Programable Logic Controller questions. PLCS

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MODULE TITLE: PROGRAMMABLE LOGIC CONTROLLERS

TOPIC TITLE: INTERFACING

LESSON 2: INPUT INTERFACE DESIGN

PLC - 3 - 2

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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published and without a similar condition including this

condition being imposed on the subsequent purchaser.

________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

The previous lesson in this topic ended by covering aspects of the external

requirements of input interfaces. In this lesson we shall examine the practical

design of a digital input interface. The electronics, of necessity, must be

considered but it is hoped that the treatment will not be too dry for 'non-

electronics' students whilst hopefully being technical enough to be acceptable

to those within the electronics industry.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

On completing this lesson you should be able to:

• appreciate the factors governing input interface design

• identify additional circuit diagram symbols

• describe the function and operation of certain electronic components

• show further knowledge about isolation and coupling

• acknowledge the differences in d.c. and a.c. input requirements

• read and understand input interface circuit diagrams provided in

manuals supplied by manufacturers.

________________________________________________________________________________________

STUDY ADVICE ________________________________________________________________________________________

Circuit diagram symbols are to BS 3939 wherever possible.

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________________________________________________________________________________________

DIGITAL INPUT INTERFACE ________________________________________________________________________________________

The previous lesson covered aspects of connecting simple switches to the input

interface terminals of a PLC. We can now consider possible circuitry which

could form the input interface and examine how some of the problems

discussed in previous lessons have been overcome.

To refresh your memory, here is a list of the functions of an input interface.

An input interface is required to:

• provide isolation between the outside world and the sensitive

microelectronics of the controller

• provide a digital-type signal with the correct voltage levels for the

input interface adaptor

• provide its own internal supply for use by simple switch inputs (d.c.

circuits only)

• provide an indication on an input panel that an input signal is present

(normally by the use of a Light Emitting Diode)

• provide for the possibility of voltage drop along the input cables.

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________________________________________________________________________________________

DESIGN CONSIDERATIONS ________________________________________________________________________________________

An engineer who can understand a manufacturer's design will stand the best

chance of analysing what is likely to be happening when things appear to be

going wrong.

One of the first design considerations is the determination of the value,or

values, of voltage which should be used for the circuitry.

The signals which will be sent from the input interface to the interface adaptor

are well defined by the manufacturer of the adaptor. In our case we are

assuming a fairly safe standard value of +5 V and 0 V.

Shoud this 5 volts be used as the supply to the switches?

....................................................................................................................................................

________________________________________________________________________________________

The answer should be NO.

There are two reasons for this.

(a) The interface needs to provide isolation between the microelectronics and

the outside world. If the same supply is used for both then common

connections will be needed and isolation would be impossible.

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(b) A 5 V supply used for the input wiring would not allow for very much

external voltage drop along the switch cables.

These two reasons point to the use of two values of voltage:

(a) a 5 V supply for the interface adaptor signals

and

(b) a larger voltage (hopefully not dangerously large) for the external wiring

to the input sensors. A 24 V d.c. supply would be suitable (and is, in fact,

quite common).

For the external circuit two parameters have now been chosen:

• a 24 V d.c. supply for the switches

• a value of switch current of about 10 mA (0.01 A).

This value of current was mentioned in Lesson 1. It seems a sensible value to

choose because the larger this current is the more voltage drop can be

expected. Remember that the PLC manufacturer does not know how long the

input cables will be but by keeping the input current small the anticipated

voltage drop should also be small.

If 24 volts exists across an input switch contacts and that switch is closed what will

happen to the 24 volts?

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________________________________________________________________________________________

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The voltage will be dropped across some other circuit component due to the

value of circuit current flowing through the switch. If the circuit has no other

components then a short circuit would exist.

What other circuit component (or components) are we talking about?

....................................................................................................................................................

________________________________________________________________________________________

It has been mentioned that the switch current flows through an LED which

illuminates to show that the switch is closed. However, an LED carrying

10 mA will have a forward voltage drop of only between 2 V and 2.4 V – this

is a long way short of 24 V! Some other components will be needed.

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________________________________________________________________________________________

LIGHT EMITTING DIODES ________________________________________________________________________________________

If you have no experience of LEDs then this section may serve to fill-in

missing detail. If you think you already know enough about LEDs then go on

to the next main heading. Questions may be asked later so if in doubt don't

skip this section!

A semiconductor diode is a component which is specifically designed so that

its current flow (or conduction) properties are different depending upon the

direction in which current is trying to flow.

