Chapter 6
Operational Amplifier
and Applications
_____________________________________________
6.0 Introduction
Operational amplifier has been used for a long time. Its initial application was in
the area of analog computation and instrumentation. Earlier day, operational
amplifier was constructed using vacuum tube and discrete components. As the
result of first development operational amplifier in integrated circuit form by
Robert J. Wildar in Fairchild Semiconductor in early 1960’s that today,
operational amplifier is constructed in integrated circuit form has much reduce
in term of price and versatility in applications.
A usual 14-lead dual in line package can contain up to quad operational
amplifiers. However, for high precision and high performance operational
amplifier, a package usually contains one operational amplifier.
Operational amplifier is basically a five terminal device which comprises
of a non inverting input denoted as '+', an inverting input denoted as '-', an
output, and two power supply terminals, which are denoted as 'V+' and 'V-'.
Operational amplifier usuallyhave additional terminals used for offset
adjustment. The circuit symbol of an operational amplifier is shown in Fig.6.1.
Figure 6.1: Circuit symbol of an operational amplifier
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6 Operational Amplifier and Applications
The operational amplifier is characterized with high input impedance, low input
current, high common-mode rejection ratio, high voltage gain, high output
current, and low output impedance.
Operational amplifier is made of three basic circuits. They are differential
amplifier, voltage amplifier, and output amplifier. The internal circuit of an
operational amplifier can be represented in Fig. 6.2. Combining these three
circuits, it would achieve the characteristics mentioned earlier.
Figure 6.2: Internal circuits of operational amplifier
6.1 Parameters of Operation Amplifier
Operational amplifier has many electrical parameters. In this section, we shall
study some important parameters of operational amplifier that are normally
utilized in the design of the circuits.
6.1.1 Input Offset Voltage
An ideal operational amplifier should have zero offset voltage. However, a real
operational amplifier has offset voltage due to the transistors of the differential
operational amplifier are not identical. A small difference in the base-emitter
voltage of transistor Q1 and Q2 would cause different in collector current. Thus,
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6 Operational Amplifier and Applications
it causes difference in output voltage. Sometime not only the base-emitter
voltage is the factor, the slight different in collector resistor RC value would also
cause non-zero offset voltage. Figure 6.3 illustrates the sources of offset
voltage.
Figure 6.3: Offset voltage of operational amplifier
Thus, input offset voltage VIO is defined as
VIO = Vout1 - Vout2
(6.1)
Alternatively, it is defined as the input voltage difference to make zero voltage
difference between Vout1 and Vout2 of the transistors. Thus,
VIO = VBE1 - VBE2
(6.2)
6.1.2 Input Bias Current
The input bias current is basically the base current of BJT transistor in the
differential amplifier. It is the gate current of FETs in the differential amplifier.
Input bias current is either positive base current I +B or negative base current I −B .
In conventional definition, input bias current Ibias is defined as
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6 Operational Amplifier and Applications
Ibias =
I +B + I −B
2
(6.3)
6.1.3 Input Offset Current
Ideally, the two input bias currents are equal so that their difference is zero. In
reality, the bias current of two transistors is not equal. Thus, input offset current
IIO is the difference between two inputs bias-current. Thus,
IIO = | I +B − I −B |
(6.4)
The relationship of input offset current and input offset voltage is defined in
equation (6.5).
VIO = I1Rin - I2Rin= ( I +B − I −B )Rin= IIORin
(6.5)
where Rin is the common-mode input impedance and I1 and I2 are base current
or gate current of the BJTs or FETs of the differential amplifier
If offset current is large, it can cause a large error at output voltage because
the offset voltage will increase. The analysis will be done in later section.
6.1.4 Input Impedance
There are two ways to specify the open-loop input impedance of an operational
amplifier. The differential input impedance is the total resistance between the
inverting and the non-inverting input. It is determined by the change of bias
current for a given change in differential input voltage. The common-mode input
impedance is measured from the common input to ground. Figure 6.4 illustrates
the difference between two specifications of input impedance.
(a) Differential input impedance
(b) Common-mode impedance
Figure 6.4: Input impedance of operational amplifier
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6 Operational Amplifier and Applications
6.1.5 Output Impedance
Ideally, the open-loop output impedance of an operational amplifier should be
zero. Typically, it is about 50Ω. Low output impedance has the purpose of
reducing attenuation of signal to next stage of the circuit.
