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Introduction......................................ACR 2The AC motor ...................................ACR 3Synchronous motors .............................. ACR 4Induction (Asynchronous) motors ............ ACR 5Stator ....................................................... ACR 5Magnetic field ........................................ ACR 6Rotor ........................................................ ACR 7Torque, slip and speed ........................... ACR 7Losses and efficiency ............................. ACR 9Improper magnetization ......................... ACR 9Equivalent diagram ............................... ACR 10AFD Speed change ............................... ACR 11Motor data ............................................. ACR 13Types of load ........................................ ACR 15The AFD..........................................ACR 17The rectifier ........................................... ACR 18Uncontrolled rectifier ............................ ACR 19Full-wave controlled rectifier ............... ACR 19The intermediate circuit ....................... ACR 20The inverter ........................................... ACR 21Transistors ............................................ ACR 23Pulse Amplitude Modulation (PAM) ..... ACR 24Pulse Width Modulation (PWM) ........... ACR 25Danfoss VVC control principle ............. ACR 27Harmonics and the rotating field ............ACR 28The control circuit ................................. ACR 30The computer in general ...................... ACR 30The computer of the AFD ..................... ACR 31Inputs and outputs of the control card ACR 31Serial communication ........................... ACR 31Operational conditions of the motor ... ACR 34Compensations ..................................... ACR 34Motor torque characteristics ............... ACR 35Choice of AFD size ............................... ACR 35Load characteristics ............................. ACR 35

Power factor of the motor .................... ACR 37Normal operational conditions............. ACR 37Operation............................................... ACR 37Motor speed control ............................. ACR 38Acceleration and deceleration ............. ACR 38Braking .................................................. ACR 38Reversing .............................................. ACR 39Ramps ................................................... ACR 39Process monitoring .............................. ACR 40Motor load and heating ........................ ACR 40Efficiencies ............................................ ACR 41Long motor cables ................................ ACR 42Intermittent operation ........................... ACR 42Parallel connection of motors .............. ACR 43Hazardous locations ............................. ACR 43Transformers and AFDs ....................... ACR 43Protection under extreme

working conditions ......................... ACR 43Electrical noise ...................................... ACR 44Ways of emission .................................. ACR 44AC line interference .............................. ACR 45Transients/Overvoltages ...................... ACR 46Radio frequency interference .............. ACR 46Shielded cables .................................... ACR 47Operational reliability............................ ACR 47Simple trouble shooting ....................... ACR 47Fault indication ..................................... ACR 47Fuses ..................................................... ACR 47Short-circuits and ground faults .......... ACR 48Influence of the supply mains .............. ACR 48Considerations before buying.............. ACR 49Appendix I: General

Mechanical theory .......................... ACR 50Appendix II: General AC theory ........... ACR 52Appendix III: Conversion table SI US .... ACR 55Subject index ............................................. ACR56

ACTechnical

Reference

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AC Technical Reference

A static Adjustable Frequency Drive(AFD) is an electronic unit for speedcontrol of AC motors.

The AFD controls the motor speed byconverting the frequency and voltageof the mains supply from fixed tovariable values.

Industry today increasingly demandsautomated plants and higherproduction speeds. Great efforts aremade to improve productionmethods.

Advantages of variablespeed controlThe AFD controlled AC motor is usedin all kinds of automated plants. Apartfrom optimizing the features of the ACmotor, the variable speed control ofAC motors gives the followingadvantages:

Energy savingEnergy is saved as the motor speedis continuously matched to themomentary demand. A good exampleof this can be seen in pump andventilation equipment where thepower consumption is reduced by thecube of the speed.

Process improvementSpeed control according to theproduction process offers severaladvantages:

Production can be increased,consumption of materials and therejection rate can be reduced.

Improved qualityThe number of starts/stops isreduced. Unnecessarily harshoperation of the machinery cantherefore be avoided.

Less maintenanceThe AFD demands no maintenance.

Improved working environmentThe speed of conveyor belts can beadapted to the working speed. Inbottling plants the bottle noise isreduced by adjusting the speed tosuit the production rate. In ventilationapplications the fan speed can bematched to suit the demand so thatnoise and drafts are avoided.

The induction motor is an importantelement of production. That is why itis so important to find the optimummethod of motor speed control.

The AFD can be designed based onvarious control principles. Thegreatest development has been seenwithin AFDs utilizing fixed DC voltageintermediate circuits, as shown inFigure 1. This technical referencesection deals primarily with this typeof AFD.

Change of A.C.motor speed

AFD withoutintermediate

circuit

AFD withintermediate

circuit

DC currentintermediate

circuit

DC voltageintermediate

circuit

Cascadecouplings

Pulseinverter

Pulseinverter

Figure 1.

Figure 2.Feedback

ControllerRegulation

Control

Introduction

Open or closed loop?With open loop speed control a signalwhich is expected to produce therequired speed is sent to the motor.

With closed loop speed control afeedback signal is returned from theprocess, as shown in Figure 2. If thespeed does not correspond to therequirements, the signal to the motoris adjusted automatically until themotor speed is as it should be.

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Rotor withsalient polesFull pole rotor

Slip ring rotorSquirrel cage

AC motors

Synchronous Induction

The AC Motor

BackgroundThe first electric motor built in1833 was a DC motor. It wassimple to control speed and tomeet the demands of variousapplications.

In 1899, the first AC motor wasdesigned, The AC motor wasmore simple and robust thanthe DC motor. However, thefixed speed and torquecharacteristics of the first ACmotors have not been suitablefor all applications.

AC motors convert electricenergy into mechanical energy bymeans of electromagnetic induction.The principle of electromagneticinduction is: If a conductor is movingacross a magnetic field, a voltage isinduced. If the conductor is part of aclosed circuit , there will be a currentinduced.

In the motor, the magnetic fieldis placed in the stationary part(stator). The conductorsinfluenced by the electro-magnetic forces are located inthe rotating part (rotor).

AC motors can be divided upinto two types: induction andsynchronous motors. Inprinciple the stator works in thesame way in both motor types.They only distinguishthemselves in the way the rotorsare built up and are movingaccording to the magnetic field.

With synchronous motors the rotorand the magnetic field are running atthe same speed; with inductionmotors the rotor and the magneticfield are running at different speeds.

Figure 3. Generator and motor principle.

S

N

SI

FF

I

N

I FF IMotor principleGenerator principle

In motors, the induction principle isutilized in “reverse order”: a liveconductor is placed in a magneticfield. The conductor is influenced by aforce, which tries to move it throughthe magnetic field., Figure 3.

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Synchronous motorsThe rotor of the synchronous motorcan be built up in different ways.

The rotor with salient poles consistsof magnets (Figure 4). The magnetscan either be permanent magnets (forsmall motors) or electromagnets. Therotor has two or more pairs of poles,therefore it can also be used for lowspeed motors. This type ofsynchronous motor cannot start byitself. This is due to the inertia of therotor and the high speed of the

0

Torque

no Speed

Tn

TK

Figure 5. Full pole rotor with stamped-out poles and its torque characteristic

The speed of the synchronous motoris constant and independent of load.

The load on the synchronous motormust be within the electromagneticforce generated between the rotorand the magnetic field. Higher loadswill break the synchronism and themotor will stall.

For example, synchronous motors areused for parallel synchronousoperation of several mechanicallyindependent machines.

rotating field. The rotor must beaccelerated up to the same speed asthe rotating magnetic field. This canbe done with a start motor or an AFD.

The full pole rotor has stamped outslots on 2/3 of the rotor surface(Figure 5). Together these stampedout slots make up one pair of poles.This motor type is often called areluctance motor. A reluctance motorcan be used for high speeds as wellas low speeds, and it can be self-starting.

0

Torque

no Speed

TK

Figure 4. Rotor with salient poles and its torque characteristic

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Induction motorsThe induction motor is the mostcommonly used motor. It requirespractically no maintenance. Astandard design ensures that asuitable supplier can always be found.There are several types of inductionmotors, but they all work according tothe same basic principle.

StatorThe stator and the rotor are the twomain parts of an induction motor.

The stator is the fixed part of themotor. Mechanically it consists of: thestator housing (1), ball bearings (2)carrying the rotor (Figure 6), bearinghousing (3) closing the stator housing,fan (4) cooling the motor and the fancover (5) protecting against therotating fan. Finally there is a housingfor the electrical connection (6).

In the stator housing is an iron core(7) consisting of thin iron sheetscoated with a thin insulation. Thethree phase windings (8) are placed inthe grooves of the iron core.

The phase windings and the statorcore must produce the magnetic fieldin a number of pole pairs. It is thenumber of pole pairs, which

-16.53

29.17

5 4 3 2 10 9 2 1

7 36

Figure 6. The physical buildup of the asynchronous motor

determines the speed of the rotatingmagnetic field (Table 1). When amotor is operated at rated frequency,the speed of the magnetic field iscalled the synchronous speed of themotor No.

The phase windings consist of severalcoils. The number is dependent onthe required pairs of poles. In two-pole motors one coil covers

= 360°

number of poles 2 360°

180°=

The interval between the startingpoints of the coils is

= 360°

no. of phase windings 3 360°

120°=

In four-pole motors the figures

are 4 360°

90°= 3 × 2 360°

60°= and

Number of Poles 2 4 6 8 12

No (RPM) 3,600 1,800 1,200 800 600

Table 1. The pole numbers of the motor vs. synchronous speed

9 110 7 8

2

10

4

Figure 7. The physical buildup of theasynchronous motor

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Magnetic fieldThe magnetic field rotates in the airgap between the stator and the rotor.

A magnetic field is induced when oneof the phase windings is connected toone of the phases of the supply voltage(Figure 8).

Figure 10. One phase gives analternating field.

Figure 9. Two phases give anasymmetrical rotational field.

When the third phase has beenconnected there are three magneticfields in the stator core (Figure 10).There is a 120° displacementbetween the three phases.

The stator has now been connectedto the three-phase supply voltage.The magnetic fields of the individualphase windings make up asymmetrically rotating magnetic field.This magnetic field is called therotating field of the motor.

The amplitude of the rotating field isconstant and equal to 1.5 x themaximum value of the individualalternating fields (Figure 11). It rotates at

I1

L1

I2

L2

N S N0

wt

I

I1

S N120 180 300 360

I2

Figure 11: The size of the magnetic field is constant.

I1

L1

I2

L2

N0

wt

I

I1

120 180 300 360N S S S N N

60 240

I2I3

L3

I3

the speed

n = f x 120

p 0

The speed is dependent on thenumber of poles of the motor (p=polepair) and the frequency of the supplyvoltage. The figure below shows thesize of the magnetic fields at threedifferent time periods.

When the magnetic field vectorrotates one revolution and is back toits starting point, the vector tip willhave traced a complete circle. Whenwe draw this field as a function oftime for any single location in thestator, we will see a sine curve forany single location in the stator.

t =1/2 max

s=1/2 max

r = max

=3/2 max

r =1/2 max

=3/2 max

t = maxs=1/2 maxt =

max3

2

r =max

32

=max3/2

N S N

1800 360wt

I

I1 1

I1

L1

Figure 8. One phase gives analternating field.

The magnetic field has a fixedlocation in the stator core, but itsdirection is varying. The rotationalspeed of the magnetic field isdetermined by the frequency of theAC line. When the frequency is 60 Hzthe field changes direction 60 timesper second.

Two magnetic fields are produced inthe stator core when two phasewindings are connected to twophases of the supply voltage at thesame time (Figure 9). In a two-polethere is a 120° displacementbetween the two fields. There is alsoa time interval between the maximumvalues of the two fields. That is how amagnetic field is created whichrotates in the stator.

The field is asymmetrical until thethird phase is connected.

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RotorThe rotor is mounted on the motorshaft. The rotor, just like the stator, ismade of thin iron sheets with slots inthem. The rotor may be a slip ringrotor or a squirrel cage rotor. Theserotors differ from each other, becausethey have different “windings” in theslots.

The slip ring rotor, like the stator,consists of wound coils, which areplaced in the slots. There are coils foreach individual phase, and they areconnected to slip rings. If the sliprings are short-circuited the rotorworks like a squirrel cage rotor.

The squirrel cage rotor has aluminiumrods cast into the slots. At each endof the rotor the rods are short-circuited with an aluminium ring.

The squirrel cage rotor is the mostcommon type. In principle the rotorswork in the same way. In the followingwe will therefore deal with the squirrelcage rotor only.

A rotor rod placed in the rotating fieldis passed by magnetic poles(Figure 12). The magnetic field ofeach pole induces a current in therotor rod. The rod is thus influenced

Figure 12. Operational field and rotor

by force, (F). The next pole passingthe rotor rod is of opposite polarity. Itinduces a current in the oppositedirection of the first one. However, asthe direction of the magnetic field haschanged the force is still affecting therod in the same direction. If the wholerotor is place in the rotating field allthe rotor rods are thus influenced byforces making the rotor rotate. Therotor speed (2) will not reach thespeed of the rotational field (1), as nocurrents are induced in the rotor rodswhere the speeds are the same(Figure 13).

Torque, slip and speedNormally, the rotor speed nn is a littlelower than the speed of the rotationalfield no.

n = f × 60

p × (1 - s)n

s, which is the difference between thespeed of the rotating field and therotor is call the slip: s = no - nn.

The slip is often indicated inpercentage of the synchronousspeed:

s = n - n 0 n

n 0× 100 [%]

Normally it is between 3 and 8percent.

The force acting upon a conductor isproportional with the magnetic field,(Φ) and the current, (I) in theconductor. In the rotor rods, voltageis induced by the magnetic field.Because of this voltage, a current, (I)can flow in the short-circuited rotorrods.

The various forces of the rotor rodsmake up torque, (T) on the motorshaft.

As the magnetic field can beconsidered to be constant, the torqueis directly proportional with thecurrent in the rotor:

T = k 1 11× Φ × I = k × I (for s/s << 1)0

The voltage induced in the rotor canbe found in the following way:

V = k 2 × (n - n ) = k × s0 n 21

In the rotor the current,V = k × s21R R

k × s22=I =

arises, where R is the resistance inthe rotor.

Figure 13. Induction in the rotor rods.

N

S

S

N

N

S

1

2

FF

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There is direct proportionality betweenthe torque output and the slip of themotor (for s/so << 1): T = k 11 × I = k × s3

From the above, it can be seen thatthe motor torque is very muchdependent on the resistance in therotor. The higher the resistance, thelower the torque.

The current heat loss (-) in the rotorincreases with the square of the slip.

(-) = V × I = R × I = k × s2 2

= R × k × s × k × s22 22

= k × s4 2

A curve shows the relationshipbetween the motor torque and thespeed (Figure 14). However, thecharacteristic is to some extentdependent on how the rotor slots aremade.

The motor torque expresses the forceor the “twist” arising on the motorshaft.

As an example, the force (F) applied toa flywheel at the radius (r) yields atorque of T = F × r. The work W doneby the motor can be calculated asfollows: W = F × d, where d is thedistance moved by force, (F).

d is the distance moved for a givenload and n the number of revolutions:

d = n × 2 × π × rThe work can also be expressed asthe power times the period where thepower is active: W = P × t.

The torque can then be rewritten to:

T = F × r = HPd

× r = P × t × rn x 2 x π x r

T = P × 5.25n

(for t = 60 seconds)

The formula shows the relationbetween the speed [RPM], the torque[lb.ft] and the motor output [HP].

The formula gives a quick surveywhen we compare n, T and P to thevalues in a fixed working point.

The working point is normally therated operating point of the motor,the formula can thus be rewritten to:

T = rPrnr

, and to P = Tr × n , where

T = rTTn

, P = rPPn

, and n = rnnn

r r

The constant 5.75 disappears in theformula.

Example: Load = 15% of rated value,rotational speed = 50% of ratedvalue.

The output is 7.5% of rated output.

Pr = 0.15 × 0.50 = 0.075.

Apart from the normal operationalrange, the motor has two brakingranges.

In the range nn > 1

0the motor is

pulled over synchronous speed by theload.

Here the motor acts as a generator.In this range the motor yields acounter torque and power istransferred back to the AC line.

Figure 15. The current and torquecharacteristic of the motor.

