Programable Logic Controller questions. PLCS

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MODULE TITLE: PROGRAMMABLE LOGIC CONTROLLERS

TOPIC TITLE: INTERFACING

LESSON 3: OUTPUT INTERFACE DESIGN

PLC - 3 - 3

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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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

The PLC output interface circuit uses signals supplied by the CPU to provide

outputs to control devices found in the outside world.

In this lesson we examine the way digital output interfacing is achieved.

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YOUR AIMS ________________________________________________________________________________________

After this lesson you should be able to:

• state the problems of using a digital logic signal to drive an external

output via an interface circuit

• identify additional electronic components from the circuit diagram

symbols

• describe the function and oper ation of the most impor tant

components typically found in output interface circuits

• understand manufacturers' specifications of output interface circuits

• select suitable output interface circuits to match the requirements of

different loads.

________________________________________________________________________________________

STUDY ADVICE ________________________________________________________________________________________

Circuit diagram symbols are to BS 3939 wherever possible.

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CPU OUTPUTS ________________________________________________________________________________________

When the CPU needs to send signals out in order that some external device

may be controlled it does so by using the logic levels at its disposal. The logic

level can have one of two possible values, i.e logic 'l' or logic '0'. By now you

should know that logic '1' can be assumed to be a +5 V d.c. signal and that the

logic '0' can be assumed to be a 0 V signal.

Whatever the logic level is it needs to be sent to an output port. When we

considered the requirements of an input signal we assumed that its port (and

others) would be part of an integrated circuit called an interface adaptor. A

similar assumption may be made about the port for output signals. That is, the

output port may be part of an interface adaptor. There are, however, other

electronic components which may be used in some manufacturers' designs.

For example, a device called an octal latch may be used instead of an interface

adaptor. In this text we shall continue to assume the use of interface adaptors.

Voltage levels are not the only consideration when dealing with output signals.

The maximum values of current which adaptors can supply are specified in

milliamperes – not tens of milliamperes or hundreds of milliamperes but ones

and twos. Of course, we could overload the adaptor and draw off more current

but this would seriously limit the life expectancy of the device.

The job of an output interface is to use this 5 V, 1 mA signal to provide a

means of controlling a load which has a much larger power requirement.

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________________________________________________________________________________________

PLC OUTPUT INTERFACES ________________________________________________________________________________________

The block diagram of FIGURE 1 shows the basic arrangement.

FIG. 1

The PLC manufacturer does not know what type of load a user will connect.

In most cases manufacturers will design a range of output interfaces which will

cope with most load requirements. The trick, then, is for the user to make the

intelligent choice and select the correct unit. In some PLCs the output

interfaces are made in interchangeable modules. The user buys the PLC void

of output circuitry and then purchases and fits the modules required for that

particular application.

In other PLCs the output interface is built into the design and is not

interchangeable. If the user makes the wrong choice then a completely

different PLC would need to be purchased.

PLC

Micro computer

Interface adaptor

Output interface circuit

External power load

Logic level

signal

Logic level

signal

Power control

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We will now look in more detail at three basic versions of output interface.

(a) Relay output.

(b) Transistor output.

________________________________________________________________________________________

RELAY OUTPUTS ________________________________________________________________________________________

The first type of output interface circuit to be examined is the relay contact

output. FIGURE 2 shows a possible circuit which we will use as an example.

Relays themselves were covered in the previous topic on the development of

programmable controllers. It is the relay coil driving circuit which is of

interest here.

The interface circuit of FIGURE 2 uses the logic signal from the output

adaptor to switch on and off the transistor (TR1). The transistor itself is acting

as an amplifier in that its collector load current is many times larger than the

current which drives it on i.e. the base current. In this circuit some of the load

current is driven through an indicating LED before all of it is passed through

the relay coil. Clearly if the LED carries a current then so does the relay coil.

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FIG. 2

This arrangement is one possible version, but it must not be assumed to be the

general case. Some PLCs allow the output LED to be illuminated whilst the

physical output of the relay is inhibited. Safety, when commissioning a new

system, may require this feature. The circuit diagram of FIGURE 2 does not

include this refinement.

You may wish to consider the design of an output interface circuit which will

allow such control.

LED

Relay coil

Logic signal

External output

terminals

0 V

+ V

TR1

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CIRCUIT DESIGN AND ANALYSIS

The following section analyses the design of the circuit shown in FIGURE 2.

The purpose of this is to show that the process is straightforward and not at all

difficult.