6.1.6 Open-Loop Voltage Gain
Open-loop gain Aol is the gain of the operational amplifier without any external
feedback from the output to input. A good operational amplifier has a very high
open-loop gain ranging from 50,000 to 200,000. This shall mean a small voltage
difference Vdiff between two inputs would drive the output to saturation easily.
This also explains how the operational amplifier configured as a comparator
works. A negative voltage difference Vdiff will swing the output to negative
saturation and a positive voltage difference Vdiff will swing the output to positive
saturation.
High open-loop gain of an operational amplifier creates an unstable
condition because the noise voltage can be amplified to a level that may drive
the operational amplifier out of its linear region and falls into the saturation
region.
A good way to stabilize and control the gain of an operational amplifier is
by mean of adding a closed-loop negative feedback circuit, which is taking part
of the output voltage and applying it back out-of-phase to the input. As the
result, the closed-loop gain is much less than open-loop gain. Thus, it stabilizes
the circuit.
6.1.7 Common-Mode Range
This is the range of input voltages applied to both inputs that will not cause
clipping or distortion of signal at the output of the operational amplifier. This is
the input voltage applied to the operational amplifier that the output of the
operational amplifier is still operating in the linear region. Most operational
amplifier has common-mode range of ± 10.5V for ± 15V dc supply. A high
performance operational amplifier circuit can achieve the common-mode range
of ± 11.5V for ± 15V dc supply. Fig. 6.5 illustrates the common-mode range of
a typical operational amplifier. As illustrated in the figure, the common-mode
range is approximately ± 11.5V for input voltage range up to approximately ±
0.27V range. The theoretical maximum output range for an operational
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6 Operational Amplifier and Applications
amplifier can be defined as ±Vsat = ± VCC m 2.0V, which is + 12.0V to -12.0V
for ± 15.0V supply.
Figure 6.5: A graph shows the typical common-mode range of an operational amplifier
6.1.8 Common-Mode Rejection Ratio
Common-mode rejection ratio CMRR is defined as the ratio of differentiatemode gain AV(dm) and common-mode gain AV(cm). i.e. CMRR =
A V ( dm )
A V ( cm )
.
Theoretically, the common mode rejection ratio CMRR of any operational
amplifier should have an infinite value but practically it is not achievable
because zero common-mode gain is not attainable due to the fact that offset
voltage is never be able to attain zero volt.
6.1.9 Power Supply Rejection Ratio
Power supply reject ratio PSRR indicates how much the offset voltage changes
when the supply voltage is changed. If an operational amplifier has a power
rejection ratio PSRR of 25µV/V, it means that the offset voltage is changed by
25µV for every one volt change of power supply. Therefore, PSRR follows
equation (6.6).
PSRR =
∆VIO
∆VDC
(6.6)
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6 Operational Amplifier and Applications
Usually, PSRR is expressed in decibel i.e. 20 log10 (∆VIO / ∆VDC ) .
6.1.10 Output Short-Circuit Current
Output short-circuit current is the maximum current that an operational
amplifier would supply when the output is shorted. The typical output short
circuit current of the µA741 operational amplifier is 20.0mA.
6.1.11 Slew Rate
It is the measurement of how fast the output voltage transition from a specified
voltage level to another specified voltage level. If the slew rate of the
operational amplifier specified to be 0.5V/µs, it means that its output can
change at the rate of 0.5V every microsecond.
Slew rate S can also be used to determine the maximum operating
frequency of an operational amplifier. If the output voltage of the operational
amplifier is Vout = VPsinωt, then maximum rate of change of output voltage is S
=
dVout
dt
= VP ω . Thus, the maximum frequency fmax follows equation (6.10).
ωt = 0
fmax =
where Vinp =
S
S
=
2πVP
2πVinp A V
(6.10)
VP
is the amplitude of the input signal. The maximum output
AV
voltage without distortion is common-mode range voltage, which is also the
saturation voltage. Thus, maximum amplitude VP of the output signal is half of
the common-mode range at output.
If the input signal is ramping at the rate of Sin voltage per second i.e. Sin =
dVin
, in order not to get distortion at the output of the operational amplifier, the
dt
gain of the amplifier should have the voltage gain AV such that its output rate of
change of voltage is not greater than slew rate S of the operational amplifier.
Since Vout = AVVin, dividing this equation with time t give rise to
Vout
V
= A V in , which is S = AVSin. Therefore, the maximum gain AV(max) of the
t
t
operational amplifier is
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6 Operational Amplifier and Applications
AV(max) =
S
Sin
(6.11)
Example 6.1
If the slew rate of an operational amplifier is 0.5V/µs and its common-mode
range is 10V and - 10.5V, what is the maximum operating frequency of this