T

I

n0n

S0S

0, TS

01

10

n 0, 0

n max ,Tmax

n N,TN

S0S

01

10

n 0 , I 0

n N , I NIS0,

IN8 x

n0n

T==

CurrentTorque

I ==

slipspeed

sn

Figure 14. The motor torque is equal to “force × radius”

In the rangenn < 0

0the braking is

called plugging.

If two phases to a motor are suddenlyswapped over, the rotating fieldchanges direction of rotation.Immediately after the rotational speed

slip ratio, nn0

will be equal to 1.

The motor which was loaded with thetorque TM will now brake with abraking torque. If the motor is notdisconnected when n = 0 it willaccelerate in the new rotationaldirection of the magnetic field.

In the rangenn0

0 < < 1 the motor will

be operating in its normal workingrange.

The operational range can be split upinto two ranges: the accelerationrange

nn0

0 < < nn0

max

and the operational range

nn0

< < 1 nn0

max

Torque

n 0n

S = slipn = speed

0 1

1 0s 0s

Fr

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The important points of the workingrange are: Ms is the starting torque ofthe motor. It is the torque which themotor produces, when it is connectedto rated voltage and rated frequencyat standstill.

Tmax is called the breakdown torque orthe maximum torque of the motor.This is the highest torque which themotor can yield, when it is connectedto nominal voltage and nominalfrequency. TN is the nominal torque ofthe motor. The nominal values of themotor are the mechanical andelectrical values for which the motorhas been designed according toNEMA standards. These values arestated on the name plate of themotor, therefore they are also calledthe rated values or rated data. Therated motor values indicate where thedesigned operational point of themotor is, when it is connected directto the AC line.

Losses and efficiencyThe motor draws electrical powerfrom the AC line, and at constantload, this power is higher than themechanical motor output to the shaftdue to various losses in the motor.The ratio of the shaft output power toelectrical power is called the efficiency

of the motor η = PPinout (Figure 16).

The value is dependent on the motorsize and is typically between 0.7and 0.9.

The motor losses consist of:

The copper loss, which is the resultof the ohmic resistance in the statorand rotor.The iron loss, which consists ofhysteresis loss and eddy-currentloss. The hysteresis losses occurwhen the iron is magnetized by analternating current.

The iron is magnetized anddemagnetized repeatedly, i.e. 120times per second at a supply voltageof 60 Hz. The magnetization and thedemagnetization requires energy.These losses increase with thefrequency. Eddy-current losses occurbecause the magnetic fields induceelectrical current in the iron core(Figure 17). These currents generateheat in the core. The currents flow incircuits at right angles to themagnetic field. The eddy-currentlosses can be reduced substantiallyby dividing the iron core up into thininsulated sheets. This divisionreduces the cross-sectional areawhere the eddy-currents flow,reducing eddy-current losses.

Figure 17. The eddy currents arereduced by laminating the motor iron.

The ventilation loss occurs due to theair resistance of the motor fan.

The friction losses result from lossesin the ball bearings holding the rotor.In practice the motor efficiency isdetermined by deducting the losses inthe motor from the electrical inputpower. The electrical input power canbe measured and the losses can becalculated or determined throughexperiments.

Improper motor magnetizationA typical motor has been designed foroperation on the fixed voltage andfrequency of the AC line. The motormagnetization is determined by thevoltage/frequency ratio. If the voltage/frequency ratio increases the motoris over-magnetized. If the ratiodecreases the motor is under-magnetized. Undermagnetizationweakens the magnetic field of themotor. Therefore, the motor cannotyield as much torque. As a resultthe motor may not start or it stalls.The starting time of the motor maybe extended to the point that itis overheated.

An overmagnetized motor isoverloaded during operation and thepower consumed for the extramagnetization is dissipated as heat inthe motor. Under worst conditionsthis can result in insulation damage.Alternating current motors andespecially induction motors are veryrobust so it is not often that loaddamages occur because of wrongmagnetization. The motor operationwill show if the magnetization is poor(falling speed at varying load, unstableor jerky motor operation etc.).

Pin

PoutShaftoutput

Copper loss

Iron loss

Ventilation loss

Friction loss

Figure 16. Losses in the motor

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Equivalent diagramIn principle the induction motorconsists of six coils. Three coils in thestator as well as the short-circuitrotor, which is magnetically acting asif it consisted of three coils (Figure18). It is possible to make anequivalent diagram using one set ofthese coils. The diagram makes iteasier to understand how the motoroperates, for example, during timeswhen the frequency of the supplyvoltage is changed.

The current in the stator coil is notlimited by the ohmic resistance of thecoil alone. In every coil connected toAC voltage there is also some ACresistance.

This resistance is called the reactanceXL = 2 × π × f × L, and it is measuredin ohm [Ω].

f is the frequency, and 2 × π × ftherefore shows the current variationper unit of time: ω [s-1]. L is theinductance of the coil and it ismeasured in Henry [H]. The reactancedoes not cause any energy losses.But as it is dependent on thefrequency it will limit the activecurrent.

The coils are loading each other withmagnetic induction G. The rotor coilinduces some additional current in thestator coil and so does the stator coilin the rotor coil. Because of thisinteraction the two electrical circuitscan be connected with a commonlink. The common link consists of Rfe,the transverse resistance, and Xh, thetransverse reactance. The currentnecessary to magnetize the statorand the rotor is flowing through these.The voltage drop over the “transverselink” is called the induction voltage.

Motor loading has not yet been takeninto account (Figure 19).

When the motor is working in itsnormal operational range the rotorfrequency is less than the frequencyof the rotational field. The rotorinductance X2' is therefore reduced bythe factor s (slip).

Figure 18. The equivalent diagram of the motor is for one phase

I1R1 X1 X’2 R’2/s

RFe Xh

L1

L2

L

L3

X’2I1

G

L1

I’1

GI’2I1

LILI

R1

X1

R’1

X’1

Figure 20. Equivalent circuit at no-load operation and blocked rotor.

I1R1 X1 X’2 R’2 I’2

RFe XhR 2

1-ss

Terminalvoltage

Induction voltage

Figure 19. Equivalent diagram for a loaded motor.

I1R1 X1 X’2 R’2 I’2

RFe Xh

R 21-ssS 1 : 0

I1R1 X1 X’2 R’2

RFe Xh

R 21-ssS 0 : 00

In the equivalent diagram (Figure 20)the effect can be described throughan increase of the rotor resistance R2’

by the factor 1s

Rs

’2 can be

rewritten to R + R × ’2

1 - ss 2

’ where

R × 21 - s

s ’ shows the mechanical

motor load. R2’ and X2’ represent therotor only.

R2’ included in the load representsthe heat loss arising in the rotor rods

when the motor is loaded. At no-loadoperation the slip s is small.

That means that R × 21 - s

s ’ is high.

Therefore no current can flow in therotor. That means that the resistorrepresenting the mechanical loadcould be removed from the equivalentdiagram under ideal conditions.

When the motor is loaded the slip

increases and R × 21 - s

s ’ decreases.

The current I2’ in the rotor increaseswith the load.

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The equivalent circuit matches motorconditions seen in practice. In mostcases it will therefore be possible todescribe the operation of an inductionmotor on the basis of this diagram.

Sometimes the induction voltage ismistaken for the terminal voltage ofthe motor. The reason is that theequivalent diagram is often simplifiedto give a better assessment of themotor conditions. It is only at no-loadoperation that the induction voltagecorresponds to the terminal voltage.The no-load current is much lowerthan the load current. At no-loadoperation, the voltage drop over R1and X1 is therefore negligible.

When the load increases, the voltagedrop must be taken into account, asI2 and I1, will be increasing with theload. This is especially importantwhen the motor is controlled by anAFD.

AFD speed changeThe motor speed (Figure 21), n isdependent on the rotational speed ofthe magnetic field no, therefore, n canbe expressed as follows:

n = n - n = 0 pf × 60

s - ns

It is possible to change the motorspeed in three ways:

• changing the number of pole pairs p

• changing the slip ns

• changing the frequency f of thesupply mains

f x 60p

Pole number FrequencySlip

n = - n8

Stator VoltageRotor

Cascade couplingResistor

Figure 21. Different ways of changing the motor speed

Pole number controlThe rotational speed of the magneticfield is determined by the number ofpole pairs of the stator (Figure 22). Ifthe motor is a two-pole motor, therotational speed of the magnetic fieldis 3600 RPM, when the supplyfrequency is 60 Hz. The speed of afour-pole motor is 1800 RPM.

Torque

Speedn2 n1

Figure 22. Torque characteristic whenchanging the pole number

Motors can be designed for twodifferent numbers of pole pairs. Thedifference is the way the statorwindings are put into the slots. Thiscan either be done as a Dahlanderwinding or as two separate windings.If a motor with three or four differentnumbers of pole pairs is needed thesewinding types are combined.

The speed change is done byswitching between the statorwindings so that the number of polepairs in the stator is changed.

The switching from the small numberof poles (high speed) to the largenumber of poles (low speed) must beconditioned with the actual motorspeed. If the change takes place tooearly the motor torque runs throughthe regenerative area, which can leadto damages of both motor andmachine.

Slip controlMotor speed control by slip variationcan be accomplished in two ways.Either by changing the supply voltageof the stator or by makingmodifications in the rotor.

Change of the stator voltageIt is possible to control the speed ofan induction motor by changing thesupply voltage without changing thefrequency. This is due to the fact thatthe motor torque falls with the squareof the voltage.

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Torque

Speed

Torque

Speedn4 n3 n2nN

ExpandedView

Figure 23. Torque characteristic whenthe stator voltage and the slip arechanged

Based on the torque characteristicsshown in Fig. 26 it is normally onlypossible to obtain stable workingpoints in the working range (nmax < n <no). When this method is used with aslip ring motor it is also possible toobtain stable working points in theacceleration range (0 < n < nmax). Thisis done by inserting resistors in therotor windings.

Rotor controlChanges in the rotor can be made intwo ways. One possibility is to insertresistors in the rotor circuits. Anotherpossibility is to connect the rotorcircuits in cascade couplings withother electrical machines or rectifiercircuits.

Change of rotor resistorsThis version of motor speed control isachieved by connecting the rotor sliprings to resistors. The motor speed ischanged by increasing the powerlosses in the rotor. Higher powerlosses in the rotor increases the slipand reduces the motor speed. Themotor torque characteristics changewhen resistors are inserted in therotor circuit.

Figure 24. The torque characteristicwhen the rotor resistors and the slipare changed

Frequency regulationWith a variable frequency supply it ispossible to control the motor speedwithout any additional losses. Therotational speed of the magnetic fieldchanges with the frequency(Figure 26). The motor speedchanges proportionally with therotational speed of the magnetic field.

To maintain the motor torque themotor voltage must change with thefrequency.

With a given load the following will

T = P × 9.55 = η × √3 × V × I × cos ϕ × 9.55 60n

(1 - s)p

f × = k × V

f

As V = k1 × f × Φ, the motormagnetization Φ must be constant.

If the ratio between the voltage supplyand frequency is held constant, themagnetization is also constant in thewhole operational range of the motor.

From Figure 24, it can be seen thatmaximum torque will remain thesame. The curve shows the speedwith different rotor resistor sizeswhere the load is the same in allsettings. A set speed is verydependent of the load. When theload is removed from the motor thespeed always increases up tosynchronous speed. The resistors arenormally variable and it is veryimportant that the size is correct forthe operational conditions.

Cascade couplingsHere the rotor circuits (Figure 25) areconnected via slip rings to DCmachines or controlled rectifiercircuits instead of resistors.

The DC machine supplies the rotorcircuit of the motor with additionalvariable voltage. In that way it ispossible to change the rotor speedand the magnetization.

If controlled rectifier circuits areconnected instead of the resistors,energy can be recovered.

M3 ~

Figure 25. Typical cascade coupling

R1 X1 X’2 R’2

Xh RFeR’ 2

1-ss

I FeI 2

I 0II 1

V1

Vs

Vq

Figure 27. Equivalent diagram of themotor

Torque

Speed

Torque

Speed

Figure 26. Motor characteristic atvoltage-frequency regulation

However, when starting and at thevery low frequencies, the magneti-zation will not be optimum. Here extraterminal voltage is required becauseof the resistance of the stator. Inappli-cations with varying load it mustbe possible to adjust themagnetization according to the load.

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Extra start voltageIt is interesting to compare the voltagedrop Vs with the voltage drop Vq.

Terminal voltage: V1 = Vs + Vq = VR1 +VX1 + Vq

Stator reactance: X1 = 2 × π × L × f

The motor has been designed for itsrated values. The magnetizing voltageVq can for example be 440 V for amotor, where V1 is 460V, and f = 60Hz. Here the motor has its optimummagnetization.

The voltage-frequency ratio is

therefore 46060

= 7.6

If the frequency is reduced to 2.5 Hzthe voltage falls to 19 V. The lowfrequency makes the stator reactanceX1 small (Figure 27). The reducedterminal voltage does not affect thetotal voltage drop in the stator. Thevoltage drop is now determined by R1alone. It is approximately the same aswith the rated values, about 25 V, asthe motor current is determined bythe load.

The terminal voltage corresponds tothe voltage drop over the statorresistor. There is no voltage tomagnetize the motor, therefore itcannot yield expected torque at thelow frequencies, if the voltage-frequency ratio is held constant in thewhole range.

It is therefore necessary tocompensate for the voltage dropduring start and at low frequencies.

Load dependent voltage boostWhen the motor is magnetizedproperly, under the starting conditionsit will be overmagnetized at lighterloads. In that case the stator current Iand the induction voltage Vq arefalling. The motor will draw too muchblind current and overheat. To achieveoptimum magnetization, automaticvoltage compensation to the motorload is required during both start andas the load varies.

VM = V1 + Vs = V1 + I1 × R1 + I1 × X1.

Motor dataThe motor has a nameplate on it(Figure 28). This plate containsimportant information about the

3-4. The stator windings can beconnected in series or in parallel(Figure 29).

If the supply voltage is 460 V thewindings are connected in series.In that case the motor current willbe 29 A per phase.

If the supply voltage is 230 V thewindings are connected inparallel. In that case the motorcurrent will be 50 A per phase.

When starting, the AC line couldbe overloaded, because here thestarting current is 5 to 8 timeshigher than the rated current.

motor.

The nameplate of a 20 hp 4-polemotor can for example contain thefollowing information:

1. It is a three-phase motor for anAC line at 60 Hz.

2. The nominal motor output is 20hp. This means that the motorcan at least yield a shaft outputof 20 hp when connected to theAC line as shown.

The rated outputs of inductionmotors are put into standardratings. This means that the usercan choose between differentmotor makes for a specificapplication.

Watts is also a unit for the outputpower of a motor. 1 hp =746 watts.

NEMA Nom. Eff. 90.2

Frame326T

4

5

13

2

7

HP 20RPM 1770Amb.40°C

TypeP

DesignB

Volts 230/460Amps 50/29Duty Cont.

Hz 60S.F. 1.15Encl. TEFC

Phase 3Code F

Ins.Class F

Identification No.

Low VoltsT4 T5 T6T7 T8 T9

T1 T2 T3L1 L2 L3

T4 T5 T6T7 T8 T9

T1 T2 T3

L1 L2 L3

High Volts

6

Figure 28. The motor plate givesmuch information

Figure 29. Motor torque and current,star and delta connection

0, 5n N

n

TnT

I nI

1

2

3

I yTy

I

T

==

TorqueSpeed

Tn

u u

I

u

I

3U3

I3

Series Parallel

HP 1 2 3 5 7.5 10 15 20 30 40 50 60 75 100

kW 0.75 1.5 2.2 4.0 5.5 7.5 11 15 22 30 37 45 55 75

Table 2. The motor power rating

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5. The motor enclosure indicatesthe degree of protection againstpenetration of liquids andextraneous matter (Figure 30).

6. The full load current IRES drawn bythe motor can be divided intotwo currents: an active currentIACT and a reactive current IREA.

Power factor is the ratioexpressing how much of thecurrent is active. The activecurrent is the one giving the shaftoutput and the reactive current isthe one providing the necessaryoutput to build up the magneticfield of the motor.

