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1S. C. Lin, EE National Chin-Yi University of Technology
Sedra/SmithMicroelectronic Circuits 6/E
Chapter 2: Operational Amplifiers
2S. C. Lin, EE National Chin-Yi University of Technology
【Outline】
2.1 The Ideal OP Amp
2.2 The Inverting Configuration
2.3 The Noninverting Configuration
2.4 Difference Amplifiers
2.5 Integrators and Differentiators
2.6 DC Imperfections
2.7 Effect of Finite Open-Loop Gain and Bandwidth on Circuit Performance
2.8 Large-Signal Operation of Op Amp
3S. C. Lin, EE National Chin-Yi University of Technology
2-1 The ideal op amp
CCV
CCV
CCV
CCV
4S. C. Lin, EE National Chin-Yi University of Technology
The properties associated with an ideal Amplifier are:
1. infinite voltage gain ( )
2. Infinite input impedance ( )
3. Zero output impedance( )
4. Output voltage when input voltages
5. Infinite bandwidth ( no delay of the signal through the amplifier)
inZ
vA
1 2V V0outZ
0outV
Low Outputstages
outZ OutputInverting input Differential
input stageIntermediate Amplifier stagesNoninverting
input
Figure 1. Block diagram of an operational amplifier
5S. C. Lin, EE National Chin-Yi University of Technology
2.1.2 Function and Characteristics of the ideal Op Amp
2v
1 0i
2 0i
1v
(virtual short circuit: 0, 0, )i i inv I R
6S. C. Lin, EE National Chin-Yi University of Technology
Some Specifications1. Open loop gain ( ): Usually several thousand.2. Input offset voltage ( ): Small, usually a few millivolts.3. Input offset current ( ): Usually between a few and several
hundred nanoamps.
4. Input resistance ( ):Typically greater than one megohm, but it can be as high as several hundred megohms.
5. Output resistance ( ): Usually less than a few hundred ohms.6. Slew rate ( S ): The maximum rate of output voltage change
given in volts per microsecond.
olA
osIosV
outR
inR
7. CMRR d
cm
AA
1BI
2BI1 2os B BI I I
7S. C. Lin, EE National Chin-Yi University of Technology
2.1.3 Differential and Common-Mode Signal
icmv
/ 2idv
/ 2idv
2v
1v
2 1
2 1
1
2
The difference input signal : The Common-mode input signal : / 2
/ 2
/ 2
id
id
icm
icm
icm id
icm id
vv v v
vv v v
v v vv v v
8S. C. Lin, EE National Chin-Yi University of Technology
2.2 The inverting configuration
11
1 1 1
1 1 1
1
1
2 1 0
(virtual short circuit: 0, 0, )
0
0
i i i
o f f
o
i i in
f
if
fo
i
v v
v
v viR R R
v v i R v i R
v RR
Rv
v vA
v I R
Gv R
fR
1R
1i
fi
iv
ov
2 1v v
2 1A v v
fR
1R
1i
fi
iv
1v
ov
9S. C. Lin, EE National Chin-Yi University of Technology
2.2.2 Effect of Finite Open-Loop Gain
fR
1R
1iiv
ovA
0
ov
11
1
1
1
1
1
1
( / )
/
/
/
1 (1 / )(2.
/
whe
5)
n , /
i o
i o
oo f
o i of
fo
i f
f
v v AiR
v v AR
vv i RAv v v A RA R
R RvGv R R A
A G R R
CCV
CCV
iv
ov
1G /fR R
CCVG
CCVG
10S. C. Lin, EE National Chin-Yi University of Technology
/o iv v2
1 2 21
2 1
23
3 1 3
24 2 3
1 1 3
0 2
0 4
4 4
1
4
22 4
1 1
2
1 3
3
0
/0
1
ix
i
xi
ii
i
i
x
i i
v Rv v i RR
v R Rv Ri v
R R Rv Ri i i vR R R
v v i R
v v RR v RR R R R
v R R Rv R R R
Example2.2: Assuming the op amp to be ideal, derive an expression for closed-loop gain =?