operational amplifier? And if the input signal is changed at the rate of 0.1V/µs,
what is the maximum gain of the operational amplifier?
Solution
The maximum frequency for the positive voltage range is fmax =
S
=
2πVP
0.5V / µs
= 7957.8Hz.
2π(10V)
The maximum frequency for the negative voltage range is fmax =
S
=
2πVP
0.5V / µs
= 7578.8Hz.
2π(10.5V)
The maximum operating frequency of the operational amplifier is 7578.8Hz.
The maximum gain of the operational amplifier is AV(max) =
0 .5
= 5.
0 .1
6.1.12 Thermal Drift
Operational amplifier is made of diodes and transistor whose parameters change
with temperature. Thermal drift is the measure of change in an offset parameter
due to a unit change in temperature. Three drift parameters are usually
measured. They are thermal voltage drift DV, thermal biasing current drift Db,
and thermal input offset current drift Di. They are respectively defined in the
following equations.
DV =
∆VIO
(V/ 0 C)
∆T
(6.12)
Db =
∆I B
(V/ 0 C)
∆T
(6.13)
Di =
∆I IO
(V/ 0 C)
∆T
(6.14)
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6 Operational Amplifier and Applications
where IB is the input biasing current assuming that I −B = I +B .
6.1.13 Other Parameters
There are many other parameters not discussed here. The examples of the
parameters are overshoot, rise time, fall time, input and output capacitance,
settling time, and etc.
6.2 Non Ideal Operational Amplifier
Operational amplifier can be modeled as shown in Fig. 6.6. It contains a finite
input impedance Ri, a finite output impedance Ro and a differential input Vi.
Figure 6.6: Operational amplifier model
It basically contains an input differentiate stage Vi = (V1 - V2) which has high
impedance Ri and a output voltage stage which is the product of open loop gain
Aol and input differentiate voltage and its low output impedance Ro.
In the earlier study of operational amplifier circuit application, the
operational amplifier is treated as the ideal type whereby its input impedance Ri
is infinity and output impedance Ro is zero.
Considering an inverting operational amplifier circuit shown in Fig. 6.7, its
ideal gain is A V = −
R2
. However, the non-ideal inverting amplifier circuit is
R1
modeled as shown in Fig. 6.8. In the circuit input resistance Ri and output
resistance Ro are added.
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6 Operational Amplifier and Applications
Figure 6.7: An inverting operational amplifier circuit
Figure 6.8: Model circuit of the non-ideal inverting operational amplifier
From the circuit shown in Fig. 6.8, the output voltage Vout is
Vout = -AolVi + IRo
(6.15)
and the input differential voltage Vi is
Vi = IR2 +Vout
(6.16)
Substituting equation (6.16) into equation (6.15), it yields Vout = IRo -Aol(IR2
+Vout). This shall mean that current I is equal to I = [(1+Aol)Vout]/(Ro –AolR2).
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6 Operational Amplifier and Applications
Applying Kirchhoff’s voltage law for the loop gives rise the sum of open
loop output voltage and input voltage Vin is equal to
Vin = I(R1 + R2 ) + Vout
(6.17)
Combining equation (6.17) for current I into equation I = [(1+Aol)Vout]/(Ro –
AolR2).
AV =
Vout
R O − A ol R 2
=
Vin
R 1 (1 + A ol ) + R 2 + R O
(6.18)
Dividing the numerator and denominator by Aol, and for very large value of Aol,
Ro/Aol → 0, R2/Aol → 0, and (1+Aol)/Aol → 1. Thus voltage gain AV is
approaching - R2/R1 which is the ideal gain equation.
Similarly the model circuit of non-inverting operational amplifier circuit
shown in Fig. 6.9 has the voltage gain AV follows expression,
AV =
A ol (R 1 + R 2 + R O )
R 1 (1 + A ol ) + R 2 + R O
(6.19)
Figure 6.9: Model circuit of the non-ideal non-inverting operational amplifier
The equation (6.19) can be derived by finding the voltage across input
resistance Ri using Superposition theorem. The voltage across input resistance
Ri Vi is the sum of voltage across input resistance by treating each voltage
source Vin and AolVi separately. Thus, Vi1 from Vin source is equal to Vi1 = Vin
by treating [R1||(R2+Ro)] << Ri. Vi2 from (-AolVi) is Vi2 = R1/(R1+R2)x(AolVi) by
treating (R1||Ri) = R1. As the result Vi = Vin + R1/(R1+R2)x(AolVi). Sorting for Vi
and substitute Vi = Vout/Aol. The gain equation (6.19) shall be obtained.
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6 Operational Amplifier and Applications
6.3 Effect of Offset Current and Offset Voltage to Output
Offset Voltage of Operational Amplifier
In this section, we will learn the effect of output offset voltage VOS due to offset
current IIO and offset voltage VIO. Considering an inverting amplifier circuit
with compensation resistor RC shown in Fig. 6.10 has its input connected to
ground and its input bias current represented by current source I +B and I −B . The
circuit can be re-drawn into Thévenin’s equivalent circuit where the inputs have
voltage source - I +B R2 and I +B RC connected in series with resistance R2 and RC to
the input Vin- and Vin+ of the operational amplifier.
Figure 6.10: Offset current compensation circuit for an inverting amplifier
Superposition principle is used to determine the output voltage of the amplifier
as shown in Fig. 6.11. Shorting Vin+ pin, it yields output voltage VO1 =
R1 −
IBR 2 =
R2
I −B R 1 .