The total current IRES (Figure 31)the motor draws from the AC lineis called the resulting current. Itcannot be calculated just byadding the active and thereactive currents. This is due tothe fact that there is a timeinterval or phase angle betweenthe two currents. It is necessaryto use vector addition. Since thephase angle between IREA and IACTis 90, the following formula maybe used:

I =RES

I ACT2 + I

REA2√

The currents can be regarded asthe sides in a right-angledtriangle, where the long side ofthe triangle is equal to the squareroot of the sum of the squares ofthe short sides.

ϕ is the angle between theresulting current and the activecurrent, and power factor is theratio between the values of thetwo currents: power factor =

IACT/IRES. Power factor can also beexpressed as the ratio betweenthe active power P and theapparent power S power factor.

(The term “apparent power”means that only a portion of theresulting current (IACT) generatesthe power to be used) .

7. The nominal motor speed is themotor speed at rated voltage,rated frequency and rated load.

I ACT

I REA

I RES

Figure 31. Connection between theresulting, the blind, and the activecurrents

On the basis of the nameplate of themotor it is also possible to calculateother important motor data.

The rated motor torque can be foundwith the formula

P = T × n 9.55

T = P × 9.55

n =

15000 × 9.552910

= 49 Nm

The motor efficiency can be found asthe ratio between the nominal activepower and the added electric powerand

NEMACode Intended Use and Description

1 Indoor use, primarily to provide protection against contact withthe enclosed equipment and against a limited mount of falling dirt.

2 Indoor use to provide a degree of protection against limitedamounts of falling water and dirt.

3 Outdoor use to provide a degree of protection against wind-blown dust and wind-blown rain; undamaged by the formation ofice on the enclosure.

3R Outdoor use to provide a degree of protection against falling rain;undamaged by the formation of ice on the enclosure.

3S Outdoor use to provide a degree of protection against wind-blown dust, wind-blown rain, and sleet; external mechanismsremain operable while ice laden.

4 Either indoor or outdoor use to provide a degree of protectionagainst falling rain, splashing water, and hose-directed water;undamaged by the formation of ice on the enclosure.

4X Either indoor or outdoor use to provide a degree of protectionagainst falling rain, splashing water, and hose-directed water;undamaged by the formation of ice on the enclosure; resistscorrosion.

6 Indoor or outdoor use to provide against the entry of water duringtemporary, limited submersion; undamaged by the formation ofice on the enclosure.

6P Indoor and outdoor use to provide a degree of protection againstthe entry of water during prolonged submersion at limited depths.

11 Indoor use to provide by oil immersion, a degree of protection ofthe enclosed equipment against the corrosion effects of corrosiveliquids and gases.

12, 12K Indoor use to provide a degree of protection against dust, dirt,fiber flyings, dripping water, and external condensation ofnoncorrosive liquids.

13 Indoor use to provide a degree of protection against lint, dustseepage, external condensation, and spraying of water, oil, andnoncorrosive liquids.

Figure 30.

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√3 × V × I × cos ϕ η =

PE = √3 × 380 × 29 × 0.9

15000 = 0.87

The motor slip can be calculated, asthe motor nameplate containsinformation about the nominal speedalong with the frequency. A 4-polemotor has a synchronous speed of1800 RPM. The slip is therefore1800 - 1725 = 75 RPM.

In general the slip is indicated in %,therefore

s = nsns

= 75

1,800= 0.03 = 3%

A motor catalogue contains some ofthe data stated on the motornameplate, but it is also possible tofind other important data:

Type Poweroutput

kW

Speed

r/min

Effi-ciency

%

cos Currentat

380 VA

I M

Nm

Momentof

inertiakgm

Weight

kg

At rated operationst st Mmax

MMMI

2

160 MA

160 M

160 L

11

15

18.5

2900

2910

2930

86

88

88

0.87

0.90

0.90

25

29

33

6.2

6.2

6.2

36

49

60

2.3

1.8

2.8

2.6

2.0

3.0

0.055

0.055

0.056

76

85

96

Types of loadWhen the motor torque output isequal to the load torque we have astable load. In such cases the torqueand the speed are constant.

Typical load types (Figure 33) arecharacterized by the following speedtorque curves:

1. Machines for winding material atconstant material tension. Thisgroup also includes veneercutting machines and machinetools.

2. Conveyor belts, different kinds ofcranes and positive displacementpumps etc.

3. Smoothing machines, calendarrollers and other machines formaterial processing.

4. Machines working withcentrifugal forces, e.g. centrifugalpumps and fans.

The stable load occurs when themotor torque is equal to the torque ofthe working machine. This isindicated at point B.

v

Torque

Speed

Power

Speed

Torque

Speed

Power

Speed

Torque

Speed

Power

Speed

Torque

Speed

Power

Speed

r

v

m1

m2

r

n

n

n

v

n

T(n)~n -4

T(n)=k

T(n)~n

T(n)~n 2

1

2

3

4

Figure 33. Typical load characteristics

Figure 32. Example from a motor catalogue

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When choosing a motor for a specificmachine the intersection point mustbe as close as possible to the motorsfull load torque to obtain optimummotor utilization.

In the range from standstill to theintersection point there must be asurplus torque. If not, the operationwill be unstable and the stationaryload may stop low speed becausethe surplus torque is used foracceleration (Figure 34).

Torque

Speed

N

B

Figure 34. The motor must have asurplus torque to accelerate

For applications as those found ingroups 1 and 2, it is necessary to payattention to the start situation. Theseload types may require a high break-away torque which could be equal tothe starting torque of the motor(Figure 35).

If the break-away torque of the loadexceeds the starting torque of themotor, the motor will not be able tostart.

Torque

Speed

Figure 35. Especially high torquesmay be required when starting

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The AFD has developedtremendously since the first unit waslaunched at the end of the ’60s.Today's advanced micro-processorsand semi-conductors have improvedthe frequency converter substantially.

The frequency converter can bedivided up into four main components(Figure 36):

1. The rectifier converts the three-phase AC voltage from thesupply mains to a pulsating DCvoltage. There are two basictypes of rectifiers: the controlledand the uncontrolled rectifiers.

2. The intermediate circuit. Thereare three different types. Onetype converts the voltage of therectifier into a DC current. The

The AdjustableFrequency Drive

The common characteristic of AFDcontrol circuits is that they transmit asignal to the semiconductors of theinverter to switch on or off. Thisswitching pattern is determined bythe design principle (Figure 37). AFDscan be grouped according to theswitching pattern controlling themotor power.

other type stabilizes the pulsatingDC voltage and sends this on tothe inverter. The third type ofintermediate circuit converts aconstant DC voltage from therectifier into a variable value.

3. The inverter controls thefrequency of the motor voltage.One type of inverter alsoconverts the constant DCvoltage into a variable ACvoltage.

4. The electronics of the controlcircuit can transmit signals toboth the rectifier, theintermediate circuit and theinverter. The parts to becontrolled are dependent on thedesign of the AFD.

Figure 36. Simplified diagram of a frequency converter

M

1. 2. 3.

4.

RectifierInter-

mediatecircuit

Inverter

Control and regulation circuit

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Rectifier

IntermediateCircuit

Inverter

1 2

543

6 7

Current Source Inverters CSI(1 + 3 + 6)

Pulse-amplitude-modulated converters PAM(1 + 4 + 7) (2 + 5 + 7)

Pulse-width-modulated converters PWM/VVC(2 + 4 + 7)

Figure 37. Different control principles

The rectifierThe supply voltage is a three-phaseAC voltage with a fixed frequency(e.g., 3 × 460 V, 60 Hz). Figure 38shows some characteristic values.

It should be noted there is a timedelay between the three phases. Thephase current changes direction allthe time based on the inputfrequency. A frequency of 60 Hzmeans that there are 60 periods (60 ×t) per second. That means that oneperiod is 16.66 msec.

The rectifier of the AFD is eitherconstructed of diodes, thyristors or acombination of these semi-conductors. A rectifier containingdiodes only is called an uncontrolledrectifier. If the rectifier consistsexclusively of thyristors it is called afull-wave controlled rectifier. A rectifiercontaining both diodes and thyristorsis called a half-wave controlledrectifier. The half-wave controlledrectifier is not used very often inAFDs.

V

wt

T

V

wt

a b 13

a b= T

Figure 38. Single- and three-phase AC voltage

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V

wt

V

wt

I

A C

Figure 39. Mode of operation of the diode

Uncontrolled rectifierThe uncontrolled rectifier consists ofsix diodes.

A diode permits current to flow in onedirection only: from anode tocathode. If any attempt is made tosend current in the opposite direction,a diode blocks current flow(Figure 39).

With a diode it is not possible tocontrol the amount of current flow asit is with other semi-conductors.When an AC voltage is supplied to adiode circuit, it becomes a pulsatingDC voltage. When a three-phase ACvoltage is connected to anuncontrolled three-phase rectifier, theDC voltage will still be pulsating.

Figure 40 shows that the uncontrolledthree-phase rectifier consists of twogroups of diodes. One group containsdiodes D1, D3 and D5; the other,diodes D2, D4 and D6 . Each diode isconducting 1/3 T (120°). The twogroups of diodes are conducting inturns. The time interval between thetwo groups is 1/6 T (60°).

In the groups of diode D1, D3 and D5

will be conducting the most positivevoltage. If the voltage in L1 is mostpositive, then terminal A will have thesame value as L1. Above the twoother diodes there are reversevoltages of the size VL1-2 and VL1-3.

The groups of diodes D2, D4 and D6terminal B will have the most negativevoltage of the phases. Where phaseL3 has the most negative voltage thendiode 3 will be conductive. Above thetwo other diodes there are reversevoltages of the sizes VL3-1 and VL3-2.

The output voltage of the uncontrolledrectifier is the difference between thevoltages of the two diode groups(Figure 41). The average value of thepulsating DC voltage is 1.35 × AC linevoltage.

V

wt

V

wt

D1 D3 D5

D2 D4 D6

L1

L2

L3

(A)

(B)

Figure 41. The output voltage of theuncontrolled three-phase rectifier

VA-B

V

wt

wt

UB

UA

Full-wave controlled rectifierThe full-wave controlled rectifier hasthyristors instead of diodes.

Just like the diode, the thyristorpermits the current to flow in onedirection only, from anode to cathode(Figure 42). There is a difference,however; a thyristor will only conductcurrent when the third terminal calledthe “gate” receives an electric signal.The thyristor will then conduct untilthe current becomes zero.

A signal on the gate cannot stop thecurrent.

Thyristors are used in both rectifiersand inverters.

The signal on the gate is the controlsignal of the thyristor and it isdesignated α. α is a time delay statedin degrees. The degree valueindicates the time delay from zerocrossing up to the point where thethyristor must start conducting.

Figure 40. The uncontrolled three-phase rectifier

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When it is between 0° and 90° thethyristor coupling is used as rectifier.When it is between 90° and 300° thecoupling is used as inverter.

The full-wave controlled three-phasecan be divided up into two groups ofthyristors containing T1, T3 and T5 andthyristors T2, T4 and T6 respectively.

α is determined from the point wherethe corresponding diode of anuncontrolled rectifier startsconducting. This point is 30° after thezero crossing of the voltage. Otherthan this, the description follows thatof the uncontrolled rectifier.

Figure 43. The full-wave controlled rectifier

V

wt

V

wt

T1 T3 T5

T2 T4 T6

L1

L2

L3

(A)

(B)

Figure 44. The output voltage of thefull-wave controlled three-phaserectifier

The rectified voltage can be varied bychanging α (Figure 44). The full-wavecontrolled rectifier supplies a DCvoltage of the following average value:1.35 × AC line voltage × cos α.Compared to the uncontrolledrectifier, the controlled rectifierproduces large disturbances andlosses in the AC line. This is due tothe fact that the rectifier drawscurrent in short intervals. Thyristorsare typically applied only in theinverter section of the AFD. Theadvantage of the controlled rectifier isthat braking energy fed into theintermediate circuit can be transferredback to the AC line.

The intermediate circuit (Bus)The intermediate circuit can beregarded as the source, from wherethe motor, through the inverter,receives its energy. The intermediatecircuit can be built up according tothree different principles. Theintermediate circuit type useddepends on the type of rectifier andinverter concerned.

The intermediate circuit shown(Figure 45) consists of a large coil.This is used with a controlled rectifierstyle design only. The coil convertsthe variable voltage from the rectifierinto a variable DC current. The loaddetermines the level of the motorvoltage.

This type of intermediate circuit hasthe advantage that braking energy is

V

t

I

t

Figure 45. Variable AC intermediate circuit

Figure 46. Constant or variable voltage intermediate circuit

V

t

V

tV

t

V

t

Figure 42. The mode of operation of the thyristor

V

wt

V

wt

I

A CG

+-

VA-B

V

wt

wt

VB

VA

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fed back to the AC line without theuse of extra components.

The intermediate circuit can alsoconsist of a filter containing onecapacitor and one coil (Figure 46).This intermediate circuit can becombined with both rectifier types.

The filter smooths the pulsating DCvoltage coming from the rectifier. Ifthe rectifier is controlled, the voltageis held constant at a given frequency.The voltage led on to the inverter isthus a smoothed DC voltage of avariable amplitude.

If the rectifier is uncontrolled thevoltage on the input of the inverterbecomes a DC voltage with aconstant amplitude. With this type ofintermediate circuit bus the loaddetermines the size of motor currentdrawn.

Finally, it is possible to insert achopper in front of a filter, as shownin Figure 47. The chopper has atransistor that alternately switches therectified DC voltage on and off. Thecontrol circuit measures the variablevoltage behind the filter andcompares it with the input signal. Ifthere is a difference, the ratiobetween ton (conducting) and toff(blocking) is regulated. The DCvoltage becomes variable and thesize Vv depends on how long thetransistor is on:

V =v

V ×t on

t on - t off

When the chopper transistor turns off,the current the filter coil will create ahigh voltage across the transistor. Toavoid this the chopper is protected bya free-wheeling diode.

When the transistor turns on and offas shown in Figure 48, the averagevoltage will be highest in situation 2.

The filter of the intermediate circuitbus smooths the square wave voltageof the chopper. The capacitor and thecoil of the filter hold the voltageconstant at a given duty cycle.

V

t

V

ttoff t off

t on

ton ton

toff

Situation 1 Situation 2

Figure 48. The chopper transistors vary the intermediate circuit voltage

The inverterThe inverter is the last module in theAFD before the motor. Here the finaladaption of the output voltage takesplace. If the motor is connected directto the AC line the ideal workingconditions will be in the nominalworking point. The AFD providesexcellent operational conditions in thewhole control range, since the outputvoltage is matched to the loadconditions. It is therefore possible tohold a constant motor magnetization.

From the intermediate circuit theinverter either receives

• a variable DC current• a variable DC voltage• a constant DC voltageThe inverter must convert the DCintermediate circuits supply into anAC supply for the motor. The invertercan have additional functions: Whenthe inverter receives a variable currentor voltage the inverter must contributethe frequency only. However, whenthe voltage is constant the invertermust control both the frequency andthe amplitude of the voltage.

The design of inverters differs, but inprinciple they are constructed in thesame way. The main components arecontrolled semi-conductors placed inthree branches.

Today most inverter thyristors havebeen replaced by transistors. Theadvantage of transistors is that theycan change from conductive to non-conductive condition at any time,whereas thyristors do not changecondition until next time the currentthrough them goes through zero.

The switching frequency range of thetransistorized inverter can thereforebe extended significantly from 300 Hzto 15 kHz.

The semiconductors of the inverterturn on and off on the basis of signalsfrom the control circuit. The signalscan be controlled according todifferent principles.

Generally inverters based uponcurrent control require morecomponents than inverters regulatingvoltage.

Figure 47. Variable voltage intermediate circuit

V

t

V

t

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This current sourced inverter consistsof six diodes, six thyristors and sixcapacitors (Figure 49).

The capacitors must include thenecessary energy to turn off thethyristors. The size of the capacitorsand thyristors must be in accordancewith the motor size. The capacitorspermit the thyristors to switch so thatthe DC current flow 120° displaced inthe phase windings. When the motorterminals periodically are suppliedwith the current in turns U-V, V-W, W-U, U-V...., an intermittent rotationalfield with the required frequency isproduced. The motor currents aresquare-waved, but the motor voltagewill be sinusoidal. However, there willbe voltage peaks each time thecurrent is switched in or out.

The diodes isolate the capacitorsfrom the motor load current.

