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CHAPTER 1: OPERATIONAL AMPLIFIER
1.1: Introduction to Op-Amp
1.2: Symbol, Packaging, Pinouts
1.3: Ideal and Practical Characteristic of IC 741 Op-Amp
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Introduction to Operational Amplifier IC
The term operational amplifier, abbreviated Op-Amp, wascoined in the 1940s as a tube-type amplifier.
In those days, it was used in the analog computers toperform a variety of mathematical operations such asaddition, subtraction, multiplication etc.
Due to its use in performing mathematical operations, it hasbeen given a name operational amplifier.
Due to the use of vacuum tubes, the early Op-Amps were
bulky, power consuming and expensive.
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Introduction to Operational Amplifier IC
(Cont) Robert J. Widlar at Fairchild brought out the popular741 integrated circuit (IC)
Op-Amp between 1964-1968.
The IC version of Op-Amp uses BJTs and FETs which are fabricated along with theother supporting components, on a single semiconductor chip or wafer which is of apinhead size.
IC Op-Amps are inexpensive, take up less space and consume less power.
The modern linear IC Op-Amp works at lower voltages.
The Op-Amp is basically an excellent high gain DC amplifier.
The differential amplifier is the basic building block of IC Op-Amp.
Key Point: Because of theirlow cost, small size, versatility, flexibility anddependability, Op-Amps are used in the fields of process control, communications,computers, power and signal sources, displays and measuring systems.
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Symbol, Packaging and Pinouts
Figure 1.1(a)-(c) show the standard symbols of Op-Amp.
Figure 1.1(a) is a symbol of a buffer Op-Amp.
Figure 1.1(b) is a symbol of a differential input, single ended output
Op- Amp. This symbol represents the most common types of op
amps, including voltage feedback, and current feedback. It is often
times pictured with the non-inverting input at the top and the
inverting input at the bottom.
Figure 1.1(c) is a symbol of a differential input, differential output
Op-Amp. The outputs can be thought of as inverting and non-
inverting, and are shown across from the opposite polarity input for
easy completion of feedback loops on schematics.
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Symbol of Op-Amp
Figure 1.1: Standard symbols of Op-Amp;(a) buffer Op-Amp
(b) a differential input, single ended output Op-Amp
(c) a differential input, differential output Op-Amp
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Packaging of Op-Amp
The Op-Amp ICs are available in various packages. The three
popular packages available for Op-Amp are:
1) The metal can package (TO)
2) The dual in line package (DIP)
3) The flat package or flat pack
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Packaging of Op-Amp
(Metal Can Package - TO)
Available with 3,5,8,10 and 12 pins.
The metal sealing plane is at the bottom, used over which the silicon
chip bonded.
This plane is effective for the dissipation of heat.
Figure 1.2 shows the 8 pin metal can package and the connectiondiagram.
The tab is used to identify pin 8 and the pins are numbered
counterclockwise when metal can is viewed from the top.
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Packaging of Op-Amp
(Metal Can Package - TO)
Figure 1.2: Metal can package (TO)
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Packaging of Op-Amp
(Dual In Line Pack - DIP)
The DIP is popular for commercial applications.
It is easy to handle, fits standard mounting hardware andinexpensive when molded in plastic.
Ceramic DIPs are used for high temperature, high performance(usually military) equipment.
Figure 1.3 shows 8 pin and 14 pins DIPs and their connection
diagrams.
For DIPs either plastic or ceramic cases are available.
The pin 1 is indicated by a notch ordot, as viewed from the top
and other terminals are numbered counterclockwise.
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Packaging of Op-Amp
(Dual In Line Pack - DIP)
Figure 1.3: Dual in line packages (DIP)
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Packaging of Op-Amp
(Flat Package)
For circuits where the space is critical, the flat pack gives a
compact package.
Flat packs are much difficult to handle than DIPs and often donot dissipate poweras well.
The metal can packages allow easy connection to heat sink, and
are chosen when heat dissipation is the main consideration.