R 

2
Shorting Vin- pin, it yields output VO2 = − 1 + 1 I +B R C . Thus, the output
R

voltage Vout of the amplifier shall be the sum of VO1 and VO2. This output
voltage Vout is also the output-offset voltage VOS( I ) since it is caused by offsetcurrent. Thus, the output voltage due to offset current is
B

Vout = VOS( I ) = I B− R 1 − 1 +
B

 +
I B R C

- 178 R1
R2
(6.20)
6 Operational Amplifier and Applications
Figure 6.11: Thévenin’s equivalent circuit of Fig. 6.10
If the bias current I −B equals to bias current I +B equals to IB and resistor RC can
effectively eliminate the output-offset voltage i.e. VOS( I ) = 0, then value of
resistor RC shall be equal to R1||R2. If bias current I −B is not equal to bias current
B
I +B and after substituting RC =
R 1R 2
, equation (6.20) becomes
R1 + R 2
VOS( I B ) = (I −B − I +B )R 1 = IIOR1
(6.21)
The effect of offset voltage VIO on output voltage Vout can be determined using
circuit shown in Fig. 6.12.
Figure 6.12: Amplifier circuit showing the effect on output due to input offset voltage
The output voltage Vout which also the output offset voltage VOS( V ) , is equal to
IO
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6 Operational Amplifier and Applications

R 

2
Vout = VOS( V ) = 1 + 1 VIO
R
IO

(6.22)
Thus, the total output offset voltage VOS i.e. due to offset current and offset
voltage shall be

VOS = (I −B − I +B )R 1 + 1 +

R1
R2

VIO

(6.23)
6.4 Bias Current and Offset Voltage Correction
The bias current and offset voltage of a non-ideal operational amplifier are not
equal to zero, thus, in a negative feedback amplifier, a small bias current would
create a small voltage at the output due to potential cross feedback resistor Rf.
When two inputs of an operational amplifier are set to zero volt, the output
exists a voltage called offset voltage. This voltage causes error in measurement.
Bias current for operational amplifier can be compensated by adding a
compensation resistor RC to the operational amplifier as it is shown in noninverting amplifier circuit in Fig. 6.13. The value of the compensation resistance
is equal to the parallel value of the feedback resistance i.e. RC = R1||R2. In order
to offset the biasing current, the compensation resistor RC is derived based on
the fact the voltage at input Vin- and Vin+ are the same. Thus, the Thévenin’s
resistance at input Vin- and Vin+ should be the same i.e. RC = R1||R2.
Figure 6.13: Bias current compensation for a non-inverting amplifier
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6 Operational Amplifier and Applications
Figure 6.14 shows the bias current compensation circuit for an inverting
amplifier. The value of the compensation resistance is equal to the parallel value
of the feedback resistance i.e. RC = R1||R2.
Figure 6.14: Bias current compensation for an inverting amplifier
Figure 6.15 shows the bias current compensation circuit for a voltage follower.
The value of the compensation resistance is equal to the parallel value of the
feedback resistance i.e. RC = R1.
Figure 6.15: Bias current compensation for voltage follower
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6 Operational Amplifier and Applications
There are many ways can be used to compensate offset error problem. One way
to compensate offset voltage is using the circuit shown in Fig. 6.16. i.e.
connecting a voltage source to adjust the input such that it produces a zero
output. The voltage at node a Va is equal to Va = VIO =
R C Vnt
, where Vnt
R nt + R b + R C
is the Thevénin’s voltage which should be equal to VCC = VEE, whilst Rnt is the
Thevénin’s resistance which shall be equal to R/4. Usually the value of
resistance is such that Rb >Rnt > Rc so that the offset voltage VIO is
approximately equal to VIO ≈
R C VCC (= VEE )
. Note that the value Rb > 10Rnt and
Rb
Rb > 1000Rc will be ideal.
Figure 6.16: Offset voltage compensation
Offset voltage can also be compensated by connecting a potentiometer to
designated offset null pin of the operational amplifier and adjusting the
potentiometer until the output voltage read zero volt. For the µA741 operational
amplifier, a 10 kΩ potentiometer is recommended to be connected between null
pin 1 and 5 and adjusting the potentiometer via V- voltage pin until the output
read zero volt. This arrangement can be used to adjust ± 15 mV of offset voltage
for the µA741operational amplifier.
6.5 Applications of Operational Amplifier
In this section, we shall discuss the applications of operational amplifier
involving negative feedback loop. We shall begin with the most common
operational amplifier applications that are the inverting and non-inverting
amplifiers and some special operational amplifier circuits such as the current
source and logarithmic amplifiers will also be studied.
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6 Operational Amplifier and Applications
6.5.1 Inverting Amplifier
A typical inverting amplifier circuit is shown in Fig. 6.17. The input signal is
connected to the Vin- pin of the operational amplifier. The feedback portion is
also connected to the Vin- pin. The Vin+ pin is either connected to the ground
point or compensation resistance before connected to ground point.
Using KCL law, the current in feedback resistor Rf is (Vout -Vin-)/Rf and
Similarly, current in input resistor Ri is (Vin - Vin-)/Ri. Thus,
(Vout -Vin-)/Rf = - (Vin - Vin-)/Ri
(6.24)
Since Vin+ is at zero volt and (Vin- - Vin+) is VIO, we can also assume that Vin- is
equal to zero volt. Thus, equation (6.24) shall be Vout/Rf = - Vin/Ri and the
closed-loop gain is
Acl = Vout/Vin = -Rf/Ri
(6.25)
where Acl is the closed-loop gain.
Figure 6.17: Inverting amplifier
6.5.2 Non-Inverting Amplifier
A typical non-inverting amplifier circuit is shown in Fig. 6.18. The input signal
is connected to the Vin+ pin of the operational amplifier. The negative feedback
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6 Operational Amplifier and Applications
portion remains the same like any other operational amplifier applications
mentioned in this chapter.
Figure 6.18: A non-inverting amplifier
The voltage at output Vout is Vout = Aol(Vin - Vf). Thus, the feedback voltage is Vf
Ri
⋅ Vout .
Ri + R f
=

