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OP-AMP CIRCUITS

Dr Collins

Hilary Term 2012

Introduction

There are at least two broad classes of systems in which it is important to

detect continuously varying electrical signals. The first of these are wireless

communications systems, including radios, mobile phones and wireless

internet. In these applications the signals that must be detected arise from the

receiver. The second class of systems is in instrumentation and control

systems. In these systems the signals arise from sensors that are used to

detect important physical quantities, including temperature, pressure and

strain. These physical quantities can only be sensed if they affect an electrical

property, such as resistance, of a component of a circuit and the physics of the

available processes mean that there is always a weak dependence of the

electrical property on the physical quantity. When the physical quantity

changes this means that the resulting changes in the electrical signal are

therefore small.

In wireless applications the ability to detect the smallest possible signal can be

used to either increase the range (distance) over which the wireless system

can operate or reduce the power needed to transmit the signal. In

instrumentation and control systems the ability to detect small electrical signals

can be exploited to detect small changes in the physical quantities of interest.

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Sophisticated electronic systems are now inexpensive. In the consumer

market this has contributed to the popularity of a range of items including

personal computers, digital cameras and mobile telephones. The equivalent

trend in engineering has been a rapid expansion in the use of electronics to

obtain data from different systems to be able to monitor and control their

performance.

One example of this latter trend is a 2.25 Km suspension bridge built for the

Athens Olympics which was constructed with approximately 300 sensors

embedded within its structure. These sensors include:

strain gauges to keep track of framework fatigue

sensors to monitor motion in the stay cables caused by cross winds

accelerometers in the roadway to measure the impact of earthquakes

This array of sensors produces a huge amount of data which must be

recorded, processed (to extract useful information) and transmitted to both the

headquarters of the bridge operating company in Athens and the headquarters

of the structural monitoring division of one of the bridge builders, which is in

France.

The recording, processing and transmission of data are all now usually

performed when the data is represented in the digital format used by

computers. In this format binary numbers represent the amplitude of the data

sampled at particular, usually regularly separated, times. In contrast the

outputs from sensors and receivers are analogue voltages that vary

continuously in both amplitude and time. Before these analogue signals can

be recorded, processed and transmitted they must be converted from this

analogue form to an equivalent digital format using a component known as an

analogue-to-digital converter (ADCs). These components will be discussed

later in the course.

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In this course we will focus upon instrumentation systems as the application

domain, however the circuits that will be discussed can be used in

communication systems. Before we discuss ADCs we need to discuss the

circuits that are needed between any sensors and ADCs. These circuits are

needed because of a combination of two factors: On the one hand there are

the small electrical signals from sensors (As a result of this the maximum

sensor output signals are always small, typically less than 10mV.) On the other

hand ADCs are standard components with a fixed maximum input voltage,

typically several volts. To achieve maximum sensitivity the maximum output

voltage from the sensor must be amplified so that it equals the maximum input

range of the ADC.

As well as amplifying the output voltage from the sensors the circuits between

the sensors and the ADCs should also remove high frequencies from the

sensor output voltage.

The circuits to amplify the output voltage and the filters needed to remove high

frequencies are usually based upon a component known as an operational

amplifier (op-amp). The starting point for the course will be op-amps and the

properties of ideal op-amps.

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Learning Outcomes

At the end of the course students should:

1. Be able to state the characteristics of an ideal op-amp

2. Be able to explain the use of an op-amp as an voltage follower

(impedance buffer)

3. Be able to perform a nodal analysis on circuits containing ideal

op-amps, including amplifiers and filters.

4. Understand the importance of the difference between the two op-amp

inputs

5. Understand the operation, specification and use of the 3 op-amp

instrumentation amplifier

6. Be able to calculate the component values needed to obtain a

required filter response

7. Be able to analyse the response of a system containing a series of

op-amp based circuits

8. Understand the function of DACs and ADCs

9. Be able to describe the operation of a DAC based upon an R-2R

ladder

10. Be able to describe the operation of flash ADCs

11. Understand the operation of a successive approximation ADC and the

need for a sample and hold circuit at its input

12. Be able to determine the effects of quantisation noise on an

instrumentation system

13. Be able to specify the characteristics of amplifiers and filters in an

instrumentation application

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Textbook

A textbook that covers the majority of the topics in this course at an

appropriate level is

Design with Operational Amplifiers and Analog Integrated Circuits

by Sergio Franco, published by McGraw-Hill

Organisation of the Course

After describing the operation of ideal op-amps the course will concentrate

upon circuits containing op-amps and resistors that can be used to add two

voltages or to amplify either a voltage or the difference between two voltages.

