of input resistance
Differential mode input resistance
ido vR
Rv
1
2 oin vRRif 1
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Instrumentation Amplifier
Not is differential mode, common mode input can be pass.
idv
idv
))(1(
))(1(
121
2
3
4
121
2
3
4
iio
iiid
ido
vvR
R
R
Rv
vvR
Rv
vR
Rv
)1(.2
.1
1
2
3
4
R
R
R
RA
R
d
in
Big differential gain
Defects:1. Common mode gain=differential mo
de gain.2. Resistance have to match.
sato vv
))(1( 121
2iiid vv
R
Rv
21 RandR
Advantages:
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Remove the point x
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DC non-ideal characters
1. Finite open loop gain ( finite CMRR)2. Finite BW3. Offset voltage4. Input bias and offset current
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Frequency response (open-loop)
bot
ob
t
b
o
A
AdBsA
sA
:01
)(
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Frequency response (closed loop)
ARRR
R
v
v
i
o
/)1(11
2
1
2
b
o
sA
sA
1)(
)1/(1
)1/()1(
11
1
2
1
2
1
21
2
0
1
2
RRsR
R
RRs
RR
A
RR
v
v
tt
i
o
1
23
1 RRt
dB
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Output voltage saturation
mAi
Vv
o
o
20
13
max
max
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Slew rate
)/(max sVdt
dvSR o
)1()()(
)(
),0(1
1
/1
11
1
12
1
2
1
2
to
tt
to
i
t
i
o
b
o
i
o
teVtvs
V
s
V
s
V
ssV
stepunitVvlet
RRsv
v
s
AA
AR
RR
R
v
v
bot A
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Offset voltage
Output = + saturation or - saturation
mVvos 5~1
From the component mismatches in the input differential stage
May be positive or negative
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1. Reduce the allowable signal swing2. When input is dc we would not know th
e output is due to vos or signals
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Solution 1: Offset-nulling terminals
Solution 2: Capacitive coupling (only ac signal be amplified)
)1(1
2
R
Rvv oso
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Microelectric Circuit by Meiling CHEN
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Offset current
Input bias currents21
21
2
BBos
BBB
III
III
Input offset currentnAInAI osB 10,100
221 RRIV Bo
Upper limit R2
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Solution : introducing R3
0
//)1(
)]1([
)(
21
1
2
23
1
232
21
1
321232
o
Bo
BBB
BBBo
V
RR
RR
RRchoose
R
RRRIV
IIIif
R
RIIRRIV
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Inverse amplifier
Non-inverse amplifier
Reduce the effect of Vos
Reduce the effect of Ios
Reduce the effect of Vos
Reduce the effect of Ios
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1
2
21
000
Z
Z
v
vG
Z
v
Z
v
I
O
OI
Integrator & Differentiator
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1202
122
221
)()()(
0)](0[)(0)(0
R
v
R
vvsC
R
tv
R
tv
dt
tdvC
dt
tvdC
R
tv
R
tv
io
ioo
ooi
220
22
1
2
122
11
2
1
1
1
RC
RsC
RR
v
v
RsCRRZ
Z
v
v
I
O
I
O
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Miller Integrator
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Consider Vos offset voltage
Consider Ios offset current
satV
satV
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Solution :
osV
tos
oso
osCR
RCosos
R
Ro
R
R
R
RR
os
o
f
oosoosos
CfR
ff
ff
f
eRC
VVtv
Vs
VVsC
v
sCsC
sC
V
v
R
vV
sC
vV
R
V
1
)(
]1[
]1[
01
1
1
1
1
1
1
1
11
tRC
VVtv os
oso )(
osCR
RC
osf
f
ossCf
o
Vs
VCsRR
R
VR
Rsv
f
)1(
)1
11(
)//
1()(
1
1
1
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Microelectric Circuit by Meiling CHEN
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fR
tCR
BBo
f
BB
f
BB
f
fB
o
fBo
f
f
oBoBB
feCIRItv
CRs
CIRI
RsC
IRI
RsC
RsC
RRI
v
RsC
RRIv
RsC
R
vRI
sC
vRI
R
RI
1
22
22
22
2
2
222
)()(
)1
()1
()1
(
]11
[
]11
[)1
(
0/1
tC
IRIv os
Bo 2
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Microelectric Circuit by Meiling CHEN
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Example 2.7 sketch output response
VV
MRR
nFC
kR
sat
FF
13
1,
10
10
Case I: FR
Case II: MRF 1
dttvRC
tvt
io )(1
)(0
F
F
I
O
RsC
RR
v
v
2
1
1
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Differentiator
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Problem *2.31
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Problem *2.43
]1248[16
]1248[80
5
]80402010
[5
00
80
50
40
50
20
50
10
50
080402010
0123
01230
01230
00123
00123
SSSSk
R
SSSSk
RV
k
S
k
S
k
S
k
SRV
R
V
k
S
k
S
k
S
k
S
R
VV
k
SV
k
SV
k
SV
k
SV
f
f
f
f
f
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Problem **2.69
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Problem *2.77
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Problem *2.78
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Problem C*2.126
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Non-ideal OP amplifiers1. Type A: Finite open-loop gain (unknown)2. Type B: Finite open-loop gain = K3. Type C:
1. Infinite input impedance2. Zero output impedance3. Zero common-mode gain4. Infinite open-loop gain5. Infinite bandwidth
Ideal OPA characters
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Type A: Finite open-loop gain (unknown)
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Type B: Finite open-loop gain =K
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Type C: iov RRA ,0,
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