Chapter   6:   Operational   Amplifier   (Op   Amp)
‐
‐
1

6.1
What   is   an   Op   Amp   (1)
• An   operational   amplifier   (op   amp)   is   a   integrated   circuit   (IC)   composed   of   perhaps   30   BJTs   and/or   FETs,   10   resistors,   and   several   capacitors.
• The   functions   like   a   voltage ‐ controlled   voltage   source .
• It   is   an   active   circuit   element   designed   to   perform   mathematical   operations   of   addition ,   subtraction ,   multiplication,   division ,   differentiation and   integration .
Pin   configuration                         Circuit   symbol                          Powering   the   op   amp
2

6.1
What   is   an   Op   Amp   (2)
(a) μ A741   IC with   has
8   connecting   pins.
(a)                                    (b)
(b) Correspondence   btw   circled   pin   numbers   of   the   IC   and   the   nodes.
(c) An   op   amp,   including   power   supplies   v
+ and   v
-
.
(c)

6.1
What   is   an   Op   Amp   (3)
The   equivalent   circuit of   the   non ‐ ideal   op   amp
Op Amp output: v o as a function of v d v d
= v
2
– v
1 v o
= Av d
= A ( v
2
–v
1
)
Typical   ranges   for   op   amp   parameters
Parameter
Open ‐ loop   gain,   A
Input   resistance,   R i
Output   resistance,   R o
Supply   voltage,   V
CC
Typical   range
10 5 to 10 8
10 5 to 10 13 
10 to 100 
5 to 24 V
|v |
 v sat
|i |
 i sat
| dt
( )
|

SR
Ideal   values
∞
∞ 
0 
4

6.1
What   is   an   Op   Amp   (4)
• Equivalent   circuit   of   the   op   amp:
5

6.2
Ideal   Op   Amp
An   ideal   op   amp   has   the   following   characteristics:
1. Infinite   open ‐ loop   gain,   A ≈ ∞
2. Infinite   input   resistance,   R i
3. Zero   output   resistance,   R o
≈ ∞
≈ 0
Example: Determine   the   value   of   i o .
when v s
 1 V
6

6.3
Nodal   Analysis   of   Circuits   with   Op   Amps   (1)
Use   node   equations   to   analyze   circuits   containing   ideal   op   amps .
There   are   3   things   to   remember:
1. The   node   voltages   at   the   input   nodes   of   ideal   op   amps   are   equal .
Thus,   one   of   these   two   node   voltages   can   be   eliminated   from   the   node   equations.
v
1
 v
2
→ v
2 can   be   eliminated   from   the   node   equations.
2. The   currents   in   the   input   leads   of   an   ideal   op   amp   are   zero .
These   currents   are   involved   in   the   KCL   equations   at   the   input   nodes   of   the   op   amp.
3. The   output   current   of   the   op   amp   is   not   zero .
This   current   involved   in   the   KCL   equations   at   the   output   node   of   the   op   amp.
Applying   KCL   at   this   node   adds   another   unknown   to   the   node   equations.
If   the   output   current   of   the   op   amp   is   not   to   be   determined,   then   it   is   not   necessary   to   apply   KCL   at   the   output   node   of   the   op   amp.
7

6.3
Nodal   Analysis   of   Circuits   with   Op   Amps   (2)
Example: Analysis   of   a   Bridge   Amplifier
The   op output
amp of
the and
resistors, bridge.
As   a
R
5 and   R
6
,   are   used   to   amplify   the   consequence,   v
1
= 0 and   i
1
= 0 ,   determine   the   output   voltage,   v o
,   in   terms   of   the   source   voltage,   v s
.
(a)                                           (b)                              (c)                                   (d)
(a)   A   bridge   amplifier,   including   the   bridge   circuit.
(b)   The   bridge   circuit   and   (c)   its   Thévenin   equivalent   of   the   bridge.
8

6.3
Nodal   Analysis   of   Circuits   with   Op   Amps   (3)
First,   notice   that   the   node   voltage   v a is   given   by   (using   KVL)   v a
  v oc