Aol  Vin −
Therefore,
the
output


Ri
Ri
⋅ Vout  . Since Vout 1 + A ol


Ri + R f
Ri + R f




Ri
Aol/ 1 + A o
Ri + R f

voltage
Vout
is
Vout
=

 = AolVin, therefore Vout/Vin =



Ri
 . The term Aol
>> 1, thus, the closed-loop gain Acl is

Ri + R f


Acl = Vout/Vin = 1 +

Rf 

R i 
(6.26)
6.5.3 Weighted Summing Amplifier
An n-input summing amplifier circuit is shown in Fig. 6.19.
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6 Operational Amplifier and Applications
Figure 6.19: An n-input weighted summing amplifier
The differential inputs are virtually at ground potential. Using Kirchhoff’s
current law KCL, the sum of the current in all n-input is equal to negative
current flowing in feedback resistor Rf.
Using Kirchhoff’s current law, a mathematical expression can be formed
such that the current in the feedback resistance is equal to the same of the
current in the input resistors. Thus,
V
Vout
V
V
V
= − IN1 + IN 2 + IN3 + ....... + INn
Rf
R2
R3
Rn
 R1



(6.27)
If R1 = R2 + R3 = …. = Rn = Rin , the Vout shall be equal to
nR f
Vout
(VIN! + VIN 2 + VIN3 + .... + VINn ) .
=−
Rf
R
Example 6.2
An op-amp circuit is shown in the figure. Given that R1 = 10kΩ, R2 = 100kΩ,
R3 = R4 = R5 = 10kΩ, R6 = 20kΩ, V1 = 20mVdc, V2 = 500mVdc, V3 = 200mVPP ac. Sketch the signal at output Vout and find the amplitude of ac voltage and
dc voltage at Vout.
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6 Operational Amplifier and Applications
Solution
The output of operational amplifier one is - 200mVdc
This - 200mVdc becomes the input to the second amplifier.
The input sum of second amplifier is
(-200mVdc + 500mVdc +200mVP-P ac)
=300mVdc + 200mVP-P ac
Since the amplification factor is 2, thus,
The output voltage shall be
-2(300mVdc + 200mVP-P ac)
= -600mVdc - 400mVP-P ac
The sketch of the output is shown in the figure below
The dc voltage at Vout shall be - 600mVdc and amplitude of the ac is
200mV.
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6 Operational Amplifier and Applications
6.5.4 Difference Amplifier
The circuit of a difference amplifier is shown in Fig. 6.20. The principle of
superposition shall be used to determine the relationship of output Vout with
input V1 and V2.
Figure 6.20: A difference amplifier
Firstly the input V2 is treated as zero and determines the relationship of Vout1
with V1 and resistive components. Secondly the input V1 is treated as zero and
determines the relationship of Vout2 with V2 and resistive components. Finally,
adding Vout1 and Vout2 will get Vout.
Figure 6.21 shows the superposition analysis of circuit shown in Fig. 6.20.
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6 Operational Amplifier and Applications
(a)
(b)
Figure 6.21: Superposition analysis of difference amplifier shown in Fig. 6.20
From the circuit shown in Fig. 6.21(a), the output one Vout1 is Vout1 = −
R2
V.
R1 1
From Fig. 6.21(b), the input Vin+ and input Vin- are approximately equal, without
considering the offset voltage VIO, thus,
Vout2 is Vout 2 = V2
R4
R3 + R4
R4
R1
V2 =
V and output 2
R3 + R4
R 2 + R 1 O2
 R2 
1 +
 . Since Vout is the sum of Vout1 and Vout2, thus,
R 1 