Figure 1.4 shows 10 pin flat package where the chip is enclosed in a
rectangular ceramic case. The terminals are taken out through the
sides and ends.
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Packaging of Op-Amp
(Flat Package)
Figure 1.4: Flat package
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Pinouts of Op-Amp
The main three pins which are normally appear in an Op-Amp circuitschematic diagram are:
1) Pin 2 (-) - inverting input
If a positive signal is sent to inverting terminal, the output
signal would be inverted and it would be negative. Conversely, if anegative signal is sent to the inverting terminal, the output would be
inverted and it would be positive.
2) Pin 3 (+) - non-inverting input
If a positive signal is sent to the non-inverting terminal, the
output signal would not be inverted and it would remain positive.
3) Pin 6 - Amplifier output
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Pinouts of Op-Amp
The other pins includes:
1) Pin 1 & 5 - The null offset
Provide a way to eliminate any offset in the output voltage of theamplifier.
The offset voltage is additive with output (pin 6 in this case), can beeither positive or negative and is normally less than 10 mV.
Because the off-set voltage is so small, in most cases we can ignore
the contribution of offset voltage and leave the null offset (pins 1&5)to be open.
Have a special functions such as fine-tuning when the Op-Amp isrequired to amplify DC signals.
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Pinouts of Op-Amp
2) Pin 7 (positive) & pin 4 (negative) power supply
In reality, the amplifier needs the power source to increase the
input signal to the strength in order for the output signal to be
useful.
3) Pin 8 (labeled NC)
Mentioned as Not Connected and not been used.
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Pinouts of Op-Amp
Figure 1.5: The schematic symbol for an Op-Amp
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Pinouts of Op-Amp
Figure 1.6: Pinout for a single Op-Amp (741 included) when housed in
an 8-pin DIP (Dual Inline Package) integrated circuit
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Ideal And Practical Characteristic of
IC 741 Op-Amp
+
AVinVin Vout
Zout
~
Zin
Practical op-amp
Sr. No Parameter Symbol Ideal Practical 741 IC
1 Open loop voltage gain AOL 200,000
2 Output Impedance ZOut 0 75
3 Input Impedance Zin 2 M
4 Input offset current I os 0 200 nA
5 Input offset voltage VOs 0 2 mV
6 Bandwidth B. W 1 MHz
7 CMRR 90 dB
8 Slew rate S 0.5 V/s
9 Input Bias Current Ib 0 80 nA
+
~
AVin
Vin Vout
Zout=0
Ideal op-amp
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Ideal Characteristic of
Op-Amp
- Infinite voltage gain: It is the differential open loop gain and is infinite for an ideal Op-Amp
-Infinite input impedance: Infinite for ideal Op-Amp. This ensures that no current can flow into an
ideal Op-Amp
-Zero output impedance: Zero for ideal Op-Amp. This ensures that the output voltage of the Op-Amp
remains same
-Zero offset voltage: The presence of the small output voltage though V1 = V2 = 0 is called an offsetvoltage. It is zero for an ideal Op-Amp. This ensures zero output for zero input signal voltage
-Infinite Bandwidth: The range frequency over which the amplifier performance is satisfactory is
called bandwidth. The bandwidth of an ideal Op-Amp is infinite. This means the operating frequency
range is from 0 to .This ensures that the gain of the Op-Amp will be constant over the frequency
range from DC (zero frequency) to infinite frequency. So, Op-Amp can amplify DC as well as AC signals
-Infinite CMRR: The ratio of differential gain and common mode gain is defined as CMRR. Thus infinite CMRR of anideal Op-Amp ensures zero common mode gain. Due to this common mode noise output voltage is zero for an ideal
Op-Amp
- Infinite slew rate: This ensures that the changes in the output voltage occur simultaneously with the changes in the
input voltage
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Practical Characteristic of
Op-Amp
- Open loop gain: It is the voltage gain of the Op-Amp when no feedback isapplied. Practically it is several thousands
- Input impedance: It is infinite and typically greater than 1M. But usingFETs for the input stage, it can be increased up to several hundred M
- Output impedance: It is typically few hundred ohms. With the help ofnegative feedback, it can be reduced to a very small value like 1 or 2 ohms
- Bandwidth: The bandwidth of practical Op-Amp in open loop configuration isvery small. By application of negative feedback, it can be increased to a
desired value
- Input offset voltage: Practical Op-Amp shows a small non zero outputvoltage
- Input bias current: The practical Op-Amps do have some input currents
which are very small, of the order of 10^-6 to 10^-14 A
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CHAPTER 1: OPERATIONAL AMPLIFIER
1.4: Differential Gain (Ad)1.5: Common Mode Gain (Ac)
1.6: Common Mode Rejection Ratio (CMRR)
1.7: Slew Rate
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Differential Gain (Ad)
Ad = differential gain
V1 - V2 = difference voltage denoted as Vd
Generally the differential gain is expressed in decibel (dB) value
Vo = Ad (V1-V2)Vo = Ad (Vd)
Ad = Vo / (Vd) Ad = 20 Log10
(Ad) in dB
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Common Mode Gain (Ac)
Assume V1 = V2, then ideally the output voltage V0 = (V1-V2) Ad must be 0.