This is followed by a short discussion of two techniques to convert a current

into a voltage.

The course then moves on to describe circuits that contain op-amps, resistors

and capacitors that can be used to amplify any component of a voltage signal

at some frequencies whilst attenuating components at other frequencies.

Circuits with a response that depends upon the frequency of the input voltage

are known as filters.

The final section of the course covers analogue to digital converters that are

needed to convert the analogue voltages from circuits into the digital format

used in computers and the digital to analogue converters that do the

conversion in the opposite direction.

The course will depend upon all the techniques that you have been taught in

P2a1 and P2a2 together with complex algebra from P1b and Laplace

Transforms from both P1d and P2a2.

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An Example Sensor: The Strain Gauge

A strain gauge is a thin, wafer-like device that can be bonded to a variety

of materials that might be subject to strain. To understand how strain

causes a change of resistance assume that the strain gauge is made from a

material with a resistivity ρ has an overall length L and a uniform cross-

sectional area A. Then the resistance R is

A

LR

When strain is applied to the strain-gauge it is either stretched or compressed.

This obviously changes the length of the strain gauge, but it also changes the

cross-sectional area. As a result the fractional change in resistance, α

R

R

where ∆R is the change in resistance R that is proportional to the applied

strain, ε. The constant of proportionality between α and ε is known as the

gauge factor

The gauge factor for a typical metallic strain gauge is 2. Once the gauge

factor is known the strain applied to a strain gauge can be calculated

from the resulting change in resistance.

To estimate the change in resistance that might occur assume a

maximum strain of 310

, i.e. an 0.1% change in length. With a typical

metallic strain gauge, G=2, the maximum fractional change in resistance,

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will be3102 . Since the unstrained resistance of a standard gauge will

be 100 Ohms, this corresponds to a change in resistance of 0.2 Ohms

(i.e. the change is 0.2% of the original value).

One method of converting this change of resistance to a voltage change is to

use a simple potential divider formed by a strain gauge and a reference

resistor. The resistance of this reference resistor is usually chosen to be the

same as the unstrained resistance as the strain gauge and a voltage is applied

across the strain gauge and reference resistor connected in series. The output

voltage from the circuit is the one shown in Figure (1).

Figure 1 A Bridge Circuit formed by one strain gauge and a standard resistor.

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With a voltage across the strain gauge and the resistor, the current

flowing through these two components is

00 1 RR

VI applied

This means that

00

00 1 RR

RVRIV applied

out

which reduces to

)2/1(22

appliedapplied

out

VVV

since α is small this can be approximated to

2

2/1 applied

out

VV

with zero strain and hence no change in resistance and so 0

a finite strain will then cause a change in the output voltage

If the gauge factor is 2 then for a maximum strain of 10-3 the fractional

change in resistance, , is 2 x 10-3. Assuming that a voltage of 10V is

applied to the resistances, then the maximum change in output signal for

a strain of 10-3 will be 5mV.

Since this is the maximum signal any accurate strain measurements can

only be made if the system is designed to reliably measure changes in

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output voltage of a small fraction of a millivolt. To obtain accurate

measurements the small changes in output voltage must be amplified.

This can be done using circuits containing operational amplifiers (op-

amps).

Characteristics of an ideal op-amp

A critical component of many circuits is the operational amplifier, usually referred

to as the op-amp. An op-amp has two inputs and one output and the symbol

used to represent an op-amp in a circuit diagram is shown in Figure (2). As this

figure shows this symbol is based upon a triangle with the output coming from

one apex of the triangle and the two inputs on the opposite side of the triangle.