R i t 1
Because   v
1
= 0 and   i
1
= 0 , v a
 v oc
Now,   writing   the   node   equation at   node   a i
1
 v a

R
5 v o  v a
R
6
 0
Because   v a
= v oc v oc and
R

5 v
o i
1

= 0 v
R oc
6
,
 0
Solving   for   v v o




1 
R
5
R
6

 o

,   we   have v oc




1 
R
R
5
6


 R
1
R

2
R
2

R
3
R

4
R
4


 v s
9

6.4
Configurations   of   Op   amp   (1)
• Inverting   Amplifier:   reverses   the   polarity of   the   input   signal   while   amplifying   it.
Negative   feedback   btw   the   inverting   input   ( v i
) &   output   ( v o
) v i is   connected   to   the inverting   input   via   R
1
To   find   the   relationship   btw   v i
&   v o
:
By KCL at node 1,
 v
1
 i v
1
2 i v i
 v
1  v
1
 v o
R
1
R f
 0 for an ideal op amp since the noninverting terminal is grounded.
Noninverting   input   is   grounded
 v o
 
R f
R
1 v i
Closed-loop voltage gain is
A v
 v v o i
 
R R f 1
Example: Find   v
0.5 V ,   R
1 o
&   i in   R
= 10 kΩ ,   &   R f
1 if   v i
=
= 25 kΩ .
10

6.4
Configurations   of   Op   amp   (2)
• Current ‐ to ‐ voltage   converter   (trans ‐ resistance   amplifier)
PD   is   adopted   to   use   an   inverting   current v o
 
Ri s v o
 
R
1


1 
R
3 
R
3
R
1
R
2

 i s
11

6.4
Configurations   of   Op   amp   (3)
• Noninverting   Amplifier :   designed   to   produce   positive   voltage   gain.
To   find   the   relationship   btw   v
By K CL at inverting terminal, i i
1 i
0 
R
1 v
1  v
1
 v o
R f
& v o
: v i connected   to   noninverting   input   terminal
 v
1
 v
2 v o


 1 
R f
R
1

 v i
Voltage gain:
A v
 v v o i
R f
R
1
• To   isolate   two   cascaded   stages:
R f
 0 (short circuit) or
R
1

  (open c ircuit) v o
 v i a s voltage follow er
Has   a   very   high   input   impedance   and
12 thus   eliminate   interstage   loading

6.4
Configurations   of   Op   amp   (4)
Circuit   1   (a)   before   and   (b)   after   circuit   2   is   connected.
(c)   Preventing   loading ,   using   a   voltage   follower.
A   voltage   divider   (a)   before   and   (b)   after   a   30 ‐ k Ω  resistor   is   added.
(c)   A   voltage   follower   is   added   to   prevent   loading.
13

6.4
Configurations   of   Op   amp   (5)
A   common   application   of   op   amp   is   to   scale   a   voltage ,   that   is,   to   multiply   a   voltage   by   a   constant,   K ,   so   that   v o

Kv in
The   input   voltage,   v in
,   is   provided   by   an   ideal   voltage   source.
The   output   voltage,   v o
,   is   the   element   voltage   of   a   100 ‐ k Ω  resistor.
Circuits   that   perform   this   operation   are   usually   called   amplifiers .
The   constant   K is   called   the   gain   of   the   amplifier .
There   are   four   cases   to   consider:   (b)   K = -5 (< 0), (c)   K = 5 (> 1), (d)   K = 1, and (e)
K = 0.8 (0 < K < 1) .
(a)                                           (b)
(c)                                                (d)                                     (e)
14

6.4
Configurations   of   Op   amp   (6)
Example:    calculate   the   output   voltage   v o
.
by v o
 
R f
R
1 v i by v o

1

R
R
1 f

 v i
15

6.4
Configurations   of   Op   amp   (7)
• Summing   Amplifier :   combines   several   inputs   and   produces   an   output   that   is   the   weighted   sum   of   the   inputs.
To   find   the   relationship   btw   v i
& v o
:
By KCL at node , i
1 i i
3
But i
1
 v
1
 v a
R
1
, i
2
 v
2

