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Chapter 4. Operational amplifiers
2015. 9. 24.
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Contents
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1. Introduction
2. Op-Amp Models
3. Fundamental Op-Amp Circuits
4. Comparators
5. Application Examples
6. Design Examples
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Why operational amplifiers ?
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1. Originally, the op-amp was designed to perform mathematical operations such
as addition, subtraction, differentiation, and integration.
2. By adding simple networks to the op-amp, we can create these “building blocks”
as well as voltage scaling, current-to-voltage conversion, and myriad more
complex applications.
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2. Op-amp model
Ideal op-amp :
Ri  , RO  0, A0  
iin  0, in  0
iin
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Example
VO  Ri   RL 

A0 
 A0  


VS  Ri  RTh1   RO  RL 
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Example : unity gain buffer
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VS  I ( RS  Ri  RO )  A0Vin
VO  IRO  A0Vin
Vin  IRi
VO
IRO  A0Vin
IRO  A0 IRi


1
VS I ( RS  Ri  RO )  A0Vin I ( RS  Ri  RO )  A0 IRi
VO  VS
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Example 4.2
Let us determine the gain of the basic inverting op-amp configuration shown in
the figure using both the non-ideal and ideal op-amp models
1   S 1 1  o
 
0
R1
Ri
R2
0  1 o  A e

0
R2
Ro
0

S
 R2 / R1
 1 1 1  1 1 
1       
 R1 R2 Ri  R2 Ro 
 1  1 A  
   
 R2  R2 Ro 

R2
R1
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Ideal model
I
I
0 [V ]
Virtual short
 S  IR1 , o   IR2 
•
•
•
o
R
 2
S
R1
Step 1. Use the ideal op-amp model: Ao → ∞, Ri → ∞, Ro = 0.
i+= i-=0, v+= vStep 2. Apply nodal analysis to the resulting circuit.
Step 3. Solve nodal equations to express the output voltage in terms of the
op-amp input signals.
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Example 4.3
Let us now determine the gain of the basic non-inverting op-amp configuration
shown in the figure.
      in
in 0  in

RI
RF
 1
1  
in     0
 RI RF  RF

0
R
 1 F
in
RI
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Example 4.5
Consider the op-amp circuit shown in the figure. Let us determine an expression for
the output voltage.
   
R4
2
R3  R4
1       o

R1
R2
o    
 R  R4
R2
R
R
(1    )   2 1  1  2 
 2  2 ( 2  1 )
R1
R1
R1
 R1  R3  R4
( R4  R2 , R3  R1 )
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Example 4.6
The circuit shown in the figure is a precision differential voltage-gain device. It is used
to provide a single-ended input for an analog-to-digital converter. We wish to derive an
expression for the output of the circuit in terms of the two inputs.
1  o 1   a 1   2


0
R2
R1
RG
 2   a  2  1  2

 0
R1
RG
R2
 R 2R 
o  (1   2 )1  2  2 
 R1 RG 
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Example 4.7
V1
Vx
V1  V2 V1  Vx

0
10k
10k
V1  V2  V1  Vx  2V1  V2 
Vx 
10k
V
Vo  o
10k  30k
4
V0
0
4
Vo  8V1  4V2
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4.4 Comparators
(a) An ideal comparator and (b) its transfer curve.
(a) A zero-crossing detector and (b) the corresponding input/output waveforms.
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Application example 4.11
An instrumentation amplifier of the form shown in Fig. 4.26 has been suggested. This
amplifier should have high-input resistance, achieve a voltage gain Vo/(V1-V2) of 10,
employ the MAX4240 op-amp listed in Table 4.1, and operate from two 1.5 V AA cell
batteries in series. Let us analyze this circuit, select the resistor values, and explore the
validity of this configuration.
Vx
2
V1
V2
Vx  V1 V2  V y V1  V2


R1
R2
R
Vx Vx
 Vo
2  2
 Vo  Vx  V y
RA
RA
Vy 
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Vx  V1 
R1
R
(V1  V2 ), V y  V2  2 (V1  V2 )
R
R
R 
R
R
R 


Vx  1  1 V1  1 V2 , V y   2 V1  1  2 V2
R
R
R
R


R  R2 

Vo  Vx  V y  Vx  1  1
(V1  V2 )
R



Vo
2R
 1  1  10 ( R1  R2 )
V1  V2
R
 R1  4.5R
ex )
R  100 k, R1  450 k
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Design example 4.14
We wish to design a weighted-summer circuit that will produce the output
Vo= - 0.9V1 - 0.1V2
The design specifications call for use of one op-amp and no more than three resistors.
Furthermore, we wish to minimize power while using resistors no larger than 10kΩ.
If
I1  I 2  I f
V1 V2 0  Vo


R1 R2
R
    0
Vo  
R
R
V1  V2
R1
R2
R
 0.9,
R1
R
 0.1
R2
R1 
R
R
, R2 
0.9
0.1
ex. R  270 [], R1  300 [], R2  2.7 [k],
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