The two inputs are known as the inverting and the non-inverting inputs of

the op-amp and in the symbol these inputs are distinguished by placing a

minus sign (-) near to the inverting input and a plus sign (+) near to the non-

inverting input. The op-amps also need to be connected to a power supply,

however for simplicity these connections are not usually included in either the

symbol or circuit diagrams.

Figure 2 The symbol often used to represent an op-amp in a circuit diagram.

Inverting Input

Output

Non-inverting Input

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The op-amp is designed to generate an output voltage many millions of

times larger than the difference between the two input voltages. Since it

acts as a voltage source the output from the op-amp can be represented by

a Thevenin circuit in which the open circuit voltage is a large multiple

(usually millions) of the difference between the two input voltages. This

multiple is known as the differential gain of the op-amp, so that

where Adiff is the differential gain and V+ is the voltage applied to the

non-inverting input and V – is the voltage applied to the inverting input. In addition

to the voltage source the Thevenin circuit also contains an output impedance Zo.

It may also be necessary to supply a small current to each of the inputs in order to

sustain, or change, the voltage applied to the two inputs. This feature of the

op-amps behaviour is represented by including an input impedance Zin between

the two op-amp inputs.

Figure 3 The equivalent input and output circuits for the inputs and output from an op-amp drawn

within the conventional circuit symbol for the op-amp.

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In this course we will deal with ideal op-amps. An ideal op-amp has three

important properties: an infinite input impedance, an output impedance of

zero and an infinite differential gain.

The first of these properties means that the current flowing into each

input of the op-amp is zero. This simplifies analysis of op-amp circuits.

The zero output resistance means that the op-amp can drive any load

impedance at its output. This means that we can analyse a circuit without

considering any other circuits that will be connected to the output of the

op-amp.

In the ideal op-amp the differential gain is approximately infinity. This

means that a negligibly small differential input will result in a finite output

voltage. Thus when the op-amp is included in a circuit with a connection

between its output and one of its inputs, a type of connection known as a

feedback connection, the output voltage will settle at the value required

to ensure that

In effect this means that an op-amp circuit with a feedback connection can be

analysed assuming that the voltage difference between the two inputs is

negligibly small so that

This means that during the circuit analysis the two inputs are indistinguishable.

However, as we will see op-amp circuits will only be stable if the feedback is

connected the inverting input. It is therefore critically important to connect the

feedback to the correct input.

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The Voltage Follower

The simplest circuit containing an op-amp is the voltage follower shown in

Figure (4). In this circuit the output from the op-amp is connected to the

inverting input of the op-amp and the input to the op-amp circuit is connected

to the non-inverting input of the op-amp. Because the output impedance of

the op-amp is zero the behaviour of this op-amp circuit is independent of the

load connected to the op-amp output and therefore there is no load shown in

the circuit diagram.

Figure 4 An op-amp connected to create a voltage follower.

The connection of the op-amp output to the inverting input means that

OUTVV

and the input connection means that

INVV

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Then since the op-amp is ideal,

VV

and so

INOUT VV

Since the output voltage is equal to the input voltage then the circuit seems

rather unnecessary. However, the infinite input resistance of the op-amp

means that this circuit draws no current into its input. If connected to an

input circuit with a large output resistance, then there will be no voltage drop

across the output resistance of the circuit connected to the op-amp input. In

addition the zero output resistance of the op-amp means that it will apply the

input voltage any circuit connected to the output, even one with a very low

resistance.

The importance of using the correct feedback connection

In the analysis so far the two inputs of the op-amp seem interchangeable. To

understand why it is critical to make the connections to the correct op-amp

input consider what happens when the input to the voltage-follower changes.

For this analysis consider the circuit in Figure (5) which includes the Thevenin

equivalent circuit for the op-amp output and a load capacitance.

Start by considering the current flow out of the output of the op-amp. The

voltage difference across the output resistance, Ro in Figure 5, is

outdOL VvA

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Figure 5 A voltage follower circuit loaded with a capacitor showing the Thevenin equivalent circuit at the output of the op-amp. Note the output is connected to the non-inverting

input.

