Presentation1.ppt

NotesSome slides and formatting are missing from this version it will be updated asap.24/11/04Stephen Norton

Mechanical Project SafetyIntroductionWe will not and cannot cover every aspect of safety in this one hour. All I hope to do is to bring safety to the forefront of you mind. You are young, and of course nothing will happen to YOU ..as youre going to live forever. A big but, I have been on sites where people have been killed or injured because of silly oversights.

Mechanical Project SafetyIntroduction: The Purpose Of This LectureThe purpose of this lecture is to remind you of the safety risks associated with projects and to try to review the correct way to approach the design and construction of various circuits.

As the old saying goes PPPPPP.

Mechanical Project SafetyThe First and Overiding CommandmentIf in doubt, leave it and ask.

Repeat DO NOT UNDERTAKE ANY WORK YOU DO NOT UNDERSTAND OR CANNOT DO IN COMPLETE SAFETY.

Do not work alone in the workshop on equipment that can be dangerous.

Do not break the above rules.

Mechanical Project SafetyUp to now we have had no accidents in the project labs. Keep it that way as this is not the kind of first you want to achieve.

RememberPPPPPP!

Definitions

PRINCIPLE FACTORS IN SHOCK OR ELECTROCUTION

The following items have been found to be the leading causes in a person being shocked or electrocuted.

State of equipmentLack of Off indicator/indication on front panelModifications to original installationNew equipmentPoor/wrong documentation (are the mods documented?)Working aloneLack of supervisionRepetitionStressVacation (before and after)Not following procedures/specsNot looking outUntrained techniciansInadequate or inoperative interlocksLack of attentionPoor maintenance of equipmentLack of proper groundingFailure to shut off equipmentFailure to short high voltage leads to groundImproper toolsLack of final inspectionIs equipment ready to be energized?

Protective Strategies and DevicesProtection against Overload;Recall circuit functioning correctly with excessive load. Fuse will protect the circuit after a long delay. Circuit breakers have a similar overload reponse to fuses.For motor circuits a separate overload is normally provided or integrated into another component.

A Short Circuit is a fault condition and the circuit needs to break the power quickly.

Short Circuit Protection

Utilisation Categories

EarthingEarthing strategies were done in second year.For project purposes, all exposed conductive surfaces should be earthed and equipotential bonded as shown across.

Structure Of Automatic Control Circuits. With Safety in mind is important to clearly understand the function of piece of equipment that you may use.

Data AcquisitionData Acquisiton consists of retrieving information on the state of external phenomena.SafetyLight curtains using pnoto-electric detectors shown be used to protect against access to machinery. Should you have this????At a push limit switches connected by string can serve the same purpose.Tied here

Data Acquisition-Applications

Man Machine DialogueThese are manual devices designed to switch low current, eg contactor coils, applications.SafetyThe emergency stop switch is the most important of these devices. If you are using mains power you circuit must have one.

ActuatorsActuators Convert electrical energy into mechanic energy [in control circuits].Linear Motion Safety Issues;Beware sudden unexpected movements; Restrict access to the motive area.Rotation Motion Safety Issues;Do not were lose clothingCheck direction of rotation before connecting up coupling to machine/drive.Test circuit with ohmmeter and lamp if applicable before powering up.

Actuators. Induction MotorThe three phase induction motor is the most common industrial machine. There are single phase induction motors also. Speeds are high so be careful around the shaft as when it is running it can look static. Overload protection needs to be provided. You will need to size cables correctly. Operation of the motor is covered in the electrical section of the presentation.Look out for excessive vibration, noise or heat.

Actuators; DC Motors

Series type DC motor is shown across.Torque Current squaredSpeed Recipricol Of CurrentApplications; hoists, trains, starter motorsShunt Type DC motor is shown across.Torque CurrentSpeed VoltageApplications; Pumps, machines, compressors.

ActuatorsCompound Type DC motor is shown across.Combination of both series and shunt and can have either characteristic used in traction systems.

Permanent magnet motors are used in robotics, toys etc.Similar to the shunt typr motor.

Actuators; Stepper MotorsStepper motors are rugged, simple and easy to control.There are three main types;Permanent magnetMost widely usedLow cost,speed, torqueVariable RelucatanceNo permanet magnet.No detent torqueNot widely usedHybrid MotorsMost widely used

Data ProcessingRelays were traditionally used to decide the action required for a particular input or sequence. Today PLC and PC Control of circuits is more common.

Data Processing

Power ControlPower control. ContactorsElectric power is supplied by means of a distribution network but cannot be permanently connected to equipment. Power switching systems must be used which allow the making and breaking of elec tric power from the mains to the load.Switchfuses, isolators, circuit breakers and contactors are used to provide this function called control.Electronic and pneumatic equipment can also carry out power control but are not dealt with in this publication.In some cases, the operator is at some distance from the power control units and remote control must be provided.A drive unit (electromagnet) serves as intermediary for this purpose and may be controlled by pilot device. (push-buttons for example).Contactors, with their electromagnets, fulfil both power switching and remote control functions. Remote operation frequently requires monitoring either by an illuminated display or by a sequence interlock controlling another component.These complementary electrical circuits, called interlock circuits, also use contacts incorporated in the contactor.

Power Control-ContactorsA contactor is a mechanical switching device controlled by an electromagnet with go-nogo operation.When the coil of the electromagnet is energised the contactor closes and, by means of contacts, completes the circuit between the main supply and the load. As soon as the coil is de-energised the contactor opens.The moving part of the contactor are detailed on the next slide.Contactors offer a great number of advantages, such as: breaking high single or poly-phase currents by means of a low current device, providing intermittent as well as continuous service, executing a remote, automatic command using cable of a small cross-section thus greatly reducing the length of the power cables. multiplying the number of control stations and placing them close to the operator. after a momentary break in supply, protect personnel against automatic restart provide easier distribution of emergency stop and interlock stations inhibiting operation of the machine until all safety precautions have been taken,

Contactor; Contact DescriptionMain Contacts are numbered 1 to 6 or 1 to 8.The second number of the auxiliary contact 1,2 and 5,6 are normally closed. 3,4 and 7,8 are normally open.Coil normally A1 and A2

Power ControlA Contactor, auxilary contact block and a timer block are shown opposite and the movements in the contactor are shown below.

Power Control; Contactor Problems

Power Control-Basic Contactor Circuits

Power Control-Induction Motor CircuitsStarting of squirrel cage motorsDirect-on-line startingWhen voltage is applied to a motor the current surge in the main supply is high and can, especially if the cross section of the supply line is insufficient, cause a voltage drop capable of affecting the operation of other loads. This method is ideal as long as the peak current is acceptable and if the initial starting torque of the motor (determined by its rotor construction and usually about 1 .5 Tn) is adequate to start the machine.The current peak when the motor is energized is very high, of the order of 4 to 8 times the rated current. Torque during starting is always higher than the rated torque especially for modern motors with complex cages. It is a maximum when the motor reaches about 80 % of its speed ; at this moment the peak current is con siderably reduced.This arrangement permits the starting of machines, even at full load.However, with a starting torque of about 1.5 Tn, this method is not recommended if starting should be smooth and progressive (elevators, hoists, conveyors, etc.).

Star-delta startingThis method is applicable only to motors having both ends of three stator windings brought out to the terminals and whose delta connection corresponds to the mains voltage (example: for a 380 V supply a motor with 380 V delta/660 V star is required). This method consists of starting the motor with star connected windings the applied voltage is equal to the mains voltage divided or about 58 % of the rated voltage. This voltage is constant during the entire first step. The torque, which is reduced by a ratio of the square of the voltage, is equal to one third of the torque.- the second step the connection is changed from star to delta. Each winding then has the full main supply voltage applied and the motor resumes its normal running characteristics. Motor torque remains low during the entire star connected period and the speed, stabilized at the end of this time, can be low. This results in high peaks in current and torque oaring transition from star to delta. Furthermore it should be noted that the current flow in the motor windings is not continuous as it is interrupted at the moment the contactor is opened and then abruptly resumed, at the full mains voltage, when the delta contactor is closed. Considering the highly inductive characteristics of the windings transition to the delta connection is accompanied by high transient current peaks.

Power Control-Induction Motor Circuits

Power Control-Induction Motor CircuitsSingle Phase Induction Motor starting.

Power Control; DC MotorsShunt and permanet magnet motors can be controlled by a H bridge. T1 and T4 are on together and for the oppposite direction of rotation T2 and T3 are on.To vary speed turn the relavent pair of switeches on and off as showm.DC motors speed is most simplycontrolled with variable reistors.

Power Control; Stepper MotorsStepper motors [Except VR stepper] step the rotor by edging the magnetic field around the poles and the rotor follows. They stay in position by virtue of the permanent magnet. It is done via a series of 1s and 0s

Design ExampleGiven the following pump house application as an example application.

Design the control and power circuits. See next slide for these.Determine the required components.Obtain your components.

Design Example

Design ExampleLay your componets out as you wish to wire them.Typically keep low voltage or dc circuits away from ac circuits.Contacts link the power and control circuits.