But the output voltage of the practical differential amplifier not only depends on thedifference voltage but also depends on the average level of the 2 inputs.
Such an average level of the 2 inputs is called common mode signal (Vc
) .
The gain with which it amplifies the common mode signal to produce the outputis called common mode gain of the differential amplifiers (Ac).
Thus, there exists some finite output for V1 = V2 due to such common mode gain(Ac), in case of practical differential amplifiers.
Vc = (V1 + V2 ) / 2
V0 = Ac Vc
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Common Mode Gain (Ac)Cont
So the total output of any differential amplifiercan be expressed as:
For an ideal differential amplifier, the differential gain Admust be infinite while
the common mode gain must be 0. This ensures 0 output for V1 = V2
But due to mismatch in the internal circuitry, there is some output available for
V1= V2 and gain Ac is not practically 0.
V0 = Ad Vd + Ac Vc
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Common Mode Rejection Ratio (CMRR)
When the same voltage is applied to both inputs, the differential amplifier is said tobe operated in a common mode configuration.
Many disturbance signals, noise signals appear as a common signal to both inputterminals of the differential amplifiers.
Such a common signal should be rejected by the differential amplifier.
The ability of a differential amplifier to reject common mode signal isexpressed by a ratio called common mode rejection ratio (CMRR).
Ideally the common mode voltage gain is 0, thus the ideal value of CMRR isinfinite
For a practical differential amplifier, Ad is large and Ac is small, thus the value ofCMRR is large
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Common Mode Rejection Ratio (CMRR)
Cont Thus, CMRR is expressed by:
Many times, CMRR is expressed in dB as:
The output voltage can be expressed in terms of CMRR as:
CMRR= =
CMRRin dB = 20 log
dB
Vo = AdVd [1 +1
.
]
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CMRR (EXAMPLE 1.1)
Determine the output voltage of a differential amplifier for the input voltages of 300 V and
