Op Amp Workbook 2014

1

Preface

This booklet on operational amplifiers has been compiled to fulfill the requirements of the Caribbean Advanced Proficiency Examinations (CAPE). It addresses a specific need in some of our high schools and is intended to augment the teachers material on the subject matter. Booklets are provided to students during the CAPE Physics Workshop offered annually by the Department of Physics at the University of the West Indies, Mona Campus. These workshops target problematic topics in the Physics Syllabus and use lectures and laboratory sessions to teach the material in a manner that enhances the students understanding.

The booklet starts with an overview of the CAPE Physics Electronics requirements then moves into the relevant subject matter. The author presents the content of each section in a way that relates to the targeted age group and facilitates quicker understanding. Worked examples are included throughout the booklet and additional CAPE type questions are added at the end.

The procedures for the practical sessions are added at the end of the booklet. In Lab 1, a chosen operational amplifier circuit is constructed and tested on the lab bench. The results of these measurements are plotted on a graph. In lab 2, the same circuit is simulated on a computer and its results are compared to that of Lab 1. This technique is intended to provide the students with a comparative feel for the hands-on experiments versus the simulated ones.

The Department of Physics hopes that the students find this booklet to be a valuable aid in their learning process and wishes each and every student success in their upcoming examinations.

Paul R. Aiken (PhD) Department of Physics UWI, Mona 2014

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Table of Contents

CAPE Requirements . 2 What are Operational Amplifiers? .. . 3

Circuit Schematic Representation . 4 Power Supply Requirements.. 4

The Ideal Operational Amplifier 5 The Typical (Real) Operational Amplifier 9 The Op Amp as a Comparator 10 Application of the Comparator 13

Potential Dividers and Variable Resistors . 13 Light Dependent Resistors (LDR) . 14 Thermistors 14

Strain gauge ... 14 Feedback in Op Amp Circuits 17

Positive Feedback Negative Feedback

The Inverting Amplifier 18 The Non-inverting Amplifier .. 21

Effect of Negative Feedback on Gain and Bandwidth 22 Voltage Follower 26 Summing Amplifier .... 27 Difference Amplifier .. 29 An Operational Amplifier Circuit Example .. 29 Op Amp Problems . 32 Experiment 1 ..... 34 Experiment 2 . 36

3

CAPE Requirements

4. Operational Amplifiers Students should be able to: 4.1 describe the properties of the ideal operational

amplifier; 4.2 compare the properties of the typical and the

ideal operational amplifier; 4.3 use the operational amplifier as a comparator; 4.4 use the fact that magnitude of the output

voltage cannot exceed that of the power supply;

4.5 discuss the effect of positive and negative feedback in an amplifier;

4.6 explain the meaning of gain and bandwidth of an amplifier;

Consider these effects in terms of whether they are advantages or disadvantages.

Typical as well as ideal values for these quantities should be discussed.

4.7 explain the gain-frequency curve for a typical operational amplifier;

4.8 determine bandwidth from a gain frequency curve;

4.9 draw the circuit diagram for both the inverting and non-inverting amplifier with a single input;

4.10 use the concept of virtual earth in the inverting amplifier;

4.11 derive and use expressions for the gain of both the inverting amplifier;

4.12 discuss the effect of negative feedback on the gain and bandwidth of an operational amplifier;

4.13 perform calculations related to single-input inverting amplifier circuits;

4.14 perform calculations related to single-input, non-inverting amplifier circuits;

4.15 describe the use of the inverting amplifier as a summing amplifier;

Include the fact that frequency is usually plotted on a logarithmic axis and explain the reason for this. Precise numerical value related to the response of the ear is not required. Students should be familiar with several representations of the same circuit.

Explain why the virtual earth cannot be connected directly to earth although it is `virtually" at earth potential. State the two "Golden Rules' of Operational Amplifier circuit analysis and show how they lead to the results required here. Mention the effect of negative feedback on other op-amp characteristics.

Include the fact that it is also possible to configure the op-amp as a difference amplifier.

4.16 solve problems related to summing amplifier circuits;

4.17 describe the use of the operational amplifier as a voltage follower;

4.18 analyse simple operational amplifier circuits; 4.19 analyse the response of amplifier circuits to

input signals, using timing diagrams.

Mention the important practical use of the voltage follower as a buffer or matching amplifier.

Refer to note 4.11

4

What are Operational Amplifiers? The term Operational Amplifier was originally used to describe an amplifier circuit which performed various mathematical operations such as differentiation, integration, summation and subtraction. Operational Amplifier, or Op Amp, is now more loosely applied to any high gain alternating current (ac) and direct current (dc) amplifier capable of operating in various configurations. Op Amps have extremely wide applications and may be found in all types of circuit and system designs.

Op amps are a member of the family of linear integrated circuits. Integrated circuits (ICs) consist of many transistors and few resistors and capacitors. Transistors are a special form of semiconductors with properties similar to those of junction diodes. ICs are usually fabricated on a specially prepared material using extremely precise process control. The final product generally occupies areas less that one square centimeter, even for the most complex ICs.

(a) LM741 Single OP amp (b) Quad LM741 op amp (LM324)

(c) Pin labels for LM741 (d) Pin labels for single packaged quad op amp

Fig. 1 The 741 Op amp single (8-pins) and quad packages (14-pins)

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There are many different types of op amps designed for varying applications. The most popular of these is the LM741 op amp developed by National Semiconductor Company. Fig. 1 shows pictures of two package types of the LM741 Op amp and their corresponding internal schematic.

