3.1 applications of op amp part 1 3.1 Introduction
The internal structure and the electronics of op amp 1s discussed in the last chapter. It
has been mentioned that op-amp is generally used in the closed loop configuration,
using feedback.
The countless simple circuits using one or more operational amplifiers, alongwith
some external resistors and the capacitors can be constructed. Such applications of
re
classified
as
linear and nonlinear type of applications.
In the linear applications, output voltage varies linearily with respect to the input
voltage. In such linear applications, a negative feedback is used from the amplifier
output
Key
to the
Point
inverting input
Some
of
the
terminal.
linear
applications
are
voltage follower,
instrumentation amplifier, integrators, differentiators etc.
In the nonlinear applications, a feedback is provided from the output to the
noninverting input terminal. The feedback may be provided to the inverting input
terminal using nonlinear elements like diodes, transistors etc. The nonlinear circuits have
highly nonlinear input
to
output characteristics.
Key Point The typical nonlinear applications are precision rectifier, log amplifiers, antilog
amplifiers, comparators,
Schmitt
trigger
circuits etc.
3.2 Realistic Simplifying Assumptions
We can make two assumptions which are realistic and simplify the analysis of
op-amp circuits to
3.2.1
Zero
a
Input
great
extent.
Current
The current drawn by either of the input terminals (inverting and non-inverting) is
zero.
In practice, the current drawn by the input terminals is very small, of the order uA or
nA.
Hence the
assumption
of
zero
input
current
is realistic.
3.2.2 Virtual Ground
This
means
the
differential input voltage V between the
input terminals is essentially zero.
non-inverting
and
inverting
This is obvious because even if
output voltage is few volts, due to large open loOp
gain of op-amp, the difference voltage Va at the input terminals is almost
zero.
e.g. if output voltage is 10 V and the AOL i.e. open loop gain is 10* then
Vo
=
Vd AOL
V
Vd
100
=
AOL
104
1 mV
, the difference voltage Va&gt;0 and realistically
assumed to be zero for analysing the circuits.
Hence
Va
is very small. As
Vo
Vd
(V - V2)
V1
AOL
AoL
= o
=
= (
V2
(1)
Thus we can say that under linear range of operation there is virtually short circuit
between the two input terminals, in the sense
trom the input terminals to
the ground. The
Fig. 3.1 shows the concept of the virtual grournd.
Rr
R
that their voltages are same. No current flows
ww
w
v
The thick line indicates the virtual short circuit
V
I
I
Virtual
short
between the input terminals.
Now
if
the
non-inverting
terminalis
Fig. 3.1 Concept of virtual ground in
grounded, by the concept of virtual short, the
an op-amp
inverting terminal is also at ground potential,
though there is no physical connection between the inverting terminal and the ground.
This is the
principle
of virtual
ground.
Key Point Thus from the equation (1), the voltage at the non-inverting input terminal
of an op-amp can be realistically assumed to be equal to the voltage at the
inverting input terminal.
The realistic
simplifying assumptions
used to analyse the
applications. The steps of analysis based on the assumptions
are
always
practical
op-amp
Step 1: Input current of the ideal op-amp is always zero. Using this, the current|
distribution in the circuit is obtained.
Step 2
The input terminals of the op-amp are always at the same potential. Thus if
one is grounded, the other can be treated to be virtually grounded. From this, the
expressions for various branch currents can be obtained.
Step 3: Analysing the various expressions obtained, eliminating unwanted variables,
the output expression in terms of input and circuit parameters can be obtained.
3.3 ldeal Inverting Amplifier
As the name suggests the output of such an amplifier is inverted as compared to the
he
input
signal
signal.
means
The
having
inverted
output
phase
shift of
a
R
180 as compared to the input signal.
R
So, an amplifier which provides a
phase shift of 180&deg; between input and
vn
ww
A I
0
Op-amp
output is called inverting amplifier.