The inverter (Figure 50) consists of sixthyristors or transistors. In principlethe function is the same regardless ofthe type of semi-conductor you see.The control circuit turns the semi-conductors on and off according todifferent principles and they are thusvarying the output frequency.

The intervals of conduction of theinverters semi-conductors form apattern, which is repeatedcontinuously.

The switching pattern of the semi-conductors is controlled by the size ofthe variable voltage. The mostcommon ones are produced by aswitching pattern of either 6 or 18pulses. A voltage controlled oscillatorwill always make the frequency followthe amplitude of the voltage. Thisprinciple of inverter control is calledPulse Amplitude Modulation(Figure 51).

Another principle applies a fixedintermediate circuit voltage. Themotor voltage is made variable as themotor windings are applied with the

I

t

I

t

Figure 49. Inverter for variable intermediate circuit current

V

t

V

t

I

t

I

t

I

t

Figure 50. Inverter for variable or constant intermediate circuit voltage

(Figure 51). Traditionally the controlcircuit establishes the turn-on andturn-off times of the semi-conductorsas the intersection points between atriangular voltage and a sine-shapedreference voltage (sine controlledPWM).

There are other ways of establishingthe turn-on and turn-off times of thesemi-conductors. In the DanfossVoltage Vector Controlled AFD theoptimum switching times for thesemi-conductors of the inverters arecalculated by means of built-in microprocessors.

V

t

V

tPAM PWM

Figure 51. Modulation of pulseamplitude or width

intermediate circuit voltage for shorteror longer periods. The frequency iscontrolled by applying positive pulsesin one half-period and negative pulsesin the next half-period. This principlealso varies the width of the voltagepulses, called Pulse Width Modulation

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TransistorsTransistors can be made for highvoltages and high switchingfrequencies. They can replace thethyristors previously used in theinverter of the AFD. Contrary to boththe thyristor and the diode thetransistor is independent of the zero-crossing of the current. The transistorcan be changed from conducting tonon-conducting condition at any time.

The upper limit of the switchingfrequency is now several hundredkilohertz. The acoustic noiseproduced because of the “pulse”magnetization of the motor can beavoided.

High switching frequency also has theadvantage that the modulation of theoutput voltage of the frequencyinverter is very flexible. A near perfectmotor current waveform is obtainedthrough a special switching patternfor the inverter transistors (Figure 52).

The switching frequency of theinverter is a compromise between thelosses in the motor due to motorcurrent distortion and the losses inthe inverter. When the switchingfrequency increases the losses in theinverter will increase by the number ofsemiconductor switchings.

The high frequency transistors can begrouped as follows:

• bipolar, including Darlingtontransistors

• MOS-FET• IGBTThe IGBT transistor is a combinationof the bipolar transistor and the MOS-FET transistors. It has the MOS-FETtransistors' desired features on the

input and the bipolar transistors' bestfeatures on the output.

Figures 53 and 54 show the mostimportant differences.

The IGBT transistors are well suitedfor AFDs. The primary benefits arethe power range, the goodconductive features, the highswitching frequency and the simplecontrol.

0

In

0

0

fp =1,5 kHz

fp =3 kHz

fp =12 kHz

t

Figure 52. How the switchingfrequency affects the motor current Figure 54. Power and frequency range of power transistors

FeatureSemi-conductor

MOS-FET IGBT Bi-Polar

Symbol

Configuration

Conductivity

Blocking voltage

Switching conditions

Control conditions

Current conductanceLosses

LowHigh

PowerMethod

LowVoltage

LowVoltage

HighCurrent

HighSmall

HighSmall

Upper limit Low High Medium

Turn-on timeTurn-off timeLosses

ShortShortSmall

MediumMediumMedium

MediumShortLarge

E G E

N+ N+PN-P+

S G S

N+ N+PN-N+

B E

N+ N+PN-P+

Figure 53. Comparison between different power transistors

KVA

kHz

Bi-Polar

IGBT

MOS-FET

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Pulse Amplitude Modulation(PAM)The amplitude of the output voltage iseither varied by the intermediatecircuit chopper or by the inputrectifier, and the frequency is variedby the inverter.

A normal output signal is made up of6 or 18 pulses per period.

A 6-pulse switching pattern meansthat each of the six semiconductorsof the inverter is controlled with onepulse per period. With an 18-pulseswitching pattern eachsemiconductor is controlled withthree pulses per period.

The voltages between the outputterminals U, V and W of the AFDdepend on the pattern controlling thesemiconductors of the inverter.

In the following diagram (Figure 55)we look at two inverter branches tofind the voltage between the terminalsU and V. The voltages between V andW as well as between W and U canbe found in the same way. It is thesemiconductors T1, T2, T3 and T4 thatresult in the voltage betweenterminals U and V. Thesemiconductors work like contactscutting the intermediate circuit voltageUm on and off.

In the example shown (Figure 56), T2and T3 are conductive. That meansthat the semiconductors are turnedon with a control signal and that thevoltage across them is zero. T1 and T4are turned off. They do not receiveany control signal. The contacts are

Figure 55.

open and there is a voltage acrossthem.

• The voltage between U and “–” iszero

• The voltage between V and “–” isequal to - Um

• The voltage between U and V is U -V = - V = Um

When we look at the 6-pulse and the18-pulse patterns this way the outputvoltages of terminals U - V will be asshown on Fig. 62.

It can be seen that the outputvoltages are pulses of the amplitudeVm. The duration of the pulses ischaracterized by two time intervals t1and t2, where t2 = 2 × t1.

The actual value of the output voltagecan be calculated as the square rootof the ratio between the area coveredby the pulses and the area of thewhole period.

VU-V VU-V

tt

t1 t1

t 2

t1 t 2 t1t 218 t1

t1t 2 t1 t 2

U T1

U T2

U T3

U T41 2 3 4 5 6 7 1 2 3 4 5 6 7

U T1

U T2

U T3

U T4

T1

T2

T3

T4

T5

T6

P1

P4

P6

P5

P3

P2

P1 P2

P3P4 P5 P6

P8

P9

P7

P12

P11P10

P13 P14 P15

P16 P17 P18

18-pulse modulation6-pulse modulation

Figure 56. The semi-conductors of the inverter work like switches

U-

t

V-

t

U-V

t

+

-

V V

T4

T2

T3

T4

VUUm

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For the 6-pulse pattern Vact can becalculated as:

V = ACT × √V m2 2 × t 13 × t 1

= V × 0.816m

For the 18-pulse pattern:

V = ACT × √V m2 24 × t 136 × t 1

= V × 0.816m

For both pulse patterns the activevalue of the output voltage is 86.6% ofthe intermediate circuit voltage. Thismeans that the intermediate circuitvoltage must be higher than the activevalue of the supply mains.

The non-sinusoidal output voltageshave some side effect on the motoroperation:

• Cogging torques• Increased heat lossesBoth side effects are due to the factthat the motor is supplied with pulsevoltages instead of sinusoidal voltages.

Every periodic voltage can be split upinto a number of sine voltages withdifferent amplitudes and frequencies(harmonic voltages).

The total torque is only slightlydisturbed by the harmonic frequenciesbecause the motor cannot effectivelyconvert the high frequency voltages totorques. This can be observed on themotor current, it is more sine-shapedthan the motor voltage.

Because of the increased heat loss themotor windings are loaded more thanthey should. The total heat loss mustnot exceed what the motor is able towithstand. Therefore the motor mustnot be loaded 100% all the time.Should an application require 100%continuously, a larger motor must beinstalled.

The difference between a 6- and 18-pulse pattern is that the 18-pulsepattern substantially reduces the side-effects from harmonic voltages andcurrents. This is clearly seen whencomparing the motor currents(Figure 57). The more sine-shaped themotor current the less the harmoniccurrents affect the motor operation.

Figure 57. The motor current is more sine-shaped with an 18-pulse signal

18-pulse modulation 6-pulse modulation

V

wt

Vs Vs Vs

V

wtV

wt

s

1

2

V

wt

1 -V2

Figure 58. The principle of the sine-controlled PWM

Pulse-Width-Modulation (PWM)The inverter varies both theamplitude and thefrequency of the outputvoltage. The controlprinciple is working with asine-shaped referencevoltage for each AFDoutput. The three referencevoltages Vs1, Vs2 and Vs3

are supplied with atriangular voltage. Thesemiconductors turn on oroff, where the triangularvoltage and the sinereference intersect eachother (Figure 58).

The electronics of thecontrol card compares theintersection points. Theoutput pulse is negativewhere the triangularvoltage is higher than thesine-shaped voltage andpositive where it is lower.The maximum outputvoltage of the AFD is thus determinedby the intermediate circuit busvoltage. The output voltage(Figure 59) is controlled by applyingthe intermediate circuit bus voltage tothe motor for shorter or longerperiods.

The output frequency is controlled byapplying positive pulses in one half-period and negative pulses in the nexthalf-period. The amplitude of thenegative and positive voltage pulsesfrom line to neutral positions withinthe motor will be equal to half of theintermediate circuit voltage.

The switching frequency affects theaudible motor noise. Semiconductors,and their high frequency switchingrates, have allowed the audible noiseto be reduced substantially. Usingthese advanced semiconductors it ispossible to achieve almost sinusoidaloutput current.

A PWM AFD using an entirelysinusoidal reference modulation canonly yield up to 86.6% of ratedvoltage.

The intermediate circuit voltage Vm isequal to √2 times the supply voltage.The line to neutral voltage seen by themotor is equal to half of the

ACR 26


intermediate circuit voltage divided by√2. It is thus equal to half of the ACline voltage. The line to line voltage ofthe output terminals is equal to √3times the line neutral voltage, that is0.866 times the AC line.

It is possible to increase the outputvoltage of the AFD to a higher valuethan that obtainable with the puresine modulation.

The traditional way of obtaining theadditional voltage is to reduce thenumber of pulses, when thefrequency exceeds about 40 Hz. Thismethod has the disadvantage thatthere is a step voltage change. Thatcauses an unstable motor current.When the number of pulses isreduced, the content of harmonics onthe AFD output increases and so dothe motor losses.

Another method is to use otherreference voltages instead of thethree sine references Vs1-3. Thesecould for example be trapezoidvoltages, step-shaped voltages orvoltages with some other waveform.

It is relatively easy to produce areference voltage which utilizes thethird harmonic of the sine reference(Figure 60). By adding some thirdharmonic voltage the voltage to themotor can be increased up to 15.5%.

1,00

0,50

0,00

-0,50

0,50

-1,00

V-WU-V W-U

0 60 120 180 240 300 360

Switching pattern for phase U

Phase voltage (0-point=half intermediate circuit voltage)

Phase-phase voltage to the motor

Motor voltage / Mains voltage

V

wt

s

1.1551

0

V

wt

s

10.866

0

V

wt

s

1

0

0.166

Figure 59. The output voltage at PWM

Figure 60. The output voltage can be increased by utilizing the third harmonic

ACR 27

AC

Tech

nic

al R

efe

ren

ceThe optimum motor magnetization is

achieved, because the AFD modelsthe motor constants R1 and X1 andadapts them to the different motorsizes. The AFD calculates theoptimum output voltage on the basisof these data. As the AFD measuresthe load current continuously, it canchange the output voltage accordingto the load.

The motor magnetization is matchedto the motor and it compensates theload changes.

Unlike the sine controlled PWMprinciple the VVC control principle isbased on digital production of thedesired output voltage.

The VVC principle is integrated in anApplication Specific Integrated Circuit(ASIC) circuit of the VLT's AFD.

The VVC designThe VVC design includes a number ofdifferent functions:

Registers including the data, whichthe micro processor of the computertransmits to the circuit over thedata bus.

The Danfoss VVCcontrol principleThe Danfoss AFD VVC inverter controlsboth the amplitude and the frequencyof the output voltage (Figure 61).

The control circuit uses a mathematicalmodel which calculates two differentfactors:

• The optimum switching times for thesemiconductors of the inverter

• The optimum motor magnetization atvarying load (see “compensationpossibilities”).

The principle for the switching timesworks as follows:

• the numerically largest phase is for a1/6 sine period held fixed on thepositive or negative potential.

• the two other phases are varied sothat the resulting output voltage isentirely sinusoidal and of the correctamplitude.

Full rated motor voltage is ensured. It isnot necessary to overmodulate toutilize the third harmonic. The motorcurrent is entirely sinusoidal and themotor performance is the same as ACline operation.

Figure 62. The buildup of the VVC design

Figure 61. The full output voltage can be obtained with Danfoss VVCcontrol principle

1,00

0,50

0,00

-0,50

-1,00

V-WU-V W-U

0 60 120 180 240 300 360

Switching pattern for phase U

Phase voltage (0-point=half intermediate circuit voltage)

Phase-phase voltage to the motor

Motor voltage / AC line voltage

Semi-conductor 0-60 60-120 120-180 180-240 240-300 300-360

T1T3T5T2T4T6

t10

t2T-t1

T

T-t2 t1

t2

0

T-t2

T

t10

T-t1

t2

T

T

T-t2T-t1

0

t2

t1

T

T-t1

T-t1

T-t2

t2

0

t20

T

T-t2

T-t2T-t1

t10

RegistersAddress

calculator Multiplicator

Timer

Sequencecontrol

Databus

T1T2T3T4T5T6

ACR 28


Address calculator, calculatingthe address for a cosine table,which is placed in a ROMmemory.

Multiplicator, which calculatesthe product of the amplitudeand the value requested fromthe cosine table. For eachcalculation interval themultiplicator calculates twovalues t1 and t2 being cut-intimes for the invertersemiconductors.

Timer converting t1 and t2 tocontrol signals.

Sequence controller distributingthe control signals on theoutputs of the circuit 1-6according to the chart inFigure 62.

From the chart it can be seenthat semiconductor T4 is heldfixed on the negative potentialwhile semiconductors T1 and T5

are modulating the sine shape.

In the next interval T1 is heldfixed on the positive potentialwhile semiconductors T4 and T6are modulating the sine shape.

t1 is the period, wheresemiconductor T1 is activatedand is switching to +.

t2 is the corresponding periodfor semiconductor T5.

An addition of the phasevoltages will show that thevoltage between the outputterminals of the AFD reachesits rated value and that it isentirely sine-shaped. It is notnecessary to over-modulateand to use the third harmonic(Figure 63).

Figure 63. VVC control gives full output voltage

Harmonics and therotational fieldAC motors have beendesigned for sine-shaped ACvoltages and currents. Thatthe motor can still be drivenby square pulse voltages dueto the fact that all periodicvoltages can be split up intoseveral sine voltages. Thesesine voltages have differentfrequencies and amplitudes.The motor will be driven bythe dominant sine-voltage(Figure 64.

If the output voltage of theAFD is not sinusoidal themotor will receiveoverharmonic voltages inaddition to the voltage of therequired frequency(fundamental frequency orthe 1st harmonic).

The harmonic frequenciesare 5, 7, 11 and 13 timeshigher than the fundamentalfrequency and theiramplitudes are decreasingwith increasing frequency(Figure 65).

The harmonic frequenciescause torque pulsationscogging, vibration, increasedaudible noise, reduced motorefficiency and increased heatlosses in the motor.

These disadvantages areespecially significant at lowspeeds. Around the ratedmotor speed the harmonicfrequencies do not havemuch influence and none atall when the speed isincreased to 1.5 times therated value. This is due to thefact that the harmonicfrequencies are so high here,that they are reduced by thereactances of the motorwindings.

wt

wt

wt

wt

wt

wt

wt

wt

0

T1 t1

T2 T-t1

T3 0

T4 T

T5 t 2

T6 T-t 2

U-O V-O W-U

0.5

U+V V-W W+U1.0

-1.0

0 60 120 180 240 300 360

ACR 29

AC

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wt

V

V1

V

V5 V71/5 V1

Figure 64. The harmonic numberindicates how many times itsfrequency is higher than the basicfrequency

With the sine-controlled PWM it isespecially important to take intoaccount that the amount of harmonicfrequencies depends on the ratiobetween the frequency of thetriangular voltage and the frequencyof the sine voltage. If the ratiobetween the two frequencies is 6then the fifth and the seventhharmonic will have a high amplitude. Ifthe ratio is 15, then the thirteenth andthe seventeenth harmonic will behigh. The ratio between thefrequencies should therefore be highand divisible by three. All harmonicswith a frequency divisible by three areeliminated in a three-phase system.