240 V. The differential gain of the amplifier is 5000 and the value of the CMRR is 100.
= 300-240
= 60 V
V1 +V2=Vc
2
Vd = V1V2
300+240=2
= 270 V
Ad=CMRR
Ac
5000=100Ac
Ac = 50
Vo = Ad Vd + Ac Vc
= (5000 x 60) + (50 x 270)
= 313 500 V
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Slew Rate (SR)
The slew rate is defined as the maximum rate of change of output voltage withtime
The slew rate is caused due to limited charging rate of the compensatingcapacitor and current limiting and saturation of the internal stage of Op-Amp,when a high frequency, large amplitude signal is applied
The internal capacitor voltage cannot change instantaneously
It is given by
=
By large charging rate, the capacitor should be small or charging currentshould be large
Hence, the slew rate for Op-Amp whose maximum internal capacitor chargingcurrent is known, can be obtained as:
S = Imax / C
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Effect of Slew Rate (SR)
Due to slew rate of Op-Amp, for a particular input frequency, output get distorted as shown in Figure 1.7
From Figure 1.7, S =
(V/sec)
The typical value of S for 741 Op-Amp or 0.5 x 10^6 V/sec or 0.5 V/ s
Figure 1.7: Effect of slew rate
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Slew Rate (SR) Equation
For distortion free output, the maximum allowable input frequency fm can beobtained as:
fm = S / 2 Vm (Hz)
Where fm = maximum allowable input frequency
Vm = peak of output waveform
S = slew rate
This is also called full power bandwidth of Op-Amp
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Slew Rate (Example 1.2)
An Op-Amp operates as a unity gain buffer with 3V (peak to peak) square wave input. IfOp-Amp is ideal with slew rate 0.5 V/s, find the maximum frequency of operation
3V of square wave=Peak to peakVp-p=Vm
2
3=2
1.5V=
S=fm2V
m
(0.5/10^-6)=
2 x 1.5
53.051kHz=
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CHAPTER 1: OPERATIONAL AMPLIFIER
1.8: Closed Loop Operation1.8.1: Basic Op-Amp
(a) Inverting Op-Amp
(b) Non-Inverting Op-Amp
(c) Voltage Follower
1.8.2: Integrator
1.8.3: Differentiator1.8.4: Summing Amplifier
1.8.5: Difference (Subtractor) Amplifier
1.9: 3 Op-Amp Instrumentation Amplifier
Cl d O i
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Closed-Loop Operation
Closed-loop configuration
reduces the gain
It involves a feedback from the
output to the input
Positive feedback: used
exclusively with oscillator circuit
Negative feedback: output is fed
back to the inverting input
through a feedback resistor or
capacitor
A. Basic Op-Amps
D. Summing Amplifiers
B. Integrator
E. Difference (Subtractor) Amplifiers
C. Differentiator
_
+
B i O A
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Basic Op-Amps
Inverting Amplifier
Non-Inverting Amplifier
Voltage Follower
I ti A lifi
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Inverting Amplifier
Important: The Vo can never exceed VCC.
Theve sign denotes a 180 degree phase shift between input and output.
What happen to ACL if R equals to Rf.
_
+
V1
Rf
R +V
-V
VoVi=0
I- =0I1= IfI1
If
V1-Vi
R=
Vi-Vo
RfV1
R =
-Vo
Rf
Vo
Vi=
-Rf
R= ACL
I1= If+I-
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Inverting Amplifier (Example 1.3)A sine wave of 0.5 V peak voltage is applied to an inverting amplifier using R1 = 10k
and Rf= 50 k. It uses the supply voltage of12 V. Determine the output and sketch
the waveform.If now the amplitude of input sine wave is increased to 5V, what will be the output? Is it
practically possible? Sketch the waveform.
Gain = Vo
Vi
= - Rf
R1
= - 50
10
= 5
i) For Vm = 0.5V
(Vo)m = (Vin)m x Gain
= 0.5 x 5 = 2.5VpeakThe input & output waveforms are
inverted with respect to each other and
shown in Figure 1.8(a).
ii) For Vm = 5V
(Vo)m = (Vin)m x Gain
= 5 x 5 = 25Vpeak
But Op-Amp saturates at 12V (supply voltage).