11
1 1
2 11
0 (virtual short)
i i
i
o oi
v v viR
v
Rvi
A
R
vv
i
1R
1iiv
2i3i
4i2R 4R
3R
xxv
1v
Solution
11S. C. Lin, EE National Chin-Yi University of Technology
1 21 2
1 2
1 21 2 3
1
1 2 1 2
2
1 21 2
, , ,
( )
If The
n
)
( )
(2.7
n on f
n f
o nf
f n
f f fo
n
nn
f o n
v vv vi i I IR R R R
v vv vI I I IR R R R
R R R R
R R Rv v v v
Rv v v
R Rv
2.2.4 An Important Application - The Weighted Summer
2R
fR
fi
1R
1i
ni
2i
1v
2v
nv
ov
nR
12S. C. Lin, EE National Chin-Yi University of Technology
2.3 The Noninverting ConfigurationfR
1R
1i
fi
iv
ov
11
11 1
(
,
1 )
i o if
f
if o i
o fi
f
v v vi iR R
v v vi iR
Rv v
RR
CCV
CCV
iv
1G fR / R
CCVG
CCVG
13S. C. Lin, EE National Chin-Yi University of Technology
11
11
1 1
1
1
1
( / )( / ) /( / ) , ,
/( / )
( ) (1 )
1 ( / ) (2. 1 ( / )
1
when , 1
)
(
1
/ )
1
i o
o i o o i oi of
f f
o i oi of
f
i f o f
fo
fi
f
v v v Av v v A v v v Av v Ai i
R R R
v v v Av v Ai iR R
v A R R v A R R
R RvG R RvA
A G R R
=
2.3.3 Effect of Finite Open-Loop Gain
14S. C. Lin, EE National Chin-Yi University of Technology
2.3.4 The Voltage Follower
CCV
CCV
iv
ov
( ) 1G
CCV
CCV
ov
, , 0o i in outv v R R
15S. C. Lin, EE National Chin-Yi University of Technology
2.4 Difference Amplifiers
icmv
/ 2idv
/ 2idv
1 2id i iv v v
1 2 / 2icm i iv v v
1 / 2i icm idv v v
2 / 2i icm idv v v
11 1CMRR
CMRR , CMR (2.14R 20log )
cm icmo d id cm icm d id d id
d id
cm icm icmd id d id
d id id
d ddB
cm cm
A vv A v A v A v A vA v
A v vA v A vA v v
A AA A
16S. C. Lin, EE National Chin-Yi University of Technology
2.4.1 A Single Op-Amp Difference Amplifier
4R
2R
3R
1R
ov
1iv
2iv
4
2
21
1 3
2 22 1
3 4 1 1
2 4
1
2 1
' ''
(2)Assume 0,
then ''
If ,
then
1
i
O i
o
a b
b bo i i id
a a
bd
i
a
O O
i
v v vR R Rv v
R R
R R R R R RR Rv v v vR R
RAR
R R
vRv vR
(2 .17)
1
2
1
4 22
3 4 1
(1)Assume 0,
then ' 1
1
i
O
i
v
Rv vR
R RvR R R
17S. C. Lin, EE National Chin-Yi University of Technology
A common-mode signal applied at the input
4R
2R
3R
1R
4
3 4icm
R vR R
2i
1i
icmv
ov4
2 23 4
34 2
3 4 1 3 4
34 2
3 4 1 4
34 2
3 4 1 4
1 3 2 4If ,
1
1 ,
.
(2.1
)0
9
o icm
icm icm
icm
ocm
ic
c
m
m
Rv v i RR R
RR Rv vR R R R R
RR R vR R R R
v RR RAv R R R R
R R R R A
41 2
1 3 4
3
1 3 4
1
1
icm icm
icm
Ri i v vR R R
R vR R R
18S. C. Lin, EE National Chin-Yi University of Technology
1 1 1
1 (2 2.20)
idid
i
id i
id
vRi
v R i R iR R
2R
1R
2i
1i
idvov
1R
2R1i
idR
19S. C. Lin, EE National Chin-Yi University of Technology
4R
4 22 1
3 1
21 ( )oR Rv v vR R
1R
3R
2R
1v
2v
3R
4R
1 2v v 1 2 1( ) /v v R
1 2 1( ) /v v R
21 2
1
2( ) 1 Rv vR
1ov
2ov
1R
1fR 1vR
2R
0V
1 2 1( ) /v v R
0V
0A
0A
ov
2.4.2 A Superior Circuit-The Instrumentation Amplifier
20S. C. Lin, EE National Chin-Yi University of Technology
2.5 Integrators and Differentiators
iV
oV
1Z
2Z
2
1
o
i
V ZV Z
21S. C. Lin, EE National Chin-Yi University of Technology
Example 2.4For the circuit in Fig.2.23, (a) derive the transfer function. ( ) / ( )o iv s v s
Solution:
(a)1
2 1
2
2
1 2
2
02 2
2
( ) 1( ) ( ) ( )
1
/
1
11
o
i
v sv s z s Y s
R sCR
R RsC R
ωC R
2R
1R
2C
iv
ov
22S. C. Lin, EE National Chin-Yi University of Technology
02 2
(c) Evaluate 3-dB frequency
the 3-dB frequency 1ωC R
2
1
(b) find the dc gain
The dc gain K RR
2 1
(d) design the circuit to obtain a dc gain of 40 dB, a 3-dB frequency of 1 kHz, and input resistance of 1 k .