You need to take note of the approach and the way that assumptions are

brought into the design. Once a value has been assumed then other values

depend upon it.

Current will flow into the interface circuit from the interface adaptor when a

logic 'l' i.e. +5 V has been sent out. The value of current will be limited by the

resistor R1 in series with the base emitter junction of TR1. If the transistor

used for TR1 is silicon then the base emitter voltage dr op will be

approximately 0.6 V. The remaining 5 V – 0.6 V = 4.4 V must be dropped

across the resistor R1 when it carries the base current.

If the base current is assumed to be limited to 1 mA, then by using OHM'S

LAW the value of R1 may be calculated as:

therefore

Unfortunately this 4.4 kΩ is not a standard preferred value and so the nearest suitable value would need to be chosen.

A value of 4.7 kΩ from the E12 range or a value of 4.3 kΩ from the E24 range could be used.

In the above analysis 1 mA has been assumed for the value of base current.

R V

4 4 4400

4400 4 4

= = =

= =

. V 1 mA

mV 1 mA

. kΩ Ω

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Explain why this value was chosen.

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

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If you can't explain then refer back to page 2.

The relay coil will require a larger current than 1 mA. Here again an assumed

value must be used. (The actual relay used must operate with this value.)

Can you suggest a value of relay coil current?

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For this design we shall choose a realistic value of 50 mA. This value of

current must flow through the transistor (TR1) when it is switched on by the

base current of 1 mA. The chosen transistor must, therefore, have a current

gain of fifty.

i.e. collector current = 50 × base current Ic = 50 × 1 mA.

therefore Ic = 50 mA.

The relay used must be a type that will energise when carrying 50 mA through

its coil.

50 mA would be adequate for the purpose of switching on the relay but would

you allow the 50 mA to flow through the LED?

From the work covered in Lesson 2 what would be a reasonable value of current

through the LED?

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

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Very few LEDs, if any, can safely carry 50 mA.

Typical values were previously stated as between 10 mA and 20 mA.

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We shall assume a value of 10 mA is adequate to illuminate the LED. If this is

the case then 50 mA – 10 mA = 40 mA must be diverted or shunted around

the LED. This is the purpose of the resistor R2.

What ohmic value should R2 have?

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

________________________________________________________________________________________

Perhaps you have been able to calculate a suitable value of R2.

If not, don't worry, we shall go through the process.

R2 is connected in parallel with the LED. What we want to happen is that

40 mA will flow through R2 when 10 mA is flowing through the LED. It was

suggested in the previous lesson that an LED carrying 10 mA will have a

voltage drop of about 2.4 volts. R2 is in parallel with the LED and so it will

also have 2.4 V across it.

Using Ohm's law the value of R2 can be calculated.

R V

I2 2 4 0

2400 0

60= = = = . V

4 mA mV

4 mA Ω

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The nearest preferred values are:

62 Ω if the E24 range is used

or 56 Ω if the E12 range is used.

Compare your value for R2 if you did your own calculation.

Were you correct?

FIG. 3

One component not mentioned so far is the diode D1. This diode is connected as

a flywheel diode. The name may not mean anything to you but it has a specific

purpose. It does not carry any of the current which flows through the transistor

because it is connected reverse bias, that is to say, the wrong way round.

D1 carries any current which is generated within the relay coil due to the

electromagnetic energy stored within its magnetic circuit. Normally this

energy holds the relay armature in its energised position but when the

transistor is switched off the stored energy could generate a high voltage which

would damage the transistor. Diode D1 prevents this happening by presenting

a low resistance flow path to the generated voltage.

In practice the value of V shown in FIGURE 2 would need to be about 9 volts.

To relay coil

R2 56 Ω

+ V

10 mA40 mA

2.4 V

50 mA

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On the load side of FIGURE 2 a resistor R3 is connected in series with a

capacitor C1 across the relay contacts. This network is not provided so that it

shares the load current with the contacts. If the relay contacts are going to

switch highly inductive loads then sparking will take place at the contacts.

Sparking and arcing seriously reduce the life of the contacts as well as there

being the possibility of sparks igniting any flammable gases which may be

present. R3 and C1 are provided to reduce the arcing. The values of R3 and C1 depend upon the characteristics of the load.

Relay outputs are often termed volt free because the manufacturer does not

provide a supply to the contacts. The manufacturer has no idea what load will

be connected or whether the supply for the load needs to be a.c. or d.c., low

voltage or extra low voltage, etc.