The ratio between the differentharmonics can be shown in a systemof co-ordinates. The X-axis shows thefrequency of the harmonic and the Y-axis shows the amplitude of theharmonic in relation to the amplitudeof the first harmonic A1.

The harmonics affect the rotationalfield of the motor. It is possible tomeasure the quality of the different

n

A nA 1

0,1

0,5

1,0

1 5 7 11 13 17 19 23 25 29 31 35 37 n

A nA 1

0,1

0,5

1,0

1 5 7 11 13 17 19 23 25 29 31 35 37

Harmonic number

Rat

io o

f fun

dam

enta

l har

mon

ic a

mps

Harmonic number

Rat

io o

f fun

dam

enta

l har

mon

ic a

mps

Figure 65. The harmonic amplitudes at 6-pulse and 18-pulse signals

switching patterns. A vector analyzergenerates a picture of the operationalfield on the basis of the stator currentand stator voltage. This picture canbe displayed on an oscilloscope.

Figure 68 shows the rotational fieldsfor a motor connected to a 6-pulsePAM-AFD, an 18-pulse PAM-AFD, aPWM-AFD and a VVC-AFD,respectively.

6-puls PAM 18-puls PAM

PWM VVC

Figure 66. The rotational field of the motor can be displayed on an oscilloscope

The diameter of the circle indicatesthe strength of the magnetic field.The uniformity of the circle indicateshow well the AFD controls themagnetization. The edges on thepicture displayed indicates how willAFD does not manage the deviatesfrom a circle. The motor operation willbe unstable and cogging torques willincrease.

ACR 30


The control circuitThe control circuit is the fourth mainblock of the AFD. The control circuitis handling two things: It controls thesemiconductors of the AFD and itreceives signals from surroundingequipment to the AFD and transmitssignals from the AFD to otherequipment (Figure 67). Such signalsmay be from an operator at a controlpanel or from a PLC control.

For many years the control of the AFDwas based on the analog technique.However, today the AFD uses microelectronics incorporating digital dataprocessing.

Today's advanced technique hasreduced the calculating functions ofthe control circuit substantially. It isnow possible to store the pulsepatterns for the semiconductors ofthe inverter in a data memory.

The microprocessor built into the AFDcalculates the optimum pulse patternfor the motor used. The figure showsa PAM controlled AFD with anintermediate circuit chopper. Thecontrol circuit controls the chopperand the inverter. This is done on thebasis of the instantaneous value ofthe intermediate circuit voltage.

The intermediate circuit voltagecontrols the address counter for thedata memory. This memory containsthe pulse pattern output sequence forthe semiconductors of the inverter.The address counting speed followsthe intermediate circuit voltage. Withincreasing intermediate circuit voltagethe sequence is run through fasterand the output frequency of the AFDincreases.

For the chopper control theintermediate circuit voltage is firstcompared with the set referencesignal. The reference signal is avoltage signal, which is expected togive the correct output voltage andfrequency. Any difference betweenthe reference and intermediate circuitsignals will cause a PI controller tochange the chopper frequency. Theintermediate circuit voltage isconstantly matched to the referencesignal.

The computer in generalThe microprocessor consists of threebasic units, each with individualfunctions (Figure 68).

Vf

PI voltageregulator

Control circuit forchopper frequency

Sequencegenerator

Figure 67. Control circuit principle for a chopper-controlled intermediate circuit

Figure 68. The principle build up ofthe computer

The microprocessor is the heart of thecomputer. If the processor is suppliedwith the right sequence of instructions(program), it can execute a number offunctions on data stored in thecomputer memory. Themicroprocessor interacts with unitsaccording to the program entered.

The memory must store both theprogram and the various data.

The program can be stored in circuitsof an EPROM (ErasableProgrammable Read Only Memory).An EPROM does not lose its contentsin case of voltage loss. Theinformation in an EPROM can only beerased by exposure to ultra-violetlight. The microprocessor may onlyread information, not programmedinformation in the EPROM. RandomAccess Memory (RAM) will not retaindata after a voltage loss.

RAM is where the microprocessortemporarily stores data duringoperation.

The third section is the I/O whichcontains the Inputs and Outputs thecomputer needs to communicatewith. Peripheral equipment I/Osprovide connections to controlpanels, printers or other electronicequipment in the system.

A bus is a number of parallelconductors linking the units togetheras a working computer. The data bustransfers data between the units. Theaddress bus signals from where thedata must be taken and to where theymust be delivered. The control businsures that the data is transferred inthe right order.

Data bus

Mic

ropr

oces

sor Address bus

Ram Rom I/O

Control bus

ACR 31

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0 D

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The AFD’s computerIn addition to the three unitsmentioned previously the computer ofthe AFD (Figure 69) also comprises amemory which makes it possible forthe user to program. This memory isan EEPROM (Electrically ErasableProgrammable Read Only Memory). Itcan be programmed andreprogrammed electrically. When theAFD must be programmed for aspecific function an EEPROM is usedso the information is not lost.

The computer of the AFD alsoincludes an ASIC (Application SpecificIntegrated Circuit). The ASIC is anintegrated circuit where its functionsare specified by the AFD's designer.An example of such a design wouldbe the Danfoss VVC control principle.

Inputs and outputs of thecontrol cardThe number of inputs and outputs isdependent on the application type.AFD in automated applications mustfor example be able to receive analogand digital control signals (Figure 70).

• Analog signals can have any valuewithin a specific range

• Digital signals can have two valuesonly (ON and OFF)

There are no set standards for controlsignals; however, some signals are sowidely used that they can beregarded as such. An example ofthese “standard” analog signalswould be 0–10 V or 4–20 mA.

The digital outputs of a PLC mustelectrically match the digital inputs ofthe AFD. Typically these digital signalsare a nominal 24 VDC.

Serial communicationIn a working process the AFD is anactive part of the equipment. It iseither installed in a system withoutfeedback (control) or in a system withfeedback (regulation) from theprocess.

A system without feedback can bebuilt up with one singlepotentiometer. A system withfeedback is more demanding andoften includes a programmable logiccontroller (PLC).

The PLC may deliver control (speed)and command signals (start, stop,and reversing).

The output signals of the AFD, e.g.motor current or motor frequency, areoften used in conjunction with panelmeter, read out display, etc.

A PLC system (Figure 71) consists ofthree basic components:

• central unit• input and output modules• programming unit

CentralunitIn Out

Figure 71. The principle build up ofthe PLC

A control program is entered into thecentral unit by means of theprogramming unit. The central unit“sorts” the input signals and activatesthe output signals according to theprogram. The central unit can onlyprocess digital signals internally(Figure 72). That means signalschanging between two values, e.g. 24V and 0 V. The high voltage can eitherbe stated as “1” or “ON” and the lowvoltage as “0” or “OFF”.

Figure 69. The computer of the AFD

V

t

V

t

Figure 70. Analog signal and digital signal

Data bus

Mic

ropr

oces

sor Address bus

RAM EPROM EEPROM

Control bus

VVC

Powersection

Digital I/O

Operationindication

Analog I/O

Alarm On

Menu

Data

+

–

Jog

StopReset Start

Fwd.Rev.

ACR 32


V

t

0

1

Figure 72. The digital signal can be“ON” or “OFF” for short or longintervals of time

Basically an AFD and a PLC can belinked together in two ways:

One method is to connect the inputsand outputs of the PLC usingseparate wires to the inputs andoutputs of the AFD. The inputs andoutputs of the PLC thus replace theseparate components such aspotentiometers, control contacts anddisplays. The other method(Figure 73) is to transfer severalsignals at different times over one pairof conductors. Information A istransferred during time interval t1 to t2and information B is transferred fromt2 to t3 etc. This form of signaltransmission is called serialcommunication.

The principle to be chosen for serialcommunication depends on the kindof communication required and thenumber of units connected(Figure 74). One principle demandsmany conductors, if each unit must

both transmit and receive data.Another principle makes it possiblefor several units to communicate overtwo wires only. Here it is possible toconnect several receivers but onlyone sender. A third principle makes itpossible for all the connected unitsboth to send and to receive data overtwo wires. The communication linkbetween them is called a bus.

To ensure that units of different makescan “pick up” the serial signal all theunits must have a common signal level.

There are various standards des-cribing the common signal levels.These standards only comply to inter-connections, the information sentover these connections is determinedby the software. Both theinterconnection and the software

Figure 73. Serial communication ensures faster signal transmission, simplifiedinstallation

A

t1 t2 t3 t4 t5

A

t1 t2 t3 t4 t5

B C D

PLC

PLC

DIA

DID

S

S

S

A A

AID

A

DID

D

S S

S S

PrincipleStan-dard

(appli-cation)

::

transmitterreceiver

Numberof unitsper setof wire

Max.di-

stanceft.

Numberof wires

Signallever

RS 232(point

topoint)

1 trans-mitter

1receiver

49.5

Duplex:min. 3

+ variousmoni-toring

signals

±5 V min.±15 V max.

±3,6 V min.±6 V max.

±2 V min.

±1,5 V min.

RS 423(point

topoint)

RS 422(point

topoint)

RS 485(Bus)

1 trans-mitter

10receivers

1 trans-mitter

10receivers

32 trans-mitters

32receivers

3960

3960

3960

Duplex:min. 3

+ variousmoni-toring

signals

Duplex:4

Semiduplex:

2

Figure 74. Standards for serial connections

must be compatible for successfuloperation.

To date, RS 232 has been the mostcommon hardware standard. Theuse of it, however, is limited becauseof the short transmission distanceand low transmission speed. RS 232is mainly used where signals must betransmitted periodically, for examplewith terminals and printers.

RS 422 and 423 are suited for longtransmission distances and highertransmission speeds. They are suitedfor process automation, where thesignal transmission is morecontinuous.

ACR 33

RD

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ce

RS 485 is the only standard where itis possible to link multiple unitstogether for operation over a commonpair of wires. The units transmit datain turns over the common wireconnection (the bus).

In the communication between PLCand AFD there are three types ofsignals (Figure 75):

• control signals (speed change,start/stop, reversing)

• status signals (motor current, motorfrequency, frequency reached)

• alarm signals (motor stopped, overtemperature)

The AFD controls the motoraccording to the signals receivedfrom the PLC. The AFD transmitsinformation to the PLC about how thecontrol signals affect the motor/process (status signals). If the AFDsstops because of abnormaloperational conditions it transmits analarm signal to the PLC.

RS 485 makes it possible to designprocess systems in different ways.For example, the PLC can be placedin a panel and control many AFDs in aremote panel (Figure 76).

PLCStatus

Control

Alarm

Figure 75. Three signal typesbetween PLC and AFD

PLC

RS 485

Figure 76. The bus provides many new possibilities for application design.

ACR 34


Operational conditions of themotor

CompensationsPrior to the VLT series it was difficultto adapt the AFD to the motor.

It is easier today however becausethe VLT AFD is able to set startvoltage, start compensation and slipcompensation automatically basedupon the motor ratings.

With most AFDs it is also possible tochange these compensation settingsmanually.

Start compensation and startvoltageThe purpose of these twocompensations is to ensure optimummagnetization and maximum torqueat start and low speeds. This is doneby adding extra output voltage. In thisway the ohmic resistance in the motorwindings at the low frequencies, iscompensated for.

Start compensation is a loaddependent voltage, whereas the startvoltage is independent of the load.

If the motor is much smaller than therecommended motor size it may needadditional starting voltage, this is setmanually.

When motors are to operate inparallel the start compensation shouldnormally not be used.

Slip compensationThe slip of an AC motor is dependentof the load and it is approximately 3-4% of the rated speed. With a four-pole motor the slip will be about 75RPM.

When an AFD is operating a motor at180 RPM (10% of rated speed), theslip makes up 50% of the desiredspeed.

When the AFD is to operate the motorat 4% of its rated frequency the motorwill stall when it is loaded. Withefficient current measuring in theoutput phases of the AFD it ispossible to compensate for all theslip. The AFD compensates for theslip by increasing the frequencyproportionally according to the activecurrent. This form of compensation iscalled active slip compensation.

AFD and motor

Load dependent output voltageThe start voltage optimizes the AFDfor low performance motors. Afterstarting, the load will normally bevarying and the motor may beovermagnetized, when the loaddecreases. The motor will then takeup too much blind current and it willbe overheated.

The Danfoss VVC control principlematches the voltage to the presentload. The motor constants R1 and X1are modeled by the AFD and theycan be modified to different motortypes. On the basis of R1, X1 andaccurate measurement of the actualmotor current the AFD continuouslycalculate the optimum output voltage.Vk1 = V1 + I1 × R1 + I1 × X1

Load dependent control of the outputvoltage is called flux control andimproves dynamic performance ofthe motor.

ACR 35

RD

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Motor torque characteristicsIf the AFD could supply a currentwhich is several times higher than thefull-load current, the motor torquecharacteristics would be like thatshown in Figure 77. High currentsmay damage both the motor and theAFD. They are not needed for normalmotor operation. The AFD limits themaximum motor current.

The current limit is variable andmakes sure that the motor current isnot sustained higher than the desiredvalue. The AFD is able to hold themotor speed independent of the load.The motor torque characteristicswould now appear as rectangularwithin the rated working range of themotor, as shown in Figure 77.

Torque (%)

Speed(%)

100755025

1007550

Figure 77. The torque characteristicof an AFD-controlled motor isrectangular

It is preferred that the AFD can yieldan overtorque of up to 160% of ratedtorque momentarily. Most frequencycontrolled motors can also operate inthe oversynchronous range up to200% of synchronous speed.

The AFD cannot supply a voltage thatis higher than the voltage from the ACline.

Therefore, the voltage-frequencyratio is reduced when the speedexceeds the rated value. The mag-netic field is weakened and the torqueyielded by the motor is reduced by

n1 .

During oversynchronous operationthe AFD maintains maximum outputcurrent. The power output will beconstant up to 2 × nN.

Torque (%)

Speed(%)

20015010050

100

160

Figure 78. The torque and over-torqueof the motor

The motor speed can be stated indifferent ways: revolutions per minute[RPM], Hertz [Hz] or in percent of thesynchronous motor speed [%]. Thebasis is always the synchronousspeed of the motor at ratedfrequency.

Choosing the AFD sizeTo choose the correct AFD size for agiven load it is necessary to know theload characteristic. Then one must

P

f [Hz][%]

n[rpm]

72120

2160

3050

900

120200

3600

100

n/n0

V [v]

f [Hz]

460

72

=6.3uf

60

Figure 81. How the speed can beindicated (this curve shows the speedof a four-pole motor)

find the AFD which can yield the rightpower output. The necessary poweroutput can be calculated in four ways.The method to be applied will dependon the amount of motor dataavailable.

Load characteristicsWe distinguish between two loadcharacteristics:

P

f [Hz][%]

n[rpm]

601001800

3050900

1202003600

100

n/n0

V [v]

f [Hz]

460

60

=7.6uf

Figure 80. The motor power

Figure 79. The motor power

A change of the voltage-frequencyratio will affect the torquecharacteristic. If it is reduced to 6.3[V/Hz] the sequence will be asfollows:

P

Speed(%)

100 200

T = 100%

T = 160%

T = Torque

Torque

Speed

Torque

Speed

Constant (CT)

Square (VT)

Figure 82. Constant and squareload torque

ACR 36


We distinguish between two loadcharacteristics for the followingreasons:

• When the pump or fan speedincreases the power neededincreases by the cube of the speed(P = n3). The speed of pumps andfans should not exceedsynchronous speed.

• The normal working range of pumpsand fans is within 30-80% ofmaximum load.

These two conditions can be drawnfor the speed/torque characteristicsfor an AFD controlled motor.

Figures 80 and 81 show torquecharacteristics for two different AFDpower sizes. Figure 81 is one powersize smaller than that in Figure 80.

The load characteristic for the samepump is represented on both torquecharacteristics. In Figure 81 theoperating range of the pump (0-100%) is within the rated motorvalues. Since the normal operatingrange of the pump is 30-80% it ispossible to use an AFD with a lowerpower output.

When the load characteristic hasbeen determined the correct powersize of the AFD can be found on thebasis of other motor data.

