Lecture 5: Linear Op Amp Applications
In the previous lecture, the near-ideal op amp and two of its more important applications
were introduced. It was established that the actual closed loop gain was very close to
what we called the ideal closed loop gain — that value of closed loop gain obtained under
the assumption that the loop gain  Av is very much larger than unity. In this lecture, we
first define the fully ideal op amp and then show that analysing a linear op amp system
on the basis of the op amp being ideal directly yields the ideal closed loop gain
expression. The analysis task is greatly simplified.
Learning Outcomes:
On completing this lecture, you will be able to:




5.1
List the key properties of the ideal operational amplifier;
Describe and apply a methodology for working out the ideal closed loop gain;
Analyse a variety of linear circuits employing ideal op amps;
Carry out appropriate design calculations for circuits to meet given specifications.
The Methodology
The approach taken in the previous lecture to analysing a circuit with a near-ideal op amp
was to put an expression for the closed loop gain in the form
Acl  Acl id
Av
1  Av
and then assume (with good reason) that
Av  1
so that Acl  Acl id .
What we are now going to tackle is essentially a short-cut route to obtaining Acl id. The
approach starts by defining an ideal op amp as having all the properties of the
near-ideal op amp but with the open loop gain now being considered to be
indefinitely large and having the following consequence:
Consider an op amp operating in the linear region:
+
eD
-
e2
vO
e1
The output voltage is given by
vO  Av eD  Av e2  e1 
5-1
Now the output voltage is finite, lying within the limits of –VCC and +VCC. Therefore the
larger Av, the smaller eD  e2  e1 . In the limit
Av   
ie
e2  e1   0
e2  e1
This property, together with the assumption that the input resistance of the op amp is
indefinitely large, constitutes the main lever by which we can analyse any linear op amp
system. We illustrate the approach in the next section by re-visiting the two examples of
the previous lecture.
5.2
Non-Inverting and Inverting Amplifiers Re-Visited
Consider first the non-inverting amplifier:
e2
+
-
e1
i2
iIN
R2
vI
vO
R1
vF
i1
e2  vI
As before
Again, with
RIN   , we have i IN  0 and therefore i2  i1 giving
e1  v F 
R1
vO
R2  R1
Negative feedback implies linear operation and if the op amp is deemed ideal then we can
take
e2  e1
This gives
5-2
R1
vO
R2  R1
vI 
Re-arranging we have
vO
R  R1
R
 Acl id  2
 1 2
vI
R1
R1
Note the same result as obtained in the previous lecture.
Now consider the inverting amplifier.
R2
i2
R1
i1
e1
vR1
e2
vI
vR2
iIN
+
vO
RIN   , we have i IN  0 and therefore i2  i1 . Next we derive expressions
Again, with
for i1 and i2.
i1 
Noting that
v R1 v I  e1

R1
R1
i2 
v R 2 e1  vO

R2
R2
e2  0 and the fact that the op amp is ideal gives e2  e1 , we have
e1  o
i1 
and
vI
R1
Finally, because
and the expressions for the currents simplify to
and
i2 
 vO
R1
i2  i1 we have
v I  vO

R1
R2
yielding for the closed loop gain
vO
R
 Acl id   2
vI
R1
5-3
Again, the result is the same as for the previous lecture.
5.3
The Differential Amplifier
i2
R2
R1
i1
i3
-
e1
e2
R3
+
i4
v1
v2
R4
vO
We now investigate a circuit having two signal inputs, v1 and v2. The objective is to
determine the relationship between the system output vO and the signal inputs on the
basis that the op amp is ideal. Broadly speaking the approach is:
Equate i1 with i2 and i3 with i4 on the basis that
RIN   and therefore no current flows
into (or out of) either op amp input. We then make use of the infinite open loop gain
property to explore the consequences of e2  e1 .
With
i2  i1 the resistors R1 and R2 form a resistor divider giving
e1 
With
R2
R1
v1 
vO
R1  R2
R1  R2
i4  i3 the resistors R4 and R3 also form a resistor divider giving
e2 
R4
v2
R3  R4
Again the ideal op amp implies
e2  e1 giving
R1
R4
R2
vO 
v2 
v1
R1  R2
R3  R4
R1  R2
With four arbitrary resistors, we cannot say a great deal further about the relationship
between vO and v1 and v2. In practice, however, the four resistors are normally chosen to
satisfy the requirement that
5-4
R4 R2

R3 R1
We now develop this relationship as follows:
R3 R1

R4 R2
Now add 1 to each side and substitute appropriately for 1
R3
R
1  1 1
R4
R2
R3 R4 R1 R2



R4 R4 R2 R2
R3  R4 R1  R2

R4
R2
Inverting this, we get
R4
R2

R3  R4 R1  R2
Substituting this back into the vO expression above, we get
R1
R2
R2
vO 
v2 
v1
R1  R2
R1  R2
R1  R2
And, finally, simplifying this we obtain
vO 
R2
v2  v1 
R1
The circuit amplifies the signal difference (v2 – v1) by the factor
R2
R1
and hence is
referred to as a differential amplifier. Note that this sensitivity to the input signal
difference is critically dependent on fulfilling the resistor ratio criterion
R3 R1
.