This means that the current flowing through the output resistance is

o

outdOL

R

VvA

Since the input impedance of the op-amp is infinity, this current flowing

through the output resistance must charge or discharge the load capacitance

when the input is changing. This means that

o

outdOLoutload R

VvA

dt

VdC

If the output is connected to the non-inverting input of the op-amp and the

input voltage is applied to the inverting input of the op-amp (as in Figure (5))

then

inoutd VVv

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The differential equation for the output voltage is then

inOLoutOLout

loado VAVAdt

VdCR 1

Assuming that the initial input and output voltages are zero, Laplace

transforming this differential equation leads to

)()(1 svAsvAsvsCR inOLoutOLoutloado

But since the open loop gain is very large, and in particular much larger than

one, this simplifies to

)()( svAsvAsvsCR inOLoutOLoutloado

which can be rearranged to give

OLloado

inout AsCR

svsv

/1

)()(

To be comparable with the standard inverse transforms this can be re-written

as

loadoOL

in

loado

OLout CRAs

sv

CR

Asv

/

)()(

If the input is a unit step then from HLT page 14

ssvin

1)(

and so

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)/(

1)(

loadoOLloado

OLout CRAssCR

Asv

To make the comparison with the standard transformation easier let

loado

OL

CR

Aa

so that

)(

1)(

assasvout

From HLT page 14

ateaass

11

)(

1

which means that

ateass

a

1)(

1

For a unit step input this means that the response of the op-amp circuit is

atout etV 1)(

Which with the definition of a means that

tCR

A

outloado

OL

etV 1)(

This shows that in response to a unit step input the output rapidly approaches

minus infinity (at least in theory). The circuit is therefore unstable to any

changes to the input. This is an example of positive feedback, which must be

avoided (in almost all situations).

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Figure 6 A voltage follower circuit loaded with a capacitor

showing the Thevenin equivalent circuit at the output of the op-amp. Note the output is connected to the inverting input.

If the output is connected to the inverting input of the op-amp and the input

voltage is applied to the non-inverting input of the op-amp, to form the circuit

shown in Figure (6), then the differential input voltage is

outind VVv

The differential equation for the output voltage is then

inOLoutOLout

loado VAVAdt

VdCR 1

Assuming that the initial input and output voltages are zero and a very large

open loop gain this transforms to

)()( svAsvAsvsCR inOLoutOLoutloado

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which can be rearranged to give

loadoOL

in

loado

OLout CRAs

sv

CR

Asv

/

)()(

In this case

loado

OL

CR

Aa

Then in response to a unit step change in the input

tCR

A

outloado

OL

etV

1)(

This shows that with the output connected to the inverting input of the op-amp

the output voltage starts at zero and rapidly tends to 1, which is the same as

the input. With negative feedback (i.e. with the output connected to the

inverting input) the circuit is stable and can follow changes in the input

voltage.

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Non-inverting Amplifier

Next consider a circuit that can be used to amplify an input voltage so that the

output voltage is larger than the input voltage. One circuit that does this is the

non-inverting amplifier shown in Figure (7).

Figure 7 An op-amp and two resistors used to create a non-inverting amplifier.

To understand how the non-inverting amplifier in Figure (7) works assume that

the op-amp is ideal. Since this means that the input impedance of the

op-amp is infinite there will be no current flowing into the op-amp inputs.

This means that the two resistors form a potential divider and hence

but

and since the op-amp is ideal so that

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then

The circuit therefore acts like an amplifier with a gain of

So the gain of the circuit is determined by the resistors used in the feedback

circuit using with the op-amp.

Inverting Amplifier

Figure 8 An inverting amplifier circuit

An alternative amplifier circuit is the inverting amplifier, shown in Figure (8).

Again to understand the circuit assume the op-amp is ideal. Since the input

resistance (impedance) of the op-amp is infinite applying Kirchoff’s law at the

inverting input leads to

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and since the op-amp is ideal and there is a feedback loop

which leads to

i.e. the circuits has a gain of

Either of these two amplifier circuits can be used to increase the amplitude of a

signal.