Design ExampleSelect a suitable enclosure to fit the component layout you have designed.Mount components on DIN Rail as shown below.

Design Example Testing.Initially Bell Out Circuit and use a lamp to turn on/off for the contactor power etc.

System testWith the main supply and all external power and control circuits connected in accordance with the circuit diagram, a system test of the equipment can be carried out. This can be done in two steps:Off load, testThe purpose of this test is to verify that all external control, safety and signalling devices have been correctly connected. To carry out the test, the supply to all loads must be isolated- either by removing the fuses protecting the power circuit;- or by placing the isolator in the test position, without closing the power circuit (isolator closed, links removed or fuse rack switched over).

Design Example Testing.With the control circuit live, operating the pilot device which authorizes starting should cause the closing of one or more contactors and, on more complex equipment validate the operation of the cycle as specified on the circuit diagram or as described in the operating manual.At this point, it is recommended that the external pilot devices are operated manually or simulated and that all possible control and operating malfunctions are tested. This will permit the effective ness of the control circuit to be checked as well as the sequence control, and the safety and signalling circuits.On load testWith the main circuit live, a system test identical to the preceding one should be carried out in order to check connections and the operation of the various loads. If the complexity of the installation justifies it, the test will be completed by a series of tests that check that the automatic equipment duly performs the different mecha nical functions of the installation.

Cables/Terminations and Cable SizingCables must be mechanicly protected.Cables must be supported, they cannot hang from the terminals as the conductors are maliable and will stretch and eventually break or worse become a fire hazard.Cables entering enclosures should be glanded to provide mechanical support for the cable.Cables should be the appropriate size. 1.5mm will do for most small power circuits. For larger power requirements use the following table as a guide. If the cable run is long voltage drop needs to be analysed also.

Step1: Define the installation method Table AA Conduit in free airB Embedded conduitC Mouldings, skirtings, door frames, slotsD Direct wall fixing by cleats or staplesEDirect fixing to ceilingFCable traysGHangersHDuctingJ Open ductingK TrunkingL3 Direct in open or ventilated trenchL4 Direct in closed trenchN AlveolesP Alveolated blocks0 Frames

Step2 Determine the current rating index Table CInstallation method Corresponding current rating column number in table D(See table A, above)InsulationPVCButyle, ethylene propylene, PRCNumber of cores3 cores2 cores3 cores 2 coresA Conduit in free air2345B Embedded conduit2345C Mouldings, skirtings, 2345door frames, slotsD* Direct wall fixing by 3456cleats or staplesE Direct fixing to ceiling3456F* Cable trays3456G Hangers3456H Ducting2345J Open ducting3456K Trunking2345L3 Direct in open or 2345ventilated trenchL4 Direct in closed trench2345N Alveoles2345P Alveolated blocks2345Q* Frames2345* For single core cables use the next highest column number.

Step 3 Determine the current carrying capacity Iz in amps (A)

Table DCurrent carrying capacity Iz in ampsNominal crossColumn numbers (selected above, table C)section of cable 2345mmCu AlCuAlCuAlCuAl1 1213,5 15171,515,5 17,5 19,5 2224272,5212426304283235406 36414652105057637116687685962589101 112 127 3511186125 97138108157 125 50134105151118168131190 15170171133192150 213166 242 19295207161232181258 200 293 232120239186 269 210 299 236 339 269150275 309 240 344 268 390 309185314 353 275 392 305 444 353240369 415 325 461360 522 415

Step 4 Cables laid together; Select a derating factor K1 depending on the number of conductors.Cables laid together in alveolated conduits or ductings.The current carrying capacity lz in Table 52 D for two core circuits must be multiplied b by factor K1. Table G Number of live conductors 4 6 8 10 12 16Multiplying factor K1 0,80 0,69 0,62 0,59 0,55 0,51

Cables laid together on cable trays.The current from Table D must be multiplied by factor K1. Table HNumber of cablesin a single layer234 or 56 to 89 and over

K1, cables laid 0,850,780,750,720,70horizontallyK1, cables laid 0,800,730,700,680,66vertically

STEP 5 Depending or the ambient temperature (if it is above 30 C) select derating factor K2 The current carrying capacity lz in Table D must be multiplied by factor K2 Table FAmbient temperature (C)35404550556065K2 for PVC Cable0,93 0,87 0,79 0,710,610,50 K2 for butyle, PRC 0,95 0,90 0,85 0,80 0,74 0,67 0,60ethylene or propylene Cable

ExamplesExample 1How to determine the continuous current carrying capacity; in a three core cable with copper conductors: PRC insulated;nominal section of each conductor: 240 mmthe cable is PVC insulated and has three copper or aluminium4 identical cables laid together horizontally on the same tray;ambient temperature 40C.Installation method Table A: F2.Current rating index Table C 53.Current carrying capacity at ambient temperature 30C (TableD): 522 A4. Influence of groupings of conductors and cables, derating factor; derating factor in table H; K1 =0.75F: K2 = 0.90.

Overall derating : 0.75 * 0.90 = 0.675.Current carrying capacity: 522 x 0.675 = 352 A.Example 2What is the minimum cable size needed to supply a permanent load of 135A.Cable is PVC insulated and has 3 conductors; copper or Al7 identical cables laid together horizontally in a single layer on the same try.Ambient temp is 40 degrees C

Derating Table H K1=0.72 Table F K2 = 0.872. Corrected current rating lb 135 = 216AK1 x K2 = 0.72 x 0.87Overall derating : 0.75 )( 0.90 = 0.675.

3. Minimum section of each conductor : This is determined directly using tableD (column 3; installation method F, PVC 3 conductors). copper (Cu) 95 mm aluminium (Al) 150 mm

Workshop SafetyYou need to have completed the six hours training on each tool you want to use.You may only use the workshop when you are supervised.You may only use the workshop when you are scheduled unless you have made special arrangements with a qualified lecturer.Follow all the safety rules displayed in the workshop.

Pneumatic SafetyCompressed air represents a source of considerable potential energy, hence precautions should be taken to prevent accidents.It must never impinge upon the body, eg blocking are with your hand.Check all connects, mountinmgs and pipework before connecting the supply.Air and electrical supplies must be disconnected before adjustments or maintenace is carried out.Never exceed pressure or force ratingsIsolate moving parts from accidental contact.Use air cushioned cyclinders. Do not adjust under pressure.

THE BASIC PNEUMATIC SYSTEM

Pneumatic cylinders, rotary actuators and air motors provide the force and movement of most pneumatic control systems, to hold, move, form and process material.To operate and control these actuators, other pneumatic components are required i.e. air service units to prepare the compressed air and valves to control the pressure, flow and direction of movement of the actuators.A basic pneumatic system, shown on the next slide, consists of two main sections: the Air Production and distribution System the Air Consuming System

Air Production System

1 CompressorAir taken in at atmospheric pressure is compressed and delivered at a higher pressure to the pneumatic system. It thus transforms mechanical energy into pneumatic energy.2 Electric MotorSupplies the mechanical power to the compressor. It transforms electrical energy into mechanical energy.3 Pressure SwitchControls the electric motor by sensing the pressure in the tank. It is set to a maximum pressure at which it stops the motor, and a minimum pressure at which it restarts it.4 Check ValveLets the compressed air from the compressor into the tank and prevents it leaking back when the compressor is stopped.5 TankStores the compressed air. Its size is defined by the capacity of the

compressor. The larger the volume, the longer the intervals between compressor runs.6 Pressure GaugeIndicates the Tank Pressure.7 Auto DrainDrains all the water condensing in the tank without supervision. 8 Safety Valve. Blows compressed air off if the pressure in the tank should rise above the allowed pressure.9 Refrigerated Air DryerCools the compressed air to a few degrees above freezing point and condenses most of the air humidity. This avoids having water in the downstream system.10 Line FilterBeing in the main pipe, this filter must have a minimal pressure drop and the capability of oil mist removal. It helps to keep the line free from dust, water and oil.

The Air Consuming System

1 Air Take-offFor consumers, air is taken off from the top of the main pipe to allow occasional condensate to stay in the main pipe, when it reaches a low point a water take off from beneath the pipe will flow into an Automatic Drain and the condensate will be removed.2 Auto DrainEvery descending tube should have a drain at its lowest point. The most efficient method is an Auto Drain which prevents water from remaining in the tube should manual draining be neglected. 3 Air Service UnitConditions the compressed air to provide clean air at optimum pressure, and occasionally adds lubricant to extend the life of those pneumatic system components which need lubrication.4 Directional ValveAlternately pressurises and exhausts the cylinder connections to control the direction of movement. 5 ActuatorTransforms the potential energy of the compressed air into mechanical work. Shown is a linear cylinder; it can also be a rotary actuator or an air tool etc.6 Speed ControllersAllow an easy and stepless speed adjustment of the actuator movement.

Safety issues same as electrical actuators

Electro Pneumatic CircuitDraw the required pneumatic actuator circuit.Design the control circuit.Implement with electro-pneumatic or PLC control circuit.

End of Safety Lecture

Pneumatics

Pneumatics

We will firstly describe the components of a pneumatic system. We will then look at basic pneumatic circuits to see how the compressed air is used and controlled to provide motion.