If you consider a length of copper wire in which current is flowing then it

makes no difference to the value of the current if it flows in one end and out of

the other or vice versa. This is because the wire has the same electrical

resistance in both directions. A diode, however, does not have the same

resistance in both directions. If current flows in one direction the resistance

may be low (a low number of ohms), if, on the other hand, the current flow is

in the opposite direction then the resistance may be many times larger (a large

number of ohms). An ideal diode would have zero resistance in one direction

and an infinitely large resistance in the other direction. It is easy for a current

to flow in the low resistance direction and therefore this value of current may

be relatively large, but it is difficult for a current to flow in the high resistance

direction and so this value of current will be small.

Well, that is all relevant to a semiconductor diode but how does it relate to an

LED?

An LED is a semiconductor diode. The semiconductor material may be

something such as gallium arsenide. When a current flows in the low

resistance direction across a junction formed in gallium arsenide an emission

of photons of light is given off. Hence the name of light emitting

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diode(LED). Different coloured LEDs are constructed with different

compositions of the semiconductor material and require different values of

voltage across their terminals to make them light up. The current required for

the emission can be as low as 2 mA and when increased will increase the

brightness of the LED. The maximum value of current specified by the

manufacturer must not be exceeded otherwise the device will glow brilliantly

and then burn out.

One difference between LEDs and other diodes is that the maximum safe value

of voltage in the reverse direction (i.e. when trying to push current in the high

resistance direction) is very low. Values of 3 V and 5 V are very common.

If these values are exceeded then the LED will again become damaged.

The safe direction of current flow is into the anode (A) terminal and out of the

cathode (K) terminal. FIGURE 1 shows the BS symbol for an LED.

FIG. 1

The current flow direction is indicated by the arrow of the symbol. To

distinguish an LED from other diodes the emission of light is shown coming

away from the symbol.

Anode (A)

Cathode (K)

Direction of current flow

Light

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________________________________________________________________________________________

OPTOCOUPLED ISOLATORS ________________________________________________________________________________________

At this point we need to consider how we can achieve the isolation between

the 24 volt supply being used for the input wiring and the 5 volt supply for the

interface adaptor signals. One common method used by PLC manufacturers

involves incorporating into the circuit an opto-isolator device. Opto-isolator

devices are available in many forms. What we will briefly look at is the

operation of an optocoupled transistor. You should note that the description of

operation given here is a basic one and is not a description of the physics of its

operation.

FIG. 2

The optotransistor represented in FIGURE 2 has two distinctly separate parts:

an LED (considered earlier) and a phototransistor. The phototransistor is used

as a switch. The simplest explanation is that any flow of current through the

transistor must be via the collector (c) terminal and the emitter (e) terminal

(this must be arranged by the connections of the supply external to the device,

i.e. the correct polarity).

A c

K e

Phototransistor

Plastic encapsulation

Currents to enter by these two pins

LED

Connecting pins

light

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If no current flows in the LED then the phototransistor is effectively switched

off and no current can flow through it. If, on the other hand, the LED carries a

current from its anode (A) to its cathode (K) then the phototransistor is

switched on and a current is allowed to flow through it in the forward

direction.

Both component parts are encapsulated within a light-proof plastic package but

there is no electrical connection between them. They are therefore electrically

isolated. The value of insulation between the two components is often

expressed, not in ohms as might be expected, but in terms of the value of

voltage which can be tolerated between the components. The value is called

the isolation voltage. Values of isolation voltage are much higher than the

maximum working voltage across the components. The coupling medium is

the beam of light which is emitted from the LED when it carries current.

Photons of light from the LED fall on the phototransistor and have the effect of

switching it on.

Opto-isolators are so well coupled that the LED requires very little current

(around 2 mA) to switch the phototransistor on. This value of diode current is

obtained with only about 1.6 V correctly polarised across the LED.

If the indicating LED were connected in series with the opto-isolator LED would this

account for the 24 V input supply?

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________________________________________________________________________________________

The answer is NO.

If the two were series connected then their combined voltage drop would be

2.4 V + 1.6 V = 4 V.

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FIG. 3

If 24 V were connected then both of these devices would burn out unless some

other component limited the current to a safe value (10 mA in this case). The

other component would need to be a resistor and this resistor would need to

drop (24 V – 4 V) = 20 V whilst carrying 10 mA. Using Ohm's law the

value of this resistance is

Part of the input circuit can now be drawn.

20 10

20 000

2000 2

= =

V mA

mV 10 mA

kΩ Ω

2.4 V

1.6 V

LED

10 mA

4 V

10 mA

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FIG. 4

There is still one problem with this part of the circuit. The switch current is

10 mA but only 2 mA is needed thr ough the opto- isolator L E D.