A practical problem

Figure 9 The inverting amplifier circuit (again)

The changes in input voltages that are interesting are usually very small,

typically less than one millivolt, which is one reason why these voltages need

to be amplified. Either an inverting or a non-inverting amplifier could be used,

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however, a more detailed consideration of these circuits uncovers a problem.

Consider the diagram for an inverting amplifier circuit in Figure (9), which

explicitly shows that the ground contact for the input voltage is different from

the ground contact for the inverting amplifier. As before assuming that the

op-amp is ideal, so that the input current into the op-amp is zero, then applying

Kirchoff's Law to the op-amp inverting input gives

Since the op-amp is ideal and there is feedback

and hence the expression for the op-amp output voltage becomes

If the non-inverting input to the op-amp is grounded correctly, that is it

shares the same ground as the input voltage, then and the

circuit acts as an inverting amplifier as expected. However, if this

assumption is not valid then changes in the voltage of the non-inverting

input will be indistinguishable from changes in the input voltage.

Grounding a circuit correctly sounds simple but consider the simple schematic

grounding scheme on the left-hand side of Figure (10). For simplicity this

diagram shows a system divided into three functional blocks, an input amplifier

(such as the one in Figure (9)), some other op-amp circuits and all the digital

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circuits. The ground connections of these three parts of the system are then

simply connected together on the circuit board before a single connection is

made to earth. A hidden danger with this simple grounding scheme is that

each connection on the circuit board has a small but finite resistance.

These resistances are represented by , and in Figure (10),

which shows that this grounding scheme creates common resistances to

ground. The problem caused by these common resistances (or more

generally impedances) is that the current flowing to ground through one

circuit can change the ground potential of other circuits.

Figure 10 Schematic diagrams of a poor earthing scheme, on the left, and a good earthing scheme, on the right.

In the input amplifier the resulting variations in the ground potential will

be equivalent to an equal, but opposite, change in the input signal. The

common resistances to ground therefore create a potential source of

interference. Since the shared resistances created by this grounding scheme

are formed by contacts and tracks that have an ideal resistance of zero, it is

understandable to think that this effect is negligible. However, the total

resistance that is shared could be 10 Ohms. In this situation it would only

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require a current of 50μA to flow through this shared resistance to change the

'ground' potential of the input amplifier by 0.5mV. This seems small compared

to the voltages used to power the circuit but it is a significant part of the

changes in the input voltage. It is therefore quite possible for changes in the

current drawn by the digital part of the circuit to create a fluctuation in the

'ground' potential of the amplifier circuit that will be interpreted as a significant

change in the input voltage.

The potential for this type of interference must be avoided by careful

design of the grounding scheme. The coupling caused by the shared

earth tracks can be avoided by using a single-point grounding scheme,

often referred to as a star connection or star grounding scheme, shown

on the right hand side of figure (10).

Whilst devising a star grounding scheme it is also important to avoid creating

multiple ground paths from a single unit. The problem is that any multiple path

could form an aerial, which could then couple with any radiated

electromagnetic fields in the environment. This can result in large time-varying

currents flowing to through the ground tracks that could interfere with the

amplifier ground.

At low frequencies it is relatively easy to avoid creating multiple ground

paths. However, at higher frequencies, above 10MHz, parasitic

capacitances between different tracks have the potential to form ground

loops.

There are two methods for reducing radiative coupling into ground loops. The

first is to carefully design the earthing scheme for a particular system to avoid

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capacitive coupling whenever possible. However, in many situations it is

easier to avoid this particular problem by enclosing the circuit in a metal,

or metal coated plastic, box to prevent external radiation from

penetrating into the space occupied by the circuit.

Once a system has been shielded from external radiation the only

remaining problem is to avoid creating any electromagnetic

radiation within the enclosed, protected space. The two commonly

occurring potential sources of radiation within the enclosure are

any mains connections and switching transients within digital

circuits. To avoid interference from any high frequency signals on

the mains supply it is advisable to fit a mains filter close to the

point at which the power lead enters the enclosed space. Any

problems with transients in the current flowing into digital circuits,

which might cause interference by radiation or via a shared earth

connection, can then be minimised by placing a small capacitance

between the two power supply connections of each digital circuit.