Introduction to PneumaticsA fluid power system is one that transmits and controls energy through the use of pressurised liquid or gas. There are two methods of achieving this hydraulically, with fluid and pneumatically.In Pneumatics, this power is air. Compressed or pressurised air is mainly used to do work by acting on a piston or vane.The correct use of pneumatic control requires an adequate knowledge of pneumatic components and their function to ensure their integration into an efficient working system.

What can Pneumatics do?The applications for compressed air are limitless, from the opticians gentle use of low pressure air to test fluid pressure in the human eyeball, the multiplicity of linear and rotary motions on robotic process machines, to the high forces required for pneumatic presses and concrete breaking pneumatic drills.

Advantages Of Compressed Air

Availability; Most factories and industrial plants have a compressed air supply. Storage; It is easily stored in large volumes if required.Simplicity of Design and ControlChoice of Movement, It offers both linear movement and angular rotation with simple and continuously variable operational speeds.EconomyLow installation and maintenance costs. Reliability, Pneumatic components have a long working life. Resistance to EnvironmentIt is largely unaffected in the high temperature, dusty and corrosive atmospheres in which other systems may fail.Environmentally Clean; It is clean and with proper exhaust air treatment can be installed to clean room standards.Safety; It is not a fire hazard in high risk areas, and the system is unaffected by overload as actuators simply stall or slip. Pneumatic actuators do not produce heat.

THE BASIC PNEUMATIC SYSTEM

Pneumatic cylinders, rotary actuators and air motors provide the force and movement of most pneumatic control systems, to hold, move, form and process material.To operate and control these actuators, other pneumatic components are required i.e. air service units to prepare the compressed air and valves to control the pressure, flow and direction of movement of the actuators.A basic pneumatic system, shown on the next slide, consists of two main sections: the Air Production and distribution System the Air Consuming System


1 CompressorAir taken in at atmospheric pressure is compressed and delivered at a higher pressure to the pneumatic system. It thus transforms mechanical energy into pneumatic energy.2 Electric MotorSupplies the mechanical power to the compressor. It transforms electrical energy into mechanical energy.3 Pressure SwitchControls the electric motor by sensing the pressure in the tank. It is set to a maximum pressure at which it stops the motor, and a minimum pressure at which it restarts it.4 Check ValveLets the compressed air from the compressor into the tank and prevents it leaking back when the compressor is stopped.5 TankStores the compressed air. Its size is defined by the capacity of the

compressor. The larger the volume, the longer the intervals between compressor runs.6 Pressure GaugeIndicates the Tank Pressure.7 Auto DrainDrains all the water condensing in the tank without supervision. 8 Safety Valve. Blows compressed air off if the pressure in the tank should rise above the allowed pressure.9 Refrigerated Air DryerCools the compressed air to a few degrees above freezing point and condenses most of the air humidity. This avoids having water in the downstream system.10 Line FilterBeing in the main pipe, this filter must have a minimal pressure drop and the capability of oil mist removal. It helps to keep the line free from dust, water and oil.

The Air Consuming System

1 Air Take-offFor consumers, air is taken off from the top of the main pipe to allow occasional condensate to stay in the main pipe, when it reaches a low point a water take off from beneath the pipe will flow into an Automatic Drain and the condensate will be removed.2 Auto DrainEvery descending tube should have a drain at its lowest point. The most efficient method is an Auto Drain which prevents water from remaining in the tube should manual draining be neglected. 3 Air Service UnitConditions the compressed air to provide clean air at optimum pressure, and occasionally adds lubricant to extend the life of those pneumatic system components which need lubrication.4 Directional ValveAlternately pressurises and exhausts the cylinder connections to control the direction of movement. 5 ActuatorTransforms the potential energy of the compressed air into mechanical work. Shown is a linear cylinder; it can also be a rotary actuator or an air tool etc.6 Speed ControllersAllow an easy and stepless speed adjustment of the actuator movement.

AIR COMPRESSIONA compressor converts the mechanical energy of an electric or combustion motor into the potential energy of compressed air.Air compressors fall into two main categories: Reciprocating and Rotary.The principal types of compressors within these categories are shown below.

Single Stage Reciprocating Piston CompressorAir taken in at atmospheric pressure is compressed to the required pressure in a single stroke.Downward movement of the piston increases volume to create a lower pressure than that of the atmosphere, causing air to enter the cylinder through the inlet valve.At the end of the stroke, the piston moves upwards, the inlet valve closes as the air is compressed, forcing the outlet valve to open discharging air into a receiver tank.This type of compressor is generally used in systems requiring air in the 3-7 bar rangeIn a single-stage compressor, when air is compressed above 6 bar, the excessive heat created greatly reduces the efficiency.

Two Stage Reciprocating Piston CompressorBecause of this, piston compressors used in industrial compressed air systems are usually two stage.Air taken in at atmospheric pressure is compressed in two stages to the final pressure.If the final pressure is 7 bar, the first stage normally compresses the air to approximately 3 bar, after which it is cooled. It is then fed into the second stage cylinder which compresses it to 7 bar.The compressed air enters the second stage cylinder at a greatly reduced temperatureafter passing through the inter-cooler, thus improving efficiency compared to that of a single stage unit. The final delivery temperature may be in the region of 120C.

Rotary Sliding Vane CompressorThis has an eccentrically mounted rotor having a series of vanes sliding in radial slots.As the rotor rotates, centrifugal force holds the vanes in contact with the stator wall and the space between the adjacent blades decreases from air inlet to outlet, so compressing the air.Lubrication and sealing is achieved by injecting oil into the air stream near the inlet. The oil also acts as a coolant to limit the delivery temperature.

Rotary Screw CompressorTwo meshing helical rotors rotate in opposite directions. The free space between them decreases axially in volume and this compresses the air trapped between the rotors.Oil flooding provides lubrication and sealing between the two rotating screws. Oil separators remove this oil from the outlet air.Continuous high flow rates in excess of 400 m are obtainable from these machines at pressures up to 10 bar.More so than the Vane Compressor, this type of compressor offers a continuous pulse-free delivery. The most common industrial type of air compressor is still the reciprocating machine, although screw and vane types are finding increasing favour.

Volume between threads decreases as you move from right to left causing the compression to occur.

Review: Identify and explain the operation of the following air production and consuming equipment.

Review: Identify and explain the operation of the following equipment

Air ReceiverAn air receiver is a pressure vessel of welded steel plate construction. installed horizontally or vertically directly downstream from the aftercooler to receive the compressed air, thereby damping the initial pulsations in the air flow.Its main functions are to store sufficient air to meet temporary heavy demands in excess of compressor capacity, and minimise frequent loading and unloading of the compressor. It also provides additional cooling to precipitate oil and moisture carried over from the aftercooler, before the air is distributed further. To this end it is an advantage to place the air receiver in a cool location.The vessel should be fitted with a safety valve, pressure gauge, drain, and inspection covers for checking or cleaning inside.

Air DehydrationAftercoolersAfter final compression the air will be hot, when it cools it will deposit water in considerable quantities in the airline system which should be avoided. The most effective way to remove the major part of this condensate is to subject the air to aftercooling, immediately after compression.Aftercoolers are heat exchangers, being either air cooled or water cooled units.We will just look at an air cooled aftercooler.

Air Cooled AftercoolerConsisting of a nest of tubes through which the compressed air flows and over which a forced draught of cold air is passed by means of a fan assembly. A typical example is shown below.The outlet temperature of the cooled compressed air should be approximately 15C above the ambient cooling air temperature.

Air DryersAftercoolers cool the air to within 10-15C of the cooling medium. The control and operating elements of the pneumatic system will normally be at ambient temperature (approx. 20C). This may suggest that no further condensate will be precipitated, and that the remaining moisture passes out with the exhaust air released to atmosphere. However, the temperature of the air leaving the aftercooler may he higher than the surrounding temperature through which the pipeline passes, for example during night time. This situation cools the compressed air further, thus condensing more of the vapour into water.The measure employed in the drying of air is lowering the dew point, which is the temperature at which the air is fully saturated with moisture (i.e. 100% humidity). The lower the dew point the less moisture remains in the compressed air.There are three main types of air dryer available which operate on an absorption, adsorption or refrigeration process.We will only look at the adsorption dryer

Absorption (dessicant) Drying

A chemical such as silica gel or activated alumina in granular form is contained in a vertical chamber to physically absorb moisture from the compressed air passing through it. When the drying agent becomes saturated it is regenerated by drying, heating, or, as here, heatless by a flow of previously dried air.Wet compressed air is supplied through a directional control valve and passes through dessicant column 1. The dried air flows to the outlet port.Between 10-20% of the dry air passes through orifice 02 and column 2 in reverse direction to re absorb moisture from the dessicant to regenerate it. The regenerating air flow then goes to exhaust. The directional control valve is switched periodically by a timer to alternately allow the supply air to one column and regenerating the other, to provide continuous dry air.Extremely low dew points are possible with this method, for example 40C.A colour indicator may be incorporated in the dessicant to monitor the degree of saturation. Micro- filtering is essential on the dryer outlet to prevent carry over of adsorbent mist. Initial and operating costs are comparatively high, but maintenance costs tend to be low.