So portion above +12V and below -12V will be
clipped off from the output. So, 25V peak output
is not practically possible. The input and output
waveforms are shown in Figure 1.8(b)
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Inverting Amplifier (Example 1.3) Cont
Figure 1.8:Inverting Amplifier
N I ti A lifi
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Non-Inverting Amplifier
_
+
V1
RfR
VoVi=0V1= V
+
=V-
R+RfVo
= R
Vo
V1
=R+Rf
R
= ACL= 1 +Rf
R
+V
-V
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Non-Inverting Amplifier (Example 1.4)
For the Op-Amp configuration shown in Figure 1.10, determine the voltage gain
Calculation:
Figure 1.10: Non-Inverting Op-Amp
Vo
Vin
Rf= 1+
R1
10 x 10^3= 1 +
1 x 10^3
= 11
V lt F ll
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Voltage Follower
VA =
Also known as source follower, unity gain amplifier, bufferamplifier or isolation amplifier
_
+
Vo
Vin
B
A
Vin + Vb ..(1)
The node B is at potential Vin . Now at node
A is also at the same potential as B
Vo = VA...........(2)
Equating the equations (1) and (2),
Vo = Vin
Bi C t C ti
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Bias Current Compensation For the three basic op-amps, the expression of Vo is considering an
ideal op-amp representation
However in practical, Vi or Vd, I+, and I- have a very small value
In order to compensate the offset voltage and bias current, those
three op-amps are used
_
+V1
RfR
Vo
+V
-V
Rc=R||Rf
_
+
V1
Rf
R +V
-V
Vo
Rc=R||R
f
Inverting Amplifier
Voltage Follower
Non-Inverting Amplifier_
+
Vo
Vin
I t t
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Integrator
_
+
V1
C
R +V
-V
VoVi=0
I- =0
IR
Ic
IR= IC
V1-V
i
R =V
i-V
o
XCV1
R=
-Vo
Xc
Vo
V1=
-1
sRC= ACL
=-Vo
1/sC
-sCVo=
The output is the integral of the input.
Integration is the operation of summing the area under a waveform
or curve over a period of time.
This circuit is useful in low-pass filter circuits and sensor.
conditioning circuits.
vo(t) = -1
RCv1(t) dt
Diff ti t
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Differentiator
_
+
V1C
R
+V
-V
VoVi=0
I- =0
Ic
IR
IC= IRV1-Vi
XC=
Vi-Vo
R
Vo
V1= -sRC = ACL
Interchanging the location of the capacitor and the resistor of the
integrator circuit results in the circuit above which performs the
mathematical function of differentiation.
The differentiator takes the derivative of the input.
This circuit is useful in high-pass filter circuits.
V1
XC
=-Vo
R
vo(t) = -RCdv1(t)
dt
S mming Amplifiers
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Summing Amplifiers
Inverting Summer Non-Inverting Summer
Inverting Summer
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Inverting Summer
_
+
V1
Rf
R1 +V
-V
VoVi=0
I- =0
I1+I2+I3 = If
V2
V3
R2
R3
I1
If
I2
I3
V1
R1+
V2
R2+
V3
R3=
-Vo
Rf
Vo= -Rf
R1V1+
Rf
R2V2+
Rf
R3V3
Non Inverting Summer
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Non-Inverting Summer
_
+
V1
Rf
R1
+V
-V
Vo
V2R2
Applying Superposition Theorem for v+
RStep 1: Let V2 = 0
V1R1+R2
v+1= R2
Step 2: Let V1 = 0
V2
R1+R2
v+2= R1
Step 3: v+=v+1+v+
2
Vo = 1 + Rfv+ R
1
v+ = v+1+v+
2
V1+R1+R2
= R2 V2R1+R2
R1 2
Step 4: 2 1
Vo = 1 + Rf
RV1+
R1+R2
R2 V2R1+R2
R1
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Difference (Subtractor) Amplifiers
V2
_
+
R1
Vo
Vo = - RfR1
V1
Case 1: With V2 = 0, the circuit acts as an inverting amplifier
Rf
Rf
V1
R2
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Difference (Subtractor) Amplifiers
V2
_
+
R1
Vo
Vo = RfR1
(V2V1)
Case 2: With V1 = 0, the circuit acts as an inverting amplifier
Rf
Rf
V1
R2
I
I
I
A
B
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3 Op-Amp Instrumentation Amplifiers
V1_
+
R1
Vo
Vo = R2
R1
(V2V1)
R2
R2
R1
1+ 2RfRG
A3
V2
Rf
RG
Rf
A2
A1
_
_
+
+
3 Op Amp Instrumentation Amplifiers
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3 Op-Amp Instrumentation Amplifiers
(Cont)
Applications of Instrumentation Amplifier:
1) Temperature Controller
2) Temperature Indicator
3) Light Intensity Meter
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