Relay outputs are very useful because they can be used to switch either a.c. or

d.c. loads. The rating of the relay contacts is, however, different for each type

of supply and so this is one of the manufacturer's specifications which must be

checked by the user against the requirements of the load.

Typical contact ratings may be, for example, 2 A @ 24 V d.c. or 2 A @ 240 V

a.c.

At this point you should refer back to the diagram of FIGURE 2 and write

down the values of, and reasons for, each component in the circuit.

Do this without reference to the lesson if you can.

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________________________________________________________________________________________

TRANSISTOR OUTPUTS ________________________________________________________________________________________

The second type of output interface that we will look at is the transistor output.

Small loads which require a low value d.c. supply are catered for by the use of

transistorised output circuits. Such outputs are ideal for the control of

pneumatic solenoid valves or indicator panels. The user must provide the

supply for the load. The maximum load voltage obviously must not exceed the

maximum voltage rating of the output transistor. In practice a manufacturer

may use a power transistor with a maximum voltage rating of about 35 volts

whilst specifying that the user should not exceed 24 volts. A wise engineer

will follow the manufacturer's recommendations.

Turn your attention now to the circuit diagram of FIGURE 4.

There are no new symbols in this diagram although one component, the

inverter (IC1), has not been explained.

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FIG. 4

CIRCUIT OPERATION

You should be able, by now, to examine the circuit, pick out parts which have

been covered, and make a reasonable decision regarding the function of each

component. Similarities between this circuit and those designed or examined

in previous lessons should be of great help.

What then is the purpose of the inverter (IC1)? This logic gate accepts a logic

input and provides a logic output in the following way.

(a) The gate accepts either of the two possible logic levels but changes

whichever input it receives to the opposite level output.

Logic signal

External load

terminals

+ 5V

TR2

D 2

Power transistor

56 ΩR2

R1 200 Ω

OP 1

10 mA

Inverter IC1

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i.e. If a logic 'l' is applied to the input the output will become a logic '0'

and if a logic '0' is applied then the output will become a logic 'l'. The

input is inverted to provide the output.

(b) A suitably chosen inverter can also act as an amplifier of current

(sometimes called a buffer stage).

In the circuit of FIGURE 4 both of these points are important.

If a logic 'l' signal is sent from the interface adaptor into the inverter then the

output will become a logic '0'. A logic '0' (previously assumed to be 0 volts) at

the inverter output will allow a current to flow through the LED, through the

R1 – OP1 combination, through the resistor R2 and into the output of the

inverter gate. This may seem strange but it happens this way. (This circuit has

been chosen to illustrate this point.)

The current flowing into the inverter input will be less than 1 mA but it

controls the flow of the larger current into the inverter output.

Now assume the value of the LED (D1) current to be 10 mA and attempt to

answer the following three questions.

1. What value of voltage do you assume to be dropped across D1?

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2. What is the purpose of R1? (You have already seen the R1 – OP1 arrangement in

the circuit of FIGURE 5, page 10 of Lesson 2.)

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3. What is the purpose of R2?

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

________________________________________________________________________________________

Answers may be found on page 28.

When the adaptor signal is logic 'l' the LED will illuminate and the

phototransistor will switch on. The phototransistor obtains its current through

the external supply but it is unlikely that a phototransistor would be able to

carry the full load current. For this reason the phototransistor of FIGURE 4 is

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being used to switch on and off another transistor (TR2). The second transistor

must be rated to carry at least the full load current otherwise it is likely to burn

out. It is, therefore, a power transistor.

When the adaptor signal is logic '0' the LED stops glowing, the phototransistor

switches off and the load current stops flowing through the power transistor.

The diode D2 has been included as a protective device to ensure that no high

voltage spikes can build up across the transistors.

Can you think of any advantages that transistorised outputs have over relay contact

outputs?

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

________________________________________________________________________________________

Solid state outputs:

• have no moving parts and, therefore, do not require the same level of

maintenance

• are likely to last longer

• do not produce sparks

• are silent in operation.

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________________________________________________________________________________________

TRIAC OUTPUTS ________________________________________________________________________________________

This is the third type of output interface to be considered. The silent,

maintenance free operation of the d.c. transistorised circuit can also be

obtained for use in switching a.c. loads. The same output circuit cannot be

used because the transistor will not operate an a.c. load. For these loads a

different circuit which incorporates thyristors, or more commonly triacs, is

used.

FIGURE 5 shows a circuit diagram which may be suitable for handling a.c.

loads.

Examine the circuit diagram carefully and note any new component symbols

or different methods of connecting components.