1. The fastest and most preciseway of finding the right AFD is tomeasure the current IM drawn bythe motor under full load. If themotor is not fully loaded thecurrent might be determined onthe basis of measurements onsimilar applications which are inoperation.

I VLT I M

Figure 86.

Based on this technical data anAFD is selected with a maximumcontinuous output current IVLThigher, or equal to 15.5 A atconstant or square torque.

Example: [10 HP] 3 × 460 Vmotor takes up 15.5 A

S = M√3V × I ×

1,000= √3460 ×15.5 ×

1,000= 10.2 kVA

Based on the data given on AFDwith a maximum continuousoutput SVLT rating of 10.2 kVA orhigher at constant or square loadis chosen.

3. The AFD can also be chosen onthe basis of the power output ofthe motor PM. However, due tothe power factor of the motorand the motor efficiency ηchanging with the load, thismethod is less precise than thosepreviously mentioned.

Torque (%)

Speed(%)

100

100

160

50

Tacc

Torque (%)

Speed(%)

100

100

50Tacc

Figure 85. The overtorque can be used for acceleration

If the load's torque is constant themotor's torque capabilities must begreater to allow for acceleration.

If the AFD allows a temporaryovertorque of 60% this would besufficient for acceleration and therequired starting torque ensures thatthe application can withstand loadfluctuations.

If the AFD does not allow anyovertorque, it must be sized so thatthe acceleration torque Tacc is withinrated torque (See Figure 85).

2. The AFD can be chosen on thebasis of the power SM requiredby the motor and the poweroutput SVLT of the AFD.

Figure 88.

S VLT P M

Figure 87.

S VLT S M

Torque (%)

Speed(%)

100

100

160

80

30

Figure 83. Larger frequency converter

Torque (%)

Speed(%)

100

100

160

80

30

Figure84. Smaller frequency converter

ACR 37

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Example: A [5 hp] motor with anefficiency η and power factor of0.80 and 0.81 produces poweras:

S = Pη × cos ϕ M = 3.0

0.80 x 0.81 = 4.6 kVA

Based on the technical data anAFD whose maximum continuousoutput SVLT is higher or equal to4.6 kVA at constant or squareload torque is required.

4. For practical reasons the powersizes of typical AFDs usuallyfollow the ratings of AC motors.AFDs are often chosen on thebasis of these values but thismay result in inaccurate sizingespecially when the motor is notfully loaded.

3

M

IACT IRES

IREA

IRES

IACT

IREA IRES =IACTcos

Figure 90. Currents in the AFD

Normally, the motor manufacturerstates the power factor of the motorat full-load current. If the power factoris low the maximum motor torqueoutput must be reduced. Thisproblem is avoided by sizingaccording to the current drawn bythe motor and the maximum outputcurrent of the AFD (method 1).

If there is a capacitor across themotor terminals the effect will be likea short-circuit making the motorcurrent increase dramatically. Thiscondition is caused by the highswitching frequency of the AFD.

Normal operational conditionsOperationSome AFDs must be set andadjusted by means of switches andpotentiometers built into the unit. It isusually necessary to open theenclosure to gain access for theseitems which is not desirable in theindustrial environment.

Undesirable elements such as dustand static electricity may damage theelectronics. Therefore, it is veryimportant that the AFD is opened aslittle as possible. For this reasonmany AFDs can be operated and setremotely by the use of a panelproviding various data and allowingthe user to alter some settings(Figure 91).

Many of those AFDs are digital unitsand are set by menus and software.

A menu indicates parameters such ascurrent limit, minimum speed, ramp

Figure 91. The operation can bebased on menus

up, etc., and the number of menusmay vary between unit types.

The number of menus do not indicatethe operational complexity, as it isonly necessary to use some of themenus, since many are often presetfrom the factory.

The convenience of an AFD can beconsidered on the basis of thefollowing:

• How easy is the display to read andunderstand?

• Can the panel be operated withoutthe use of tools?

• Are the menus well arranged?• What information does the display

indication give during operation?

Alarm On

Menu

Data

+

–

Jog

StopReset Start

Fwd.Rev.

P M

Figure 89.

When selecting an AFD on the basisof the power output (method 2-4)make sure the calculations are basedon the same voltages as those statedin the technical data.

When choosing an AFD on the basisof current the voltage level is of noimportance, as it is the output currentof the AFD, which is determining theother values.

Power factor of the motorThe current which magnetizes themotor comes from the capacitor inthe intermediate circuit of the AFD(Figure 90). The magnetizing currentis a blind current flowing from thecapacitor to the motor and backagain.

ACR 38


• Can the menu indications replaceall measuring instruments duringinitial set-up?

• Minimum amount of programmingto start AFD?

• Are the values indicated in theappropriate units, current in Amp.,voltage in Volt etc?

Motor speed controlThe output frequency of the AFD andthe motor speed can be controlledwith a signal called speed reference.Typically as the speed referenceincreases, the motor speedincreases.

If the loads torque is less than thatobtainable by the motor within thecurrent limit setting (point A Figure93) the speed will be the desiredvalue.

If the torque curve intersects thecurrent limit setting (point B) thespeed will not continuously be able toexceed the corresponding value.

It is possible to set the AFD not to tripwhen exceeding the current limit(point C) for a predetermined periodof time.

V

t

Torque

Speed

Figure 92. Relationship between the reference signal and the motor torquecharacteristic

Figure 94. Acceleration and deceleration times

Speed

ttacc

n2

n1

Speed

ttdec

n1

n2

Since the motor always follows theoutput frequency of the inverter, it ispossible to switch directly fromdeceleration to acceleration.

The acceleration and decelerationtimes can be calculated if themoment of inertia on the motor shaftis known:

t = J ×accn - n2 1

T + T × 9.55acc fric

t = J ×decn - n2 1

T + T × 9.55dec fric

J is the moment of inertia of the loadas applied to the motor shaft.

Tfric is the friction torque of the load.

Tacc is the starting torque used foracceleration.

Tdec is the braking torque occurringwhen decreasing the speedreference.

If the AFD cannot produce anovertorque for a limited period oftime, the acceleration and thedeceleration torque can be put equalto the rated motor torque TN. Inpractice the acceleration time willtherefore be equal to the decelerationtime.

Example: J = 0.997 lb ft2, n1 = 500RPM, n2 = 1000 RPM,

Tfric = 0.05 × TN, TN = 19.9 ft lbs

If the AFD cannot produce anyovertorque, it is necessary to knowmore about Tacc and Tdec.

BrakingWhen the speed reference decreases,the motor acts as a generator andbrakes the degree of braking isdetermined by the output power ofthe motor.

A motor connected directly to the ACline can generate braking power backto the line. With an AFD this is not thecase because the intermediate circuitabsorbs the braking power.

When the braking power is higherthan the power loss of the AFD theintermediate circuit voltage can rise.The intermediate circuit voltage canrise until the AFD trips out for reasonsof inverter protection. It may benecessary to load the intermediatecircuit with an external resistor inwhich the braking power can bedissipated as heat. By using adynamic brake resistor, heavy loadscan be slowed down very rapidly(Figure 95).

I

Speed

160

n, VLT

(%)I

I LIM

B 1

B 2A

C

Figure 93. The motor current canexceed the current limit

Acceleration and decelerationAcceleration indicates at what ratethe speed increases in time to thedesired speed. The value is called theacceleration time tacc.

Deceleration expresses at what ratethe speed is falling. The time until thespeed is down to the new desiredspeed is called the deceleration timetdec.

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If the AFD is rectifier controlled, thebraking power can be sent back tothe AC line. This is done through aninverter connected in antiparallel withthe rectifier (Figure 96).

U V W

L L L1 2 3

U V W

L L L1 2 3

Figure 97. The motor speed directionchanges when the phase sequence ischanged

ReversingThe shafts operating direction of ACmotors is determined by the phasesequence of the supply voltage. Thedirection is changed by inverting twophases, causing the motor to reverse.In most motors the shaft is turningclockwise, when the connection is asshown on the Figure 97. The phasesequence of the output terminals ofmost AFDs follows the same principle.

The AFD can change the motorspeed direction by changing thephase sequence electronically.Reversing is initiated by a negativespeed reference or through a digitalinput signal. If the motor must have aparticular speed direction on start upcheck the factory setting of the AFD.

Since the AFD limits the motorcurrent to a specific set value, theAFD controlled motor can bereversed more frequently than motorsconnected directly to the AC line(Figure 98).

RampsTo ensure smooth motor operationmost AFDs are supplied with rampfunctions. These ramps areadjustable and they ensure that thespeed reference can only increase ordecrease by the set value.

If the set ramp times are too short,the motor current can increase untilthe current limit is reached.

If the ramp down time is too short thevoltage in the intermediate circuit mayincrease so much that the protectiveelectronics trip the AFD.

Speed

t

Speed

t

Figure 99. Variable ramping times

Figure 98. The braking torque of the AFD during reversing

Torque

Speed

n = speedT = torque

n T Tn

n T Tn

Figure 95 Brake resistor

Figure 96 Inverter connected inantiparallel

An AC motor can also be braked byapplying a DC voltage between twomotor phases. This produces astationary magnetic field in the stator.The braking power stays in the motorwhich could cause overheating. Thatis why DC braking is primarilyintended for frequencies below 2 Hz.

ACR 40


n

tt down

n ref

n N

Figure 100. Setting of ramp times

The optimum ramp times (Figure 100)can be calculated by means of theformulas below:

t = J ×upn

(T - T ) × 9.55 [s]n fric

t = J ×downn

(T + T ) × 9.55 [s]n fric

Normally the ramp times aredetermined on the basis of the motorsrated speed.

Process monitoringThe AFD monitors the application'sprocess and takes action in case ofoperational disturbances.

There are three types of monitoring:

Application monitoring. The AFDmonitors the application on the basisof the output frequency, current andmotor torque. On the basis of thesevalues it is possible to set a number oflimit values for the control, such asminimum allowable speed ormaximum allowable motor current.The AFD can be programmed toexecute special functions when theselimits are exceeded. For example, itcan be programmed to provide analarm signal to increase motor speedor to brake the motor as quickly aspossible.

Motor monitoring. The AFD monitorsthe motor based on a calculation ofthe thermal conditions. Similar to athermal overload relay the AFD helpsprevent motor overloading. The AFDalso takes the output frequency into

account which ensures that the motoris not overloaded at low speedswhere the self ventilation of the motoris reduced.

Unit monitoring. The AFD may be setto trip in case of overcurrent. SomeAFDs can yield a momentaryovercurrent.

The fast microprocessors used in theAFD can monitor the motor currentand time, which ensures optimumutilization without overloading theAFD.

Motor loading and heatingMotors that are connected to AFDsshould be adequately cooled.

There are two factors to take intoconsideration:

• The amount of cooling air is reducedwith lower motor speed.

• The motor generates additional heatif the applied current is not entirelysinusoidal.

At low speeds the motor fan cannotsupply sufficient amounts of coolingair. This problem arises when the loadtorque is constant in the overalloperational range. The reducedventilation determines the maximumacceptable torque under constantload.

A motor that is to run constantly at aspeed being less than half of its ratednameplate speed requires extracooling (grey area of Figure 101).

The problem can be solved byselecting from the following:

• an inverter rated motor• motor with high service factor• energy efficient motor, or• larger motorThe motor receives harmoniccurrents if the applied current is notsinusoidal. The harmonic currentsdissipate additional heat in the motor,which is dependent on the size of theharmonic currents (Figure 102). If themotor current is not sinusoidal themotor must not continuously beloaded 100%.

Speed (%)

Torque (%)

100

50

100 200

Speed (%)

Torque (%)

100

50

100 200

Figure 102. Extra heat is dissipated inthe motor if the current is not entirelysinusoidal

Figure 101. Need for additionalventilation when using a rated motorsize and an oversize motor

Torque (%)

Speed (%)

Torque (%)

Speed (%)

100

100

100

100

Oversize motor

Rated motor

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EfficienciesThe efficiency of a unit (Figure 103) isdefined as the ratio between the poweroutput P2 and the power consumptionP1:

PPη = 1

2

The difference between P1 and P2 iscalled the power loss PT, which isdissipated in the unit as heat.

The efficiency can be calculated for theAFD alone, the motor alone, or for theAFD and the motor (system efficiency).

The efficiency of the AFD is calculated

as PP2

1

The motor efficiency is calculated as

PP3

2

The system efficiency is calculated as

PP3

1

From the curves it can be seen that themotor efficiency has significantinfluence on the system efficiency. Theefficiency of the VLT AFD is high in theentire control range and during highand low load.

The curves also show that theefficiency is lowest at low speeds. Thatdoes not mean that the absolutepower loss is highest at low speeds.

Example from Figure 104c:

1. n = 800 RPM, P3 = 9628 W, η = 77%

Pη3P =1 = 12504 W, P = P - P = 2876 WT 1 3

2. n = 500 RPM, P3 = 1500 W, η = 77%

Pη3P =1 = 2143 W, P = P - P = 643 WT 1 3

The high efficiency of the VLT AFD hasseveral advantages:

• The higher the efficiency, the less theheat loss to be removed from theinstallation. This is highly advanta-geous for panel mounted AFDs.

• The less heat loss that is dissipatedin the semiconductors and coils ofthe AFD, the longer their expectedlifetime.

Figure 103. Outputs and efficiencies

600 1200 1800 2400 300000

20

40

60

80

100

rpm

%

A

B

600 1200 1800 2400 300000

20

40

60

80

100

rpm

%

BA

B

600 1200 1800 2400 300000

20

40

60

80

100

rpm

%

B

A

1

2

Figure 104a. Efficiency forVLT-type 3016 at 100%(A) and 25% (B) load

Figure 104c. Efficiency fora frequency converterand motor at 100% (A)and 25% (B) load

Figure 104b. Efficiency fora typical motor at 100%(A) and 25% (B) loadwhen fed from afrequency converter

P1

PT

P2 M

P1 P2 P3

ACR 42


Long motor cablesAFDs are designed for motor cablesof a certain maximum length and of acertain wire gauge. These cablevalues vary greatly between AFDtypes. Other specifications of the AFDwill be affected if the maximum cablelengths are exceeded.

Motor cable length required for theappropriate placing the AFD and themotor must be considered in eachcase.

The longer the motor cable is, themore heat will be generated in theAFD. Always check what cablelengths and wire gauge the AFD willallow, as this is of great importance toits thermal conditions.

If the length or the gauge of the motorcables exceeds the maximum valuesthe maximum allowable continuousoutput decreases.

The longer the cable length or thelarger the gauge, the lower thecapacitive reactance. High capacitivereactance will increase the losses inthe cable. The resulting outputcurrent must be reduced by about5% for each step the wire gaugeincreases (Figure 105). The current isreduced linearly, when the cablelength exceeds the maximum forwhich the AFD has been designed.

The typical mode of operation for theAFD causes short voltage rise times inthe motor cable. This may damagethe insulation of the motor windings.The problem intensifies as theswitching frequency of the inverterincreases.

The problems of dv/dt and thecapacity of the motor cable can besolved by installing a motor filter in theoutput of the AFD. Always check if theAFD incorporates an effective motorfilter, or if such a filter is available asan option.

Intermittent operationTo describe this kind of operation weapply the current consumed by theAFD.

IH is the current it consumes duringhigh torque situations and IL is thecurrent at low torque situations.

When both currents are lower orequal to the rated input current of theAFD no problems will occur. When IHexceeds the rated input current thetime intervals tH and tL, the duty cyclebetween IH and IL must be taken intoaccount. The temperature in the AFDincreases in the period tH and falls inthe period tL.

cable length

max. output current

nominallength

nominal cablecross section

increased cablecross section

Figure 105. The maximum outputcurrent of the AFD depends on thelength and gauge of the motor cable

Example 1 (see Figure 107):

IH = 160% (corresponds to a 60%overtorque).

At IL = 100% tL may last 600 sec. andtH 30 sec.

When IL is changed to 80%, tL can bereduced to 100 sec, or tH beincreased to 108 sec.

Example 2 (see Figure 107):

IL = 100%

If IH is reduced to 140%, tL can bereduced from 600 sec. to 300 sec., ortH can be increased from 30 to 50 sec.

t

I N, VLT

t H t L

A

B

I H

I L

Figure 106. Intermittent operation

Referring to Figure 106, area A canbe increased when area B isincreased. When B is reduced it maybe necessary to reduce A.