output voltage Vout is
Vout = −
R2
1 + R 2 / R1
V1 +
V2
R1
1+ R3 / R4
(6.28)
If R1 = R3 and R2 = R4 then equation (6.28) becomes,
Vout =
or Vout =
R2
(V2 − V1 )
R1
(6.29)
R4
(V2 − V1 ) . If R1 = R2 = R3 = R4 then
R3
Vout = V2 - V1
(6.30)
The significance of difference amplifier is that if there is a common-mode input
V1 and V2, then output Vout is equal to zero.
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6 Operational Amplifier and Applications
6.5.5 Comparator
Operational amplifier is often used as the comparator to compare the amplitude
of one voltage with another. This is probably the only application that using the
open-loop configuration with voltage at one input and a reference voltage on the
other. This shall mean a small voltage difference Vdiff between two inputs would
drive the output to saturation. A negative Vdiff will swing the output to negative
saturation and a positive Vdiff will swing the output to positive saturation.
6.5.5.1 Zero-Level Detection
A basic application of comparator is to determine when the input voltage is less
than or exceed the zero level such as the one shown in Fig. 6.22 as zero level
detector. When the input voltage is greater than the voltage at the V- input, the
output of the op-amp will swing to positive saturation voltage. Likewise, when
the input is smaller than V- voltage, the output of the op-amp will swing to
negative saturation voltage. The output waveform normally will have a 50%
duty-cycle waveform.
(a) The comparator circuit
(b) The input and output waveform
Figure 6.22: Zero-level detector
6.5.5.2 Nonzero-Level Detector
Nonzero-level detector uses a fixed reference voltage such as the circuit shown
in Fig. 6.23. The reference level is defined as Vref =
R2
( + V) . Similarly a
R1 + R 2
negative reference level can be provided voltage of the divider circit to be
negative. The duty-cyle of nonzero-level detector usually has different dutycycle. For the case of positive reference, the duty-cycle is less than 50% and for
the case of negative reference, the duty-cycle is more than 50%.
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6 Operational Amplifier and Applications
(a) The comparator circuit
(b) The input and output waveform
Figure 6.23: Nonzero-level detector
6.5.6 Integrator
In mathematical calculus, integration is the process of finding the area under the
curve for a given interval of x-axis values.
The operational amplifier can be configured as an integrator that can
perform integration as the circuit shown in Fig. 6.24.
Figure 6.24: An operational amplifier integrator
Since Vin is virtually ground, using KCL, I = Vin/R, where I is the current
flowing in resistor R. This current is used to charge the capacitor C until its
voltage reaches negative Vin. Also current is equal to I = - C
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dVout
. Thus,
dt
6 Operational Amplifier and Applications
dV
Vin
= - C out
dt
R
(6.31)
Thus, the output voltage Vout is Vout = −
dVout
1
Vin dt + cons tan t .
is also known
∫
dt
RC
as rate of change of output voltage Vout.
Example 6.3
An integrator circuit as shown in Fig. 6.24 has R = 50kΩ, C = 0.01µF, and input
voltage Vin = 5sin3000πt. Find the amplitude and phase angle of the output.
Solution
=
−
5cos3000πt/(3000πx50kΩx0.01µF). Thus, the amplitude
x50kΩx0.01µF) = 1.06V, the phase difference is 900.
is
The
integration
of
5sin3000πt
will
yield
Vout
1
Vin dt
RC ∫
=
0.5/(3000π
6.5.7 Differentiator
An operational amplifier differentiator is shown in Fig. 6.25. Differentiator will
process the rate of change of a curve for a given interval, which is the inverse of
integrator. The derivative of a curve at a point is its instantaneous rate of
change. Therefore, the output of an operational amplifier differentiator is
proportional to the derivative of the input.
Figure 6.25: An operational amplifier differentiator
The equation of differentiator shall be
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6 Operational Amplifier and Applications
Vout = − RC
dVin
dt
(6.32)
Differentiation of a triangular signal will yield a square signal at the output of
differentiator as shown in Fig. 6.25.
Example 6.4
An differentiator circuit as shown in Fig. 6.25 has R = 50kΩ and C = 0.01µF.
The input voltage Vin is a triangular wave of amplitude 1.0V and frequency
1.5kHz. Find the type and amplitude waveform expected at the output of the
differentiator.
Solution
The expected waveform at the output is a square wave of frequency 1.5kHz
since differentiation of a linear slope line would yield a constant line. The
ampliftude of the square wave can be determined from equation (6.32), which is
Vout = − RC
dVin
. The rate of change of input voltage Vin is 2V/(T/2), where T is
2dt
the period of the triangular wave, which is 1/1.5kHz = 0.666ms. Thus, the
amplitude of the square wave is
50kΩx 0.01µF
2
⋅
= 1.5V.
2
0.333x10 −3
6.5.8 Voltage Follower
The voltage follower circuit is shown in Fig. 6.26. The input voltage Vin is
transferred to the output without changing of the phase.
Figure 6.26: A voltage circuit
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6 Operational Amplifier and Applications
The circuit operates likes an emitter or source follower except the gain is equal
to one i.e. Vout equals to Vin.
6.5.9 Current Amplifier
Small current value can be measured using a current amplifier circuit as shown
in Fig. 6.27.
Figure 6.27: Current amplifier circuit design
The unknown small load current IL can be calculated by reading the amplified
current I from the ammeter. The voltage difference (Vout – V1) = ILR2, which is
also equal to IR1. Thus, the load current IL is equal to
IL =
R1
xI
R2
(6.33)
The ratio of R2/R1 can be set conveniently to allow ease of reading. Note that
the power supply is used to sink current, which is normally not the case.
6.6 Analysis on Effect of Negative
Operational Amplifier Impedances
Feedback
on
We shall discuss the effect of output and input impedance of the operational
amplifier due to negative feedback for both non-inverting and inverting
amplifier. Indeed the feedback network of certain configuration will result in
increasing the input impedance and decreasing the output impedance. This is in
fact, what is desired by a designer.
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6 Operational Amplifier and Applications
6.6.1 Input Impedance of a Non-Inverting Amplifier
Consider a non-inverting amplifier shown in Fig. 6.28, its closed-loop gain is