The basic circuit diagram of an
inverting amplifier using
shown in the Fig. 3.2 (a).
op-amp is
Fig. 3.2 (a) Inverting amplifier
Derivation of closed loop gain:
As node B is grounded, node A is also at ground potential, from the concept of
virtual ground, so VA = 0.
I
=
Vin-VA
Ri
I
= in
(1)
R1
Now from the
output side, considering the direction of
we can
write,
VA-Vo
I
I
current I
Rf
=
-Vo
RE
Entire current I passes
(2)
through Rf
as
op-amp
input
current is
zero.
Equating equations (1) and (2) we get,
Vin
R1
AvF
The
output
R
is
is the
opposite
Vo
Rf
Vo
(Gain
R
V in
gain of
the
to that of
emplifier
input.
with feedback)
while
Hence it is called
.
(3)
that the polarity of
polarity
inverting amplifier.
The input and output waveforms are shown in the Fig. 3.2 (b).
Vin
(Input)
Timet
Phase shift
of 180&deg;
Vo
(Output)
Timet
Fig. 3.2 (b) Waveforms of inverting amplifier
Observations
1. The
output
is inverted with
respect
2. The
voltage gain is independent
assumed to be large.
3. The
to
input,
of open
which is indicated
by
loop gain
op-amp, which
of the
minus
the ratio of the two resistances. Hence
and R1, the required value of gain can be easily obtained.
voltage gain depends
on
sign.
is
selecting R
4. It Rf &gt; Ri, the gain is greater than 1.
If Rf &lt; Ri, the gain is less than 1.
If
Rf
=
R1, the gain
is
unity
Thus the output voltage
voltage, in magnitude.
5. If
the ratio of
Rf
and
R
can
be
greater than,
less than
is K which is other than
or
equal
to
the circuit is called
one,
changer while for R/ R1 = 1 it is called phase inverter.
6.
The closed
loop gain
is
denoted
as
Ayp or AyCL
.e.
gain
the
with feedback.
input
scale
3.3.1 Sign Changer
then
if Rs
Ri
In the ideal inverting amplifier
the input
is s a m e as that of
of
output
magnitude
=
but its
is
gain
the
sign
is
AcL
-1. Thus the
=
opposite to that of the
input.
V
=-Vin
This circuit is called
for
R
sign changer
=
or
R1
phase
inverter.
3.3.2 Scale Changer
the
In the ideal inverting amplifier if R&cent; # R1 then
K=R/R1. Thus the circuit is used to multiply input by
gain
a
is
AcL
constant K
=
-
K where
called scaling
factor.
Vo
=KV in
The resistances must be
precision
This circuit is called scale
resistors to
changer.
the
scaling
factor K
precisely.
3.4 ldeal Non-inverting Amplifier
An amplifier which amplifies the
input without producing any phase
shift between input and
called
basic
non-inverting
+
outputis
amplifier.
The
circuit
diagram of a
non-inverting amplifier using op-amp
is shown in the
is
R
R
ww
applied
Fig.
the
to
(a).
3.5
B
Op-amp
Vo
The
input
non-inverting input
terminal of the op-amp.
Derivation of closed loop gain:
Fig. 3.5 (a) Non-inverting amplifier
The node B is at potential Vin, hence the potential of point A is same as B which is
Vin, from the concept of virtual share.
VA
From the
=
VB
= Vin
output side
I
we can
(1)
write,
Va
Rf
I
= o-Vin
(2)
Rf
At the inverting terminal,
I
A -0
R
I =in
(3)
RI
Entire current passes through R 1 as input current of op-amp is zero.
Equating equations (2) and (3),
Vo-Vin
Vin
R1
Rf
Vip in
Vo
R
Rf
Vo
R
=
+
R
(R +Rf)
inR
R
(R +R) Rf_R +R
R, R
Vin
AyF
R
Vlt R
R
in
(4)
The positive sign indicates that there is no phase shift between input and output.
The input and output waveforms are shown in the Fig. 3.5 (b).