To reduce the amount of radiation which may occur this capacitance,

which is typically 10-100nF, is placed as close as possible to each digital

circuit.

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The Summing Amplifier

In some situations, including within the digital-to-analogue converters described

later in this course, an ability to add two voltages is useful. This can be achieved

using the inverting summing amplifier shown in Figure (11).

Figure 11 An inverting summing amplifier

To understand how this circuit operates assume that the op-amp is ideal. Since

the op-amp is ideal , but, in this circuit and so the current

flowing into this circuit node from the two inputs is

Since no current flows into the inverting input of the ideal op-amp (because it has

an infinite input impedance) all this current must flow around the feedback loop

through resistor .

This will only happen when the op-amp output voltage is

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which becomes

Note that the inverting amplifier is a special case of this circuit which has only

one input.

The differential amplifier

Another useful circuit is one that amplifies the difference between two voltages.

A simple op-amp circuit that can amplify the difference between two voltages,

usually known as a differential voltage, is shown in Figure (12).

Figure 12 Single op-amp differential amplifier Again, to obtain an expression for the output voltage, assume that the op-amp

is ideal. The current flowing through the resistors connected to the inverting

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input of the op-amp is

The right hand pair of equations can be re-arranged to give

The non-inverting op-amp input is connected to the second input via a

potential divider and since the op-amp input has an infinite input resistance

However, the infinite gain of an ideal op-amp and the feedback path to the

inverting input mean that and the above two equations led to

which then gives

The output voltage is therefore proportional to the differential input

voltage, , with a gain of

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The most surprising aspect of this result is that the output of this circuit

is independent of the average input, usually referred to as the common

mode input.

To see how this arises consider a situation in which the two inputs are

equal. With a feedback loop the op-amp will generate the output voltage

that ensures that its own two input voltages are the same. Since the two

resistors connecting the circuit input to the op-amp inputs are identical and the

voltages at the two ends of each resistor are also identical, the same current

will flow through both input resistors, i.e. the two resistors labelled . Now the

input impedance of the ideal op-amp is infinite and hence the current in each

branch of the circuit will flow through the resistors labelled . With the same

current flowing through identical resistors Ohms Law means that there will be

an identical voltage drop across both these resistors. Since the two op-amp

inputs are at the same voltage, this means that the output voltage is zero,

i.e. the output is independent of any common-mode input. This analysis

clearly shows that this situation arises because two sets of identical

resistors have been used.

In the above analysis of the ideal circuit it has been assumed that there are

two sets of identical resistors. However, no two resistors are identical, and

any difference between two ideally ‘identical' resistors will destroy the

symmetry that ensured that the output is independent of the common-mode

input. In this situation the output voltage will depend upon both the difference

between the two input voltages and their average value, usually referred to as

the common mode voltage. In general this means that the output voltage from

the circuit can depend upon both the differential and the common mode

voltages, so that

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where is the differential gain of the amplifier circuit and is the

common mode gain. For the ideal differential amplifier the common mode

gain is zero, , so that the output is independent of the common

mode signal. Non-ideal effects, such as differences between the resistances of

‘identical' resistors, mean that the common-mode gain of a real differential

amplifier is finite.

The ability of a differential amplifier to reject any common mode signal

whilst amplifying the differential signal, is usually characterised as the its

common-mode rejection ratio (CMRR)

Since ideally the common-mode gain is zero, the common mode rejection ratio

of an ideal differential amplifier circuit is infinity. Despite non-ideal effects the

CMRR of a real circuit is usually large and it is therefore often quoted in

decibels

In principle, individual resistors can be individually selected to achieve a

particular CMRR. However, even a circuit which has been hand-crafted in

this way has a potential problem: In particular, an ideal voltage

measuring system draws no current whilst performing a measurement,

i.e. it has an infinite input resistance. Unfortunately, the input resistance

of this differential amplifier is finite, because a finite current flows

through the two input resistors.