In order to obtain a dc gain of 40 dB, we select R /R 100.
Solu
For
tion:
an in
1
2 0 2
0 2 3 32 2 0 2
put resistance of 1 k , we select R 1 k , and thus R 100k , for a 3-dB frequency 1kHz, we select C from
1 1 1 1.59 F. 2 1 10 100 10
f
ω C nC R ω R π
23S. C. Lin, EE National Chin-Yi University of Technology
2.5.2 Inverting Integrator
RC
ov
1i
1i
iv
Cv
10
10
1( ) ( )
1( ) ( ) ( )
t
C C
t
o C C
v t V i tC
v t v t V i tC
24S. C. Lin, EE National Chin-Yi University of Technology
12
We can be described alternatively in the frequency domain 1by subtituting ( ) , and ( )
( )Z s R Y s sC
Z s
1
1
1
1
1 ,
(
/ 90
the unity gai
)
n frequency as1
1( )
( ) 1
( )
o
o
oo to
t
t
v
v v vv
sv s sRC
v jv j j RC
RC
RC
1RC
20dB/decade
(dB)o
i
VV
25S. C. Lin, EE National Chin-Yi University of Technology
1
2
1
( )1( ) 1 1
( ) /1( ) 1
1the Corner frequency as ,
the dc gain as /F
F
F
F
F
F
o FF
F
Z s RRZ ssR
R
CscR
Rv s R RsR Cv s C
R
sR
R
R
C
Fig2.42 The Miller integrator with a large resistance RF connectedin parallel with C in order to provide negative feedback and
hence finite gain at dc
FR
1R
C
( )iv t
( )ov t
26S. C. Lin, EE National Chin-Yi University of Technology
i oi C C
o
R
i oR C
i
dv vdQQ Cv , i i C , Idt dt R
dv vi
dvv RC
dti C
dt R
iv
R
C
ov
Ri
Ci
2.5.3 The Op-Amp Differentiator
27S. C. Lin, EE National Chin-Yi University of Technology
1 2
The frequency domain transfer function of the differentiator circuit 1can be found by subtituting ( ) , and ( )Z s Z s R
sC
1
1
1
1
,
/ 90the unity gain frequency as
1
( ) ( )( ) ( )
o
t
oo
t
o
t
o
v R
v s v jsRC j RCv s v
Cv
v v
RC
j
1/ RC
20dB/decade
/ (dB)o iV V
28S. C. Lin, EE National Chin-Yi University of Technology
2.6.1 Offset Voltage
fR
1R
osV
Offset-free op amp
2
1
1o osRV VR
oV
2.6 DC Imperfections
+V
V
29S. C. Lin, EE National Chin-Yi University of Technology
2R
1RC
( )iv t
( )ov t
2R
o osV V
osV
30S. C. Lin, EE National Chin-Yi University of Technology
2.6.2 Input Bais and Offset Currents
1BI
2BI
1 2
1 2
2B B
B
OS B B
I II
I I I
31S. C. Lin, EE National Chin-Yi University of Technology
2R
1R
1BI
2BI
0
1BI
1 2o BV I R
1 2 2O B BV I R I R
32S. C. Lin, EE National Chin-Yi University of Technology
2R
1R
3R
1BI
2BI
31 2
1B B
RI IR
2 3
1
BI RR
2BIoV
2 3BI R
2 3 2 1 2 3 1
1 2
2 3 2 1
3
2 1 23 1 2
2 1 1 2
/Consider first case , which results in
1 /
Thus we can reduce to zero by selecting such that
//1 /
O B B B
B B B
O B
O
V I R R I I R RI I I
V I R R R R
V RR R RR R RR R R R
33S. C. Lin, EE National Chin-Yi University of Technology
2.7 Effect of Finite Open-Loop Gain and Bandwidth on Circuit Performance
2.7.1 Frequency Dependence of the Open-loop Gain
0 0 0
0
t 0
t t
( ) ( ) ( )1 / 1 /
( )
unity gain frequency
( ) ( )
b
b b
b
b
t