Main Line FilterA large capacity filter should be installed after the air receiver to remove contamination, oil vapours from the compressor and water from the air.This filter must have a minimum pressure drop and the capability to remove oil vapour from the compressor in order to avoid emulsification with condensation in the line. A built-in or an attached auto drain will ensure a regular discharge of accumulated water.The filter is generally a quick change cartridge type.

Air Consuming System

Air DistributionThe air main is a permanently installed distribution system carrying the air to the various consumers. There are two main layout configurations: DEAD END LINE and RING MAINDead End LineTo assist drainage, the pipework should have a slope of about 1 in 100 in the direction of flow and it should be adequately drained. At suitable intervals the mains can be brought back to its original height by using two long sweep right angle bends and arranging a drain leg at the low point.

Ring MainIn a ring main system main air can be fed from two sides to a point of high consumption.This will reduce pressure drop. However this drives condensate in any direction and sufficient water take-off points with Auto Drains should be provided. Isolating valves can be installed to divide the air main into sections. This limits the area that will be shut down during periods of maintenance or repair.

Sizing Compressed Air MainsIn a closed loop ring main system, the supply for any particular take-off point is fed by two pipe paths. When determining pipe size, this dual feed should be ignored, assuming that at any time air will be supplied through one pipe only.The size of the air main and branches is determined by the limitation of the air velocity, normally recommended at 6 m/s, while sub-circuits at a pressure of around 6 bar and a few metres in length may work at velocities up to 20 m/s. The pressure drop from the compressor to the end of the branch pipe should not exceed 0.3 bar. The nomogram at the end of this section allows us to determine the required pipe diameter.Bends and valves cause additional flow resistance, which can be expressed as additional (equivalent) pipe lengths in computing the overall pressure drop. The table below gives the equivalent lengths for the various fittings commonly used.

Sizing Distribution/Compressor Air Mains. Example 1Example 1:Determine the size of pipe that will pass 16800 l/ min of free air with a maxinium pressure drop of not more than 0.3 bar in 125 m of pipe. The 2 stage compressor switches on at 8 bar and stops at 10 bar: the average is 9 bar.Solution to Example 1:1 bar is 300Kpa therefore 0.3 bar is 30Kpa.30 kPa pressure drop in 125 m of pipe is 30Kpa/125m = 0.24 kPa / m.Draw a line from 9 bar on the pressure line through 0.24 kPa / m on the pressure drop line to cut the reference line at X.16800 l/min = 16800/60 l/s = 280l/s = 280/1000 m3/s = 0.28 m3/sJoin X to 0.28 m /s and draw a line to cut the pipe size line at approximately 61 mm.Pipe with a minimum bore of 61 mm can be used, a 65 mm nominal bore pipe (would satisfy the requirements with some margin.

Standard Gas Pipe Nominal Sizes Air Hose Pipe Nominal Sizes

Sizing Distribution/Compressor Air Mains. Example 2Example 2:Determine the size of pipe that will pass 16800 l/ min of free air with a maxinium pressure drop of not more than 0.3 bar in 125 m of pipe. The 2 stage compressor switches on at 8 bar and stops at 10 bar: the average is 9 bar as before. This circuit has a number of fittings in the line, e.g., two elbows, two 90 bends. six standard tees and two gate valves, will a larger size pipe be necessary to limit the pressure drop to 30 kPa?Solution to Example 2:1 bar is 300Kpa therefore 0.3 bar is 30Kpa.two elbows:2 X 1.4 m = 2.8 mtwo 90 bends:2 X 0.8 m = 1.6 msix standard tees:6 X 0.7 m = 4.2 mtwo gate valves:2 X 0.5 m = 1.0 mTotal= 9.6 m The twelve fittings have a flow resistance equal to approximately 10 m additional pipe.30 kPa pressure drop in 135 m of pipe is 30Kpa/135m = 0.22 kPa / m.Draw a line from 9 bar on the pressure line through 0.22 kPa / m on the pressure drop line to cut the reference line at X.16800 l/min = 16800/60 l/s = 280l/s = 280/1000 m3/s = 0.28 m3/sJoin X to 0.28 m /s and draw a line to cut the pipe size line at approximately 61 mm.Referring again to the nomogram.: The pipe size line will now cut at almost the same dia; a nominal bore pipe of 65 mm will be satisfactory.Note: The possibility of future extensions should be taken into account when determining the size of mains for a new installation

Air Take OffsUnless an efficient aftercooler and air dryer are installed, the compressed air distribution pipework acts as a cooling surface and water and oil will accumulate throughout its length.Branch lines are taken off the top of the main to prevent water in the main pipe from running into them, instead of into drainage tubes which are taken from the bottom of the main pipe at each low point of it. These should be frequently drained or fitted with an automatic drain.Auto drains are more expensive to install initially, but this is off-set by the man-hours saved in the operation of the manual type. With manual draining neglect leads to compound problems due to contamination of the main.

Float Type Automatic DrainIn the float type of drain, the tube guides the float, and is internally connected to atmosphere via the filter, a relief valve, hole in the spring loaded piston and along the stem of the manual operator.The condensate accumulates at the bottom of the housing and when it rises high enough to lift the float from its seat, the pressure in the housing is transmitted to the piston which moves to the right to open the drain valve seat and expel the water. The float then lowers to shut off the air supply to the piston.The relief valve limits the pressure behind the piston when the float shuts the nozzle. This pre-set value ensures a consistent piston re-setting time as the captured air bleeds off through a functional leak in the relief valve.

Air TreatmentFiltering, Pressure Regulation and Air Lubrication are carried out to suit particular requirements. You can look these up if you need to in the future.We will not deal with these but will go on to look at actuators and directional control valves so you can control a pneumatic system with a PLC.

Review: Identify the two distribution systems on the left and do you recall how we size air mains?

Review: Recall the air take off strategies and how a float type auto drain operates.

ActuatorsThe work done by pneumatic actuators can be linear or rotary . Linear movement is obtained by piston cylinders, reciprocating rotary motion with an angle up to 270 (standard) by vane or rack and pinion type actuators and continuous rotation by air motors.There are two basic types of linear actuators from which special constructions are derived: Single acting cylinders with one air inlet to produce a power stroke in one direction. Double acting cylinders with two air inlets to produce extending and retracting power strokes.Single Acting CylinderA single acting cylinder develops thrust in one direction only. The piston rod is returned by a fitted spring or by external force from the load or spring. It may be a push or pull type.The Single acting linear actuator is shown belowSingle acting cylinders are used for clamping, marking, ejecting etc. They have a somewhat lower air consumption compared with the equivalent size of double acting cylinder. However there is a reduction in thrust due to the opposing spring force, and so a larger bore may be required. Also, accommodating the spring results in a longer overall length and limited stroke length.

Double Acting Linear Cylinder

With this actuator, thrust is developed in both extending and retracting directions as air pressure is applied alternately to opposite sides of a piston. The thrust available on the retracting stroke is reduced due to the smaller effective piston area, but is only a consideration if the cylinder is to pull the same load in both directionsCylinder ConstructionThe construction of a double acting cylinder is shown. The barrel is normally made of seamless tube which may be hard coated and super-finished on the inner working surface to minimise wear and friction. The end caps may be aluminium alloy or malleable iron castings held in place by tie rods or, in the case of smaller cylinders, fit into the barrel tube by screw thread or be crimped on. Aluminium, brass, bronze or stainless steel may be used for the cylinder body for aggressive or unsafe environments.

DIRECTIONAL CONTROL VALVESControl valves are used to control the movement of actuators.

Valve FunctionsA directional control valve determines the flow of air between its ports by opening, closing or changing its internal connections.The valves are described in terms of the number of ports, the number of switching positions, its normal ( not operated ) position and the method of operation. The first two points are normally expressed in the terms 5/2, 3/2, 2/2 etc. The first figure relates to the number of ports (excluding pilot ports) and the second to the number of positions.The main functions and their ISO symbols are explained in the next few slides.

Explaination Of The Valve Symbol

Control Valve Operator Symbols

Putting The Symbols Together

Control Valve Port IdentificationThe identification of the various ports are not uniform; it is more of a tradition than a respected standard.Originally, the codes previously used the older hydraulic nomenclature. P for the supply port comes from pump, the hydraulic source of fluid energy.The outlet of a 2/2 or 3/2 valve has always been A, the second, antivalent output port B.The exhaust is invariably R from Return (to the oil tank). The second exhaust port in 5/2 valves was then named 5, or the former Ri and the latter R2.The pilot port initiating the power connection to port A was originally coded Z (the two extreme letters in the alphabet belong together) and the other Y.After 20 years bargaining about pneumatic and hydraulic symbols, one of the ISO work groups had the idea that ports should have numbers instead of letters, delaying the termination of the standard ISO 1219 by another 6 years. Supply should be 1, the outputs 2 and 4, the pilot port connectingl with 2 is then 12 etc. Table 7.2 shows the four main sets of port identifications in use. The preferred option is now numbers.