- 31 -

The input resistance (impedance) of the circuit can be increased by

simply inserting a unity-gain op-amp voltage follower circuit on each

input of the differential amplifier as shown in figure (13). The input to the

circuit is then directly connected to one input of an op-amp that has a very

high, if not infinite, input impedance. However, if the op-amps are ideal

which means that the input signals are applied to the differential amplifier as

before. Adding op-amp buffers to the standard differential amplifier therefore

simply increases its input resistance (impedance). However, the circuit still has

to be hand-crafted, by carefully selecting the resistors, to achieve an

acceptable CMMR.

Figure 13 A differential amplifier with buffered inputs to ensure that the input

impedance is very high (ideally infinity).

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Instrumentation Amplifiers

The ideal differential amplifier circuit has a high input resistance, a large

common-mode rejection ratio and an easily controlled differential gain.

These characteristics are achieved in the standard 3 op-amp

instrumentation amplifier shown in Figure (14).

Figure 14 The standard 3 op-amp instrumentation amplifier.

To understand the circuit first consider the current flowing vertically through the

resistors connecting to . Starting from the top of this chain the three

different resistors give three different expressions for the current. Since the

input impedance of the two op-amps is infinity the current flowing through

these resistors must be the same and so

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The middle two equations can be re-arranged to give

and the right hand pair give

Then subtracting the first of these two equations from the second gives

i.e. a differential gain of

To calculate the common mode gain consider the situation in which

. In this situation symmetry means that there will be no

current flowing in the resistors connecting to and hence

. This means that the common-mode gain is of this circuit

is 1.

The input stage can provide a high differential gain (depending upon the

correct choice of resistors) and unity common-mode gain. This is

achieved with only one pair of identical resistors that are supplied by the

manufacturer in the same package as the three op-amps. The

manufacturer therefore ensures that these two resistors are identical and the

user then achieves the required differential gain by selecting a single, resistor

.

- 34 -

The output from this input stage is fed to a differential amplifier circuit.

This last stage is often given a fixed low differential gain (unity, for

example). The advantage of this is that all the resistors within this part of the

circuit can be integrated within the instrumentation amplifier package. Again

matching to the required accuracy is then achieved as part of the

manufacturing process. With a low differential gain the main function of the

differential amplifier is to provide a single output whilst rejecting any common-

mode input signal.

Now the CMRR of a circuit in dB is

In this circuit there are two stages so the overall CMRR of the instrumentation

amplifier is the product of the two CMRR’s. If the common-mode gain of the

first stage and the differential gain of the second stage are both unity then

i.e. the CMRR of the output stage multiplied by the differential gain of the

input stage. This shows that the overall high CMRR is achieved by combining

a high differential gain in the first stage with a relatively modest rejection

(attenuation) of the common-mode signal in the second stage.

Instrumentation amplifiers have very high input impedances (because the

input stage consists of two cross-coupled followers). With a single external

component to set its differential gain the three op-amp instrumentation

amplifier is also a convenient component that is very easy to use and

therefore very popular.

- 35 -

Current-to-Voltage Conversion

The inputs to the circuits that have been described are all voltages. In some

applications the interesting signal is a current rather than a voltage. In these

situations a circuit that converts a current to a voltage is useful.

Figure 15 A current to voltage conversion circuit

One circuit that can be used to convert a current to a voltage is shown in

Figure (15). If the op-amp is ideal then the input resistance will be infinity

and so all the input current flowing into the inverting input must flow

through the resistance LOADR . This means that

LOAD

OUTIN R

VVI

Again since the op-amp is ideal and there is feedback between the

op-amp output and an input

VV

- 36 -

and since the non-inverting input is grounded

0V

This means that 0V and so

LOADINOUT RIV

which shows that the output voltage is proportional to the input current.

This simple circuit has a limitation. The largest resistor value that is used

in a typically circuit is 1MΩ=106Ω. If the input current is 1μA=10-6A then

the output voltage will be -1V, which is acceptable. However, there are

some applications in which the current that needs to be detected can be

smaller than 1pA=10-12A. For a current that is this small the output

voltage would be -10-6 V=-1μV, which is far too small. In these situations

the circuit in Figure (16) should be used.

Figure 16 A circuit to convert current to voltage by integration.