A A A ωA s A jω A jωs ω jω ω jω
A ωA jωω
ω A ωω ω fA jω A jωjω ω f
34S. C. Lin, EE National Chin-Yi University of Technology
210 310 410 610 710510
0A( )A dB
bf tf
6dB/Octave or
20dB/decade
3dB
35S. C. Lin, EE National Chin-Yi University of Technology
2.7.2 Frequency Response of Closed-loop Amplifiers
1 1
1 1
1
1
0
1
1
0
0
/ /( )( )1 ( / ) 1 ( / )
1 1
/1 ( / ) 1 ( / )
1
/
1 ( /
( ) / 1 ( / )
)1
f fo o
i if f
f
f b
f
f
bA s A
R R R Rv v sv v sR R R R
R RR R s
AR R
R
s
RA
10
0
1
where 1, ( / ) 1 ( / )
f
b fs R RA
RA
R
The inverting amplifier transfer function
36S. C. Lin, EE National Chin-Yi University of Technology
1
1
1
1
3dB1
1 /
1 /1
( )
1 /( )( ) 1
1 /
1 /
fo
i f
f t
f
o
i
t
f
R Rvv R R
A s
R Rv ssv s
R R
R R
Similarly, the noninverting amplifier transfer function:
13dB
1 1
0
0
/,
1 ( / ) 1 ( / )1
fo t t
fi f
b
b
R RvR Rv R RA
As
37S. C. Lin, EE National Chin-Yi University of Technology
2.8 Large-Signal Operation of Op AmpExample 2.7: Consider the noninverting amplifier circuit shown in Fig. below. The op amp is specified to have output saturation voltages of ±13V, And output current limits of ±20mA.
(a) Find VP =1V and RL=1k,specify the signal resulting at the output of the amplifier.
2
1
Sol:
1+ =10
10V 10mA1kΩ
the feedback current will be 10V =1mA
(9+1)kΩthe total output current is 11mA,well under its limit of 20 mA.
L
F
RGR
i
i
2 9kR
1 1kR
LR
1I
fI
iv
ov
0PV
oI LI
38S. C. Lin, EE National Chin-Yi University of Technology
(b) Find VP =1.5V and RL=1k, specify the signal resulting at the output of the amplifier.
Sol: is increased to 1.5V , will saturate at 13V
13V 13V 13mA, =1.3mA1kΩ (9+1)kΩ
= 14.3 mA, well under its limit of 20 mA.
p o
L F
o
V V
i i
i
ov
ccV
ccV
13V
15V
13V
15V
39S. C. Lin, EE National Chin-Yi University of Technology
(d) Find Vp=1V, what is the lowest value of RL for which an undistorted sine-wave output is obtained?
(max)min
min
Sol: 1.5V ,
10V 10V 20mA=9k +1k
=526 .
p
oL
L
V
iR
R
Sol: The maximum value of for undistoted sine-wave output 1.3V. The output will be a 13-V peak sine-wave. The op-amp output current at peak will be 14.3mA.
PV
(c) Find RL=1k, what is the maximum value of VP for which an undistorted sine-wave output is obtained?
40S. C. Lin, EE National Chin-Yi University of Technology
2.8.3 Slew Rate
iv
ov
iv
t
t
ov
t
ov
V
V
V
Slop V SRt
Slop SR
( )d
( )b
( )cmax
11 /
( ) (1 )t
o
o
i tω t
o
dvSRdt
vv s ω
v t V e
41S. C. Lin, EE National Chin-Yi University of Technology
2.8.4 Full-Power Bandwidth
max
max
max
ˆ sin
ˆ cos
2
The Maximum amplitude of the undistortedoutput sinusoild is given by
i i
ii
M o
Mo
Mo o
v V tdv V tdtSR V
SRfV
V V
42S. C. Lin, EE National Chin-Yi University of Technology
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