lesson op amps

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Operational Amplifiers Lesson 06

Operational Amplifiers

(Op-amps)

6.1 Introduction

Voltage amplification is a main application of an operational amplifier. It can also be used to

perform mathematical operations. Considering these two main applications, this electronic

device is termed as operational amplifier(Op-amp).Operational amplifiers can be

constructed from discrete components, mainly transistors, which provide a stable and high

voltage amplification. But commonly they are available as monolithic integrated

circuits(ICs) as shown in Figure 6.1.

Figure 6.1: Top view of an Op-amp IC

Op-amps can be used in electronic circuits to perform a number of linear and non-

linear mathematical operations such as addition, subtraction, integration and differentiation.

They are also used as video and audio amplifiers, oscillators, etc. Because of their versatility,

Op-amps are widely used in all branches of electronics; both digital and analogue circuits.

One of the most common Op-amp IC is CA 741. Figure 6.2 shows the symbol for an

Op- amp. It has two inputs named as inverting (V-) and non-inverting (V+) and one output.

Figure 6.2: Symbol of an operational amplifier

69

V_

Inputs

V+

Vo Output

Lesson 06


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The Op-amps originally behave as differential amplifiers, which provide a high linear

gain on the voltage signal, applied to the non-inverting (V+) input terminal with respect to the

inverting signal voltage(V-). The linear voltage gain of this behavior is known as differential

gain (Ad ) or open loop voltage gain.

An Op-amp shows this linear voltage amplification under certain conditions. The low

frequency of the input signal and the low input voltage difference (V+ - V- ) are prominent of

them. An Op-amp IC is activated by applying a dual DC power supply (approximately 15V

and +15V). In the symbol it is not usually mentioned.

6.2 Open loop and closed loop amplifiers

Figure 6.3 shows a complete diagram of an open loop operational amplifier circuit. On the

diagram +VCC = +15V and -VCC = -15V.

Figure 6.3 Complete diagram of an open loop operational amplifier circuit

For a linear amplifier the output voltage (V0) is Ad x ( V+ - V- )

When non inverting input (V+) is grounded(= 0V)V0 = - Ad x V- ie; the sign of the output signal is the inversion of V- .

When both input terminals bear the same voltage signal

V0 = 0V

When inverting input (V-) is grounded(= 0V)

V0 = Ad x V+ ie; the sign of the output signal is same as the input signal (V+)

70

Ad = Vo /( V+ - V- )

+VCC

V_

Inputs

V+

Vo Output

-VCC


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The above relationships between the input and output voltage signals of an open loop

amplifier is graphically represented by Figure 6.4.

Figure 6.4: Voltage amplification of an open loop Op-amp

A Linear amplifier is one that has an output voltage (Vo) which is directly

proportional to the input voltage (V+ - V-). According to the above graph, the linearamplification is valid within the input voltage range of -V1 to +V2 and out of that the output

voltage saturates around +VCC and -VCC . The voltage range of the input signal corresponding

to the linear amplification of an Op-amp is typically a few Vs . However in most of the

applications, Op-amps are used in closed loop amplifier mode which consists of negative

feed back(a portion of output voltage is negatively fed back to the input voltage signal) and

reduces the voltage gain.In closed loop amplifiers, the input voltage can take up to few mV

s in the linear amplification region. Figure 6.5 shows the general form of a closed loop

(negative feed back) amplifier circuit.

Figure 6.5: Closed loop operational amplifier circuit

71

Vo

+VCC

Saturated region

V+

- V-

-V1

+V2

Saturated region-V

CC

Linear region

V_

Inputs

V+

Vo Output

Closed

loop


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In amplifier circuits the feed back path of the closed loop mode consists of a resistor

which controls and stabilizes the amplifier gain. The closed loop gain depends on the

components connected externally and not on the amplifier characteristics .