Vin
(Input)
Timet
0
V
No
(Output)
Phase
shift
Timet
O
Fig. 3.5 (b) Waveforms of non-inverting amplifier
3.4.1 Comparison
The Table 3.1
provides
the
comparison
of the ideal
amplifier op-amp circuits.
Sr. No.
Ideal inverting amplifier
2
The
output
input.
3.
is inverted with
4.
respect
to
No
phase
shift between
input and
output.
The
The
The
input impedance is Ri.
non-inverting
Voltage gain = 1 +(R\$/ R1)
The voltage gain can be
greater than, equal to or less than one.
tkmme
and
Ideal non-inverting amplifier
Voltage gain = - Rr/ R1
1.
inverting
One.
voltage gain
is
always greater than
input impedance
is
extremely large.
3.10
Summer
or
As the input impedance of an op-amp is extremely large, more than one input signal
can e
appliei to the inverting amplifier. Such circuit gives the addition of the applied
gnals at the output. Hence it is called summer or adder circuit. Depending upon the
sign
of the output, the summer
circuits are classified
as
summer and
inverting sumnmer
non-inverting summer.
3.10.1 Inverting Summer
In this circuit, all the input
signals to be added are applied to
the
inverting input terminal
point
B is
grounded, due
to
virtual ground concept the node A
is
also
at virtual
w
w
of the
oPamp. he circuit with two input
signais is shown in the Fig. 3.24
As
R
R1
R2
V2
ww
A
B
oVo
2
ground potential.
0
(1)
Fig. 3.24 Inverting summer
Now trom the input side,
Vi-Va . V
R
2
RI
V
R2
R2
.(2)
(3)
Applying KCL at node A and as input op-amp current is zero,
I
= I1 + I2
.. (4)
From the output side,
I
=A-V,
RE
Vo
Yo
RE
Substituting (5), (2) and (3) in (4),
Vo
RRR
.
(5)
Vo
6)
If the three resistances
are
Vo
equal, R1
R2
=
=
Rf
-(Vi + V2)
.. (7)
By properly selecting Rf, R1 and R2, we can have weighted addition of the input
signals like aV1 +bV2, as indicated by the equation (6).
Infact in such a way, n input voltages can be added.
Key Point
Thus the magnitude of the ouput voltage is the sum of the input voltages and
hence circuit is called summer or adder circuit.
3.10.2 Non-inverting Summing Amplifier
The
circuit
discussed
above
R
is
ww-
inverting summing amplifier, which can
-I
R
equation (6). But a summer that gives
non-inverted
I
of the
input signals is, o
non-inverting
summinng
sum
w
called
amplifier. The circuit is shown in thev,
Fig. 3.25.
A
R1
oVo
ww=2
R2
Fig. 3.25 Non-inverting summing amplifier
Let the voltage of node B is Vg. Now
the node A is at the same potential as that of B, due to virtual ground.
(8)
VA =VB
From the input side,
V2-VB
V1VB and I2
.(9)
R2
R
But as the input current of op-amp is zero,
I1 +I2
- VB2-VB
R2
R
R1
=
0
=
0
..(10)
V
R2
VB
(R2 Vi +R, V2)
(R +R2)
..(11)
Now at node A,
I
VB
RR
=
R
as
VB
=
VA
(12)
I =o Va = VoVg
and
Rf
(13)
Rf
Equating the two equations (12) and (13),
Rf
Vo
RE
=
Vo
=
[R +R]
VBR
(14)
Substituting equations (11) in (14) we get,
1.e.
v.
R2 V +Ri V2)[R +R1
Vo
R
R2 (R
R (R+Rf) V
+
(R+Kf
V1
R (R +R2)
R (R +R2)
R (R+R2)
(15)
The equation (15) shows that the output is weighted sum of the inputs.
If R
= R2 = R
Vo
Key
Point
As
= Rf, we get
.
= V1+ V2
there is
no
phase difference
between
input
and
output,
it is called
non-inverting summer amplifier.