Circuit Schematic Representation An op amp is represented by the symbol shown in Fig 2. For the LM741 op amp, pins 2 and 3 are what we called the inverting and non-inverting inputs, respectively, because of the way the input signals are acted upon. Pin 6 is the output and pins 7 and 4 are the pins to which the positive and negative power supply inputs are connected. In some circuits, pin 4 may be connected to the circuit ground (or common) instead of a negative voltage supply.

Fig. 2 The Op amp schematic symbol

Power Supply Requirements All ICs have a minimum and maximum power supply voltage rating. The LM741 op amp may be operated from a dual-rail supply voltage of 5 V dc to 18 V dc, or a single-rail supply of 10Vdc to 36 V dc with respect to ground. The op amp will not work properly if smaller voltages (than the minimum in this range) are applied and will be damaged if greater voltages (than the maximum in this range) are applied. Before using

positive power supply voltage

Negative power supply voltage

Output

Inverting input

Non-inverting input

2

3

6

7

4

-

+

positive power supply voltage

Negative power supply voltage

Output

Inverting input

Non-inverting input

2

3

6

7

4

positive power supply voltage

Negative power supply voltage

Output

Inverting input

Non-inverting input

2

3

6

7

4

-

+

6

any IC in a circuit always check its manufacturer data sheets for its maximum ratings. This may be easily found by doing a Google search of the part number.

Fig. 3 show examples of single and dual rail power supplies. At this time, the most common type available to high school students are those made from connecting batteries in series aiding arrangements (Fig.3(c)). Other types of DC power supplies are available.

(b) A 9-Volt battery (a) Single rail 9 V supply (c) A dual rail 9 V supply

Fig. 3 Power supply configuration

The Ideal Operational Amplifier The output voltage of an op amp is proportional to the difference of the voltages at its inverting (V-) and non-inverting (V+) inputs. This is represented by Equation 1

)(0 + = VVAVout (1) where A0 is called the Open-Loop gain of the op amp.

Oklets pause here for a minute. What are these open-loop and gain things we have

been talking about???. .. Well, the typical connection for an op amp in a circuit is one where some other component (usually a resistor) is connected between the output pin and the inverting input pin. This causes the signal that goes into the op amp to get loop back from its output to its input. This setup is called a closed-loop configuration. Therefore,

if there are no components connecting the output back to the input, then we can say that

+

9V-

+ 9 Vdc

Ground

+

9V-

+ 9 Vdc

Ground

++

-

-

9V

9V

+ 9 Vdc

Ground

- 9 Vdc

++

-

-

9V

9V

+ 9 Vdc

Ground

- 9 Vdc

7

we have an open-loop configuration. So Eq.1 describes the gain of the op amp when the loop is open. OK so here we go with that gain word again.

The gain of an amplifier is the amount of amplifications that is given to the input signal

to get the output signal. In other words, it is how many times the input signal gets multiplied to equal the output signal. The gain is always found by dividing the voltage

value of the output signal by that of the input signal. The gain is always just a number and has no units. So open-loop gain means how many times the input signal gets

amplified when there is nothing connected between the output and input pins. By the way, this connection we are talking about, the one between the output and the input, it is

called feedback. That is, a portion of the output signal gets fed back to the input whenever we have the connection in place. We will talk some more on this later on.

We will now attempt to describe some of the main properties of an ideal op amp. Dont

be frightened by all the new terminologies (weird words). We will list the properties first then go through line-by-line and try to provide additional explanations. So here goes

the ideal operational amplifier may be assumed to have the following properties:

(a) An infinite open-loop gain. The slightest difference in V+ and V- will caused the output to go to saturation. Saturation voltage cannot exceed the power supply

voltage.

(b) An infinite input impedance (resistance). This ensures that no current flows into the input terminals (V+ and V-). However, voltages may be present.

(c) An infinite bandwidth. This assumes that it amplifies any input range of frequencies.

(d) Zero output impedance (resistance). This ensures that the amplifier is unaffected whatever output circuit it is connected to.

(e) An infinite slew rate. The means that the input and output frequency changes are always exactly in synch.

(f) Zero voltage and current offsets. This ensures that when the input signal voltages are zero the output will also be zero regardless of the input source resistance.

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If your head is spinning at this point, just stop, take a deep breath, get some water or something. Now, read these explanations below, then go back and re-read the

properties. It is very important that you understand these concept and terminologies.

Lets start with the first one,

(a) An infinite open-loop gain. The slightest difference in V+ and V- will caused the output to go to saturation. Saturation voltage cannot exceed the power supply voltage.

By now, we all understand this open-loop gain thing. If not, re-read the top of this

section. The difference now is that we put the word infinite in front of it! Infinite just means very, very large, countless So we are just saying that when the op amp is configured without any feedback it has a very, very, very large gain.

The next word is saturation, what does it mean for the op amp to be saturated? What is it saturated with? Suppose we were to plot the gain of the op amp from its definition, i.e.

in

out

VV

gain = , where Vout and Vin are the output and input signal voltages, respectively.

We will get the plot shown in Fig. 4(a). Just a nice straight line with a constant gradient! Agreed!!! Ok.. lets move on.

(a) Constant gain (b) saturation limit Fig. 4 Op amp gain and voltage saturation

Vin (mV)

Vout (Volts)

ConstantGain

0 Vin (mV)

Vout (Volts)

ConstantGain

0Vin (mV)

Vout (Volts)

Region ofConstant gain

0

Vsupply

Region ofreducing gain

Region ofFixed output

Vin (mV)

Vout (Volts)

Region ofConstant gain

0

Vsupply

Region ofreducing gain

Region ofFixed output

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What is the maximum possible output voltage of the op amp?? No idea!!! Think about it for a while lets ask another question then. Where does the op amp get its voltage

from? Yes, this is easier The op amp gets its voltage from the voltage of the power supply that is connected to it. Therefore, it stands to reason that the maximum output voltage of the op amp cannot exceed its power supply voltage. In other words, the power supply voltage sets the voltage output limit of the amplifier. Now, lets get back to Fig. 4.