6.3 Ideal Op-amp approximations

An ideal operational amplifier has the following characteristics:

Infinite input impedance (any signal can be supplied to the Op-amp without

loading problems)

Zero output impedance (the power supplied by the Op-amp is not

limited),

Infinite voltage gain (the output voltage exceeds the power supply voltage)

Flat frequency response ( voltage gain does not depend on the input

voltage frequency)

Two Golden rules can be derived by following the above approximations.They are ,

72

Q1: The output of an Op-amp saturates at +15 V and 15V. The open loop

voltage gain (Ad) is 105. What would the output voltages be when the non

inverting input (V+) and the inverting input(V-) terminals are connected to thefollowing voltage signals?

(i). V+ = 5.0 V V- = 2.0V Answer : 0.3V

(ii). V+ = 10.0mV V- = 5.0 mV Answer : 15V

(iii). V+ = -20 V V- = -100 V Answer : 8V

(iv). V+ = 2.0 mV V- = 5.0 mV Answer : -15V

Q2: What are the disadvantage of an open loop amplifier ?

Rule 1: The voltages at the inverting and non inverting terminals are equal; V- = V+

Rule 2: The input bias are zero; I - = I + = 0A


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To apply the 1st rule requires negative feedback. The external negative feed back

network (a resistor connected across an input teminal and output terminal )brings the input

voltage difference to zero (if possible).Deu to high input impedance of the Op-amp, the 2

nd

rule indicates that the input current through the inverting and non inverting terminals can be

negligable. These two rules provide a simple way to find the approximations of Op-amp

circuits like resultant voltage gain.

6.4 Op-amp applications

Op-amps are used as basic building blocks to build different linear and non linear analogue

electronic systems. The linear circuit applications include amplifiers, voltage to current andcurrent to voltage converters, inverters (dc to ac or dc to ac) , active filters, sample and hold

circuits, logarithmic amplifiers etc. In addition, they are used to perform various

mathematical operations like addition, subtraction, multiplication, division, differentiation

and integration. Some of these applications are discussed in the following paragraphs.

6.4.1 Inverting amplifiers

Figure 6.6 shows a basic inverting amplifier circuit using a negative feed backresistor (Rf).

Figure 6.6: Inverting amplifier circuit.

The input signal is applied to the inverting input through a resistor (Rin) and a fraction

of the output is fed back into the input through the feed back resistor (Rf), limiting the

73

Rin

I

f

Rf

VoutVin

Iin


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voltage gain. The input current (Iin) through the resistor; Rin and the current Iif through the

feed back resistor are equal because of the high input impedance of the Op-amp.

ie Iin = If

f

out

in

in

R

VV

R

VV =

Kirchoff's current law, Iin = I- +If

Iin= If ( Because I- = 0, Golden rule 2)

f

out

in

in

R

VV

R

VV =

(I= V / R)

f

out

in

in

R

V

R

V = (Because V- = V+ = 0 , Golden rule 1)

The negative sign indicates that the polarity of the output signal is the opposite of the

input signal ( out of phase; 180 phase change ). Therefore the amplifier is called an inverting

amplifier. The magnitude of the voltage gain of an inverting amplifier is defined by the ratio

of two resistors (Rfand Rin)..

74

Q3: A sinusoidal voltage signal with an amplitude of 8 mV is connected to a

inverting amplifier circuit. The values of input and feed back resistors are

50k and 100k respectively. Draw the circuit diagram and input, output

voltage signals in scale.

Vout = - Rf ( Vin)Rin


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6.4.2 Summing amplifiers

Figure 6.7 shows the circuit arrangement of an Operational amplifier used as a

summing amplifier or an adder.

Figure 6.7 : Summing amplifier circuit

A summing amplifier circuit is a modification of an inverting amplifier circuit.

Kurchoff's current law; If= I1 + I2 +I3

f

out

R

VV

=1

1

R

VV

2

2

R

VV

+

3

3

R

VV

+

31

3

2

2

1

1

f

out

R

V

R

V

R

V

R

V++=

(Because V+ = V- = , Golden rule 1)

)R

V

R

V

R

V(RV

31

3

2

2

1

1fout ++=

if R1= R2= R3 =R

It is seen from the above equation that the amplifier out put is proportional to the sum

of the input signal voltages. Therefore it is known as summing amplifier(adder ) circuit.