Standard Control Valves

Speed Control Of An ActuatorA speed controller, sometimes referred to as a Uni-Directional Flow Control Valve, consists of a check valve and a variable throttle in one housing.A typical example with the flow indicated is shown below.In (a), air flows freely to the cylinder, in (b) it flows back to the exhaust port of the valve with a restricted flow.

Rest PositionMechanically operated valves, controlling the rest positions of the cylinder driven parts, are operated in rest and have to be drawn accordingly: with the external connections drawn to the square on the operator side. In a normally closed 3/2 valve, the output is then connected with the supply and therefore under pressure. Equally, if the signal line to a monostable air operated valve is under pressure, it has to be drawn in the operated position.Further rules are:As the rest position is 0, all end-of-stroke valves operated in the rest position have an index zero. Those operated in the opposite end (work position) have an index 1.

Circuit LayoutIn a circuit diagram, the flow of the working energy is drawn from the bottom to. the top and the sequence of the working cycle from the left to the right. Consequently, the air supply (FRL[With a Filter, Regulate and Lubricate on the supply line] ) unit is situated in the lower left corner, the cylinder that performs the first stroke of the cycle, in the upper left corner etc.The power valves are drawn directly below their cylinders; they form a Power Unit that is coded with a capital letter (see Nomenclature). Between them and the power units, there may be additional valves to ensure the correct sequence (memory function) and sometimes additional valves to realise certain interlocks by logical functions..

Basic Rules

A circuit diagram is drawn in the rest position of the controlled machine, with the supply under pressure, but in the case of mixed circuits, without electrical power. All components must be drawn in the positions resulting from these assumptions. The diagram across illustrates this:

ReviewIn the last class we reviewed the operation of actuators and control valves and we are now going to put this into practice to read and understand the operation pneumatic circuits.

Circuit 1Initially, in the circuit across, everything is shown in the de-activated position as shown. The spring return in valve 2 ensures that the unactivated box is used as shown because there is no air to activated switch 2, this line is vented. This results in the pressure operated lamp turning on.When the push button 1 is actuated,the second box in the control valve is how the circuit now looks. Air can then actuate Valve 2. This means that the box shown in the lower circuit is activated and the pressure activated light is vented, so it is not on.

Circuit 2Take five minutes to work out the operation of the following circuit.

The Operation Of Circuit 2Figure 1 shows the circuit at rest. The red pressure operated light is on.Figure 2 shows the block changing as soon as the pushbutton is pressed. Figure 3 shows block 2 switching to the activated block immediately pushbutton 1 is pressed.When the pushbutton is released, the pushbutton block and block 2 return to the rest position as the are both spring operated. This is illustrated in figure 4.

Basic Circuit 3. MemoryAgain take five minutes to work out the operation of the following circuit.

The Operation Of Circuit 3Pushbutton 1 is activated in figure 1. Valve 3 activates as shown and the green light is on.Releasing pushbutton 1, valve 3 maintains its position as shown i.e. memory.Pushbutton 2 is activated in figure 3. Valve 3 de-activates as shown and the red light is on.Releasing pushbutton 2, valve 3 maintains its position as shown i.e. memory.

Application Of Speed ControllersCircuit 4 and 5Circuit 4 across uses a speed controller to delay the activation of valve 2. For long delays a reseviour is added as shown, as valve 2 cannot activate until the reseviour is full.Circuit 5 across shows the speed controller limiting the speed of the movement of the actuator.

Circuit 6What is the logic function the following circuits are executing?Both circuits execute the and function.

Circuit 7; MemoryCircuit 7 operates the same as circuit 3, except that here an actuator is controlled by the memory block and the actuator is speed controlled.

Circuit 8. Form into groups of 3 and take five minutes to work out the operation of this circuit.Consider the following application, the simple simulation of the application and the control circuit for the same system. Explain the operation of the circuit?

The Operation Of Circuit 8co is locate at point 2 and c1 is located at point 4.Valve 1 is shown actuated because it the home position. Valve 4 is not activated because it is in the final position.Press start lever 1. Air goes through to C+(valve 3) if the actuator is at the home position.Valve 3 activates and air goes through to the acuator and forces it out. When it extends valve 4 operates and a C- signal is received. This reverses valve 3 position and air retracts the actuator. When it reaches home the cycle repeats until the start is turned off.

Circuit 9. Explain the operation of circuit below.

The Operation Of Circuit 9Press 1, air appears at 3, which results in the actuator extending. This trips valve 2, which gives a 3- signal. The actuator retracts and stays in the home position.Press 1 to start again.

Circuit 10. Explain the operation of the circuit below.

The Operation Of Circuit 10

The Operation Of Circuit 11;a0M+A+a1B+b1M-B-b0A-a0

Circuit 12 Explain the operation of the circuit?

The Operation Of Circuit 12;a0M+A+a1B+b1M-B-b0A-a0

The Operation Of Circuit 13;startA+a1B+b1B-b0 end of cycle

The Operation Of Circuit 13;start1+A+2B+b1B-b0 4+1- end of cycle

Electro-pneumatics

Electro-PneumaticsElectro-pneumatics is the application of electrical and electronic equipment to control pneumatic circuits

Why Use Electro-Pneumatics?Electronic/Electrical control of pneumatics is the most efficient method for compex circuits as can be seen from the graph.Purely pneumatics equipment is extremely reliable and is still used for relatively simple circuits.The graph etr indicates the cost with the additional maintenance involved with electro-pneumatics

Electro-Pneumatic Equipment; The Solenoid ValveBelow is a 3/2 Solenoid Valve. When the coil is energised the seal disk is retained against the inner rim of the chamber as shown in figure c, this allows the pressurised air to go through to port a.

Electro-Pneumatic Equipment; The Pilot Operated Bistable Spool ValveThe solenoids at the end of the spool valve operate like the solenoid valve we have just seen. The difference is that on operation the valve allows air in to operate the spool valve. This is called pilot operation. Pilot operation makes the overall circuit more efficient.

Electro-Pneumatic Equipment; Magnetic Cyclinder SwitchesMagnetically operated cyclinder switches operate by closing when the valve seal passes it.If the switch, which is mounted on the cyclinder, was located here, everytime the seal passed this point the switch closesMotion Of Spool Valve Seals

Electro-Pneumatic Equipment;Electrical RelaysThe basic function of a relay is to open or close contacts as required to control the circuit.In figue a, no power is applied to the coil, and a contact is closed. This is called the normallly closed contact, as electrical equipment is named when it is at rest.When power is supplied to the coil, the normally closed contact opens and the normally open contact closes as shown in figure b.Some real relays are shown across.

Electrical SymbolsElectrical circuits are represented by electrical symbols. These symbols are detailed on the next few slides. You do not need to know all of the electrical symbols, the circuit on the left covers most of the components you will be dealing with.Some of the common pneumatic switches are indicated below.

Single Cyclinder Circuits: A Basic CircuitConsider the circuit above. The valve is controlled by a solenoid, as described in electromagnetic equipment section. When the coil is operated, the valve actuates. On turning off the power, the spring which is compressed forces the valve to revert to its rest position. It is shown on the next slide.

The circuit across is the rest position of the circuit. Pressurised air retracts the cyclinder and extension part of the cyclinder is vented.The cyclinder is at home or the A- position.In the second circuit the push button is pressed and the solenoid activates the valve as shown. This extends the cyclinder and vents the home side of the cyclinder.The cyclinder is at the extended or A+ position.On removing the power, i.e. releasing the push button, the compressed spring forces the cyclinder back to the home position.Single Cyclinder Circuits: A Basic Circuit

Single Cyclinder Circuits: Controlling the pushbutton circuit from two locations.This circuit behaves exactly like the previous circuit except that it can be controlled from two locations.

Single Cyclinder Circuits: Control of Bistable Valve.The valve shown here is stable in both the S- ans S+ positions so it can remember the last pushbutton pressed.The linear slide actuator is shown with S- actuated, so the last pushbutton to be pressed is pb2.On pressing pb1 operates the bistable valve and the linear slide actuator go to the S+ position and remains there until pb2 is pressed.

Single Cyclinder Circuits: Rotational Motion, one cycle circuitPress the start push button;A+ coil is energised. This switches the position of the valve and the direction of rotation is changed. When A+ activates and closes magnetic proximity switch a1, coil A- is activated. This reverses the position of the valve to the status shown.Note:Some problems with this circuit;If you hold in the pushbuttonb A+ and A- will activate together.The rotational actuator will continue to rotate when the electrical power is off. How would you fix this?

Single Cyclinder Circuits: Automatic CyclingPress the start push button;(Assumption b0 is actiavted)B+ coil is energised. This switches the position of the valve and the linear slide actuator is extended.

When B+ activates and closes magnetic proximity switch b1, coil B- is activated. This reverses the position of the valve to the status shown.

The cycle repeats, as b0 now closes activating B+, until the start switch is released.

Single Cyclinder Circuits: Hold In ContactsCircuit a: Push button 1 or pb 2 activates A and extends the actuator.Circuit b: Push button 1 turns on the relay R1. R1 contact turns on the lamp and solenoid A and extends the actuator. If pushbutton 2 is pressed the solenois A is turned on directly.

