### R 1

```Operational Amplifiers
1
Introduction
(The first IC op amp was introduced in mid-1960s.)
Why op amp is so popular ?
 We can do almost anything with op amp, such
as, summation, subtraction, amplification,
differentiation, integration, ….
 Op amp has almost ideal characteristics.
 Op amp works almost same as it is expected.
What’s inside of op amp ?
Operational Amplifier
You should be able to design nontrivial
circuit by the end of this chapter !
Chap. 9, complex but useful for high level engineer.
2.1 The Ideal Op Amp
2.1.1 The Op-Amp Terminals
No terminal of op-amp package
is physically connected to
ground.
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2.1.2 Function and Characteristics of the Ideal Op Amp
• Op amp is a differential-input single ended-output amplifier.
• Op amp is a directly-coupled or dc amplifier.
- useful, but can cause some serious practical problems
o  A(2  1 )
A: open-loop gain, differential gain
2.1.3 Differential and Common-mode Signals
differential input signal:  Id  2  1
Common-mode input signal:  Icm 
TABLE 2.1 Characteristic of the ideal Op Amp
1. Infinite input impedance
2. Zero output impedance
3. Zero common-mode gain or, equivalently,
infinite common-mode rejection
4. Infinite open-loop gain A
(How are we going to use it?)
(Feedback, closed loop)
5. Infinite bandwidth
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1   Icm   Id /2
2   Icm   Id /2
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(2.1)
1
(   2 ) (2.2)
2 1
(2.3)
(2.4)
3
2.2 The Inverting Configuration
Negative feedback
Figure 2.5 The inverting closed-loop configuration.
(Output is 180o phase-shifted w.r.t the input.)
Figure 2.12 The noninverting configuration. p.77 Sect.2.3
(Output is in phase with the input.)

Let's find G 
I
2.2.1 The closed-Loop Gain
Ideal op amp: open loop gain A = ∞

If there is a finite output   ,  2  1    0
1
1  2  0
o
2
A
The terminal 1and 2 are virtually shorted.
The terminal 1 is a virtual ground !
The virtual ground is not an actual ground. Do not short
the inverting input to ground to simplify analysis !
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Procedure to find output voltage υo.
①
Closed Loop gain: G 
0
R
 2
I
R1
• υo will be 180o phase-shifted with respect to the input wave. Inverting amplifier !
• We can make the closed-loop gain as accurate as we want by selecting passive components
of appropriate accuracy.
• The closed-loop gain is (ideally) independent to the op-amp gain.
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2.2.2 The Effect of Finite Open-Loop Gain (A≠∞)
i1 
 I  ( 0 / A)
R1

 I  0 / A
R1
0  
 2  1  0  1  1  0 / A
①
1  0 / A
G
0
A
 i1 R2  
0   I  0 / A 
A


R1
 R2

0
 R2 / R1

 I 1  (1  R2 / R1 ) / A
(2.5)
EXAMPLE 2.1 p.72
(a ) R1  1 k , R2  100 k , A=10 3 ,104 ,105 , % error of G to ideal G= ?
 I  0.1 V
(b) A changes from 100,000 to 50,000 (50% reduction),
% change of G = ?
A
3
103
10
4
10
104
5
10
105
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G  ( R2 / R1 )

 100
( R2 / R1 )

G
90.83
99.00
99.00
-9.17%
-1.00%
-0.10%
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υ11
V
-9.08mV
-0.99mV
-0.10mV
6
The production spread in the value of open-loop gain A between opamp units of the same type (part number) is very wide.
To minimize the effect of the finite open-loop gain A,
R
we have to make 1  2
A
R1

 R2 / R1
G 0 
(2.5)
 I 1  (1  R2 / R1 ) / A
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2.2.3 Input and output Resistance of Inverting Amplifier (not op-amp)
What would happen if R1 is too small ?
R
①
R2
G
( R  R1 )
Ideal input resistance
I
I
R


 R1
of inverting amplifier, i
i1  I / R1
not op-amp.
For high gain (- R2 / R1), we need small R1,
otherwise, R2 would be impractically large.
Therefore, input resistance R1 should be much
larger than output resistance of previous stage.
Unfortunately, the internal resistance of most
sensors are large.