75

R1 I1

R2 I2

R3 I3

If

Rf

Vout

V1

V2

V3

Vout = -Rf( V1 + V2 + V3 )

R


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6.4.3 Non-inverting amplifiers

It is possible to operate the Op-amp as a non inverting amplifier by applying the input

voltage signal to the non inverting input terminal. A basic non-inverting amplifier is shownin Figure 6.8.

Figure 6.8: Non-inverting amplifier circuit.

In this circuit the feedback resistor is still connected to the inverting input to limit the

voltage gain of the Op-amp. Two resistors; Rin and Rf which are connected across the

ground and output terminal behave as a voltage divider.

Iin =inR

V

and If =f

out

R

VV

But Iin = If

TheinR

V

=f

out

R

VV

in

in

R

V =

f

outin

R

VV ( Because V- =V+ = - Vin )

Therefore

76

Vout = ( 1+ Rf )Vin

Rin

Rin

If

Rf

VoutI

in

Vin


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According to the above equation the polarity of the output signal is the same as the

input signal ( in phase). Therefore the amplifier is called a non inverting amplifier. The

voltage gain of a non-inverting amplifier is 1+Rf/Rin.

The following circuit diagram shows a special case of a non inverting amplifier. Its

Rin is infinite and Rr is zero.

Figure 6.9: Unity-gain follower (Buffer) circuit

According to the above equation the voltage gain of this non inverting amplifier is

one unit and the out put follows the input. Therefore it is known as a unity-gain follower

(Vout = Vin). However this circuit is very useful as a buffer. When a voltage source with ahigh internal resistance has to be loaded with a low resistor, this buffer circuit can be used as

a interface to prevent the drain of high current from the source as shown in the following

figure.

Figure 6.10 : Buffer circuit as an interface

According to the Golden rule 2, no current flows into the positive input, and therefore

the source resistance is not loaded at all.

77

Vin

Vout

Q 4: What are the main differences between an inverting and a non-inverting

amplifier circuit.

Q 6:Give a suitable circuit diagram of an amplifier with a gain of +10, using anOp- amp.

Low resistive

loadHigh resistive

source


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6.4.4 Differential amplifiers

The voltage difference at the two inputs of the circuit is amplified by a finite value in

the differential amplifier circuit. Figure 6.11 shows a simple one Op-amp differentialamplifier circuit.

Figure 6.11: One Op-amp differential amplifier circuit

The current, I2 flows from V2 through R1 and R2 to the ground. By Golden rule 2,

current does not flow into the non inverting terminal of the Op-amp. Here resistors R1 and R2

act as a simple voltage divider. Then the voltage at the non inverting terminal (V+);

By Golden rule 1, what ever voltage that appears at the non inverting terminal (V+) also

appears at the inverting input terminal(V-).Therefore the currents through the resistors R1 and

R2 from V1 to V+ ;I1 and V+ to Vo ; I3 are given by the following equations

Where I1 = I3 (Golden rule 2 and Kirchoffs current law)

By solving the above three equations

78

V+ = V2 x R2 = V3R

1+ R

2

I1

= V1

V

+and I

3= V

+

Vo

R1

R2

VO

= (V2- V

1)

x R

2

R1

I1

I3

I2

I2


11/15


The relationship between the input and the output voltage signals of an one Op-amp

differential amplifier is given by the above equation. The one Op-amp differential amplifier

is quite satisfactory for low resistance sources ( V3 and V4). But when the input currents ( I1

and I2) are consumed by the circuit , the sources which have high resistance drop theiractual voltages internally. Therefore high resistive sources should be connected with the high

impedance differential amplifiers which has zero input bias currents. Figure 6.12 shows the

three Op-amp differential amplifier circuit.

Figure 6.13 shows the use of an Op-amp with a feed back resistor replaced by a diode.