Single Cyclinder Circuits: Timer Delay Closing Circuits: Circuit APress start, Assuming co is closed the timer relay T1 turns on. The contact, T1, is delayed moving from left to right and then closing.This extend the actuator [C+] and c1 will then close and retract the actuator [C-].

Single Cyclinder Circuits: Timer Delay Closing Circuits: Circuit BPress start, Assuming co is closed the timer relay T1 turns on. The contact, T1, is delayed moving from left to right and then closing.This extend the actuator [C+] and c1 will then close the timer relay T2 turns on. The contact, T2, is delayed moving from left to right and then closing.This will close after the set time and retract the actuator [C-].

Single Cyclinder Circuits: Timer Delay Opening CircuitsAssumption a0 will trip in unactuated state and a1 will trip in the actuated state.Assume initially a0 is on and a1 is off. Press start, TR turns on, R1 turns on, this closes the R1 contacts, changing the actuators direction[A]. After the set time TR opens, when a1 actuates, power is supplied to R2, which de-energises the R1. The cycle repeats like this until the start switch is turned off.

Two Cyclinder CircuitsWe will go through this circuit step by step to show you how it works.

Two Cyclinder Circuits

PracticeTry the following circuits yourself.

Two Cyclinder Circuits

More Complex Circuits

Operational AmplifiersAmplification

AmplifiersAn Amplifier is an electronic device or circuit which creates a larger {more powerful} copy [amplifies]of a voltage or current input signal (or both). The quality of an amplifier is determined by seeing how accurately the input signal was replicated during amplification.The additional power the output signal has is obtained from the external DC Power Supply.

Amplification [Voltage example]Vout is a larger copy of Vin

VinVout

Operational AmplifiersGainGain is a unitless quantity because it is a ratio.Gains can go from 1 to over a million. The amount of zeros can cause transcribing errors so decibels are sometimes used to express gain in a more convient way.Voltage Gain is Vout/Vin = AvCurrent Gain is Iout/Iin = AiPower Gain is Pout/Pin = Ap

Operational AmplifiersDecibelsGain is a unitless quantity because it is a ratio.Gains can go from 1 to over a million. The amount of zeros can cause transcribing errors so decibels are sometimes used to express gain in a more convient way.

Decibels (dBs)An amplifier provides power gain.Since gains can range from 1 to 1,000,000 the more convenient decibel notation is preferred by electronic engineers. For audio frequencies and above it is more convenient to use the following expression for power: P = V2/RPower Gain = Pout/Pinif Rin = Rout = RAp = Pout = Vout2/R Pin Vin 2/R

So Ap = Vout2 Vin 2A bel format is obtained by getting the log of this gain.Ap = logAp Bels= log[Vout2/Vin 2] Bels= 2log[Vout/Vin ] BelsWe bring it to decibels, which is 1/10th of a bel by multiplying by 10, because of the deci prefix, to obtain;

Ap = 20 log [Vout/Vin ]dBsAp = Av = 20 log Av dBs The indicates gain in decibels

ExamplesWhat is an amplifier.State the equation for converting a voltage gain to decibels.State the equation for converting a decibel voltage gain to a voltage gain.For the following input and outputs calculate the gain and the gain in decibels for the following examples.(i)Iin DC = 2mA, Iout DC = 200mA (ii)Iin AC = 4mA, Iout AC = 40mA (iii) Vin DC = 4mV, Vout DC = 200mV (iv) Vin AC = 2mV, Vout AC = 50V (v) Pin = 10W, Pout = 40W Convert the following decibel gains to gains;(i)Av = 100dB(ii)Ap = 66dB (iii)Ai = 20dB

Solutions To ExamplesWhat is an amplifier.

State the equation for converting a voltage gain to decibels. Av = 20 log [Av]State the equation for converting a decibel voltage gain to a voltage gain.Av = log-1[Av/20]For the following input and outputs calculate the gain and the gain in decibels for the following examples.(i)Iin DC = 2mA, Iout DC = 200mA Ai = Iout/Iin = 200mA/2mA = 100 Ai = 20 log 100 = 20 * 2 = 40 dB(ii)Iin AC = 4mA, Iout AC = 40mA Ai = Iout/Iin = 40mA/4mA = 10 Ai = 20 log 10 = 20 * 1 = 20 dB

Solutions To Examples(iii) Vin DC = 4mV, Vout DC = 200mV Av = Vout/Vin = 200mV/4mV = 50 Av = 20 log 50 = 20 * 1.699 = 33.98 dB

(iv) Vin AC = 2mV, Vout AC = 50V Av = Vout/Vin = 50V/2mV = 25000 Av = 20 log 25000 = 20 * 4.398 = 87.96 dB

(v) Pin = 10W, Pout = 40W Ap = Pout/Pin = 40W/10W = 4 Ap = 20 log 4 = 20 * 0.602 = 12.04 dB

Convert the following decibel gains to gains;(i)Av = 100dB Av = log-1[Av/20] = log-1[100/20]= 100,000(ii)Ap = 66dB Ap = log-1[Ap/20] = log-1[66/20] = 1995.26

(iii)Ai = 20dBAi = log-1[Ai/20] = log-1[20/20] = 10

OPERATIONAL AMPLIFIERS

Transistors Transistor circuits are the basis for all types of analog amplifiers and also the fundamental building block of most digital circuit components. e. g microprocessor memory

The Single Transistor, The Simplest Amplifier.The transistor revolutionised electronics and our world. It is a simple device which could amplify the power of a signal to produce a larger copy of the input signal.This basic circuit requires a lot more components to eliminate problems and increase amplification. We will look at some of these circuits later in the year.A number of basic amplifier strategies were developed each with their own applications but the most widely used transistor amplifier circuit is the operational amplifier in a chip form

Operational Amplifiers Introduction

741 CIRCUIT DIAGRAMThe actual circuit for a 741 is shown below. It is a combination of three different types of amplifier all based on a transistor amplifier.

The Operational AmplifierPin numbers used are for the 741 Op AmpThe schematic diagram of an Operational Amplifier is shown across.It is common for the power supply circuit not to be included for simplicity.This circuit is formed on a silicon chip and encapsulated as shown later to form an integrated circuit on a chip. Integrated circuits are more reliable, simpler to use and cheaper to manufacture than the discrete version of the circuit.

741 DIL PackageAfter the integrated circuit is made, it is encapsulated into a Dual in Line Package.The chip and the pin-out are shown across.Different op-amps have different characteristics and are designed for different applications. The 741 is a general purpose op amp and we will base most of our circuits on it, even though their might be better chips for the application.Each different type of op amp will have its own pin-out although this arrangement is not uncommon.

What The Operational Amplifier Does!The Operational Amplifier has two input terminals and one output terminal.There is a non-inverting input terminal and an inverting input terminalExternal power is supplied to the Op-Amp via pins 4 and 7. Pin 4 has the negative supply voltage, -Vee, attached to it [or O in some cases] and Pin 7 has the positive supply voltage Vcc attached to it.Pin 1 and 5 are for zeroing the op amp and removing any offsets.Pin 6 is the output and pin 8 is not used.

What The Operational Amplifier Does!The equation for the operational amplifier, (when not saturated), is Vout = AOL * (V+ - V-)The operational amplifier is differential amplifier. That is, it amplifies the difference between the non-inverting and inverting input terminals. The differential input is represented in the operational amplifier equation by (V+ - V-)AOL is the open loop gain of the operational amplifier.AOL is typically 100dB or 100,000

What The Operational Amplifier Does!The output voltage is limited between the supply voltage, i.e. -Vee < Vout

What The Operational Amplifier Does!When the inverting terminal voltage is more positive than the non-inverting terminal the output is negative.When the non-inverting terminal voltage is more positive than the inverting terminal the output is positive.These can be deduced from the Op Amps equationVout = AOL * (V+ - V-)An op amp can be therefore be used as a comparator in this form but it is not much good as an amplifier as we shall see now.


Open Loop Gain Open loop gain means that the output of the circuit is not connected to the input.

Linear Operation Of An AmplifierLinear in this context means that there is only one output for each input.Simply it can be described as;Ouptut = Input * Constant [Number]This is what you would expect from an amplifier as it to make an exact larger copy of the input.

The Operational Amplifier Open Loop GainConnect a circuit up as shown.An amplifier is only functionalas an amplifier in its linear region. This is the region where there is a different output for every input becauseVo = Constant * Vinso for every value of Vin a different output should be obtained to have an exact copy of the input voltage.The results are graphed after the exercise on the next page showing how small the actual linear region is; in this example it is 200 microvolts.

Class Open Loop Exercise. You have five minutes. Off You Go.Let Vcc = 10 V, Vee = -10V, andAOL = 100,000.Calculate Vo for the values below of Vin [(V+-V-) as shown on the diagram] from -200V Tto 200 V in steps of 50 V.Plot Vout versus Vin.Determine the linear region Q. What is max and min output from the op amp?Note: = 10-6

Linear RegionSaturation; one output for all inputs.Saturation; one output for all inputs.Output voltage in voltsDifferential input [V+-V-] in microvolts

The Operational Amplifier Open Loop GainMost inputs are bigger than this so to use this circuit as an amplifier, the input range should be reduced or use feedback to effectively reduce the input voltage.Most of the circuits we will work with this year use feedback and we will have a brief look at it before we move on.