Look especially at Fig.4(b) Whenever the output voltage starts getting close to the value of the power supply voltage, Vsupply, something strange starts happening. The gain

starts decreasing, i.e. Vout/Vin is getting smaller. As a result, the gradient of the slope starts decreasing and is getting flatter and flatter. By the time it reaches the power supply

voltage value, it becomes a flat straight line which will never exceed the power supply voltage value. Just as it reaches this point, the op amp is said to be saturated. That is, the output signal remains at a constant voltage irrespective of any increases in the input

signal voltage.

Moving on to the second property:

(b) An infinite input impedance (resistance). This ensures that no current flows into the input terminals (V+ and V-). However, voltages may be present.

What is impedance anyway? It is the term used to describe the combined resistances of

all the circuit elements, including elements that you will not study at the CAPE level. Impedance values depend on the frequency of the signal. Its unit is the Ohm, same as that

for resistances. So, for the sake of simplification, lets think of impedance as resistance.

Why is it that no current flows into an infinitely high resistance? It goes back to Ohms law, which state that V = IR. If V is 10 volts and R is 10 M, then the current I = 10V

10,000,000. Therefore, I = 1 A or 10-6 A. In these kinds of circuit, this current is considered negligible. This is the same principle behind the operation of an ideal

voltmeter. It measures the voltage while having negligible current flow through it.

Moving on ..

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(c) An infinite bandwidth. This assumes that it amplifies any input range of frequencies.

Bandwidth is the range of frequencies over which the op amp operates with a constant

gain. This will be explained in more detail later on.

(d) Zero output impedance (resistance). This ensures that the amplifier is unaffected whatever output circuit it is connected to.

This is self explanatory by now. It just means there is no output resistance when the op amp is connected to a load. In this state it is capable of driving any load.

Points (e) and (f) are self explanatory. Also, understanding these features is not a CAPE requirement at this time.

So lets now restate the assumptions of an ideal Op amp:

(a) An infinite open-loop gain. (b) An infinite input impedance (resistance). (c) An infinite bandwidth. (d) Zero output impedance (resistance). (e) An infinite slew rate. (f) Zero voltage and current offsets.

The Typical (Real) Operational Amplifier Real op amps have characteristics that approach those of ideal op amp, but never quite attained them. They deviate from the ideal op amp in the following ways:

(a) The open loop gain is usually in the range of 105 106. Although this is high, it is not infinite.

(b) They have large but finite input impedances usually in the range of 106 1012. Thus, drawing very small, but measurable currents at their input terminals.

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(c) They have a finite bandwidth which is dependent on the gain. The higher the gain the smaller the bandwidth. This is usually described in its frequency response

characteristics or the Gain-Bandwidth product. (d) The output impedance is usually about 100 . (e) They have finite slew rate and voltage and current offsets.

Note that while the ideal op amp does not exist, its properties serve as a valuable starting point for preliminary circuit analysis.

The Op Amp as a Comparator As discussed earlier, an op amp has an inverting (V-) and a non-inverting (V+) input. These inputs may be connected as single-ended inputs or as a differential input. In single-

ended input mode, only one of the inputs has a voltage signal while the other is grounded. In differential mode, both inputs have voltages with respect to ground. Equation 1 may be

applied to both of these cases to create these three open-loop gain scenarios:

1. += VAVout 0 Input signal (Vin) is on V+ input and V- terminal is grounded

2. )(0 = VAVout Input signal (Vin) is on V- input and V+ terminal is grounded

3. )(0 + = VVAVout for cases where a differential input signal is applied

This may be represented schematically as shown in Fig. 5.

(a) Case 1 (b) Case 2

-

+

- Vsupply

+ Vsupply

ground

Vin )(0 = VAVout

-

+

- Vsupply

+ Vsupply

ground

Vin )(0 = VAVout

-

+

- Vsupply

+ Vsupply

ground

Vin+

= VAVout 0

-

+

- Vsupply

+ Vsupply

ground

Vin+

= VAVout 0

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(c) Case 3 Fig. 5 Open-loop gain scenarios

Some important observations: 1. In case 1, the output signal will always be of the same polarity (or phase) of its

input signal.

2. In case 2, the output signal will always be of the opposite polarity (or phase) of its input signal.

3. In case 3, the polarity of the output will depend on which of the inputs has the

larger voltage:

a. If + < VV then polarity will be opposite to that of V-

b. If + > VV then polarity will be same as that of V+

4. A negative output voltage can only be obtained if the op amp is connected to a dual rail supply (i.e Vsupply).

A Comparator may be made up of any of these three configurations. The signals at the

input terminals are compared and their difference is multiplied by the open-loop gain of the op amp to produce the output voltage. Lets look at these three examples:

-

+

- Vsupply

+ Vsupply

ground

VinVin )(0 + = VVAVout

-

+

- Vsupply

+ Vsupply

ground

VinVin )(0 + = VVAVout

13

Example 1: What is the output Voltage for an op amp circuit with the following

characteristics?

V+ = 1V V- = 0 Volt (or grounded) A0 = 105 +Vsupply = +12Volts -Vsupply = - 12 Volts

Solution: (Use circuit in case 1 of Fig. 4)

000,10011050 +===+VAVout Volts

But stop right hereRemember, you cannot get an output voltage that is greater that your power supply voltages. Therefore, Vout cannot exceed +Vsupply, i.e.

Vout = + 12 Volts

Whenever this happens, the op amp is said to be SATURATED. Saturation

voltages cannot exceed the power supply voltages.