Solution: example 2.2
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EXAMPLE 2.2 p73
(a)
(a) Find
0 /  I .
 0  0

0
A

    I  0  I
i1  I


R
R
R
(b) Design the inverting amplifier with gain of
100 and an input resistance 1 MΩ.
 
i2  i 
I
R
 x    i2 R2  0 
I
R
R2  
0 x
R2


R3
R2 R3 I

R
i4  i2  i3  I  2  I
R R1 R3
R2
I
R
i3 
 0   x  i4 R4
=


R2
R
 I   I  2  I  R4
R1
 R R1 R3 
0
R
 2
I
R1
 R4 R4 
 
1 
R
R3 
2

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(b)
input resistance 1 MΩ
R1=1 MΩ
maximum resistance in practical circuits: 1 MΩ
R2, R4=1 MΩ
gain of -100
R3=10.2 k Ω
(If we adopt a typical inverting amplifier and R1=1 MΩ,
R2=100 MΩ, impractically large !)
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Transresistance (transimpedance) amplifier and current amplifier
• Photodiode (photodetector) generates electronhole pairs proportional to the incident light power.
• Therefore, photodiode is a current source (I).
• υo = - IR2 means current input, voltage output !
Transresistance amplifier !
• Input resistance is R2/A, very small. (Try this!)
• Since the photodiode has capacitance inside, if a
large resistor for high output voltage is used
instead of op-amp, time constant would be large,
which means it can not be used in high frequency
circuit.
• Transresistance amplifier is currently used as a
part of receiver in high speed fiber-optic
communication systems.
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Figure 2.9 A current amplifier based on
the circuit of Fig. 2.8. The amplifier
delivers its output current to R4. It has a
current gain of (1 + R2/R3), a zero input
resistance, and an infinite output
resistance. The load (R4), however, must
be floating (i.e., neither of its two
terminals can be connected to ground).
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2.2.3 An Important Application-The Weighted Summer.
i1 
1
R1
,
i1 
2
R2
,
,
in 
i  i1  i2    in
n
Rn
(2.6)
  0  iRf  iRf
 Rf
   
 R1
Figure 2.10 A weighted summer.
1 
Rf
R2
Rf
Rn

n 

(2.7)
Mom, virtual grounds are extremely handy !
 Rc 
 Ra   Rc 
 Ra   Rc 
 Rc 






  2 
  3   4  
R
R
R
 1  b 
 2   Rb 
 R4 
 R3 
   1 
 2   
(2.8)
Figure 2.11 A weighted summer capable of implementing summing coefficients of both signs.
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2.3 The Non-inverting Configuration
2.3.1 The close-Loop Gain
The inverting configuration.
Figure 2.12 The noninverting configuration.
Let’s derive the expression of the voltage gain.
0
R
 1 2
I
R1
(2.9)
 Another derivation
υo is divided to R1 and R2.

R1 

 R1  R2 
1   0 
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 the Role of Negative feedback
1. Let υI increase.
2. υId is increased.
3. υo is increased.
4. Fraction of υo will be fed back to (-) terminal.
5. This feedback will be counteract the increase in υId ,
driving back to zero. – degenerative feedback
(2.10)
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2.3.2 Characteristics of the noninverting Configuration
0
R
 1 2 > 0
I
R1
- Input and output signals have same phase.-noninverting.
- Input impedance is infinite.
- output impedance is zero.
2.3.3 Effect of Finite Open-Loop Gain
- Follow the procedure used for inverting amplifier.
- denominators are identical !
How come?
G
(2.11) - non-inverting amplifier
0
 R2 / R1

 I 1  (1  R2 / R1 )
(2.5)
A
G
identical negative feedback
(think about 0 V input !)
0
1  ( R2 / R1 )

 I 1  1  ( R2 / R1 )
- inverting amplifier
A
- numerators are different !
How come?
different amount of feedback
(closed loop gain)
In order to minimize the effect of the finite open-loop gain,
A
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1+
R2
R1
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(2.12)
13
2.3.4 The Voltage Follower
- Input impedance is infinite.
- very
desirable feature.
0
R
 1 2 = 1
I
R1
- Buffer Amplifier !
to connect a source with a high
Usually, gain = 1 (Sect. 1.5)
R1  , R2  0
Why do we need this?