Think About ThisFor the previous explaination. Draw the output signal from the op amp for the following input waveforms;

A 200V peak to peak ac signal



Non Linear Operation Of An AmplifierNon linear in this context means that there is not one output for each input.What you would expect from an amplifier as it to make an exact larger copy of the input. In non linear mode the output waveform is not an exact copy of the input wave because the op amp saturates.We will look at the comparator as the most basic non linear circuit.

OPERATI ONAL AMPLIF IERS

Comparator Circuit A comparator can distinguish between the magnitude of two voltages. It is the simplest of the operational amplifiers circuits. It is a non linear circuit.

The Operational Amplifier As A ComparatorConsider the circuit shown above. When the non inverting input goes above zero volts the output is positive[ Vcc]. When the non inverting input goes below zero volts the output is negative[-Vee]

The Operational Amplifier As A ComparatorThe disadvantage of the extremely narrow linear region of the amplifier can be turned to an advantage if the op amp is used as a comparator.As soon as one input becomes about 0.1mV more positive than the other input the output will reach saturation [maximum or minimum output, Vcc or -Vee]If V+ > V-, Vout = Vcc If V+ < V-, Vout = -VeeIf V+ = V-, Vout = 0V

The Operational Amplifier As A ComparatorConsider the circuit shown above. When the inverting input goes above zero volts the output is negative[-Vee] When the inverting input goes below zero volts the output is positive[ Vcc].

The Operational Amplifier As A ComparatorConsider the circuit shown above. When the non inverting input goes above the reference voltage the output is positive[ Vcc]. When the non inverting input goes below reference voltage the output is negative[-Vee]

Think About ThisFor the comparator described,can you work out the output voltage for all possible inputs.Can you draw the circuit for the comparator.What would the output look like if Vee = 0


Feedback Feedback occurs when there is a feedback path[Connection] between the output and the input. There are two types of feedback, negative and positive.

When there is feedback, the circuit gain is called the closed loop Gain, ACL.It is obtaine by dividing the circuit output by the circuit input. In the following examples the amplifier A would be the op amp.

Positive Feedback

Negative Feedback

Class Feedback Exercise. Again you have five minutes. Off You Go.Determine the gain of the negative feedback circuit shown on the previous slide.Tip. Start with calculating in - out and work your way to the output, out. Then re-arrange the equation to obtain out/ in to determine the closed loop gain of the circuit, ACL.The Circuit is closed loop because the output is connected to the input.

Negative Feedback. The Maths. How it Works.in-[B*out] = error

error * A = out

So {in-[B*out]} * A = out

A.in-B.out.A = out

A.in= out +B.out.A

A.in= out [1+B.A]Therefore out = A. in [1+B.A]If AB>>1out = A. = 1 in B.A B

This is the advantage of the op amp having a very large gain that is AB>>1, so the overall gain of the circuit just depends on the feedback network, B.The feedback network in our initial circuits is normally resistors. Therefore the overall gain of the circuit can be controlled by selecting the value of the resistors in the circuit.This gives us a more useful linear range in our circuits than that available from an op amp alone.

Think About ThisConsider the circuit across that you have just analysed.The amplifier, A, is an op amp in a circuit and the feedback network is a series of resistors with value B.We have seen that the gain of the circuit, ACL = 1/B. What happens the circuit gain, ACL , if the op amp is replaced with an op amp with a different gain?


Characteristics of the operational amplifier. We will now review three of the most important characteristics of operational amplifiers, input resistance [Zin] output resistance [Zout] and Open Loop Gain [Aol] and see how the ideal characteristics arise.

Operational Amplifier GainAs already discussed in the open loop gain section the gain of an operational amplifier is circa 100dB .In reality this means that the output voltage is 100,000 times bigger than the input and any difference in the input basically drives the output to saturationIdeally the open loop gain, Aol, of the operational amplifier is infinity.

Think About ThisWhy do we assume that the ideal gain of the op amp infinity. By assuming the gain is infinity, what characteristic of the op amp do we ignore?Can you work out what a gain of 100dB would mean if it was applied to the law of the lever? Assume the short end is 1cm.

Operational Amplifier InputImpedance

Because the input to the operational amplifier is through the base of a transistor, the current is very low. In addition it also has to flow through a large resistance Re which contributes to ib being very small, typically in the low micro range to the pico range for specialized op amps. If the input current is low, the input impedance is considered high.Because the input current is normally a thousand or million times smaller than the other circuit currents we can consider the input current to be zero because the input impedance is ideally infinity.

Output Impedance of anOperational Amplifier. The output circuit of the op amp is shown across. There is only a transistor between Vcc and the output, so the resistance is relatively low, about 75 Because this is much lower than most electronic circuit resistances we can idealise it to be zero.So Zout=O

What we are about to doWe will determine how we can use the characteristics of an op amp to analyse an op amp circuit.We will review the ideal characteristics and determine a procedure for analysing linear op amp circuits

Review Of The Ideal Characteristics Of An Operational Amplifiers

Aol, the open loop gain of the operational amplifier is ideally infinity. It can therefore be deduced that the difference between the inputs to the operational amplifier is zero. V+-V- = Vout/ = 0Zin, the ideal input impedance to an operational amplifier, is infinity. We can therefore say that the current entering the op amp is also ideally zero. I+=I-=oZout, the ouput impedance is ideally zero.This means that the output that is predicted should appear at the output of the op amp.


Ideal Characteristics of the operational amplifier.The input resistance, Zin, is infinity so no current enters the op amp.The output resistance, Zout, is zero so the desired output voltage is obtained at the output.The Open Loop Gain, Aol,is infinity so the difference between the two inputs is zero.

Deriving the gain of an operational amplifier circuit

Step1. Make the current entering the op amp input zero. I+=I-=oStep2. Draw in the circuit currents. Assume directions if required. Remember current flows from positive to negative. Step3.Find the fixed voltage. Normally on the opposite side to Vout. The fixed voltage is normally Vin or 0V.Step4. Make V+=V- and bring the fixed voltage to the other side of the op amp circuit, normally the output side.Step5. Solve for Vout/Vin using Ohms law, KVL or KCL as required.

Most of our circuits this year can be arranged as shown shown. This circuit can easily be altered to form inverting, non-inverting or summing amplifiers.The basic format of most basic operation amplifier circuits.

OPERATIONAL AMPLIFIERS The inverting operational amplifier circuit. The output voltage is the opposite polarity to the input voltage, as the name of the circuit suggests. We will now use the steps derived from the op amps characteristics to determine the closed loop gain of the complete circuit, i.e. Acl = Vout/Vin

The circuit for an Inverting Operational Amplifier Circuit

OPERATIONAL AMPLIFIERS The non-inverting operational amplifier circuit. The ouput voltage is in phase with the input voltage to the input voltage, as the name of the circuit suggests. We will now use the steps derived from the op amps characteristics to determine the closed loop gain ofthe circuit, i.e. Acl = Vout/Vin

The circuit for the Non-Inverting Operational Circuit is as shown below;

ExamplesDesign an Operation amplifier circuit with a voltage gain of 50.Design an Operational amplifier circuit with a voltage gain of 40dB.An inverting amplifier has a gain of 30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals.(i)0.3V DC(ii)0.5 VDC(iii)-0.2VDC(iv)0.6V p-p sine wave(v)1V p-p sine wave

Design an Operation amplifier circuit with a voltage gain of -50.

A voltage gain is 50. The circuit required is the non inverting amplifier which is shown below. The closed loop voltage gain ofthe circuit is given by; ACL = R1+R2 R1

Acl = R1+R2 = 50 R1

thereforeR1 + R2=50*R1OrR2 = 49*R1Let R1 = 10kWhich gives a value for R2

R2 = 49*lOk= 490K

Design an Operation amplifier circuit with a voltage gain of -40dB.

A voltage gain is -40dB. The circuit required is the non inverting amplifier which is shown below. The closed loop voltage gain ofthe circuit is given by; ACL = R1+R2 and R1Acl = 20Log([R1+R2]/R1) dBs

Acl =20Log(R1+R2/R1) dB40 = 20Log(R2/R1) dBs2 = Log(R2/R1) dBsAcl = Log -1 2 = 100So 100= R1+ R2 R1thereforeR2=99*R1

Let R1 = 10kWhich gives a value for R2

R2 = 99*lOk= 990k

An inverting amplifier has a gain of 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (i) O.3V DC Vo = 3OVin = 30 * O.3VDC = 9VDC

The input and output signals are in phase as you can see, if you compare the input and output waveforms

An inverting amplifier has a gain of 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (ii) O.5V DC

Vo = 3OVin = 30 * O.5VDC = 15VDC

The calculated ouput is 15V DC. This cannot be supplied as the maximum output is Vcc. Here it is 10V DC. This circuit is saturated at 10V DC.