R
2 v a , i
3
 v
3
 v a
R
3
, i
 v a
 v o
R f
 v a
0 v o
 


R f
R
1 v
1

R f
R
2 v
2

R
R
3 f v
3


Example:    calculate   v o
&   i o
.
16

6.4
Configurations   of   Op   amp   (8)
• Differential   Amplifier :   amplifies   the   difference   between   two   inputs   but   rejects   any   signals   common to   the   two   inputs.
v
1
 v a
 v a
 v o
R
1
R
2
 v o



R
2
R
1
 1

 v a

R
2
R
1 v
1
Example: Design   an   op   amp   circuit   with   v
1
&   v
2 such   that   v o
=  ‐ 5 v
1
+ 3 v
2 v
2
 v b
 v b
R
3

R
4
0
 v b

R
3
R

4
R
4 v
2
B ut v a
 v b
 v o

R
R
2
(1
(1


R
1
/
/
R
2
)
) v
2

R
2 v
1
1 3 4
R
1
To reject a signal common to the two inputs,
 v o

R
2
 v
2
 v
1

, if
R
1
 v o
 0 when v
1
 v
2
To become a subtractor,
R
1 
R
R
2
R
3
4
 v o
 v
2
 v
1
, if
R
2 
R
R
1
R
3
4
 1
17

6.4
Configurations   of   Op   amp   (9)
18

6.4
Configurations   of   Op   amp   (10)
19

6.5
Cascaded   Op   Amp   (1)
• Cascaded   Connection :   a   head ‐ to ‐ tail   arrangement   of   two   or   more   op   amp   circuits   such   that   the   output   of   one   is   the   input   of   the   next.
• Op   amp   circuits   have   the   advantage   that   they   can   be   cascaded   without   changing   input ‐ output   relationships.
This   is   due   to   the   fact   that   each   (ideal)   op   amp   circuit   has   R i
≈ ∞ &
R o
≈ 0 .
• Note   that   to   design   an   actual   op   amp   circuit   must   ensure   that   the   load   due   to   the   next   stage   in   the   cascade   does   not   saturate   the   op   amp .
• Overall   gain:
20

6.5
Cascaded   Op   Amp   (2)
Example   :     If   v
1
=   1   V   &   v
2
=   2   V,   find   v o
21

6.6
Op   Amp   Circuits   &   Linear   Algebraic   Eqs (1)
A   voltage   or   current   that   is   used   to   represent   something   is   called   a   signal.
Symbolic   representations   of   (a)   & (c)   multiplication   by   a   constant   and   (e)   addition   whereas   (b),   (d),   and   (f)   show   the   corresponding   operational   amplifier   circuits.
‐ To   obtain   a   2V   signal
22

6.6
Op   Amp   Circuits   &   Linear   Algebraic   Eqs (2)
‐ To   implement v z
 4 v x
 5 v y
 2
‐ An   improved op   amp   circuit
| v x
|  v sat 
15
4 4
 and | | z
 v sat

3.75V, |
15V v y
|  v sat 
15
5 5
 3V,
23

6.7
Applications   (1)
• The op amp is a fundamental building block in modern electronic instrumentation.
• Digital ‐ to ‐ Analog Converter (DAC) : transforms digital signals into analog form.
• Four ‐ bit DAC :
- block diagram

V
0

R f
R
1
V
1

R f
R
2
V
2

R f
R
3
V
3

R f
R
4
V
4
‐ binary   weighted   ladder   where
V
1
– MSB,   V
4
– LSB
V
1 to   V
4 are   either   0   or   1   V   (voltage   level)
Note: The   bits   are   weights   according   to   the   magnitude   of   their   place   value,   by   descending   value   of   R f
/ R n so   that   each   lesser   bit   has   half   the   weight   of   the   next   higher.

