Deke 2443 Analogue Electronics (Chapter 1_op Amp)

7/28/2019 Deke 2443 Analogue Electronics (Chapter 1_op Amp)

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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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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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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

Documents

Deke 2443 Analogue Electronics (Chapter 1_op Amp)