An inverting amplifier has a gain of 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (iii) -O.2V DC Vo = 3OVin = 30 * -O.2VDC = -6 VDC

For DC signals the minus sign means the input is inverted as you can see if you compare the input and output waveforms

An inverting amplifier has a gain of -30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals. (iv)0.6V p-p sine wave

Vo = 30 Vin = 30 * 0.6 V p-p= 18 V p-p

For ac waves the minus sign means the wave is inverted as you can see if you compare the input and output waveformsInput 0.6V p-pOutput 18V p-p inverted

An inverting amplifier has a gain of -30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals. (v)1V p-p sine waveVo = 30 Vin = 30 * 1 V p-p= 30 V p-p

The Black is the actual output from the circuit which is chopped when it reaches the supply voltage. The grey is the calculated output for the gain and input.Input 1V p-p Output 30V p-p

ExamplesDesign an Operation amplifier circuit with a voltage gain of -50.Design an Operational amplifier circuit with a voltage gain of -40dB.An inverting amplifier has a gain of -30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals.(i)0.3V DC(ii)0.5 VDC(iii)-0.2VDC(iv)0.6V p-p sine wave(v)1V p-p sine wave

Think About ThisHow do you Invert an AC Wave?

Design an Operation amplifier circuit with a voltage gain of -50.

A voltage gain is -50. The minus, -, indicates that the circuit required is the inverting amplifier which is shown below. The closed loop voltage gain ofthe circuit is given by; ACL = -R2/R1

Acl = -R2 = 50R1

So 50=R2 R1

thereforeR2=50*R1k


R2 = 50*lOk= 500K

Design an Operation amplifier circuit with a voltage gain of -40dB.

A voltage gain is -40dB. The minus, -, indicates that the circuit required is the inverting amplifier which is shown below. The closed loop voltage gain ofthe circuit is given by; Acl = -R2/R1 and Acl = 20Log(R2/R1) dBs

Acl = 20Log(R2/R1) dBs40db = 20Log(R2/R1) dBs2 = Log(R2/R1) dBsAcl = Log -1 2 = 100So 100=R2 R1thereforeR2=100*R1k


R2 = 100*lOk= 1M

An inverting amplifier has a gain o f - 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (i) O.3V DC Vo = -3OVin = -30 * O.3VDC = -9VDC

For DC signals the minus sign means the input is inverted as you can see ifyou compare the input and output waveforms

An inverting amplifier has a gain o f - 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (ii) O.5V DC

Vo = -3OVin = -30 * O.5VDC = -15VDC

The calculated output is - 15V DC. This cannot be supplied as the minimum output is Vee. Here it is -10V DC. This circuit is saturated at -10V DC.

An inverting amplifier has a gain o f - 30 and a supply voltage of 1OV. Calculate and draw the outputs for the following input signals. (iii) -O.2V DC Vo = -3OVin = -30 * -O.2VDC = 6 VDC

For DC signals the minus sign means the input is inverted as you can see if you compare the input and output waveforms

An inverting amplifier has a gain of -30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals. (iv)0.6V p-p sine wave

Vo = - 30 Vin = - 30 * 0.6 V p-p= - 18 V p-p

For ac waves the minus sign means the wave is inverted as you can see if you compare the input and output waveformsInput 0.6V p-pOutput 18V p-p inverted

An inverting amplifier has a gain of -30 and a supply voltage of 10V. Calculate and draw the outputs for the following input signals. (v)1V p-p sine waveVo = - 30 Vin = - 30 * 1 V p-p= - 30 V p-p

The Black is the actual output from the circuit which is chopped when it reaches the supply voltage. The grey is the calculated output for the gain and input.Input 1V p-pOutput 30V p-p inverted

summing

Nonlinear Operational Amplifier CircuitsNon Linear OpAmp CircuitsFeedback not always present or not always present for all inputs.Input and output waveforms are not normally the same shape.Circuits routinely operate in saturation.Op amp equation is not always true.Typical applications are waveshaping complex mathematical operations and comparison of signals.Linear OpAmp CircuitsNegative feedback always present.Benefits of feedback always present i.e. reduced interference, offsets, gain, output impedance, increased input impedance.Input and output waveform shapes are the same.Circuits are not normally operated in saturation.Op amp equation is always true.Typical applications are amplification or completing mathematical operations.

A nonlinear operational amplifier is a circuit where;There is not a different output for every input signal and/orThe output waveform is not the same shape as the input waveform

The inverting Operational Amplifier Comparator Circuit. (A non inverting comparator is obtained by swaping the inputs to the op amp.)The basic function of a comparator is to compare two voltages and produce an output voltage to shown which input voltage is more positive;If V+ > V- then Vout is +Vcc or +Vsat and If V+ < V- then Vout is Vee or -VsatThe minimum difference in the input voltages required to produce a saturated output voltage is;Vout = AOL(V+ - V-)Typically AOL = 100,000 and Vout saturated is ideally 15v so Difference in input signals = Vout = (V+ - V-) = 150VAOL

V- InputV+ Input is zero here but it can be any reference voltageVout is +Vsat when V- < V+Vout is -Vsat when V- > V+The output voltage from the comparator does not change very quickly as can be noticeably seen from the slope when it changes stateA non inverting comparator would have the opposite output waveform to that shown

Noise a major problem for the basic comparator circuit. (Noise is simulated by adding a small sine wave to the previous input as can be seen on the next slide.)Noise causes the false triggering of the output and since the comparator is commonly used as an input to digital circiuts this will cause inaccurate outputs in the overall circuit because of the extra changes of comparator output state due to noise.

There are two basic problems with the basic comparator circuit;It changes state relatively slowlyIt is prone to false triggering of the output due to noise.

False output triggering caused by the noise on the input signalThe input is the original signal with noise

The Schmitt Trigger Operational Amplifier Circuit.The Schmitt Trigger is basically a fast comparator with hystersis. It is fast because it uses positive feedback to increase the speed of switching between +Vsat and Vsat and vice versa.It can be designed to be immune from the effects of noise by making the Hysteresis larger than the noise present.Note: Hysteresis basically means the circuit switches at one voltage and will only switch the opposite way at a higher or lower voltage than the original trip voltage.

The input signal to the circuit is connected to the inverting input terminalThis voltage here is the trip voltage. It is connected to the non-inverting terminal.The ouput voltage can have two possible values i.e. +Vsat or VsatThe voltage at the trip point can there have two values as shown belowTP = UTP(upper trip point) or LTP(lower trip point)TP = R1 * Vout so the UTP = R1 * Vsat and the LTP = R1 *- Vsat R1 + R2 R1 + R2 R1 + R2Hystersis = UTP LTP = R1 * Vsat -R1 *- Vsat = 2*R1 * Vsat R1 + R2 R1 + R2 R1 + R2

+Vsat-VsatLTP+UTPV- Input to the op ampV+ Input to the op ampThe output swaps when V+>V- i.e. when the input is less than the LTPThe output swaps when V+

TPVoutVinIt can be seen that the schmitt trigger output voltage changes extremely quickly. It goes from Vsat to +Vsat in 0.064 of a ms.This overcomes one of the limitations of a comparator. We can see how the positive feedback generates this faster swiching on the next page.0.064msVoutVin+Vsat-VsatLTP = -BVsatUTP = BVsatSchmitt Trigger Hysteresis

VinTP = UTPTP = LTPTP Vin or V+ - V- i.e. The input to the op ampThe positive feedback can be seen clearly here. As soon as the op amp switches the difference between the op amps inputs jumps from zero to the magnitude of the hystersis. This drives the schmitt trigger circuit to saturation extremely quickly. Also this means that any noise or interference has to be at least the same magnitude as the Hystersis to cause erroneous tripping as in the comparator circuit

Schmitt Trigger with Speed Up CapacitorC1 is stray capacitance. This means that the TP voltage cannot change until C1 charges. To speed this charging up C2 in included in the circuit to provide the charge needed to charge C1 and switch the TP voltage more quickly. This is only required if very fast switching is required.C2 = R2 * C1 R1Typically this value may be doubled and approximate values are between 10pF and 100pF

Schmitt Trigger Relaxation OscillatorThe period for the square wave is T = 2RC ln[(1+B)/(1-B)] and the freuency is f = 1/TThe circuit is based on the schmitt trigger. The RC circuit enables the circuit to oscillate and generate a regular square wave output. When the output is at Vsat the capacitor charges towards Vsat. When the capacitor voltage builds up to be more positive than UTP the op amp output switches to Vsat and the capacitor starts to discharge towards this voltage. When it drops below LTP the op amp output switches to +Vsat and the cycle begins again.

There seems to be no input voltage to start the oscillations. To start oscillations a small bit of noise present in all resistors is all that is required. If R Load has noise of +1mv the signal slowly builds up as shown hereT = time period of oscillations

UTP+VsatLTPVc, Discharging Towards -Vsat-VsatWhen Vc < LTP Vout changes state to +VsatWhen Vc > UTP Vout changes state to -VsatVc, Charging Towards +VsatT = Time period

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