Example 2: Same as Example 1 except V+ = 0, and V- = 1 Volt.

Solution: Vout = -100,000 Volts But by now we know this cannot exceed the -12 Vsupply. Therefore the real

answer is: Vout = -12 Volts.

Example 3: Same as Example 1 except V+ = 1.3 V, and V- = 1.1 V

Solution: Applying Case 3 equation,

( )1.13.110)( 50 == + VVAVout = 0.2 x 105 = 20,000 Volts Again the actual value is:

Vout = +12 Volts

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Application of the Comparator The comparator is used in many circuit applications where two states need to be

compared to produce a desired output signal, which is usually used to control some other circuit or to switch some state. A not so obvious use is that of a voltage-level-shifter. That is, use can be made of the fact that an output voltage that is equal to the power supply voltage can always be obtained, irrespective of how small the input voltage difference is.

Usually, when the op amp is used as a comparator, its input terminals are connected to

some kind of potential divider circuit. One of these potential dividers creates a fixed voltage that is used as a reference, while the other varies in accordance with some other

physical or electrical property that may either be internal or external to the circuit. It may be important at this point to say a little about potential dividers and variable resistors.

Potential Dividers and Variable Resistors A Potential Divider or voltage divider is a simple arrangement of two resistances across one voltage source. From Ohms and Kirchhoffs laws, we know that the sum of the

voltage drops across each resistance will be equal to the supply voltage. In other words, the voltage drop across each of the resistances is always less than that of the voltage source. By varying the resistances, we can vary the amount of voltage across them. We

can collect an output voltage at the point just between the two resistances. In most cases this output is taken with respect to the ground. Note that we are referring to two

resistances and not two resistors. There is a reason for this! While two resistors will work perfectly well as voltage dividers, there are a set of resistors that has variable resistances.

They have three terminals, one of which is a slider that is used to vary the resistance that is seen at the output. Terminals 1 and 3 (see Fig. 6) may be connected across a voltage source and the output taken from terminal 2. An illustration of these resistors is shown in Fig. 6.

15

Fig. 6 Fixed and variable resistors

Fig. 7 shows actual voltage divider examples using these two types of resistors. The value

of the output voltage (in Fig. 7) can always be determined from Equation 2.

21

sup2

RRVR

V plyout +=

(2)

Note that in Fig.7 (b) the variable resistor is divided into two resistances, the one above the slider, R1, and the other below, R2. Eq. (2) applies in all cases where the output is connected to a very high resistance (or impedance), for example, the input terminal of an op amp.

(a) Two fixed resistor divider (b) Variable resistor divider

Fig. 7 Voltage Dividers

Fixed resistor1 2

Variable resistor

1 3

2

Variable resistor

1 3

2

Fixed resistorFixed resistor1 2

Variable resistor

1 3

2

Variable resistor

1 3

2

Variable resistor

1 3

2

Variable resistor

1 3

2

+

Vsupply

R1

R2 Vout

+

Vsupply

Vout

R1

R2

+

Vsupply

R1

R2 Vout

+

Vsupply

Vout

R1

R2

16

Example 1: If Vsupply =12 V, R1 = R2 = 10 k, what is the value of Vout?

Solution: Using Eq.2;

62

12000,10000,10

12000,10==

+

=outV

= 6 Volts

Example 2: If Vsupply =12 V, R1 = 10 k, what value of R2 is required to let Vout =

4V?

Solution: Modifying Eq.2 so that R2 becomes the subject:

==

=

= kVV

VRRoutply

out 55000412

4000,10sup

12

Other forms of variable resistors exist that have only two terminals, instead of three.

Their resistances usually depend on some external physical condition, like temperature, light or strain. Resistors with resistances that depend on the amount of light present are

called Light Dependent Resistor (LDR) and those that depend on temperature are called Thermistors.

An LDR is made by sandwiching two metal electrodes by a film of cadmium sulphide. In complete darkness, it has a resistance of about 10M, but in bright

sunlight, its resistance falls to about 100. Therefore, by varying the amount of light shining on the LDR, we can vary its resistance. In Example 1 above, Replace R2 with a LDR and calculate Vout for both darkness and sunlight cases.

A thermistor is a temperature dependent resistor which is manufactured from the

oxides of various metals. They are made in all kinds of shapes and sizes. Negative temperature coefficient types have resistances which becomes smaller as

temperature increases.

A strain gauge is made by sealing a length of very fine wire in a small

rectangular of thin plastic sheet in such a way that if the plastic is stretched (i.e. under strain), the wire will be stretched, which in turn increases its resistance.

17

So lets get back to our discussion of the op amp as a comparator. Fig 8 shows an example of a comparator application. Resistors R3 and R4 fixed the voltage at the non-

inverting input to half the +Vsupply voltage. The voltage at the inverting terminal will

depend on the value of the resistance of the LDR, R2. In the presence of light (such as daylight) the LDR will have a resistance that is less than R1. This makes V+ less than V- and the output to be equal to Vsupply. In this output state, no current will flow through

the Light Emitting Diode (LED) and it will remain off.

Fig. 8 Comparator application: A light sensitive circuit

In the absence of light (such as at night) the LDR resistance will exceed that of R1, creating a voltage at V+ that is greater than that at V-. The output of the op amp will then

switch to the positive saturation voltage which equals +Vsupply. In this state, a current

will flow through the LED to turn it on. That is, the LED will be lit. Resistor R5 is required to limit the current through the LED. Usually, LED draws a current of 10 mA

and has1.8 V drop across it, therefore R5 value is =+ 72001.0

8.1sup plyV.