FIG. 5

Refer to the point in FIGURE 5 where the logic signal from the interface

adaptor enters the circuit. The symbol for IC1 is that which is often used for a

OP 1 C1

R 3

Output

IC1

Logic signal

0 V

Opto-triac 1

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buffer amplifier. The circuit of FIGURE 4 used a buffer amplifier but the two

symbols are different. The buffer of FIGURE 4 was an inverter whereas the

buffer of FIGURE 5 is a non-inverter. This device gives out the same logic

level as it receives at its input but a current amplification is possible through

the buffer.

We shall now consider the design of FIGURE 5. You will be expected to

calculate the values of some of the components. Notice that in this design the

two LEDs are not connected in series. Each LED is itself in series with a

current limiting resistor; R1 in series with the indicating LED (D1) and R2 in

series with the LED inside the opto device. These two circuit branches are

connected in parallel across the output of the buffer amplifier. The buffer

accepts the low current logic signal from the output adaptor and provides a

larger current to supply the LEDs. The logic levels at the buffer input and

output are the same i.e. logic 'l' in gives logic 'l' out.

When a logic 'l' (+5 V) is available at the buffer output, current will be

supplied to the LEDs.

You should, by now, be confident about being able to calculate the values of

components R1 and R2 if certain other required or assumed values are

specified. (In other lessons Ohm's law has been used for this purpose.)

Consider first the branch with the indicating LED.

Assume that a suitable value of current through this LED is 10 mA and that

while carrying this current the voltage drop across the LED is 2.4 volts. (This

is shown by FIGURE 6 overleaf.)

What value of resistance should R1 have?

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To answer this question work through steps 1 – 3 below.

FIG. 6

Step 1. How much voltage needs to be dropped across R1?

Step 2. This value of voltage (V1) is dropped while _____ mA of current is

flowing through R1.

Step 3. Use these values of voltage and current to determine the resistance of

R1.

Ω

The value of resistance calculated will not be a standard preferred value.

Choose a suitable resistor from the range given below.

180 Ω, 220 Ω, 270 Ω, 330 Ω, 390 Ω

Chosen value = Ω

Solutions given on page 29.

R V

I1 1= = = =

Logic '1'

0 V

2.4 V

+5 V

10 mA

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What value of resistance should R2 have?

In this case we will assume that 2 mA of current will be sufficient and that the

LED inside the opto device will drop 1.6 volts when this current is flowing.

Provide your own sketch of the R2 branch and then calculate the value of R2 using the same method as for R1.

Step 1.

Step 2.

Step 3.

Choose a preferred value from:

1 kΩ, 1.2 kΩ, 1.8 kΩ, 2.2 kΩ, 2.7 kΩ

Chosen value = Ω

Solution given on page 29.

Well, having got some of the design calculations out of the way we can go

back to the diagram of FIGURE 5 to consider the rest of the circuit.

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Look at the opto device. This symbol is different from the optotransistor

covered in Lesson 2. It is, in fact, the symbol for an opto-triac.

OPTO-TRIAC OPERATION

Remembering back to Lesson 2, the phototransistor operation was briefly

explained as follows.

When switched on by the LED light beam a current was allowed to flow into

the collector terminal and out of the emitter. This device can only allow

current to flow in one direction and can therefore only be used for d.c. loads.

The opto-triac, as you might imagine, again requires the light from the LED to

enable conduction through the triac component. The difference here is that the

current through the triac is alternating i.e. changing direction at regular

intervals of time. The operation of the triac must allow conduction in both

directions but at different points in time and only when the LED is carrying

current. The symbol for the triac shows two arrows to signify conduction in

both directions through what is called an inverse parallel pair.

In the circuit of FIGURE 5 the current to the external load flows through the

triac when the LED is conducting. Sometimes the external load requires more

current than the opto-triac can safely carry. When this is the case the circuit

would be changed so that the opto-triac controls (i.e. turns on and off) a power

triac. The power triac should then be the device carrying the external load

current.

The components shown in the dotted box do not form part of the load circuit

but are often provided for the protection of the semiconductor triac. The R3 in

series with C1 forms a snubber circuit which will suppress unwanted switching

interference and the varistor will suppress a.c. transients. The values of these

components depend upon the size and type of load being switched.

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If you have understood this lesson then attempt the following Self-Assessment

Questions.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

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

FIGURE 7.

FIG. 7

(a) (b)

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2. State the characteristics of the signals obtained from an output interface

adaptor.