A more precise correlation betweenthe values of IH, IL, tH and tL willappear from the specifications on theindividual AFD.

5 10 432 100 432 1000 43234

6

10

2

3

6

100

4

2

34

6

5 10 432 100 432 1000 43257

10

2

34

6

100

2

34

6

1000

H [s]†

H [s]†

L [s]†

L [s]†

I

I

= 140%

= 160%

H

H

I L = 0 %

100

90

80

50

I L = 0 %

100

90

80

50

Figure 107. Load degrees influencethe intermittent operation

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Parallel connection of motorsAn AFD can control several parallel-connected motors. If the motorspeeds must be different, motors ofdifferent rated speeds must beapplied. The motor speeds can bechanged simultaneously and the ratiobetween the motors is maintained inthe entire control range.

The total current consumed by all ofthe motors may not exceed themaximum output current of the AFD.

The individual motors can be switchedand reversed an unlimited amount onthe output of the VLT AFD.

If the total starting current of themotors is higher than the maximumoutput current of the AFD, the outputfrequency falls. The output current ofthe AFD can exceed the rated currentof the individual motor, making itnecessary to protect each motor as ifit were connected to the AC line.

If the motor sizes deviate very much,problems may arise during startingand low speed operation. This is dueto the fact that small motors have arelatively large ohmic resistor in thestator, therefore they demand morecompensation voltage during startingand low speeds.

Often it will be possible to increase thestart voltage and find an acceptablestart condition for all the motors. If thisis not possible it may be necessary toreplace the small motor with a largerone. This does not necessarilydemand a bigger AFD, as themechanical power output of the motoris unchanged.

Explosion hazardous areasAC motors are available in versionsdesigned specially for use in explosionhazardous areas, i.e. areas wherethere are inflammable gasses, steamsor dust.

AFDs are not intended to be placeddirectly in the hazardous area. Theyare only typically available inenclosures up to NEMA 12.

Explosion-proof motors have beensubjected to several tests byauthorized organizations according to

At high loads the start problem can besolved by using a transformerspecially designed for a high voltage/frequency ratio. Another possibility isto replace the motor.

Protection under extremeworking conditionsExtreme working conditions include allkinds of abnormal disturbances on theinput and output sides of the AFD.

On the input side such disturbancemight be overvoltages and transientsand on the output side overcurrentsfrom short-circuits, ground faults,switching motor cables andregenerative operation.

The AFD protects its electronics by“predicting” the extreme conditions.This is possible when the AFD isdesigned with the appropriate logicsuch as preprogrammed ASIC.

ASICs can allow a fast and precisecurrent measuring. It is possible forthe AFD to take action against anextreme situation before theelectronics are damaged.

SafetyOften it is necessary to place anemergency stop near the motor. It isimportant that it is possible to place aswitch in the motor cable and that theAFD is not damaged no matter howoften the switch is activated.

The control inputs of the AFD must beisolated from the power section andthe AC line. If not, the control leadswill have the same voltage in relationto ground as the AC line. In that caseit would be highly dangerous to touchthe control leads and other equipmentmay be damaged.

To take precaution against fire it isimportant that the AFD has a built-inthermal relay cutting out its operationwhen the cooling is inadequate.

Sometimes an AFD controlled motormay start without warning after cutoutfor thermal overload or if the motor isnot switched off while operationalparameters are altered. All localHealth and Safety precautions mustbe complied with to prevent danger topersonnel.

specific standards. When the motorhas been approved it can then bemarked according to the standard inquestion.

A motor's certification and approval istypically based on measurementsmade on the fixed voltage andfrequency of the AC line.

If such a motor is to be controlled byan AFD they will test and measure theheat generation in the whole rangeduring operation.

Explosion-proof motors rated for usewith AFDs are available from severalmotor manufacturers.

Transformers and AFDsA transformer can be placed ahead ofthe AFD or between the AFD andmotor.

Isolating transformers or autotransformers designed for rectifieroperation can be used ahead of theAFD. An isolating transformer can beused under any circumstances. Anauto transformer can only be usedwhen the transformer is star-connected and the star pointgrounded. The AFD loads thetransformer like an ordinary three-phase rectifier and the transformercan be selected on the basis of thevoltage of the supply mains and therated voltage of the AFD, maximuminput current, cos ϕ and power factor.

A transformer between AFD andmotor is used to adjust the outputvoltage, to 48 V motors for example,that have been chosen to protectpersonnel or give galvanic isolation.

Transformers are usually intended fora specific frequency. There will be avoltage drop as a consequence of theohmic resistance in the transformerwindings. This has the same influenceas the corresponding voltage drop inthe motor windings. When the AFDmust compensate for the ohmicvoltage drop in both transformer andmotor the transformer will receive atoo high voltage-frequency ratio atstart and at low speed.Overmagnetization will result and themotor will be unable to start.

ACR 44


GroundingThrough a protection wire the groundterminal of the AFD is connected toan earth electrode. This form ofprotection demands that theimpedance of the earth electrode issufficiently low.

Electrical noiseElectrical noise is electric disturbancethat affects a unit or that is emittedfrom a unit.

The electrical noise can be split upinto three main groups (Figure 108).One group is the thermal interferencevoltage coming from all components.The limit for a unit's sensitivity isdetermined by physical laws.

The second group is atmosphericnoise, for example, voltage peaks onthe supply voltage during lightningstorms. The atmospheric noisecannot be damped and itsdisturbance of electric installationscan only be limited by taking differentmeasures.

The third group is the noise comingfrom switches, radio transmitters andother electric devices. This man-made noise can be controlled.

EMC is a term that is often used inconnection with electrical noise. EMC

AC line interference and radiofrequency interference (RFI).

Emission typesThe electrical noise can be spreadover the AC line (conducted) and bythe air (coupled/radiated noise).

CouplingThe coupling is dependent on howthe electric circuits have beendesigned. The coupling can begalvanic, capacitive or inductive.

is an abbreviation of Electro MagneticCompatibility, i.e. a unit's ability toresist electrical noise and not to emitelectrical noise to other surroundingequipment.

Emission is the electro-magneticenergy emitted from a unit.

Immunity is the unit's ability to resistelectro-magnetic disturbances. Inregards to AFDs, electrical noise like

Figure 110. Capacitive coupling

Figure 108. Different kinds of electrical noise

Thermal noise Human-made noiseAtmospheric noise

Electrical noise

Figure 109. Galvanic coupling

Z

Z L1

Z 0

M

The galvanic coupling may occurwhen two electric circuits have acommon impedance.

In Figure 108, line impedance and theground impedance are common forthe AFD and another electric device.Dependent on the impedanceconditions a noise voltage can betransferred to the unit over the twocommon impedances ZL1 and Zground.

The capacitive coupling can happen,when two electric circuits have acommon ground. A typical exampleis, where the motor cables are placedtoo close to other cables connectedto sensitive devices.

The capacitive noise current dependson the switching frequency of theinverter and how far the motor cableis from other cables. The highfrequency of the output voltage givesa low capacitive resistance in themotor cable and results in acapacitive noise current.

The noise current may flow as shownin Figure 110.

Inductive coupling may occur whenthe magnetic field around a live wireaffects another wire or another unit(Figure 110). The strength of themagnetic field depends on thecurrent, the wiring and the distance tothe live wire. AC voltages mayespecially induce noise into anotherwire loop. The size of the inducedvoltage depends on the frequency anamperage of the induced voltage(Figure 108).

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Figure 111. Inductive coupling

M

Conductednoise

Over-couplednoise

Unit

Figure 112. Ways of emission of electrical noise

TransmissionElectrical noise can also be spread viathe wires of the AC line. This takesplace when the sine-shape of thesupply voltage is deformed or whenthe line is disturbed by high frequencynoise.

AC line interferenceAC line interference is a distortion ofthe sinusoidal curve shape of thesupply signal. The distortion goesfrom the AFD and back through theAC line. A humming sound can oftenbe heard from other units connectedto the same AC line.

The rectifier of the AFD sends apulsating DC voltage to theintermediate circuit. Its capacitor ischarged at each voltage peak. Duringthese chargings AFD draws upcurrents of relatively largeamplitudes. The AFD can become apulsating load, distorting the sine-shape of the supply voltage. Thedegree of distortion depends on theimpedance of the AC line and thesize of the loads current.

The AC line interference consists ofharmonics of the basic frequency ofthe supply voltage, and for eachindividual harmonic, it is possible tocalculate the amplitude of the noisevoltage (An) in percentage of thenominal voltage.

√(R + n × X 2A [%] = 100 × 1I ×

Un

2 2

n is the ordinal of the harmonic

I is the size of the load current

R is the ohmic part of the lineimpedance

X is the reactive part of the lineimpedance

V is the nominal voltage

The formula applies to units wherethe zero conductor is not connected.Line interference can be reduced bylimiting the amplitudes of the pulsecurrents. In practice, this is done byinserting coils in the intermediatecircuit of the AFD.

If the AFD does not include thesecoils as standard, they must bepurchased and mounted separately.

ACR 46


Transients/OvervoltageMost industrial AC lines are disturbedby line transients which can be shortovervoltages of up to 1000 V.

They arise when high loads are cut inand out elsewhere on the AC line. Alightning strike directly to the supplywire causes a transient wave of highvoltage. The transient may damageinstallations in a distance of up to 4miles from where the lightning strikes.Short-circuits in the supply lines canalso cause transients. High currentsdue to short-circuits can result in veryhigh voltage in the surrounding cablesbecause of inductive coupling.

Radio frequency interferenceAny current or voltage deviating fromthe sine curve will contain componentsof higher frequencies.

The frequencies will depend on howsteep the sequence is.

When a switch is activated the currentincreases very quickly from zero torated current. In that case thesequence is very steep. In a radio youwill hear a crackling.

One single noise pulse will not do anyharm, but as the semiconductors of theAFD are acting as switches they emitnoise disturbing the surroundingelectronic equipment.

Radio frequency interferenceRFI) is defined as electric oscillations offrequencies between 150 kHz and 30MHz.

The RFI degree depends on differentconditions:

• the impedance conditions of the ACline

• the switching frequency of the inverter• the frequency of the output voltage• the mechanical buildup of the AFD• the power level of the AFDRadio frequency interference can beconducted and coupled noise.

There are different standards for themaximum allowable radio noise from aunit. An example is the German VDE

Figure 114. EEC’s limits for emissionof radio frequency interference

Curve NRFI filter(extra)

Curve GRFI filter(integrated)

Mains supplyinterference Transients

Currentmeasuring

dudt

Figure 115. How the frequency converter is damping electrical noise

Figure 113. Limits for emission ofradio frequency interferenceaccording to VDE 0875

0.1 0.2 1 2 5 10 30

MHz0

20

40

60

80

100

dB V

0.1 0.2 1 2 5 10 30

MHz0

20

40

60

80

100

dB V

standards (Figure 113). VDE 0875gives the level for the acceptable RFIemission over the AC line. VDE 0875,curve G) is the acceptable limit forindustrial equipment. Curve N gives thelimit for ordinary household equipment.

The EEC directive EEF 82/499European Standard) gives like, VDE0875, the levels for how much radionoise a unit is allowed to emit overthe AC line. From this directive it canalso be seen how much radio noise aunit is allowed to emit on the outputside. In regards to AFDs the outputside is the motor cable. As of 1992EEC 82/499 will be standard for allEuropean countries (Figure 114).

Effective suppression of the radiofrequency interference can only beobtained by means of a filter. Thisfilter is called an RFI filter and itconsists of coils and capacitors.Some suppliers of AFDs offerintegrated RFI filters as standard, ifnot, this filter must be purchased andmounted separately (Figure 115).

In the motor cable the radio noise canbe suppressed by means of an RFIfilter or by using a shielded cable. Thehigh frequencies means:

• that the capacitors of the filter drawhigh currents that may make theAFD cut out.

• that the filter coils must be very bighigher expenses and increasedacoustic noise).

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Shielded cablesFor a shielded cable a switchingimpedance is stated. It is theimpedance of the shield in thelongitudinal direction. To be sure thatthe noise current returns to the shieldthe value of the switching impedancemust be as low as possible.

To provide effective suppressionagainst emission of high frequencyradio noise both ends of the shieldmust generally be connected toground. It is important that thecontact between the shield and theground/chassis terminal is good. Abad connection will increase theimpedance of the shield and reducethe suppression of radio noise(Figure 116).

Control cables should not beconnected to ground on both endsbut only to the ground terminal of theAFD. This is partially due to the factthat any noise current in the cabinetof the control case could act as acurrent loop that will have adisturbing effect on the control.

Before you buy an AFD you shouldexamine how the electrical noise canbe limited.

Operational reliabilitySimple trouble shooting

Supply voltageIf a functional fault is observed, checkthe following:

Has the AC line been connectedcorrectly?

Has a prefuse blown?

Is the AC line within the permissiblevariation?

MotorHas the motor been connectedcorrectly?

Has the emergency switch beenactivated?

Control signalsIs the AFD receiving the controlsignals?

Are the signals of the proper type andvalues?

Menu settingsAre all settings correct?

Fault indicationSome AFDs also provide faultindication; this is of great help for themore advanced trouble shooting(Figure 117).

Here any faults arising because ofmotor overload, over- or undervoltagein the AC line, short-circuits or groundleakages etc. are either indicated on adisplay or by means of LEDs.

Figure 116. The cable shield must be connected properly

Figure 117. The AFD can be of greathelp for trouble shooting

Alarm On

Menu

Data

+

–

Jog

StopReset Start

Fwd.Rev.

FusesThe AFD may not be supplied withbuilt-in fuses, as the fast currentmeasuring and electronics provideeffective protection for the unit.

However, the whole installation mustbe protected by means of branchprotection ahead of the AFD.

The fuses must be dimensioned tocarry the maximum intermittent inputcurrent of the AFD and to protectcables and contractors. The fusesmust not be dimensioned accordingto the normal starting current of themotor, since with an AFD startingand reversing produce no inrushcurrent on the AC line.

ACR 48


Short-circuits and ground faultsShort-circuits and ground faults mayoccur

• on the supply side• on the motor side• in control leads

M

Figure 118. Where there is risk ofshort circuits

Any short-circuits or ground faults onthe supply side will cause theprefuses in the installation to blow.The AFD itself will seldom causeshort-circuits and it will not bedamaged because of faults on thesupply side.

As a rule motor faults arise because ofmissing insulation that causes short-circuits between two phases orbetween phase and ground. A short-circuit will act as an overload on theAFD which may then trip out.

Grounding can also cause the AFD totrip out. A short-circuit of the controlleads of the AFD may overload theinternal voltage supply. The internalvoltage supply is therefore protectedby a fuse.

Grounding of a control lead will notdamage AFDs with input isolation.

Insulation measurementHigh voltage tests on an AFDinstallation may damage theelectronics; therefore, the input andoutput terminals must be short-circuited. If the AFD is supplied withan RFI-filter the filter capacitors mustbe disconnected.

U

V

W

L

L

L

1

2

3

The maximum voltage drop allowedfrom the transformer of the AC line tothe AFD voltage varies from country tocountry. It is therefore necessary tofollow the local regulations(Figure 120)

The acceptable voltage drop willtypically be 4% and whendimensioning the cable the followingmust be taken into account:

• ambient temperature• set-up• cable impedance• load degreeFinally the wire cross section of thecable must be dimensioned accordingto the current consumed by the AFD.There will be no current inrush on theAC supply during motor start andreversing; however, if the motorproduces intermittent overtorque theAFD will draw an intermittentovercurrent. The duration of thiscurrent may have influence on howsmall the wire cross section may be.

Figure 120. The demand for the impedance of the AC supply must be inaccordance with the local regulations

A B

VBVA

I VLT Z

(I VLTVB = VA - Z )

VA VB-

VA100 4%

The influence of the AC lineNormally the AFD specifications aremaintained by a supply voltage thatvaries +10% from the rated value.

Generally, the impedance of the ACline is so low that it does not affectthe function of the AFD.

Figure 119. Short circuits andswitching off before high voltage testin the installation

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Considerations to be madebefore buyingTo find the right AFD for yourapplication the following points mustbe considered:

• Does the AFD automatically matchthe output voltage to the actualload dynamic flux control)?