3.10.3 Average Circuit
If in the inverting summer circuit, the values of resistance are selected as,
R1 R2 =R
and
R
R
Then from the equation (6) we get,
(16)
3.11
wwww.
Subtractor
or
Difference
Dec.-11
Amplifier
Similar to the summer circuit, the subtraction of two input voltages is possible with
the
help
of op-amp circuit, called subtractor
or
difference
amplifier
circuit.
R
The circuit diagram is shown in the
Fig. 3.28.
R
To find the relation between the
inputsand
output
let
us
N
use
Superposition principle.
Let
Vo
be the
R2
output,
Vacting, assuming V
And
Vo2
be the
with input
to be zero.
V,
acts
as
an
zero,
inverting
B
L
R
2
acting, assuming V to be zero.
V,
Vo
ww
+0
output, with input V2
Case 1: With
A
the circuit
amplitier,
Fig. 3.28 Subtractor circuit
as
shown in the Fig. 3.29. Hence we can
Write,
R
ww
R
V,
ww
Vo1
R2
ww
R
Fig. 3.29
Vol
RV
R
..(1)
Case 2: While with V1 as zero, the circuit reduces to as shown in the Fig. 3.30.
R
I
R1
V
A
B
V
L
R2
R
Vo2
Let
potential of
node
R
VB
And
I
R
=
I
A is
VR: The potential of node
R
same as
Bi.e.
Va
=
VH
input V2 loop,
divider rule to the
Applying voltage
Now
B is
(2)
V2
(3)
R1
R
=2-Va Vo2
R
-VB
(4)
R
Equating the equations (3) and (4),
VB
Vo2-VB
Rf
R1
RR
VB
R
Vo2
.
(5)
Substituting V&szlig; from (2) in (5) we get,
Vo2
(6)
RIJR +Rr
Hence using superposition principle,
Vo
=
Vol +Vo2
RV1RRR
R
(7)
Now if the resistances are selected as R1 = R2,
V
RVi *|
R,RV2 V R
Vo=+ R V-V1)
R
Key Point
(8)
Thus the output voltage is proportional to the difference between the tvo 1put
voltages. Thus it acts as a subtractor or difference amplifier.
If
Ri = R2 =
Rf is selected,
Vo =V2V1
But
by
selecting
proper values
9)
of
R1, R2
inputs with appropriate strengths like
Vo
= aV2
and
Rf,
we can
have the subtraction of
two
- bV1
Thus using adders and subtractors, various mathematical equations like
V
=(aV1 +bV2)-(cV3 +dV4) can be solved. In such equation, two adder circuits are
used and the outputs of these circuits are used as input to a subtractor to obtain the
required equation.
Example 3.12 Design the op-amp circuit which can give the output as
V
2
V1-3 V2 +4 V3 -5 V4
Solution: The positive and negative terms can be added separately using two adders
and then subtractor can be used.
2 V +4 V3, let Rf|
For
RL
V,l v
R1
Vol
R
R
and
=100 k
2
=
RFL
hence R
=
50 kS2
4 hence R3 = 25 k2
R3
For
(3V2 +5 V4),
Rf2
=
120 k2
R1V1*
R2 1RRa *3
Vo2
R2
=
RR2
and
let
R2
=
3 hence
R2
=
40 kQ
5 hence R4 = 24 k2
R4
Use the subtractor with all the resistances of same value ofR = 100 kQ.
Hence the output of the subtractor is Vo = Vo2 - Vol where V2 and Voi are the two
inputs of the subtractor, derived from the previous adders
The output
voltage
is,
Vo2 - Vol = 3 V2 -5 V4 -(-2V -4 V3)
V
=
2V
-3
V2
+4 V3 -5 Va
This is the
required output.
W
V,o
Rn
R
R
Vo1-2V1 -4V3
w-
Vow
R
V30
V2
V4O
R3
VoN
-
R
o2o1)
R2
Vo2-3V2-5V4
R4