R
Ri
i
 and the feedback voltage is V f =
Vout/Vin = Aol/  1 + A ol ⋅
Vout .
Ri + R f 
Ri + R f

Since Vout = Aol(Vdiff) and Vdiff = Vin – Vf = Vin 
Vdiff 1 +

Vout
Ri
⋅
Vdiff R i + R f


Ri
 = Vdiff  1 + A ol ⋅

Ri + R f


Ri
Vout then Vin =
Ri + R f

 . Vdiff is also equal to ZinxIin, where

Zin is the open-loop input impedance and Iin is the input current. Therefore, the
closed-loop input impedance Zin(cl) is

Zin(cl) = Vin/Iin =  1 + A ol ⋅

Ri
Ri + R f

 Zin

(6.34)
From equation (6.34), it shows that the closed-loop input impedance is very
much higher than the open-loop input impedance.
Figure 6.28: Impedance of a non-inverting amplifier
6.6.2 Output Impedance of a Non-inverting Amplifier
From Fig. 6.28, the output voltage is Vout = AolVdiff - ZoutIout. Usually AolVdiff >>

ZoutIout, then Vout ≅ Aol  Vin −


Ri
Ri
Vout . Thus,
Vout  . Vout ≅ AolVin - Aol

+
R
R
Ri + R f
i
f

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6 Operational Amplifier and Applications


 . The closed-loop output impedance Zout(cl) =



R i 
= IoutxZout(cl) then AolVin/Iout ≅ Zout(cl) 1 + A ol
.
R i + R f 

R
i
AolVin ≅ Vout 1 + A ol
R
+
Rf
i

Vout/Iout. Replacing Vout
Open-loop output impedance is also Zout ≅ Vout/Iout. Also for open-loop, output
voltage is Vout = AolVin. Thus, AolVin = IoutZout and Z out =
A ol Vin
. This implies
I out
that output impedance is

Zout(cl) ≅ Z out /1 + A ol

Ri
Ri + R f




(6.35)
Equation (6.35) shows that the closed-loop output impedance Zout(cl) for noninverting amplifier is very much smaller than the open-loop output impedance.
The factor β =
Ri
is also known feedback portion.
Ri + R f
6.6.3 Input Impedance of an Inverting Amplifier
Reference Fig. 6.29, and according to Miller's theorem, amplifier with feedback
resistance Rf, there is a Miller's input impedance parallel with open-loop input
impedance Zin of the operational amplifier. The Miller input impedance Zin(Miller)
= Rf/(Aol +1). Correspondingly, there is a Miller output impedance Zout(Miller) =
Rf
A ol
parallel with open-loop output impedance Zout of the operational
A ol + 1
amplifier. Thus closed-loop input impedance Zin(cl) and closed-loop output
impedance Zout(cl) of the operational amplifier can be represented in Fig. 6.28.
Figure 6.29: Inverting op-amp circuit showing Miller resistance and open-loop impedance
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6 Operational Amplifier and Applications
The closed-loop input impedance Zin(cl) shall be
Zin(cl) = Ri +
Rf
A ol + 1
(6.36)
|| Z in
Typically, Rf/(Aol +1) is much less than Zin and also Aol >> 1 then the above
equation can be simplified to Zin(cl) ≅ Ri +
Rf
A ol
Also Ri >> Rf/Aol then
Zin(cl) ≅ Ri
(6.37)
The closed-loop output impedance Zout(cl) is
Zout(cl) =
A ol
⋅ R || Z
A ol + 1 f out
Normally Aol >>1 and Rf >> Zout, so the closed-loop output impedance Zout(cl)
can be simplified to
Zout(cl) ≅ Zout
(6.38)
6.7 Effect of Finite Open-Loop Gain and Bandwidth
Between zero frequency to critical frequency, -3dB frequency point, the gain of
the operational amplifier is not effected. This is because there is no external RC
network that would act as high-pass filter.
As frequency increases until the critical frequency fc, the gain of
operational amplifier begins to decrease. This is because the internal RC
network of the operational amplifier becomes dominant. The gain would
steadily decrease to a point where it is equal to 1.0V/V or 0dB. The value of
frequency is called the unity-gain bandwidth or transition frequency.
The operational amplifier represented by gain amplifier and its internal RC
network elements is shown in Fig. 6.30. The gain function is AV(s) =
Vout (s)
=
Vin (s)
1
1
1
=
⋅ exp(− jφ) =
∠ − φ . The RC low-pass network
2
2
2
1 + jωRC
1 + ω2 R 2 C 2
1+ ω R C
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6 Operational Amplifier and Applications
is responsible for roll-off gain as frequency increases. From gain function, the
magnitude of the function is |T(s)| =
=
Vout
=
Vin
1
, which is also equal to AV
1 + ω 2 R 2C2
1
where reactance χC =
. Thus, as frequency increases, the
ωC
χC
R + χ 2C
2
gain attenuation due to RC low-pass network can be expressed as
Av =
=
Vout
1
=
Vin
1 + R 2 / χ C2
1
(6.39)
1 + ω 2 R 2C2
where χC = 1/ωC.
Figure 6.30: An operational amplifier showing its internal RC network
The critical frequency fc of the network occurred at -3dB gain, which
AV/ 2 .This implies that is (ωRC)2 = 1 . Thus, critical frequency fc is
fc =
1
2πRC
(6.40)
Dividing both sides with frequency f and substituting χ C =
fc/f =
χ
1
= C
2 πfRC R
1
, it gives
2 πfC
(6.41)
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6 Operational Amplifier and Applications
Substitute equation (6.40) into equation (6.41) yields equation (6.42).
Vout
1
=
Vin
1 + f / fC
(
(6.42)
)
2
Since an operational amplifier is represented by a voltage gain element and a
RC low-pass network as shown in Fig. 6.30, then the total open-loop gain is the
product of mid-range frequency open-loop gain AV(mid) and the attenuation of
the RC network. Thus,
AV =
Since AV =
A V ( mid )
(
1+ f / fC
A V ( mid )
1 + jωRC
=
(6.43)
)
2
A V ( mid )
1 + ω2 R 2 C 2
⋅ exp(− jφ) , the phase φ is
 f
 fC
φ = − tan −1 (ωRC ) = − tan −1 