• Does the AFD automaticallycompensate for the loaddependency of the slip dynamicslip compensation)?

• Does the AFD allow intermittentovertorque?

• How great a part of operation andsetting is done inside the AFD?

• Is the AFD easy to operate?• Can the AFD control the braking of

large moments of inertia?

• How extensive are the monitoringfunctions of the AFD?

• Does the motor generate extra heatbecause of the wave shape of themotor current?

• How high is the efficiency of theAFD in the whole control range)?

• Is the specified length and crosssection of the motor cablesufficient?

• Is it possible to mount emergencystop or switches in the motorcable? How often can they beactivated?

• Is the AFD fitted with built-in motorfilter or is it available as option?

• In what enclosures can you get theAFD?

• Does the AFD suit the AC line?• Is the isolation of the control leads

effective? What standard has beenfollowed?

• Is the AFD protected againstthermal overload?

• What precautions have been madeagainst electrical disturbance?Does the AFD meet any recognizedstandards UL, CSA, etc.)

• Is the AFD protected againstoverheating?

• Is the AFD protected against short-circuits and grounding? How doesthe AFD react?

• Has the AFD fault indication? Howadvanced is it?

ACR 50


Rectilinear movementIn a rectilinear movement, a body willlie still or maintain its rectilinearmovement until it is actuated by aforce.

The force “F” can be expressed asthe product of the mass of the bodyand the change per time unit of thevelocity of the body. The velocitychange per time unit is the same asacceleration “a”.

Appendix 1General mechanical theory

T = J

Angular velocity:(1) unit of measurement [ ]radians

s

(1) ; n measures in [ ] 2 n60

rmin.

Angular acceleration:radians

sα = ; unit of measurement [ ] dωdt

Moment of inertia: J; unit of measurement [kg m ]2

Figure 123.

r

FT = F r

Figure 122.

To maintain constant movement abody must be actuated all the time.This is necessary because of forcesof friction and gravity attracting thebody in the opposite direction.

The moment of inertia has, like themass, a damping effect on theacceleration. The moment of inertiadepends on the mass and form of thebody according to the axis of rotation.

When the torque and accelerationconditions of a plant are to becalculated it is advantageous to relateall masses and inertias to a totalmoment of inertia on the motor shaft

J;

J , J etc:

ω :

ω , ω etc.

Moment of inertia of the motor

Individual moments of inertia

Angular velocity of the motor

Angular velocity of the various

J = J + J1 1 ( )2

1

2+ J 3 ( )3

1

2+...

ωω

ωω

2 3

1

2 3

of the system

rotating bodies

Figure 121. Calculation of differentmoments of inertia

Rotary motionIn rotary motion a body can be forcedto rotate or to alter its rotary velocity,if it is influenced by a torque aroundits mass center.

Like the force the torque can beexpressed through its effect:

The product of the moment of inertiaof the body J and the change of thevelocity of the body per time unit, theangular acceleration α.

F = m a

Mass:

Acceleration:

Force:

“m” unit of measurement: [kg]

“a” unit of measurement: [ ]

“F” unit of measurement: [N]

ms2

d

d

r

d

r1

r2

2r

Solid cylinder

Hollow cylinder

Solid ball

J = m r 2

2

J = m r 2

4+ m l 2

12

J = m12

J = 2 m r 2

5

(r + r )22

2

Figure 125.

Figure 124.

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Work and powerThe work “W” performed by themotor in rectilinear motion can becalculated as the product of the forcein the direction of motion “F” and thelength of the movement “s”.

In rotary motions the output iscalculated as the product of thetorque “T” and the angular motion ϕ.One revolution = 2 × π [rad].

W = T

Length of motion: s Unit of measurement: [m]

Output: W Unit of measurement: [W × s]

ϕ

W = F s

Angular motion: ϕ Unit of measurement: radians

One resolution = 2 × π [rad]

The output of a conveyor system isincreasing with the time. It has nomaximum value and therefore itcannot be applied for sizingcalculations.

The power “P” express the work pertime unit and therefore it has nomaximum value.

In a rectilinear motion the power iscalculated as the product of themotion in the direction of motion andthe length of motion per time unit, thevelocity “V”.

In rotary motions the power iscalculated as the product of thetorque and the length of motion pertime unit, angular velocity.

Figure 128.

P = T ω P = F VUnit of measurement: [W]

Figure 127.

Figure 126.

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General alternating currenttheoryAC voltage alternating current) issymbolized with ~. It changesamplitude and direction.

The number of periods per second iscalled frequency, which is indicated inHertz. 1 Hz = one period per second.The duration of one period is theperiod time and it is found as T = 1/f.At a frequency of 50 Hz the periodtime is 0,02 s (Figure 129).

Contrary to DC voltage and current,AC voltage and current can havedifferent values.

Appendix II

1 period

1 rotation(of four-pole rotor)

Current one way

Current theother way

time

Figure 129. Different values at AC voltage

90

270

180 0/36045 90 135 180 225 270 315 360

med

ium

med

ium

max

.m

ax.

activ

eac

tive

peak

to p

eak

Current/voltage

Figure 130. AC voltage

Figure 131. The direction of thevector is anti-clockwise

In general, it is the actual value that isapplied and an alternating currentvalue of 1 A generates the same heatin a given resistor as a direct currentof 1 A.

Vectors are most useful in connectionwith alternating currents and voltages.They clearly show the connectionbetween current, voltage and time.

A vector is characterized by its lengthand its direction of rotation. It isrotating anti-clockwise (Figure 131).

When the magnetic field vectorrotates one revolution and is back toits starting point, the vector tip willhave traced a complete circle, i.e.360°.

The time of one revolution is equal tothe period time of the sine curve. Thevector velocity per second is calledthe angular velocity and is indicatedby the Greek letter ω. ω = 2 × π × f.

There are three forms of AC loads.

When the load consists of coils withiron cores like motors the load willprimarily be inductive. Here thecurrent will be delayed in timecompared to the voltage.

The load can be capacitive and thecurrent will be time-wise in advanceof the voltage. The load can beentirely ohmic and there will be no

displacement between current andvoltage (Figure 130).

The displacement between voltageand current is called the phasedisplacement angle and it isdesignated with the Greek letter ϕ.

By multiplying the associated valuesof current and voltage the powercurve for the three loading forms canbe drawn.

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t

CLR

Capacitive loadInductive loadOhmic load

I

U

I

U

I

U

I

V

P

I

V

P

I

V

P

0 90

270 360

I max

V max

I max

V max

0 90

270 360

0 90

270 360

V max

I max

0 90

270 360

P=O P=O

Figure 132. Current, voltage and power at entirely ohmic load

The loading forms are only theoreticalquantities when there is a matter ofAC circuits. A load will either beohmic-inductive, or ohmic-capacitive(Figure 132).

Power factorThe power factor is defined as theratio between the active power andthe apparent power.

It is often called cos. ϕ; but cos ϕ isonly defined for sinusoidal currentsand voltages.

With nonlinear loads such as AFDsthe load current is not sinusoidal. Wemust therefore distinguish betweencos ϕ and the power factor λ.

Power

Voltage

Current

Phasedisplacement

Formulasign In general Unit

P V x I x cos = S cos W or kW=

Q =

S =

V =

I =

I =

I =

cos =

RES

ACT

REA

sin

V x I x sin = S sin

V x I = Pcos = Q

sin

PI x cos = Q

I x sin = SI

PV x cos = Q

V x sin = SV

=V

S x cosPV

=I

S x sinQI

=PV x I

PS

=QV x I

QS

VAr or kVAr(Sin or ks)

VA or kVA

V

A

A

A

abstractnumber

abstractnumber

Figure 133.

P λ = I × V

where P is the active power and I andV actual values.

ϕ designates the phase differencebetween current and voltage. With anentirely sinusoidal current and voltageCos ϕ corresponds to the ratiobetween the active power and theapparent power.

ACR 54


The voltage between a phase wireand neutral wire is called the phasevoltage Vph and the voltage betweentwo phase wires is called the mainsvoltage VN.

Three-phase alternating currentIn a three-phase voltage system andvoltages are displaced 120°according to each other. The threephases are usually shown in the samesystem of co-ordinates (Figure 134).

V1

V2

V3

V

-V

120

120

180 360

270

90

180 360

270

90

V1 V2 V3

L1

L2

L3

UN UN

V

I 1

UN Uph

I n

Uph

Uph

I 2

I 3

W

U L1

L2

L3

UN UN

UN

I n

Uph

I 2

I 3

W

U

UphUph

Figure 135. Mains and phase values in a star and delta connection

The ratio between VN and Vph is √3

Figure 134. A three-phase AC voltageconsists of three individual time-displaced AC voltages

Star or delta connectionWhen the three-phase supply mainsis loaded with a motor the motorwindings are star or delta connected.

In a star connection one phase isconnected to one of the ends of themotor windings whereas the otherends are short-circuited star point).

The voltages above the variouswindings are:

V V = phN

√3

For the currents the following applies:I1 = I2 = I3 = INIn a delta connection the three motorwindings are connected in series andeach link is connected to a phase.

The voltage above the variouswindings are: Vph = VN

For the currents the following appliesI I = I = I = 1

N√32 3

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SI Prefixes

symbol prefix valuea atto 1018

f femto 1015

p pico 1012

n nano 10_9

µ micro 10_6

m milli 10_3

c centi 10_2

d deci 10_1

d deca 10h hecto 102

k kilo 103

M mega 106

G giga 109

T tera 1012

P peta 1015

E exa 1018

SI Derived Units

symbol name quantity dimensionsC coulomb charge A . sF farad capacitance A2s4/kg . m2 same as C/V)H henry inductance kg . m2/A2.s2 same as Wb/A)Hz hertz frequency l/sJ joule energy kg.m2/82 same as N.m)N newton force kg.m/82

Pa pascal pressure kg/m.s2 same as N/m2)Ω ohm resistance kg.m2/A2.s3 same as V/A)S siemens conductance A2.s3/kg.m2 same as A/V)T tesla flux density kg/A.s2 same as Wb/m2)V volt potential kg.m2/A.s3 same as W/A)W watt power kg.m2/s3 same as J/s)Wb weber magnetic flux kg.m2/A.s2 same as V.s)

Conversion factors

multiply by to obtain multiply by to obtainacre 43,560 ft2 J 0.73756 ft-lbfangstrom 1x10_10 m kg 2.20462 lbmatm 1.01325 bar kg 0.06852 slugatm 29.92 in Hg kip 1000 lbfatm 14.696 lbf/in2 kJ 0.9478 BTUbar 1x105 Pa kJ 737.56 ft-lbfBTU 778.17 ft-lbf kJ/kg 0.42992 BTU/lbmBTU 1.055 kJ kJ/kg.K 0.23885 BTU/lbm-°RBTU/h 0.293 W km 3280.8 ftBTU/lbm 2.326 kJ/kg km/h 0.62137 mi/hrBTU/lbm-°R 4.1868 kJ/kg.K kPa 0.14504 lbf/in2

cm 0.3937 in kW 737.6 ft-lbf/seccm3 0.061024 in3 kW 1.341 hpeV 1.602x10_19 J l 0.03531 ft3

ft 0.3048 m l 0.001 m3

ft3 7.481 gal lbf 4.4482 Nft3 0.028317 m3 lbf/ft2 144 lbf/in2

ft-lbf 1.35582 J lbf/in2 6894.8 Pagal 0.13368 ft3 lbm 0.4536 kggal 3.7854 x 103 m3 lbm/ft3 0.016018 g/cm3

gal/min 0.002228 ft3/sec lbm/ft3 16.018 kg/m3

g/cm3 1000 kg/m3 m 3.28083 ftg/cm3 62.428 lbm/ft3 m3 35.3147 ft3

hp 2545 BTU/hr mi/h 1.6093 km/hhp 33,000 ft-lbf/min micron 1.10_6 mhp 550 ft-lbf/sec N 0.22481 lbfhp 0.7457 kW Pa 1.4604x10_4 lbf/in2

in 2.54 cm slug 32.174 lbmin3 16.387 cm3 torr 133.32 PaJ 6.2415x1018 eV W 3.413 BTU/hr

ACR 56


AAcceleration 36,38Acceleration range 8Active current 14Address bus 31Address calculator 28Alarm signals 33Alternating field 6Analog control signals 31ASIC 31Asynchronous motor 5BBasic circuit of the AFD 17Blind current 19,35Braking 38Braking ranges 8Break-away torque 16Bus 32CCapacitive over-coupling 44Cascade coupling 12Chopper 21, 24, 30Cogging torque 25, 28, 29Computer 30Conducted noise 44, 45Constant current inverter (CSI) 18Constant voltage intermediate circuit 20,22Control bus 31Control circuit 17, 30Control signals 33Cooling forms 14Cooling 40Copper loss 9Cos ϕ 14, 37Current heat loss 8Current limit 35, 38, 40DDanfoss VVC control principle 18, 22, 23,29, 31, 34Data bus 30DC braking 38DC motor 3Deceleration 38Digital control signals 31Diode 19EEarthing 44Earth leakage 48Eddy current loss 9EEPROM 31Efficiency 9, 15, 29, 41Electrical noise 44EMC 44Emission ways of 44Enclosure 14EPROM 31Equivalent diagram 10-11Extra protection 43FFault indication 47Flux control 34Frequency control 12Friction loss 9Full-wave controlled rectifier 19Fuses 47

GGalvanic isolation 43Gate 19Generator 8Generator principle 3HHarmonics 25, 27-29Hazardous location 43Heat loss 10, 29Horse power 13Hysteresis loss 9IIGBT transistor 23Inputs and outputs 31Inductive over-coupling 45Intermediate circuit 17, 20Intermittent operation 42Insulation measurement 48Inverter 17, 19, 21-25, 39Iron loss 9LLoad dependent magnetization 13Load dependent output voltage 34Load torque 35, 38Load types 15Losses 9, 23, 26MMagnetic field 3, 4-7Magnetization 12, 27Mechanical load 10Menus 37Microprocessor 30Moment of inertia 15, 38Motor cables 42Motor catalogue 15Motor data 13, 35Motor filter 42Motor heating 40Motor principle 3NName plate 13OOperation 47Operational field 7-8, 29-30Operational range 8-9Output voltage 26, 27, 29Over-coupled noise 44Overmagnetization 9, 13, 43Overtorque 35, 36, 38PParallel connection of motors 34, 43Phase sequence 39PLC 30-33Pole pairs 5, 6Pole number control 11Power 8Power rating 13Pulse Amplitude Modulation (PAM) 18, 22,24, 25, 30Pulse Width Modulation (PWM) 18, 22, 25,26, 27RRAM 30Radio frequency interference RFI 44, 46,48Ramps 39Rated torque 9Rectifier 17-20

Registers 27Reluctance motor 4Remote ventilation 40Reversing 39Ripple 45Rotor 4, 7Rotor control 12Rotor resistors 12RS 232/422/423/485 32, 33SScreens 47Sequence controller 28Serial communication 31Short-circuit 48Short-circuit rotor 7Simple trouble shooting 47Sine-controlled Pulse Width Modulation(PWM) 22, 25-26, 29Sine reference 25Size of AFD 35Slip 7, 10, 15, 34Slip compensation 34Slip control 11Slip ring motor 11Slip ring rotor 7Slots 5, 7Speed 7Speed change 11Speed reference 38Star connection 13Start compensation 34Start current 13, 15Start magnetization 12Start torque 8, 15Static electricity 37Stator voltage change of 11Status signals 31Supply interference 45Supply mains 48Switching frequency 23Switching pattern 6- and 18-pulse signal22-25, 29Synchronous motor 4Synchronous speed 5TTime delay (α) 19Timer 28Torque 7, 8, 11, 35Transformer 43Transients/overvoltage 46Transistors 21, 23Trip-out torque 9, 12Thyristor 19, 21UUndermagnetization 9Uncontrolled rectifier 19VVariable direct current intermediate circuit20, 21Variable voltage intermediate circuit 20, 39VDE 0875 46Ventilation loss 9Voltage-frequency ratio 12, 13, 35, 36Voltage rise times du/dt 42VVC control principle 18, 23, 27-29, 31, 35WWire cross section 48Work 8

Subject Index