R4 R2
Achieving this criterion is best approached by means of integrated circuit technology
when all the resistors would be manufactured together and if there was an error in one
resistor then the likelihood would be that there would be a similar error in the others.
The differential amplifier is very important for measurement and instrumentation
systems.
5-5
5.5
The Integrator
We now wish to explore the operation of the following op amp circuit which incorporates a
capacitor.
C
iC
iR
vC
R
e1
e2
vR
vI
+
vO
The approach taken will be very similar to that adopted for the inverter but first we need
to consider a suitable current-voltage relationship for the capacitor. Simply saying that
the capacitor represents an impedance of
1
restricts our consideration to pure
jC
sinusoidal signals; we would be interested in a more fundamental perspective and
consequently we characterise the current-voltage relationship of the capacitor by
iC
vC
iC  C
C
dvC
dt
where iC is the current through the capacitor and vC is the voltage across the capacitor.
Similar to the inverter
i R  iC
and this may be converted to
d e1  vO 
v I  e1
C
R
dt
Noting the infinite open loop gain giving
e2  e1 and e2  0 , we have e1  0 and so
dv
vI
 C O
R
dt
5-6
dvO  
1
v I dt
RC
Integrating both sides and denoting the initial value of output voltage by VO 0 , we have
t
1
vO  
v I dt  VO 0
RC 0
That is, the output voltage is proportional to the time integral of the input signal voltage.
Suppose, for example, that vI is constant at some negative level –E and that the initial
output voltage is zero, then the output voltage as a function of time is given by
vO  

1
RC
t
  E dt
0
E
t
RC
E
volts per sec.
RC
ET
This is illustrated below. After T seconds, the output voltage will have reached K 
RC
That is, the output voltage increases linearly with time at the rate of
volts.
vI
0
T
t
-E
vO
K
0
T
5-7
t
Example 5.1:
An integrator uses a resistor of value R = 10kΩ and a capacitor of
value C = 0.01µF. The input signal is a square wave of 2V amplitude and 1kHz frequency
as shown below. Determine the output signal waveform.
vI
+E
0
T/2
T
2T
T/2
T
2T
t
-E
vO
+K
0
t
-K
Basically what happens in the circuit is the following:
For 0  t 
T
the system integrates the positive constant level E  2V and thus
2
produces a negative-going linear waveform;
For
T
 t  T the system integrates the negative constant level E  2V and thus
2
produces a positive-going waveform.
Joining up these two half-cycles results in a triangular output waveform as shown above.
We might also note that since the DC or average value of the input is zero and the
integral of zero is zero, the output triangular waveform will be symmetrical about zero. It
thus remains to compute the amplitude K of the output waveform.
Consider the interval 0  t 
vO  
1
RC
T
2
  E dt
T
2
 VO 0
gives
0
5-8
K 
2K 
K
E T
K
RC 2
ET
2 RC
ET
4 RC
With the data as supplied and noting that T 
K
210 3 
 
4
4 10 10
8

1
 10 3 s , we obtain
3
10
 
 0.5 101  5V
That is, the triangular waveform oscillates between -5V and +5V with a 1ms period.
5.6
A Logarithmic Amplifier
Consider the following op amp circuit:
D
iD
iR
vR
vI
vD
R
e1
e2
+
vO
The lay out of the circuit is very similar to that of the inverting amplifier or the integrator
except that, in place of the feedback resistor R2 (of the inverter) or the capacitor C (of the
integrator), we have a new component labelled D. In principle, and assuming the op amp
to be ideal, in order to work out the relationship between vO and vI all we need is a
mathematical expression for the current-voltage relationship of device D. The same circuit
analysis approach can be taken as was employed for the analysis of the inverter or the
integrator.
Device D is a semiconductor diode (or, as it is sometimes called in the textbooks, a pn
junction diode). From semiconductor physics, it may be established that the currentvoltage relationship for the device is closely approximated by
5-9
iD
D
vD
v
i D  I S exp  D
 VT



where IS and VT are (temperature-dependent) constants.
Typical values of these constants for room temperature are I S  10
15
A and VT  26mV
Resulting in an iD versus vD characteristic as follows:
iD
3.5mA
0.75V
vD
Assuming that electronic circuits typically operate with currents of the order of milliamps,
the plot is seen to rise very sharply (exponentially, in fact) in the vicinity of 0.75V.
Returning now to the task of analysing the op amp circuit at the beginning of this section,
we note as before:
iR  iD
e v
v I  e1
 I S exp  1 O
R
 VT
or
As before, infinite open loop gain gives
v
v I  RI S exp  O
 VT
e2  e1 and e2  0 , so that e1  0 resulting in



Taking the natural logarithm of each side
ln v I   ln RI S  



vO
VT
5-10
Re-arranging we get
vO
 ln RI S   ln v I 
VT
vO  VT ln RI S   VT ln v I 
That is, the output voltage is proportional to the log of vI and this is what gives the circuit
its title of logarithmic amplifier.
5.7
Concluding Remarks
We have consistently employed two key features of an ideal op amp,
RIN   and
Av   , for the analysis of a variety of op amp circuits. Each time we directly get an
expression for what we previously termed the ideal closed loop gain or for the output
voltage that would result from Acl id . If greater accuracy is required, the appropriate
expression can always be multiplied by
Av
1  Av
. However, the intent here has been to
introduce the range of possible applications of the op amp and to focus on the dominant
mode of operation associated with the op amp being considered ideal.
5-11