Therefore 10 mA – 2 mA = 8 mA will need to be diverted around this

device by the use of another resistor of value:

With the input switch closed the circuit values will be as shown by FIGURE 5.

1 6 8

1600 8

200

. V mA

mV mA

Ω

Common

Input conductors

External wiring

Interface

R1 2 kΩ

+ 24 V

10 mA (When switch is closed)

0 V

Input switch

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FIG. 5

The next part of the design is the 5 V circuit. The phototransistor is merely

required to switch on and off a 5 V supply. The load for this circuit can be

resistive. A connection is tapped off the collector terminal so that an input

signal can be sent, via a current limiting resistance to the interface adaptor port

pin.

Common

External wiring

Interface

0 V

10 mA

+ 24 V 10 mA (When switch is closed)

20 V

2.4 V

Input switch closed

1.6 V

10 mA

R1 2 kΩ

2mA 8 mA

R 2

200 Ω

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FIG. 6(a) FIG. 6(b)

The operation of this part of the circuit is such that the adaptor receives a +5 V

(logic '1') supply via the resistors in series when the phototransistor is not

switched on. When the phototransistor is switched on (by the light beam from

the input circuit) the voltage at the collector drops well below 5 V. The voltage

level reaching the interface adaptor is interpreted by the CPU as being a logic

'0'.

A capacitor may be necessary across the adaptor pins to filter out any

fluctuations in the switching signal.

Such a capacitor introduces a delay between the switching action of the

transistor and the associated change in the logic level of the interface adaptor

signal. Notice in this arrangement the interface adaptor receives a logic 'l'

when the external switch is open and a logic '0' when the external switch is

closed. FIGURE 6(a) and FIGURE 6(b) are equivalent in operation.

0 V Equivalent

circuits

+5 V

CSwitch

+5 V

0 V

To Input Interface Adaptor

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________________________________________________________________________________________

TRANSISTORISED INPUT SENSORS ________________________________________________________________________________________

The input switching devices (defined as 'simple' and listed in Lesson 1) have

all been of a mechanical operating type, that is to say, they are all types which

require some physical contact to move the switching mechanism.

e.g. A push-to-make switch requires someone to push it.

A limit switch requires a lever or bar to move the pawl to operate the

switch.

Sometimes, however, a switching action is required without any physical

contact taking place.

Consider, for example, items passing along a conveyer belt travelling past a

point where they are required to be counted. The movement of the items must

not be impeded by any physical contact. In this type of application a switching

detector is necessary. This detector could be a photosensitive device or an

inductive or capacitive proximity detector.

Such devices detect items passing by their sensitive face and do not require

any physical contact.

An examination of commercially available detectors reveals that the output

stage is quite often a switching transistor. That is to say, a transistor which acts

like a switch.

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FIG. 7

Well, if our input interface circuit works with a switch then it should work with

a switching transistor.

Can we expect it to work with any switching transistor?

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Unfortunately the answer is NO.

If we re-examine the circuit design of FIGURE 5 we will notice that when the

switch closes 10 mA of current can be expected to flow from the input

terminal, through the switch and back into the common terminal. (Shown

again in FIGURE 8.)

Input detector

Switching transistor

Input terminal

Common

External wiring

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FIG. 8

If a transistor is to replace this switch then the following conditions must be

met.

(a) It must allow a current to flow in this direction.

(b) The transistor must be conducting well enough for the resistance

between the collector and the emitter to allow sufficient current to

flow. Remember that this current is used to illuminate both LEDs

which are connected in series. If the conducting resistance is too

large then insufficient current will flow and the detector signal will

not be recognised by the PLC.

When the input switch of FIGURE 8 opens no current will flow but when a

switching transistor is turned off a small value of current may still get through.

In this case it is important to ensure that this small current is not enough to be

detected as an input signal.

For the circuit design considered any transistor switching device should have a

recommended 'off' resistance which is greater than 10 000 Ω (>10 k) and an

'on' resistance which is less than 1000 Ω (< 1 k).

Capacitive and inductive proximity detectors are not cheap items. You will

appreciate the waste of time and expense in sending for devices which are not

suitable for the PLC interface being used. It is important to check the PLC

manufacturer's requirements with the detector manufacturer's specifications

before the detectors are purchased.

Input switch closed

Input terminal

Common

10 mA

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________________________________________________________________________________________

A.C. SWITCHED INPUTS ________________________________________________________________________________________

The points covered so far in this lesson have concentrated on the requirements

of d.c. switched inputs. It is also possible, by suitable interface design, for the

PLC to recognise an a.c. input signal.

The design of FIGURE 5 cannot be used to accept a.c. A different design,

specifically for a.c., within an acceptable voltage and frequency range needs to

be examined.