This is the basic principle of operation of a photosensitive light detector, like those installed outside your houses. Note that the LDR can be replaced by a thermistor to

produce an ice-warning LED circuit.

-

+

ground

++

-

-

9V

9V

R110k

LDRR2

R310k

R410k

R5

+ 9 V

-9 V

LED

-

+

ground

++

-

-

9V

9V

R110k

LDRR2

R310k

R410k

R5

+ 9 V

-9 V

LED

-

+

ground

++

-

-

9V

9V

R110k

LDRR2

R310k

R410k

R5

+ 9 V

-9 V

LED

18

Feedback in Op Amp Circuits Feedback, as the word implies, is the process of taking some, or all, of the output signal

of an amplifier and adding it back to its input. The basic arrangement for this is shown in Fig. 9.

Fig. 9 Basic feedback in amplifier

A fraction of the output voltage is fed back and added to the input applied voltage. By

inspection of Fig. 9, it is seen that the fraction of signal fed back to the input is outV . This gets added to the input, Vin, to give outin VV + . This combined input is amplified by the gain A0 of the amplifier to give an output of ( )outinout VVAV += 0 . This may be rewritten as:

( )

00

00

00

1 AVVA

VAVVA

VAVAV

outin

outoutin

outinout

=

=

+=

The overall gain of this feedback arrangement may now be expressed as:

( )00

1 AA

VV

in

out

= (3)

AddVin

FeedbackFraction,

AmplifierGain, A0 Vout

Vout

Vin+VoutAddVin

FeedbackFraction,

AmplifierGain, A0 VoutAdd

Vin

FeedbackFraction,

AmplifierGain, A0 Vout

Vout

Vin+Vout

19

There are two kinds of feedback that depends on the polarity of . They are positive

feedback and negative feedback.

Positive Feedback

Referring to Eq. 3 above, if is positive, then the 0A term can be made to be equal to 1, making the denominator term ( )01 A equal to zero. That is, the overall gain will be infinite. So we now have an amplifier system with an infinite gain, even without any

inputs. This is the basis of the principle of operation of oscillator circuits. Note that positive feedback is only possible when the output signal is fed back in phase with the input signal, so in the case of the op amp, the feedback signal must be sent to the non-inverting input terminal.

Negative Feedback If the fraction is negative, then the denominator term in Eq. 3 is greater than unity. The overall gain of the feedback amplifying system is now much smaller than the open-loop gain. At first glance, this may seem pointless, but there are some very great advantages for doing this:

a. Lowering the gain significantly increases the bandwidth. That is, it allows amplification over a greater frequency range.

b. There is less distortion of the output signal. The op amp doesnt have to saturate anymore.

c. Overall improvement in the operational stability of the circuit.

The Inverting Amplifier The inverting amplifier is shown in Fig. 10. For simplification, the power supply

connections are not shown. Also, it is assumed that the op amp is not saturated. An input signal Vin is applied to the resistor R1, which is connected to the inverting input terminal. Resistor Rf is connected across the output and the inverting input terminal to provide negative feedback.

20

Fig. 10 The inverting amplifier

The non-inverting input is connected to ground so its potential is at exactly zero volt. The behavior of an op amp is such that when any of its input terminals is grounded it causes a virtual ground condition to exist at the other terminal. This point is labeled VG in Fig.10. This is an important point and needs restating. When the amplifier output is fed back to the inverting input terminal, the output voltage will always take on that value required to drive the signal difference between the amplifier input terminals to zero. Another point to note is that because of the very large input impedance no current flows into the input terminals of the op amp.

Let us now derive an expression for the gain in terms of the R1 and Rf. The circuits of Fig.11 (a) and (b) show the directions of current flow in an inverting amplifier. Since no currents flow into the input terminals, all currents must flow through the external

resistors, R1 and Rf.

-

+Vin Vout

Rf

R1 VG-

+Vin Vout

Rf

R1 VG

21

(a) +ve flowing (b) -ve flowing

Fig. 11 Direction of current flow in the inverting amplifier

The same current, I, that flows through R1, also flows through Rf. Therefore, Ohms law

may be used to derive a simple expression of the voltage gain.

Current through R1 = Current through Rf

From Ohms law (V=IR), we can express the current in terms of voltage and resistance;

Therefore, f

RR

R

VR

V f=

1

1

Where, VR1 and VRf are the voltage drops across resistors R1 and Rf, respectively. But, since the voltage at the inverting input is 0-volt (at the point VG) then:

01

= inR VV and outR VV f = 0

Substituting,

-

+-ve output

Rf

R1 VG

+ve inputCurrent, I

Current, ICurrent, I

0-volt

-

+-ve output

Rf

R1 VG

+ve inputCurrent, I

Current, ICurrent, I

-

+-ve output

Rf

R1 VG

+ve inputCurrent, I

Current, ICurrent, I

0-volt

-

++ve output

Rf

R1 VG

-ve input

Current, I

Current, I

Current, I-

++ve output

Rf

R1 VG

-ve input

Current, I

Current, I

Current, I

22

1

1

1

00

RR

VV

RV

RV

RV

RV

f

in

out

f

outin

f

outin

=

=

=

That is, the voltage gain of an inverting amplifier is equal to the negative of the ratio of

its feedback resistance to its input resistance.

The Non-inverting Amplifier The non-inverting amplifier is shown in Fig.12. An input voltage Vin is applied directly to the non-inverting input terminal, V+. Negative feedback is applied by means of resistors Rf and R1, as shown.

Fig. 12 The non-inverting amplifier

-

+

Rf

R1

VoutVin

0 V

-

+

Rf

R1

VoutVin

0 V

(4)

23

Assuming that the amplifier is not saturated, then the voltages at the two input terminal must effectively be the same. That is V+ = V-. In this setup, V+ is equal to Vin and V- is equal to the voltage dividing of Vout by Rf and R1. This may be expressed as:

f

outin RR

VRVVV+

===+

1

1

By rearranging, the overall voltage gain of the amplifier is given by

1

1RR

VV f

in

out += (5)

Exercise: Verify Eq.5 by using the current flow technique used to derive the gain equation for the inverting amp.

Important Observations: 1. The non-inverting amplifier produces an output that is in phase with the input

signal, hence the name non-inverting.

2. The inverting amplifier produces an output that is out-of-phase (usually by 180 degrees) with its input. Look back at the circuits of Fig. 10 and Fig. 12.

3. The input signal of the inverting amplifier goes through the input resistors. This means that the impedance seen by the input signal is dependent on the values of the resistors.

4. In the case of the non-inverting amplifier, however, the input signal goes directly to the non-inverting input of the amplifier, which has infinitely high input

impedance. Because of this, the current drawn from the signal source is negligible and this circuit configuration is extremely suitable for BUFFER AMPLIFIER applications. That is, it does not load down the signal driving circuit.

Effect of Negative Feedback on Gain and Bandwidth Fig.13 shows a circuit that could be used to measure the open-loop gain of a real op amp at a number of different frequencies. During measurement of gain, the amplifier must not

24

be saturated. Because the open-loop gain is about 105, if Vsupply is 12V then the maximum output voltage would be 12V. This means that the maximum input signal must be 12V 105 = 0.12V = 120mVpeak.

Fig. 13 Measuring the gain and bandwidth of an open loop op amp

The sinusoidal input signal may be obtained from a frequency generator and an oscilloscope may be used to measure the input and output voltages (or waveforms). The frequency may be varied from 0 Hz (dc) to 1 MHz. A plot of the gain versus frequency for this open loop amplifier is shown in Fig.14. This is called the frequency response curve of the op amp.

Fig. 14 Frequency response of the open-loop op amp circuit

-

+

- 12V

+ 12V

ground

Vac Vout

-

+

- 12V

+ 12V

ground

Vac Vout

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gain

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gain

25

The plot shows that in the open-loop mode, the op amp does not amplify all frequencies equally. The range of frequencies for which the gain is more or less constant is called the bandwidth of the amplifier. For this op amp in open loop mode, its bandwidth is about dc to 10 Hz, or just 10 Hz. An important observation from the plot is that if a larger bandwidth is desired, the gain must be reduced. How can we reduce the gain? By using negative feedback of course! We just did that in the previous sections.

Using the circuits of Fig. 10 and Fig. 12, shown together in Fig. 15, and applying a variable frequency sinusoidal signal, Gain-Frequency plots may be obtained when both input and output are measured with an oscilloscope.

(a) Inverting setup (b) Non-inverting setup

Fig. 15 Circuits to determine frequency response of negative feedback amplifiers

Note that the gain of both the inverting and non-inverting amplifiers may be easily adjusted by varying the values of one are both resistors (R1 and/or Rf). They may also be adjusted to be the same value. They have same frequency response, but different resistor values. Fig. 16 shows the frequency response curves for gains of 10, 100 and 1000.

The main observation from these plots is that as the gain decreases, the bandwidth increases. That is, as more and more negative feedback is applied, the bandwidth increases. Therefore, negative feedback improves bandwidth while reducing gain. But look more closely at the relationship between the product of the Gain and the bandwidth. Yes, it seems always to be constant. Lets look at the three cases of Fig. 16 again.

-

+

Rf

R1

Vout

0 V

-

+

ground

VacVout

Rf

R1

Vac

-

+

Rf

R1

Vout

0 V

-

+

ground

VacVout

Rf

R1

Vac

26

Fig. 16 Gain-Bandwidth for variable gain feedback amplifier

When the gain is 1000 the bandwidth is 103 making a Gain-Bandwidth product of 106. The second case has a gain of 100 and a bandwidth of 104 giving a gain-bandwidth

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 1000

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 1000

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 100

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 100

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 10

10 102 103 104 105 106 1

10

106

105

104

103

102

Frequency (Hz)

gainWithout feedback

With feedback

Overall gain = 10

27

product of 106. Likewise the third case! Go back to Fig. 14 for the case without feedback. Its gain is 105 and its bandwidth is 10 making the gain-bandwidth product of 106. This is a very important property of the op amp and is used in many design considerations.

Important observation: Sometimes you may be required to design circuits with large gain and large bandwidths. For example, suppose a gain of 100 and a bandwidth of 105 are desired. From the plots in Fig. 16, this cannot be obtained from one op amp circuit. So why not cascade (join in series) two op amp circuits, each with a gain of 10.

Voltage Follower What is a voltage follower? It is also called a unity gain buffer, that is, it has very high input impedance and low output impedance and is very applicable in cases where signal sources must not be loaded down. To load down a circuit means to connect a load that will cause too much current to be drawn. Fig. 17 shows the configuration of an op amp as a voltage follower.

Fig. 17 A voltage follower

Vout

-

+

ground

Vin

+Vs

-Vs Vout

-

+

ground

Vin

-

+

ground

Vin

+Vs

-Vs

28

In this configuration negative feedback is provided by directly connecting the output to the inverting input. Vout will adjust itself in such a way as to ensure than the voltage at V- is the same as that of V+. The voltage at V+ is Vin, therefore, Vout will always be the same as Vin. Some characteristics of the voltage follower are:

1. Infinite input impedance 2. Low output impedance 3. Input and output are always in phase 4. Output voltage is the same as the input voltage. That is, Gain = 1

Question: What is the bandwidth of an op amp voltage follower?

Answer: Since the gain is always 1, look at the frequency response plots. At

gain=1, we always have maximum bandwidth. In Fig. 14, the bandwidth at gain = 1 is 1 MHz.

Summing Amplifier The inverting amplifier configuration of the Op amp may be configured to operate with more than one input signals. Fig. 18 shows such a configuration where 3 input signals, V1, V2, and V3 are connected to resistors R1, R2 and R3, respectively. The other ends of the resistors are connected together at the inverting input terminal.

Fig. 18 The summing amplifier

-

+

R fR 2

V ou

R 1

R 3

V 1

V 2

V 3

I1

I2

I3

-

+

R fR 2

V ou

R 1

R 3

V 1

V 2

V 3

I1

I2

I3

Itotal = I1 + I2 + I3

29

The analysis we did earlier for a single input inverting amplifier may be extended for this 3-input configuration. From Kirchhoffs current law, we know that the algebraic sum of the currents flowing into a junction must be equal to the algebraic sum of the currents flowing out of that junction. Since the V+ input terminal is grounded, then the V- terminal must be at a ground potential as well, i.e. 0 volts. Also, no current will flow into the terminals of the op amp (because of their infinitely high input impedance). Therefore, the currents I1, I2 and I3 produced by the input voltages V1, V2, and V3 can only combine to flow through the feedback resistor, Rf, creating a voltage drop across Rf that (from Ohms law) is equal to Rf (I1+I2+I3). But, the voltage drop across Rf is equal to Vout. Therefore, Vout = - Rf [I1+I2+I3]. Note the minus sign!! If current flowing into the junction is positive, then current flowing out must be negative.

We can express the currents in terms of the input voltages. Once again, because of the

virtual ground (or earth) at the inverting input terminal,1

11 R

VI = ,

2

22 R

VI = and

3

33 R

VI = .

Substituting in expression for Vout:

++=

3

3

2

2

1

1

RV

RV

RV

RV fout (6)

This expression may be extended for many inputs system. Likewise, if there were only 2

inputs then the term with 3

3

RV

should be removed.

Summing circuits are used in the design of audio mixers. Input 1 may be from a microphone, input 2 from a CD player, and input 3 from a turn table (phono player), etc. The input resistors may be made variable for separate volume adjustment, while the feedback resistor may be made variable for the master volume adjustment. In most applications, however, the output of the summer is fed to a different amplifier circuit for master volume adjustments.

30

Difference Amplifier The op amp may also be configured as a difference amplifier (or subtractor). Fig. 19 shows such a configuration. Notice the use of the same resistances, R1 and R2!

Fig. 19 The difference amplifier

( )211

2 VVRRVout = (7)

An Operational Amplifier Circuit Example In this section we will examine a number of op amp circuits operations with respect to dc and ac signal inputs. Wherever appropriate, waveform diagrams (or timing diagrams) will be used to show the relationship between the input and output signals.

Response to a variable voltage 1 KHz input signal: Fig. 20 shows a typical inverting amplifier with supply voltages 10V. The input signal is obtained from a 10Vpeak, 1Kz signal generator and is varied (from 0 -10V) by adjusting the slider arm of the variable resistor.

Fig. 20 An inverting amplifier circuit

-

+

100 k

10 k+10 V

-10 VVout

SignalGenerator(10V, 1kHz

100kVariableresistor

Slider-

+

100 k

10 k+10 V

-10 VVout

SignalGenerator(10V, 1kHz

100kVariableresistor

Slider

-

+

R2

VoutR1V1

V2R1

R2

-

+

R2

VoutR1V1

V2R1

R2

31

Try to answer the following:

1. What is the voltage gain of the amplifier?

2. What is the maximum peak voltage output?

3. Draw the output waveform corresponding to input voltages of a. 0.1 volt b. 1 volt c. 5 volts d. 10 volts

4. How can the gain of the amp be increased by a factor of 10? 5. What is the maximum input voltage that will not saturate the amp?

Solutions:

1. Gain is in

f

RR

=

kk

10100

= - 10

2. Max. peak voltage = Vsupply = 10 V (or -10V) 3. In drawing waveform diagrams. Always draw the input and the output on the

same plot.

a. Note the 180 degree phase change

b. 1 volt input will give 10 volts output. Waveform looks like that of (a).

time

Voltage

input0.1V

1 V

output

time

Voltage

input0.1V

1 V

output

32

c. 5 volts input will saturate the amplifier. The output signal will be clipped at the saturation voltage of 10 V.

d. A similar waveform is obtained as in (c) when input is 10 V. Waveform will look more like a square wave.

4. Change the 100 k resistor to 1 M., or the 10k to 1k, or use any other resistor

combination that gives a gain of 100.

5. The maximum input signal that will not saturate the amp is the Vsupply divided by the gain.

11010sup == V

gainV ply

volt

time

Voltage

input5V

10 V

output

-10 V

time

Voltage

input5V

10 V

output

-10 V

time

Voltage

input5V

10 V

output

-10 V

33

OP Amp Problems

1. The circuit of Fig. 21 was used as a buffer amplifier for an audio signal.

Fig. 21 Buffer Amplifier

Answer the following questions:

(a) What amplifier configuration is this (inverting or non-inverting)?

(b) What is the theoretical gain?

(c) Sketch of graph of the output vs input voltage when the input voltage is varied from 100mV to 5V.

(d) What is the saturation voltage of this circuit? Show this on your sketch in (c) above.

(e) What is the maximum input before saturation? Show this on your sketch in (c) above.

(f) Are there any advantages of using this buffer amp over other configuration? Explain.

2. Using the circuit of Fig. 21, but replacing the audio input with variable frequency generator. It was found that the gain was different for different frequencies. Explain!

-

+ VoutVaudio

R1 = 100k+12V

-12V

R2 = 400k

-

+ VoutVaudio

R1 = 100k+12V

-12V

R2 = 400k

34

3. Explain the effect of negative feedback on

(a) The gain of an op amp

(b) The bandwidth of an op amp

(c) The gain-bandwidth product

(d) The saturation voltage

4. Design a 2-input audio mixer circuit that has gain of 10.

5. Fig. 22 shows a 2-stage op amp circuit.

Fig. 22 Two stage amplifier

Determine the following:

(a) The value of the voltage Va.

(b) The output voltage, Vout

(c) The total gain of the circuit.

(d) Which of the two op amp configuration above determines the bandwidth of the circuit?

6. Use any amplifier configuration to explain the concept of virtual ground.

-

+

100k +12V

-12V

600k

Vout-

+

+15V

-15 V

0.2V Va-

+

100k +12V

-12V

600k

Vout-

+

+15V

-15 V

0.2V Va

35

EXPERIMENT #1 Inverting Amplifier

AIM: Determine the effect of negative feedback on the gain and bandwidth of an Operational Amplifier.

APPARATUS: Dual voltage power supply ( 12 V), 741 Operational Amplifier, Cathode Ray Oscilloscope, Resistors (1 k, 10 k, 100 k) and function generator (0 1 MHz).

DIAGRAM:

1. Choose two resistors Rf = 10k and Ri = 1k, such that the Gain of the amplifier is 10 and connect the circuit as shown in the diagram. The pin labels for the Op amp may be obtained from page 3 of this workbook. Place the input signal on Channel 1 and the output signal on channel 2 of the scope.

2. Set your signal generator to produce a sinusoidal waveform and set voltage amplitude to any value between) 0.5V and 1.0 V.

3. Adjust the frequency of signal generator to 100 Hz. Remember the Cathode Ray Oscilloscope must be used to check/verify the output (frequency and voltage) of the signal generator.

36

4. Measure the output voltage of the Operational Amplifier using the cathode ray oscilloscope.

5. Repeat steps 3 & 4 at frequencies 1 kHz, 10 kHz, 100 kHz, 1 MHz.

6. Tabulate your results in the table below (in Part 2).

7. Plot graph of log of Gain vs. log of frequency.

PART 2

1. Repeat the experiment in Part 1, but set Rf = 100k and Ri = 1k to produce voltage gains of 100. Tabulate your results below:

Input Voltage (Vin)

Output Voltage (Vout)

Volatge Gain (Vout/Vin)

Log of Gain

Frequency (Hz)

Log of Frequency

10 10010 100010 10000

10 100000

10 1000000

100 100

100 1000

100 10000

100 100000100 1000000

2. Make a plot of log(Gain) vs. log(frequency) on the same graph created in Part 1.

3. Determine the Gain-Bandwidth product and the maximum bandwidth of the Op-Amp..

DISCUSSION: Discuss your result as they relate to the aim of the experiment.

37

EXPERIMENT #2 Simulation of Experiment 1

AIM: Using PSPICE simulator to determine the effect of negative feedback on the gain and bandwidth of an Operation Amplifier.

APPARATUS: Computer with student version of PSPICE loaded.

DIAGRAM: Draw the circuit of Experiment 1 in PSPICE Schematic Editor as shown:

Procedure:

The lab instructor will guide you through the simulation setup and measurement via ac analysis and transient analysis. You will use PSPICE to obtain the output voltages and make the plots as required by Experiment 1.

38

1. Click START , select ALL PROGRAMS, then PSPICE STUDENT, then Schematics (as shown).

2. The PSPICE window should open up and looks like

39

3. You will now draw the circuit inside the SPICE window by selecting and placing the components. Click on the looking glass icon (part browser) to open the component selection window. Select and place each component as follows: uA741 (for the op amp), r (for resistor x 3), V dc for voltage supply times 2, Vsin for signal, and GND_EARTH for the ground connections.

4. Place the components to match the layout in the circuit diagram (Exit 1) then make all wire connections. This is done by clicking on the THIN PENCIL, place the mouse at one component, make one click, then place the mouse at the other component, and make another click. Repeat this procedure the components are connected as shown:

40

5. Double click on each resistor to set its value.

6. Double click on the V dc parts and set values to 12V

7. Double click on Vsin to open a window. Set values as

DC = 0 AC = 0.1 (this is the desired input voltage level: 100mV) Voff = 0 Vamp = 0.1 (same as AC) Freq = 100 (this set it to 100 Hz)

Then click ok to save values. During the experiment you must re-select this window to change the frequency value to 1k, 10k, 100k, and 1MHz. :

8. Insert Vin and Vout labels by double clicking on the input and output wires:

41

9. You are now ready to start your simulations. Click on Analysis then Setup and select Transient box. Enter the values shown in the transient window.

10. Close Analysis Setup Window.

11. Click on Analysis and select Simulate to start your simulations. A percentage completion bar will be displayed on the bottom right of the screen.

12. Display the input and output waveform (Vin and Vout) and take the relevant measurements. The lab instructors will show you how to do this.

13. Record your data in the table below:

42

Inpuit Voltage (Vin) Output Voltage (Vout) Voltage Gain (Vout/Vin) Frequency (Hz) Log of Frequency

100 2

1,000 3

10,000 4

100,000 5

1,000,000 6

14. Make a plot of Gain vs. log of frequency on a graph paper (or in EXCEL).

15. Determine the Gain-Bandwidth product and the maximum bandwidth of the circuit.

DISCUSSION:

Discuss your result as they relate to the aim of the experiment, and compare your results for the simulations to that of the hands on measurements.