Figure 6.13: Logarithmic amplifier

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Figure 6.12. Three Op-amp differential amplifier (Instrumentation amplifier)

This amplifier circuit is regularly used in most of the analytical and bio-medical

instruments. The first two unity-gain follower circuits provide the high input impedance for the

differential amplifier circuit. The amplification factor of this instrumentation amplifier is the

same as the one Op-amp differential amplifier given above.

6.4.5 Logarithmic amplifiers

RVoutVin


12/15


This circuit is employed when an output voltage is desired to be proportional to the

logarithm of an input signal. Considering the relationship between the variation of the

current through a P-N junction diode and the voltage applied across it , the following

equation can be derived for the above logarithmic amplifier circuit.

Where k1 and k2 are constants

6.5.7 Integrators

The Op-amp can also be used to perform the mathematical operation of "integration" by

using a capacitor in the feedback path. The basic circuit of an integrator is shown in Figure

6.14.

Figure 6.14: Integrator circuit

By applying Op-amp Golden rules and equations related to capacitors it is possible to

derive the following equation.

The output of this electronic circuit is the integration of input voltage signal with respect to

time as shown in above equation.

80

Vin

C

Vout

R

Vout = k1 Log(k2 Vin)

t

Vout = - 1 0

Vin dt

RC


13/15


6.5.8 Differentiators

Differentiation is the opposite of integration. Therefore, the differentiator circuit is

obtained by interchanging R and C components of the integrator circuit.. The basic

component connection of a Differentiator circuit is shown in Figure 6.15.

Figure 6.15: Differentiator circuit

By applying the Op-amp Golden rules the following equation can be derived for the

above differentiator circuit.

Thus the out put voltage is the differentiation of the input signal.

81

Q7: Starting from a DC voltage signal, how do you obtain a voltage signal which

automatically increases linearly with time?Answer: By connecting a DC signal to the input of an Op-amp integrator

circuit. Then the output voltage increases gradually, because the integrationof a constant(C) with respect to time(t) gives C x t

Vout = - RC d (Vin)

dt

Vin

R

Vout

C


14/15


6.6 Details of Op- amp IC-CA741

CA 741 is the most widely used operational amplifier IC which is constructed using more

than 25 bipolar junction transistors. It is also one of the cheapest you can buy. CA 741 hasthe following characteristics.

input impedance 2M

output impedance 50

voltage gain 105 at DC

Input bias current 500 nA

Flat frequency response from DC to several MHz.

CA 741 possesses 8 - pins in a dual line integrated circuit package as shown in Figure 6.16.

Pin No: 1 Off set null

Pin No: 2 Inverting input

Pin No: 3 Non inverting input

Pin No: 4 Negative power supply

Pin No: 5 Off set null

Pin No: 6 Output

Pin No: 7 Positive power supply

Pin No: 8 No connection

Figure 6.16: Pin configuration of the CA 741 Op-amp

Op-amps are often powered by both a positive and a negative supply so that the

output can swing above and bellow ground. For the CA 741 operational amplifier,

the dual power supply should be in the voltage range of +/- 5 and +/- 18 . Negative

voltage must be connected to Pin No: 4 and positive voltage to Pin No:7.

The two inputs; non-inverting and inverting are Pin No:2 and Pin No:3 respectively.

The output is taken at Pin No:6.

In addition , there are "offset null" inputs at Pin No: 1 and Pin No:5 which is used to

calibrate the constructed Op-amp circuit. End terminals of a variable resistor are

connected to Pin No: 1 and Pin No:5.The mid terminal is connected to the negative

voltage (Pin No: 4)

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4 5

3 6

2 7

1 8

CCA7 411


15/15


A sample circuit of an inverting amplifier to illustrate the pin connection of CA 741 IC is

shown in the following figure.

Figure 6.17: Inverting amplifier circuit

83

Q4: An inverting amplifier circuit is designed selecting 2.2k, 220 resistors

as feedback and input resistors respectively. This amplifier circuitgives 52mV as the output voltage when 5mV is applied as the input.

Calculate the error of the amplifier circuit .

10k

1k

150k

-15V 0V +15V

Vout

Vin

lesson op amps

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