(6.44)
Tutorials
6.1.
Find the differential input and input current of an operational amplifier
that has open-loop gain of Aol = 2x105, input resistance Ri = 600kΩ, dc
supply are ± 12V.
6.2.
Find the maximum output voltage of an operational amplifier that that has
open-loop gain of Aol = 2x105, input resistance Ri = 600kΩ, output
resistance Ro = 75Ω, dc supply are ± 12V, and differential input Vd = 75µV.
6.3.
Calculate the total output offset voltage VOS for the circuit shown in the
figure with op-amp specified input offset voltage VIO = 4.0mV and input
offset current IIO = 150nA.
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6 Operational Amplifier and Applications
6.4.
Calculate the input bias current at each input of the op-amp having
specified values of IIO = 5nA and Ibias = 30nA.
6.5.
An inverting amplifier has slew rate 0.5V/µs and gain AV = 33.
Determine if the amplifier is capable of amplifying the following input
signal without distortion at its output.
(i)
(ii)
(iii)
(iv)
V1 = 0.01sin(1.0x106t)
V2 = 0.05sin(3.5x105t)
V3 = 0.1sin(2.0x105t)
V4 = 0.2sin(5.0x104t)
6.6.
The maximum PSRR of µA741 operational amplifier is 150µV/V. If the
dc voltage supply of an operational amplifier changes from ± 15V to ±
12V, what is the change of its offset voltage?
6.7.
The operational amplifier circuit shown in the figure has R1 = 22kΩ and
R2 = 10kΩ.
(i) Determine the value of compensation component for offset current.
(ii) Redraw the amplifier circuit with the compensation component.
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6 Operational Amplifier and Applications
6.8.
Calculate the closed-loop output voltage Vout of this operational amplifier.
You may use Superposition theorem to help you.
6.9.
Given that the common mode range of an operational amplifier is +10V
and -10.5V.
(i) Calculate the output voltage Vout of this amplifier circuit.
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6 Operational Amplifier and Applications
(ii) Calculate the output voltage Vout of this amplifier circuit.
6.10. Find the peak value of the output of the ideal integrator shown in figure
for input Vin = 0.55sin(100t)V.
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6 Operational Amplifier and Applications
6.11. Design a practical integrator that integrates signals with frequencies down
to 100Hz and produces a peak output of 0.1V when the input is 10V peak
sine wave using frequency 10kHz. Find also the dc component at output
when there is 50mVdc at input.
6.12. An suming integrator amplifier circuit shown in the figure. Write down
the expression of output voltage.
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6 Operational Amplifier and Applications
6.13. The op-amp of a non inverting amplifier shown in the figure has output
impedance Zout equal to 50Ω, input impedance Zin equal to 500kΩ, and
open-loop gain Aol equal to 100,000. Calculate the feedback portion and
closed loop impedances of the amplifier.
6.14. Design an op-amp circuitry that can be used to solve the second order
non-homogeneous differential equation
boundary conditions that
d 2V
dV
+2
+ 3V − 2 = 0
2
dt
dt
d2V
dV
= 0 and
= 0 at t = 0.
2
dt
dt
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with
6 Operational Amplifier and Applications
References
1. Adel S. Sedra and Kenneth C. Smith, "Microelectronic Circuits", fourth
edition, Oxford University Press, 1998.
2. Robert L. Boylestad and Louis Nashelsky, “Electronic Devices and Circuit
Theory”, eighth edition, Prentice Hall, 2002.
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