6.7
Applications   (2)
Note: Each   bit   has   a   value   of   0.125 V .
A   voltage   btw   1.000
and   1.125
cannot   be   represented.
For   greater   accuracy,   a   word   representation   with   greater   number   of   bits   is   required.
25

6.7
Applications   (3)
• Instrumentation   Amplifier   (IA)   :   to   amplify   low   level   signals   used   in   process   control   or   measurement   applications   and   to   be   packed   in   a   single   unit.
i flows   through   R
3
,   R
4
, & R
3 in   series
But i
 v a
 v
R
4 v
1
 v
2 b & v a
 , b
 v
2
R
4
 v o

R
2
R
1


1 
2
R
3
R
4


 v
2
 v
1

• A   typical   example   is   the   LH0036 ,   developed   by   National   Semiconductor.
26

6.7
Applications   (4)
• IA   with   an   external   resistance   to   adjust   the   gain.
v
0


 1 
2
R
R
G


 v
2
 v
1

Three   major   characteristics :
1.
Voltage   gain   is   adjusted   by   one   external   resistor   R
G
.
2.
The   input   impedance   of   both   inputs   is   very   high   and   does   not    vary   as   the   gain   is   adjusted.
3.
Output   v o depends   on   the   difference   btw   the   inputs   v
1 the   voltage   common   to   them   ( common ‐ mode   voltage ).
&   v
2
,   not   on
27

6.8
Characteristics   of   Practical   Op   Amps   (1)
In   contrast,   the   op   amp   model   shown   in   (d)   accounts   for   several   nonideal parameters   of   practical   op   amp,   namely:
• Nonzero   bias   currents
• Nonzero   input   offset   voltage
• Finite   input   resistance
• Nonzero   output   resistance
• Finite   voltage   gain
(a) An op amp, (b) the offsets model of an op amp, and (c) the finite gain model of an op amp. (d) The offsets and finite gain model of an op amp.
28

6.8
Characteristics   of   Practical   Op   Amps   (2)
Based on the nonideal model of an op amp, the followings should be accounted: bias currents , input offset voltage , and finite gain .
29

6.8
Characteristics   of   Practical   Op   Amps   (3)
• Example:   Offset   Voltage   and   Bias   Currents
(a) An inverting amplifier and (b) an equivalent circuit that accounts for the input offset voltage and bias currents of the op amp. (c) – (f) Analysis using superposition.
30

6.8
Characteristics   of   Practical   Op   Amps   (4)
In (b), the op amp has been replaced by the offsets model of an op amp. Notice that the op amp in (b) is the ideal op amp that is part of the mode1 of the op amp used to account for the offsets.
The circuit in (b) contains four inputs that correspond to the four independent sources, v in
, i b1
, i b2
, and v os
. The input v in is obtained by connecting a voltage source to the circuit. In contrast, the
“inputs” i b1
, i b2
, and v os are the results of imperfections of the op amp.
Superposition can be used to good advantage in analyzing this circuit. (c) – (f) illustrate this process. In each of these figures, all but one input has been set to zero, and the output due to that one input has been calculated.
In (c), the circuit is used to calculate the response to v in other inputs, i b1
, i b2
, and v os
, have all been set to zero.
alone. The
31

6.9
Summary
Several   models   are   available   for   op   amps.
Simple   models   are   easy   to   use.
Accurate   models   are   more   complicated.
The   simplest   model   of   the   op   amp   is   the   ideal   op   amp .
The   currents   into   the   input   terminals   of   an   ideal   op   amp   are   zero ,   and   the   voltages   at   the   input   nodes   of   an   ideal   op   amp    are   equal .
It   is   convenient   to   use   node   equations to   analyze   circuits   that   contain   ideal   op   amps.
Op   amps   are   used   to   build   circuits   that   perform   mathematical   operations .
The   output   of   the   voltage   follower   faithfully   follows   the   input   voltage.
The   voltage   follower   reduces   loading   by   isolating   its   output   terminal   from   its   input   terminal.
Practical   operational   amplifiers   have   properties   that   are   not   included   in   the   ideal   op   amp.
These   include   the   input   offset   voltage,   bias   current,   dc   gain,   input   resistance,   and   output   resistance.
51