3. Why is it necessary for a PLC to have output interface circuitry?

4. State typical contact ratings for relay output circuits.

5. What kinds of load can be supplied from a volt-free relay output?

6. Transistorised outputs are designed to handle certain types of load. What

are the characteristics of such loads?

7. What is the purpose of using a non-inverting buffer?

8. A manufacturer's leaflet states that the output circuitry of a PLC employs

power triacs. What types of load can this PLC drive and for what types

is it unsuitable?

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (a) INVERTER or INVERTING BUFFER

(b) DIODE

(d) OPTO TRIAC

2. Logic levels of voltage are available from the interface adaptor i.e. a logic

'l' being +5 V and a logic '0' being 0 V. The current available is very

limited and should be kept to a low value of around 1 mA.

3. The purpose of the output interface circuitry is to take in the logic level

signals and use these to enable control over much larger power loads.

The PLC itself cannot control these power loads directly because it is

limited to microelectronic values of power.

4. The exact ratings of any output device can only be obtained by reference

to the manufacturer's specifications. However, typical values for output

relays would be 2 A @ 24 V d.c. and 2 A @ 240 V a.c.

5. Virtually any kind of load can be supplied from volt-free relay contacts so

long as the ratings provided by the manufacturer are observed. It would

be very dangerous to exceed the values of voltage and current for either

a.c. or d.c. loads – more information on this point will be covered in

Lesson 4.

6. Transistorised outputs are designed to drive low voltage d.c. loads only.

Typically loads requiring around 1A @ 24V d.c. would be quite

acceptable. This covers indicator lighting panels and the pneumatic

solenoid valves mentioned in Lesson 3.

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7. A non-inverting buffer accepts a logic signal at its input and gives the

same level of logic signal at its output. It does not change the level of the

signal in any way. Buffers do have the advantage of drawing in a very

small current from the circuit which feeds it whilst having the capacity to

provide a much larger current at its output. In the circuit of FIGURE 5

the buffer will take in less than 1 mA whilst providing 12 mA output (10

mA for one LED and 2 mA for OP1).

8. Power triac outputs are specifically designed for control of a.c. loads at

voltages and currents dictated by the size and rating of the triac used and

the insulation and ventilation properties of its enclosure.

Triacs are not designed to operate d.c. loads of any type.

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________________________________________________________________________________________

ANSWERS TO QUESTIONS ON PAGES 14 AND 15 ________________________________________________________________________________________

1. Following the assumptions previously made a value of between 2 V and

2.4 V would be acceptable.

2. R1 diverts some of the current which flows out of LED (D1) around the

LED which is inside OP1. This is because the LED is carrying 10 mA (in

this case) but the LED in OP1 does not require this much current.

3. The voltage drop across the two LEDs in series would be a total of about

4 volts. 5 volts is being applied but only 4 volts is being dropped. The

difference in voltage, i.e. 5 V – 4 V = 1 V needs to be dropped across

the resistor R2.

The actual voltage drop across R2 will be:

In practice this would be acceptable because when the 10 mA flows into

the output of IC1 its actual '0' voltage will rise to about 0.4 V.

V I R

= × = ×

10 56

560

0 56

Ω

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________________________________________________________________________________________

ANSWERS TO QUESTIONS ON PAGE 19 ________________________________________________________________________________________

Step 1. 5 V – 2.4 V = 2.6 V

Step 2. 10 mA of current

Step 3.

Chosen value = 270 Ω

________________________________________________________________________________________

ANSWERS TO QUESTIONS ON PAGE 20 ________________________________________________________________________________________

Step 1. 5 V – 1.6 V = 3.4 V

Step 2. Current = 2 mA

Step 3.

Chosen value = 1.8 kΩ

R V

I2 3 4 3400

1700= = = = . V

2 mA mV

2 mA Ω

2 6 2600 260

. V 10 mA

mV 10 mA

= = Ω

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

This lesson commenced by looking at the reasons for, and problems associated

with, output interface circuits. An analysis of types of output circuit made it

possible to examine important points relevant to the user and to see how

manufacturers cater for the users' needs.

The actual circuits used for analysis were varied and not representative of any

one manufacturer. Components readily available and commonly used have

been chosen and assembled in diagram form to indicate the range of

possibilities available to a designer. Understanding design limitations enables

prospective users to realise the reasons for particular ratings. Lesson 4

concludes the present work on interfaces but more will be covered later when

analogue inputs are analysed.

Teesside University Open Learning (Engineering)

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