- 37 -

The voltage across the capacitance in the feedback loop is proportional

to the charge on the capacitance, so that

fbout C

QVV

where Q is the charge stored on the capacitor. Again the non-inverting input is grounded and the op-amp is ideal so that

0V and therefore

fbout C

QV

The output voltage is therefore proportional to the charge on the

capacitor. To relate this to the input current, remember that

dt

dQI

which means that

dtIdQ Assuming that the charge (and hence the output voltage) starts from zero at a time t=0 then

dItQt

0

)(

Applying this relationship to the circuit in Figure (16) gives

- 38 -

fb

t

in

fbOUT C

dI

C

QtV

0

which shows that the output voltage at a particular time is the

integral of the input current since the time at which the output

voltage was zero. Not surprisingly this circuit is therefore known as

an integrator.

If the input current is constant then

fb

inOUT C

tItV

assuming that the capacitance is 1pF=10-12F and that the input current is

1pA=10-12A, then after 1 second the output voltage will be 1V.

In principle the output voltage will continue to increase with time. In order

to measure the current at different times it is necessary to force the

output voltage to zero when required. So that this can be achieved a

switch is put in parallel with the capacitor, see Figure (17). When the

switch is closed the output is shorted to inverting input, and since the

inverting input of this ideal op-amp is grounded then the output voltage

will be zero. Integration will then start when the switch is opened and the

output is no longer shorted to the inverting input so that the op-amp

begins to integrate the current onto the feedback capacitor.

- 39 -

Figure 17 An integrator with a switch to allow the output

voltage to be zeroed.

- 40 -

Circuits with more than one op-amp

To achieve the required functionality complete systems often contain a

series of op-amp based circuits connected in series. A simple example,

containing only component circuits that you know, is shown below. The

critical first step to understanding this circuit is to identify the points at

which you can split the system to identify the constituent parts. Can you

identify the constituent parts of the circuit below, including the passive

circuit between the two op-amp based circuits?

- 41 -

Summary

There are at least two broad classes of applications, wireless communications

and instrumentation and control systems, in which it is important to detect

small continuously varying electrical signals.

These small continuously varying signals must be amplified and filtered so that

they can be sampled and converted to the digital format commonly used to

record, transmit and store information.

The circuits used to amplify and filter analogue circuits are often based upon

operational amplifiers (op-amps). These circuits have two inputs and one

output. One input is known as the inverting input and the other input is known

as the non-inverting input. The output voltage is proportional to the difference

between the two voltages applied to these inputs.

An ideal op-amp has three important properties. It has an infinite input

impedance, so that no current flowing into the op-amp inputs. It has an infinite

gain. It has a zero output resistance, so that it can provide a large current

without losing any output voltage.

When the output from the op-amp is connected to its inverting input the infinite

gain means that the op-amps output voltage will ensure that the voltages at the

two inputs are identical. If the output is connected to the non-inverting input the

circuit is unstable.

An op-amp and resistors can be used to create circuits that amplify an input

voltage or the sum of several input voltages.

- 42 -

The difference between two voltages is amplified by a differential amplifier. The

ability of a differential amplifier to amplify the differential voltage whilst rejecting

the average (or common) voltage is characterised by the common-mode

rejection ratio (CMRR).

A differential amplifier can be made with a single op-amp however the CMRR

of the circuit is limited by the fact the resistors that should be identical will

actually have slightly different values. A more popular differential amplifier is

the three op-amp instrumentation amplifier. In this two stage circuit the first

stage amplifies the differential input whilst the second stage rejects the

common-mode signal. The user only needs to calculate the value of one

resistor. This simplicity makes the circuit very simple for users.

There are two methods that can be used to convert a current into a voltage. An

op-amp and resistor can convert the signal continuously if it is large enough.

Small currents need to be converted by integrating the current onto a capacitor

for a finite time. The output voltage then represents the average current during

this time.

When analysing a large circuit containing more than one op-amp the circuit

should be divided into its component parts. The first stage of this is to divide

the circuit at the output of each op-amp. Each component circuit can then be

analysed separately. For op-amp circuits connected in series the overall

response will be the product of the responses of each circuit.