In this examination, we shall take a look at a PLC manufacturer's design noting

the points and components which are common to the d.c. circuits of FIGURE 5

and FIGURE 6 and explaining those items which are different.

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FIG. 9

With reference to FIGURE 9 make a list of any components which you

recognise from the lessons so far.

...............................................................................................

The correct list is given on page 23.

85 V – 132 V a.c.

Input switch

Input terminal

Interface

External wiring

C B

D2 D1 RL

OP1

+5 V d.c.

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________________________________________________________________________________________

CIRCUIT OPERATION ________________________________________________________________________________________

The manufacturer of this input interface has not provided the supply for the

input switch.

This is probably because a.c. supplies are so readily available within plant that

it is unnecessary for an internal supply to be used.

In this country that supply is likely to be obtained from a step down

transformer which has its primary side supplied from the public mains. The

operating frequency would therefore be 50 hertz. This means 50 cycles of

alternating current take place every second. The current flow is in one

direction for 0.01 seconds and then in the opposite direction for 0.01 seconds.

The direction of current is thus alternating by changing its direction at regular

times.

In the d.c. input interface circuit the direction of the current flow was in one

direction only.

Let us examine how this changing direction (or alternating) current affects the

circuit design.

The manufacturer has connected a capacitor (C1) in series with two resistors

(R1 and R2).

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When the input switch closes, which components will carry the same value of current

as the switch?

....................................................................................................................................................

________________________________________________________________________________________

With the switch closed and for half a cycle of a.c. the current from the input

switch can be assumed to be flowing into the input terminal. This current will

pass through C1 and R1. However, at the junction (or node) point A this

current will split. Some of the current flows through R2 and some flows

through D3 in series with D2. The two currents will combine at junction point

B to flow out into the external circuit. With the current flowing in this

direction it does not flow through either LED and so neither is illuminated at

this time. More importantly, however, the diodes D3 and D2 are fitted so that

the reverse voltage stress across the LEDs is kept to a very low value of

voltage. (If D3 and D2 are silicon diodes this reverse voltage will be the diode

forward voltage drop of about 0.6 V across each diode.) Remember that LEDs

which are made from a semiconductor material, such as gallium arsenide, can

only withstand about 5 V in the reverse direction before irreparable damage

takes place.

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During the next half-cycle the current reverses direction and will flow into

terminal C. The current splits at junction B, some of it will flow through R2 and some through the LED (D1) in series with the LED of OP1. The two

currents will recombine at point A and flow through R1 and C1 to complete the

current path back to the external circuit. For the duration of this half cycle the

current flow would last for approximately 0.01 seconds and during this time

the LED would glow and the phototransistor would be switched on.

The behaviour of the LEDs with the input switch closed is that they will

illuminate for about 0.01 seconds and be extinguished for about 0.01 seconds,

i.e. they will flash on and off. This flash rate is too fast to be detected by the

human eye and the LED will appear to be ' on' all the time. T he

phototransistor, carrying the same current will switch on and off. If the CPU

looks at this signal level when the phototransistor is switched off then it will

interpret this as meaning that the input switch is open. The R-C network of RS and C2 must filter or smooth out this on/off transistor switching action so that

the CPU does not read an incorrect signal level.

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________________________________________________________________________________________

A.C. OR D.C. INPUT CIRCUIT ________________________________________________________________________________________

It is possible to design an input interface which will accept either a.c. or d.c.

external signals within suitable voltage ranges.

The block diagram of FIGURE 10 shows a possible arrangement.

FIG. 10

The a.c. or d.c. signal is fed into the bridge rectifier which ensures that

unidirectional current is passed to a smoothing filter. A proportion of this d.c.

is applied to a stabiliser which gives a constant voltage level output. This

voltage level is used as the input to the isolation device. The digital signal

being obtained from the isolated switching components is as in the previous

explanations.

Smoothed d.c.

Rectifier

Filter

Potential divider

& stabiliser

d.c. Low voltage

d.c.

Isolation device

Logic output 5 volts d.c.

Logic signal to

microelectronics

a.c.

Input

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This lesson has covered three basic versions of interface circuitry: d.c. input,

a.c. input, and a.c. or d.c. input. It is very probable that the majority of people

who work with PLCs will know almost nothing about this circuitry and yet, as

long as the manufacturer's instructions have been followed their use of the

PLC will be satisfactory. You must not make the mistake of thinking that these

three versions are the only possibilities. No standard exists as each

manufacturer will make his own design. Hopefully you will, should the need

arise, be able to understand a little better the circuit diagrams supplied with the

PLCs you are required to use. You should also appreciate that sometimes PLC

users add on external interface circuits to cover those circumstances where

their signals, from existing plant, do not match the requirements of the PLC

being used. Coverage of such possibilities is outside the scope of this lesson.