- 100% feedback !
- Used as impedance transformer
or power amplifier.
- Elegant in simplicity !
The unity-gain buffer
or voltage follower.
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equivalent circuit model.
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2.4 Difference (differential) Amplifiers (very important application)
0  Ad Id  Acm Icm
differential gain differential input
Common mode input
(2.13)
Common mode gain≈0
• The efficacy of a differential amp is measured by the degree of its rejection
of common-mode signals in preference to differential input signals.
Common-Mode Rejection Ratio (CMRR) = 20 log
Acm
(2.14)
Transducer (sensor)
0.001sin s t
Noise (Electromagnetic Interference, EMI)
from motor, spark, lightning…
We want to amplify the sensor output only.
(we want to reject common-mode noise signal.)
1sin  N t
2.4.1 A single Op-Amp Difference Amplifier
Gain of the inverting amplifier
= -R2/R1
Gain of the non-inverting amplifier = 1+R2/R1
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Combine these two amp.
matched gain
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When I2 is zero,
O1  
R2

R1 I 1
When  I 1 is zero,
O 2   I 2
Figure 2.16 A difference amplifier.
R4 
R2 
1

R3  R4 
R1 
To make gain R2/R1,
R4 
R2  R2
1

R3  R4 
R1  R1
O 2 
Superposition !
O 

R4 R2

R3 R1
R2

R1 I 2
R2
R
 I 2   I 1   2  Id

R1
R1
(2.16)
R2
R1
(2.17)
For easier matching, R3=R1, R4=R2
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Differential input Resistance Rid
R3  R1
and
R4  R2
Rid 
 Id
iI
 Id  R1iI  0  R1iI
Rid  2R1
(2.20)
Drawback 1: for large gain, very small R1 is required.
Drawback 2: to vary the gain, two resistances should be changed.
2.4.2 A Superior Circuit-The Instrumentation Amplifier
The Goal: High differential gain with high input resistance and easy gain control
HOW? Voltage follower with gain + Difference amplifier
R4 
R2 
R4 
R2 
1
 ( I 2   I 1 ) 
1

R3 
R1 
R3 
R1  Id
R 
R 
Ad  4  1  2 
(2.21)
R3 
R1 
O 
1: in the first stage, common-mode gain = differential gain may
causes saturation of op-amp.
2: in the first stage, two amplifiers have to be perfectly matched.
3: to vary the gain, two resistances (two R1) should be changed.
There is a simple solution !
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Instrumentation Amplifier
1
2
6
4
5
8
7
3
9
1
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EXAMPLE 2.3
Design The instrumentation amplifier circuit to provide a gain that can be varied over the range of
2 to 1000 utilizing a 100 kΩ variable resistance ( a potentiometer or “pot” for short).
(Sol.)
R4 
R2 
1



R3 
R1 
(2.21)
- It is usually preferable to obtain all the required gain in the
first stage. (Low Noise)
- The second stage (difference amp) is usually designed for a
gain of 1.
- We select all the second-stage resistors to be 10 kΩ,
practically convenient value.
1

R 
Ad   1  2 
R1 

2 R2
 2 to 1000
R1 f  R1
1
1
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2 R2
 1000
R1 f
2 R2
2
R1 f  100 k
R1f =100.2 Ω,
R1f =100 Ω, 1%
R2 = 50.050 kΩ,
R2 = 49.9 kΩ, 1%
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2.5 Effect of Finite Open-Loop Gain and Bandwidth on Circuit Performance
A circuit designer has to be thoroughly familiar with the characteristics of practical op amps and the effects of such
characteristics on the performance of op-amp circuits.
2.5.1 Frequency Dependence of the Open-Loop Gain
- Internally compensated op amps are unit that have a
network (usually a single capacitor) within the IC chip.
- Capacitor’s function is to cause the op-amp gain to have
the STC low-pass response.
- This process of modifying the open-loop gain is termed to
frequency compensation.
- The purpose of frequency compensation is to ensure that
the op-amp will be stable (as oppose to oscillation: Ch. 8).
A( s ) 
Figure 2.22 Open-loop gain of a typical generalpurpose internally compensated op amp.

b , A(j )
A0b
j
(2.26)
A( j ) 
A(j ) =
A 0b

A0
1  s / b
A0
1  j / b
(2.24)
(2.25)
(2.27)
When A  1, t  A0b
(2.28) - ft = ωt/2π is specified on the data sheet as unity-gain band width.
t
t
ft
 b , A(s )
(2.30)
A(
j

)
=
(2.31)
s

f
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A(j )
t
f
= t

f
(2.31)
- If ft is known, one can easily determine the magnitude of the op-amp gain at a given frequency when f >> fb.
The production spread in the value of ft between opamp units of the same type (part number) is much
smaller than that of A0 and fb.
ft is preferred as a specification
parameter.
- An op amp having this -6 dB/octave (= - 20 dB/decade) gain roll off is said to have a single-pole model.
A( s ) 
A0
1  s / b
pole: s=-b
(2.24)
- Since this single pole dominates the amplifier frequency response, it is called a dominant pole.
(for more on poles and zeros, refer to Appendix E.)
2.5.2 Frequency Dependence of the Closed-Loop Gain
The effect of limited op-amp gain and bandwidth on the closed-loop transfer functions.
We know
Vo
 R2 / R1

Vi 1  (1  R2 / R1 ) / A
and A( s ) 
Then,
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Vo ( s )

Vi ( s )
A0
1  s / b
 R2 / R1
R2 
1 
s
1
1

A0 
R1  t /(1  R2 / R1 )
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(2.24)
(2.33)
21
for A0
1
R2
,
R1
Vo ( s )
Vi ( s )
 R2 / R1
s
1
t /(1  R2 / R1 )
(2.34)
corner frequency, 3dB 
We know,
Similary,
Low-Pass STC Network !
Vo
1  R2 / R1

Vi 1  (1  R2 / R1 ) / A
Vo ( s )
Vi ( s )
1  R2 / R1
s
1
 t /(1  R2 / R1 )
t
1+R2 / R1
(2.35)
(2.36)
(2.37)
EXAMPLE 2.4
ft = 1 MHz, find 3-dB frequency of closed-loop amp with gain
of 1000, 100, 10, 1, -1, -10, -100, -1000

Gain-bandwidth product = 1000 V/V-kHz
= 60dB-kHz
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R1
R1 R2
Non-inverting amplifier
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2.6 Large-signal Operation of Op Amps
2.6.1 Output Voltage Saturation
Rated output voltage (output saturation voltage)
= ± L (L=VDD – α V )
2.6.2 Output Current Limits
For example, maximum output current of 741 is ± 20 mA.
EXAMPLE 2.5
Output saturation voltage = ± 13 V, output current limits = ± 20 mA
(a) Output voltage for Vp =1 V, RL= 1 kΩ
υo/ υI = (1+R2/R1): Vop = 10 V, Iop = 10 mA
(b) Output voltage for Vp =1.5 V, RL= 1 kΩ
Vop = 15 V, Iop = 14.3 mA
(c) for RL= 1 kΩ, Vpmax = ? for undistorted output.
13/10 = 1.3 V ( 14.3 mA)
(d) for Vp =1.5 V, RLmin = ? for undistorted output.
iO max  20 mA 
10 V
10 V

RLmin 9 k + 1 k
RLmin  526 
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2.6.3 Slew Rate - Another nonlinear distortion due to the large output signal
Slew Rate : maximum rate of change possible at the output of a real op amp
SR =
dO
dt
(V/s) (2.38)
max
Why this happens ? : Chap. 9
SR is distinct from the finite op-amp bandwidth.
- The limited bandwidth : linear, no distortion, but reduced gain at higher frequency.
- The limited Slew-rate : nonlinear distortion, even at low frequency when output is too large.
When V is sufficiently small, the output can be
the exponentially rising ramp.
d ( A sin  t )
= A cos  t
dt
Vo ( s )
Vi ( s )
1  R2 / R1
s
1
 t /(1  R2 / R1 )
When R1  , R2  0,
(2.37)
VO
1

Vi 1  s / t
(2.39)
This is a low-pass STC response.
Output from capacitor of RC network !
O (t )  V (1  e t )
t
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(2.40)
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2.6.4 Full-Power Bandwidth

 I  V i sin  t ,


d I
  V i cos  t
dt
MVomax  SR
SR
fM 
2Vo max
If SR <  V i ,
(2.41) full-power bandwidth
Figure 2.27 Effect of slew-rate limiting on
output sinusoidal waveforms.
At which an output sinusoid with amplitude equal to the rated output voltage of the op amp
begins to show distortion due to slew-rate limiting.
 
Vo  Vo max  M 
  
(2.42) Maximum amplitude of the undistorted output sinusoid.
2.7 DC Imperfection
2.7.1 Offset Voltage
Op amps are direct-coupled devices.
They are prone to dc problems.
• With inputs being zero, the amplifier
output rests at some dc voltage level
instead of zero. The equivalent dc input
offset voltage is
V
V  o 10 mV(0.25 mV w/high cost)
OS A
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R1 =1.2 kΩ, R2 = 1 MΩ,
To include effect of offset voltage,






ic
o  A  
id CMRR
V
OS
V  3 mV
OS






υO = ?

1 MΩ 
V  1
 (0.003)  2.5V
O  1.2kΩ 
If υid =0,
o  A





ic





V
 A( )
OS
CMRR OS
CMRR 
ic
V
When an input signal is applied to the amplifier, the
corresponding signal output will be superimposed
on the 2.5 V dc.
Then the allowable signal swing at the output will
be reduced.
If signal is dc, we would not know where the output
is due to VOS or the signal.
 V/V
OS
(another interpretation of CMMR)
•
* To overcome dc offset problem
Thus, CMRR is a measure of how total offset
voltage vOS changes from its dc value VOS
when common-mode voltage is applied.
* The gain will fall off at the
low- frequency.
Figure 2.28 Circuit model for an op amp with input offset voltage VOS.
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2.7.2 Input Bias and Offset Current
* To find the dc output voltage of the closed-loop
amp due to the input bias current
In order for the op amp to operate, its two input
terminals have to be supplied with current,
termed the input bias current. Why?
Bias currents ( base currents in BJTs or gate
currents in MOSFETs or JFETs) are similar in
value with directions depending on internal
amplifier circuit type.
VO  I B1 R2
I B R2
(2.44)
How to reduce the dc output voltage due to the input
bias current
IB 
I B1  I B 2
: input bias current in data sheet
2
IOS  I B1  I B2 : input offset current
VO   I B 2 R3  R2 ( I B1  I B 2 R3 / R1 )
(2.45)
if I B1  I B2  I B , VO  I B  R2  R3 (1  R2 / R1 )
For minimum VO , R3 
R2
RR
 1 2 (2.46)
1  R2 / R1 R1  R2
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With this R 3 and I B1  I B 2 ,
VO  IOS R2 < I B R2
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(2.47)
27
Conclusion: To minimize the effect of the input bias current,
Place in the positive lead a resistance equal to
the dc resistance seen by the inverting terminal.
• There should be a continuous dc path between each input terminal and ground.
This circuit will not work.
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This circuit will work.
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2.8 Integrators and Differentiators
2.8.1 The Inverting Configuration with General Impedances
Vo ( s )
Z ( s)
 2
Vi ( s )
Z1 ( s )
(2.48)
Example 2.6, p106
Sol)
Vo ( s )
1

Vi ( s )
Z1 ( s )Y2 ( s)
Vo ( s )  R2 / R1

Vi ( s ) 1  sC2 R2
Vo ( s )
 ? * Is this STC circuit?
Vi ( s )
* dc gain=?, 3 dB frequency=?
* Design a circuit of dc gain=40 dB,
3 dB frequency=1 kHz, input
resistance = 1 kΩ
* Frequency of gain=1
and phase at this frequency
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Vo ( s )
1

R1
Vi ( s )
 sC 2 R1
R1
K 
R2
R1
0 
1
C2 R2
R2
 100, R1  1 k, R2  100 k
R1
1
0  2  1  103 
C2  1.59 nF
C2  100  103
-20dB roll off  f t  100 f0  100 kHz
40 dB = 100 V/V
from Bode plots phase shift = -270o or +90o
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2.8.2 The inverting Integrator
Figure 2.39 (a) The Miller or inverting integrator.
(b) Frequency response of the integrator.
C ( t )  VC 
1
C

t
i ( t ) dt
0 1
1
O ( t )  C ( t )  
RC

t
0
 I ( t ) dt  VC (2.49)
CR : integrator time-constant.
Vo ( s )
Z ( s)
 2
Vi ( s )
Z1 ( s )
(2.48)
Vo ( s )
1

Vi ( s )
sCR
Vo ( j )
1

Vi ( j )
jCR
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Vo
1

Vi  CR
int 
  90 ,
1
: integrator frequency
CR
* An integrator behaves as a low-pass filter with a corner
frequency of zero.
* At f=0, the magnitude of the integrator transfer function is
infinite and the op amp is operating with an open loop.
* Any tiny dc component input signal will theoretically
produce an infinite output ! Serious problem !!!
* In practice, the output saturates at power supply voltage.
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30
* Deleterious effect from the presence of the op-amp
input dc offset voltage
* Deleterious effect from the presence of the op-amp
input dc offset current
( R is added in (+) to reduce the effect of bias current.)
Place in the positive lead a resistance equal to
the dc resistance seen by the inverting terminal.
Assuming  C =0 at t  0,
 O  VOS 
VOS
t
CR
(2.55) (lineary increasing)
Op amp will saturate very soon !!!
Assuming C =0 at t  0,
The dc offset current produces a similar problem !!
O   I B 2 R 
Solution
I OS
t
C
(lineary increasing)
Let’s provide a dc current (VOS and IOS) path !
dc component of O  VOS (1  RF / R)  IOS RF Low RF is better for dc of  O .
Vo ( s)
R /R
 F
Vi ( s )
1  sCRF
 Vo ( s )
1 

High RF is better for ideal integrator. 

sCR 
 Vi ( s )
Selecting a value for RF present the designer with a trade-off between dc
performance and signal performance. (example 2.7, p110)
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EXAMPLE 2.7
Find the output by the Miller integrator due to  I ( R1  10 k, C  10 nF).
 I ( RF  1 M).
Find the output by the Miller integrator due to
Sol)
1 t
1. dt ,
0  t  1 ms
CR 0
( R1  10 k, C  10 nF) O (t )  10t ,
O (t )  
0  t  1 ms
Constant current (1 V/10 kΩ= 0.1 mA) to capacitor.
( RF  1 M, C  10 nF)
Using Eq.(1.29) p.46, O (t )  O ()  O ()  O (0)e t / CRF
O ()   IRF  0.1 103  1 106  100 V
O (0 )  0, initial value
O (t )  100(1  e t /10 ),
0  t  1 ms
O (1 ms)  100(1  e t /10 )  9.5 V
Important Applications: square-wave input to triangular output
Exercise 2.27,
(Function Generator !!),
: Active Filter !! (Ch. 12)
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2.8.3 The Op-Amp Differentiator
C (t )   I (t )
d I ( t )
Vo
 1 at   1/ CR
Vi
i(t )  C
dt
d ( t )
O ( t )  CR I
dt
CR: differentiator time-constant
(2.56)
* Differentiator is an STC highpass filter with a corner
frequency at infinity.
* Differentiator is a noise magnifier.
Z1 ( s )  1/ sC
Z2 ( s)  R
Input
Vo ( s )
  sCR
Vi ( s )
(2.57)
Vo ( j )
  j CR
Vi ( j )
Vo
  CR (2.59)
Vi
(2.58)
  90
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(2.60)
Output
* Spikes in output could cause EMI problems.
* A small resistor in series with the capacitor might solve this
problem, but the circuit becomes non-ideal differentiator.
* Due to the EMI and stability (Ch.8) problem, differentiator is
seldom used in practice.
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2.9 The SPICE Op-Amp Model and Simulation Examples
Figure 2.45 A linear macromodel used to model the finite gain and bandwidth of an internally compensated op amp.
The differential gain : A0d of the voltage-controlled voltage source Ed
Frequency response : Rb-Cb low-pass filter with corner frequency fb  1/ 2 RbCb
Eb with gain of 1 is used to isolate the low-pass filter from any load at output.
(2.61)
Figure 2.46 A comprehensive linear macromodel of an internally compensated op amp.
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