Now attempt the Self-Assessment Questions which begin on page 24.

________________________________________________________________________________________

LIST OF COMPONENTS FROM PAGE 18 ________________________________________________________________________________________

D1 – Light emitting diode.

D2 & D3 – Diodes.

OP1 – Opto-isolated transistor.

R1, R2, RL & RS – Resistors.

C2 – Capacitor.

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. Identify, by naming, each of the circuit diagram symbols shown below.

FIG. 11

2. Explain how two circuits can be coupled and yet remain electrically

isolated.

3. State a good reason why two different values of voltage should be used

within a d.c. input interface circuit.

4. Specify a typical value of input current required by a d.c. input interface

circuit.

5. Explain the term 'isolation voltage'.

(a)

(c)

(b)

(d)

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6. If a transistorised proximity detector is to be used as an input sensor

instead of a simple switch, then list the factors which need to be

considered before the detector is chosen.

7. FIGURE 12 shows, in block diagram form, an arrangement of

components which will act as an a.c. or d.c. input interface. Briefly

explain the significance of the operation of blocks A, B, C and D.

FIG. 12

Smoothed d.c.

Rectifier

Filter

Potential divider

& stabiliser

d.c. Low voltage

d.c.

Isolation device

Logic output 5 volts d.c.

Logic signal to

microelectronics

a.c.

Input A C

B D

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (a) LIGHT EMITTING DIODE OR LED.

(b) OPTO-ISOLATED TRANSISTOR.

(d) PUSH-TO-MAKE SWITCH.

2. If the two circuits are said to be electrically isolated then this means that

there is no electrical connection between the circuits. If electrically

isolated circuits are coupled then this means that the operation of one

circuit will have an effect of some kind upon the second circuit. The

lessons covered so far have used electromagnetic coupling of relays and

photons from a light beam as examples of coupling mediums.

3. One voltage needs to be matched to the signal requirements of the input

interface adaptor. This is most often a +5 V for logic 'l' and 0 V for logic

'0'. The other voltage could also be 5 V but a larger value (typically 24 V

d.c.) gives more flexibility in allowing for external voltage drop and

simpler input interface circuit design.

4. In theory almost any value of current could be stated. The input interface

would need to be designed to suit whatever the stated value is. However,

a large value of current for each input will require the manufacturer to

provide a power supply large enough and this will add to the cost of the

PLC. Additionally, large values of input current flowing in the external

cables may produce unacceptable voltage drops and power wastage. A

small value of current such as 10 mA or 20 mA is therefore much more

acceptable.

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5. The term 'isolation voltage' is used by manufacturers of electrically

isolated, coupled devices to specify the maximum value of voltage which

the two isolated components may have between them. Typical values

would be in hundreds or even thousands of volts. This is a manufacturer's

indication of the 'quality' of the insulation between the components. It in

no way implies that these values of voltage should be considered for

practical use.

6. The detector must have operating characteristics which are matched, as

closely as possible, to the requirements of the PLC interface.

Factors to be considered include the following.

(a) When switching, the direction of current must be correct.

(b) The 'on' resistance must allow sufficient current to flow so that the

input to the PLC will be recognised.

small to be recognised by the input as an 'on' signal, which means

that it will be recognised as an 'off' signal instead.

(d) When the detector is off the value of d.c. voltage from the interface

circuit must not be so large that the transistor becomes damaged.

7. A is shown as a bridge rectifier. This block accepts either a.c. or d.c.

inputs and ensures that whatever input was used it becomes a d.c.

output with a specific polarity.

B is entitled a filter block. The filter block is provided in an attempt to

filter out variations in the wave form from the rectifier. In the case of

an a.c. input signal the filter is designed to reduce the ripple

amplitude and to provide a smoother d.c. level.

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C stabilises the d.c. and reduces (or potential divides) the level to make

it correct for the input to the coupled isolator device.

D is a block signifying some form of coupled isolation device which

separates the input circuit from the logic levels which are supplied to

the microelectronics.

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson we have attempted to identify and define typical problems

encountered when interfacing input signals with the microelectronics of the

PLC. Once defined, these problems were used in a step-by-step design

exercise. The object of this was to provide an appreciation of each of the

component parts and the reasons for their use. Prospective users of PLCs need

to be able to follow the specifications of input interfaces stated by their

manufacturers. By careful examination of typical designs similarities emerge

and, once learned, these can be recognised universally.

Teesside University Open Learning (Engineering)

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setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice