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The humble motor, consisting of nothing more than an arrangement of copper coils and steel laminations, is clearly rather a clever energy converter, which warrants serious consideration. By gaining a basic understanding of how the motor works, we will be able to appreciate its potential and its limitations, and (in later sections) see how its already remarkable performance can be further enhanced by the addition of external electronic controls.
This section deals with the basic mechanisms of motor operation, so readers who are already familiar with such matters as magnetic flux, magnetic and electric circuits, torque, and motional e.m.f can probably afford to skim over much of it. In the course of the discussion, however, several very important general principles and guidelines emerge. These apply to all types of motors and are summarized in Section 1.8. Experience shows that anyone who has a good grasp of these basic principles will be well equipped to weigh the pros and cons of the different types of motor, so all readers are urged to absorb them before tackling other parts of the guide.
Wiki Series: Laboratory Manual for Electronics
Before beginning the study of three-phase transformers, it's appropriate to discuss three phase power connections and basic circuit calculations. This unit may be review for some students and new ground for others. Whichever is the case, a working knowledge of three phase circuits is essential before beginning the study of three-phase transformers.
Most of the power generated in the world today is three-phase. Three-phase power was first conceived by a man named Nikola Tesla. There are several reasons why three-phase power is superior to single-phase power.
1. The kVA rating of three-phase transformers is about 150% greater than for a single phase transformer with a similar core size.
2. The power delivered by a single-phase system pulsates. The power falls to zero three times during each cycle. The power delivered by a three-phase circuit pulsates also, but the power never falls to zero (ill. 2). In a three-phase system, the power delivered to the load is the same at any instant.
ill. 1 Single-phase power falls to zero three times each cycle.
ill. 2 Three-phase power never falls to zero.
3. In a balanced three-phase system, the conductors need be only about 75% the size of conductors for a single-phase two-wire system of the same kVA rating. This helps off set the cost of supplying the third conductor required by three-phase systems.
A single-phase alternating voltage can be produced by rotating a magnetic field through the conductors of a stationary coil as shown in ill. 3. Since alternate polarities of the magnetic field cut through the conductors of the stationary coil, the induced voltage will change polarity at the same speed as the rotation of the magnetic field. The alternator shown in ill. 3 is single-phase because it produces only one AC voltage.
If three separate coils are spaced 120° apart as shown in ill. 4, three voltages 120° out of phase with each other will be produced when the magnetic field cuts through the coils.
This is the manner in which a three-phase voltage is produced. There are two basic three phase connections, the wye or star, and the delta.
ill. 3 Producing a single-phase voltage.
The wye or star connection is made by connecting one end of each of the three-phase windings together as shown is ill. 5. The voltage measured across a single winding or phase is known as the phase voltage as shown in ill. 6. The voltage measured between the lines is known as the line-to-line voltage or simply as the line voltage.
ill. 4 The voltages of a three-phase system are 120° out of phase with each other.
ill. 5 A wye connection is formed by joining one end of each winding together.
ill. 6 Line and phase voltages are different in a wye connection.
In ill. 7 ammeters have been placed in the phase winding of a wye connected load and in the line supplying power to the load. Voltmeters have been connected across the input to the load and across the phase. A line voltage of 208 volts has been applied to the load. Notice that the voltmeter connected across the lines indicates a value of 208 volts, but the voltmeter connected across the phase indicates a value of 120 volts.
In a wye connected system, the line voltage is higher than the phase voltage by a factor of __ 3 (1.732). Two formulas used to compute the voltage in a wye connected system are:
E_PHASE = E_LINE/ sqr_rt (3)
E_LINE = E_PHASE x sqr_rt (3)
ill. 7 Line current and phase current are the same in a wye connection.
Notice in ill. 7 that there is 10 amps of current flow in both the phase and the line.
In a wye connected system, phase current and line current are the same.
Helpful Hint: In a wye connected system, the line voltage is higher than the phase voltage by a factor of sqr_rt 3 (1.732).
Helpful Hint: In a wye connected system, phase current and line current are the same.
Relationships in a Wye Connection ill. 8 Single-phase transformer with grounded center tap.
Many students of electricity have difficulty at first understanding why the line voltage of the wye connection used in this illustration is 208 volts instead of 240 volts. Since line voltage is measured across two phases that have a voltage of 120 volts each, it would appear that the sum of the two voltages should be 240 volts. One cause of this misconception is that many students are familiar with the 240/120 volt connection supplied to most homes. If voltage is measured across the two incoming lines, a voltage of 240 volts will be seen. If voltage is measured from either of the two lines to the neutral, a voltage of 120 volts will be seen. The reason for this is that this connection is derived from the center tap of an isolation transformer, as shown in ill. 8. If the center tap is used as a common point, the two line voltages on either side of it will be in phase with each other. Since the two voltages are in phase, they add similar to a boost connected transformer, as shown in ill. 9. The vector sum of these two voltages would be 240 volts.
Three-phase voltages are 120° apart, not in phase. If the three voltages are drawn 120° apart, it will be seen that the vector sum of these voltages is 208 volts, as shown in ill. 10.
Another illustration of vector addition is shown in ill. 11. In this illustration two-phase voltage vectors are added and the resultant is drawn from the starting point of one vector to the end point of the other. The parallelogram method of vector addition for the voltages in a wye connected three-phase system is shown in ill. 12.
ill. 9 The two voltages are in phase with each other.
ill. 10 Vector sum of the voltages in a three-phase wye connection.
ill. 11 Adding voltage vectors of two-phase voltage values.
ill. 12 The parallelogram method of adding three-phase vectors.
In ill. 13 three separate inductive loads have been connected to form a delta connection. This connection receives its name from the fact that a schematic diagram of this connection resembles the Greek letter delta (?). In ill. 14, voltmeters have been connected across the lines and across the phase. Ammeters have been connected in the line and in the phase. In the delta connection, line voltage and phase voltage are the same. Notice that both voltmeters indicated a value of 480 volts.
ill. 13 Three-phase delta connection.
Helpful Hint: In the delta connection, line voltage and phase voltage are the same.
ill. 14 Voltage and current relationships in a delta connection.
ill. 15 Division of currents in a delta connection.
Notice that the line current and phase current are different, however. The line current of a delta connection is higher than the phase current by a factor of __ 3 (1.732). In the example shown, it's assumed that each of the phase windings has a current flow of 10 amperes. The current in each of the lines, however, is 17.32 amperes. The reason for this difference in current is that current flows through different windings at different times in a three-phase circuit. During some periods of time, current will flow between two lines only.
At other times, current will flow from two lines to the third, seen in ill. 15. The delta connection is similar to a parallel connection because there is always more than one path for current flow. Since these currents are 120° out of phase with each other, vector addition must be used when finding the sum of the currents (ill. 16). Formulas for determining the current in a delta connection are:
Students sometimes become confused when computing values of power in three-phase circuits. One reason for this confusion is because there are actually two formulas that can be used. If LINE values of voltage and current are known, the apparent power of the circuit can be computed using the formula:
If the PHASE values of voltage and current are known, the apparent power can be computed using the formula:
VA = 3 x E_PHASE × I_PHASE
Notice that in the first formula, the line values of voltage and current are multiplied by the square root of 3. In the second formula, the phase values of voltage and current are multiplied by 3. The first formula is the most used because it's generally more convenient to obtain line values of voltage and current since they can be measured with a voltmeter and clamp-on ammeter.
Watts and VARs
Watts and VARs can be computed in a similar manner. Watts can be computed by multiplying the apparent power by the power factor:
3 PHASE × PHASE
Note: When computing the power of a pure resistive load, the voltage and current are in phase with each other and the power factor is 1.
VARs can be computed in a similar manner, except that voltage and current values of a pure reactive load are used. E.g., a pure capacitive load is shown in ill. 17. In this example, it's assumed that the line voltage is 480 volts and the line current is 30 amperes.
Capacitive VARs can be computed using the formula:
VAR c = 29,097.6
ill. 16 Vector addition is used to compute the sum of the currents in a delta connection.
ill. 17 Pure capacitive three-phase load.
Three-Phase Circuit Calculations
In the following examples, values of line and phase voltage, line and phase current, and power will be computed for different types of three-phase connections.
Example #1. A wye connected, three-phase alternator supplies power to a delta connected resistive load, as shown in ill. 18. The alternator has a line voltage of 480 volts.
Each resistor of the delta load has 8 ohm of resistance. Find the following values:
E_L(LOAD) - Line voltage of the load
E_P(LOAD) - Phase voltage of the load
ill. 18 Computing three-phase values: Example Circuit #1.
I_P(LOAD) - Phase current of the load
I_L(LOAD) - Line current to the load
I_L(ALT) - Line current delivered by the alternator
E_P(ALT) - Phase voltage of the alternator P - True power
Solution: The load is connected directly to the alternator. Therefore, the line voltage sup plied by the alternator is the line voltage of the load.
The three resistors of the load are connected in a delta connection. In a delta connection, the phase voltage is the same as the line voltage.
Each of the three resistors in the load comprises one phase of the load. Now that the phase voltage is known (480 volts), the amount of phase current can be computed using Ohm's law.
L (LOAD) 480 volts =
P LOAD () I LOAD ()
P LOAD () 480 volts =
In this example the three load resistors are connected as a delta with 60 amperes of current flow in each phase. The line current supplying a delta connection must be 1.732 times greater than the phase current.
The alternator must supply the line current to the load or loads to which it's connected. In this example, there is only one load connected to the alternator. Therefore, the line current of the load will be the same as the line current of the alternator.
The phase windings of the alternator are connected in a wye connection. In a wye connection, the phase current and line current are equal. The phase current of the alternator will, therefore, be the same as the alternator line current.
L LOAD () P LOAD () 1.732 × =
L LOAD () 60 1.732 × =
L LOAD () 103.92 amps =
LALT () 103.92 amps =
P_ALT () 103.92 amps =
The phase voltage of a wye connection is less than the line voltage by a factor of the square root of 3 (__ 3 ). The phase voltage of the alternator will be:
In this circuit the load is pure resistive. The voltage and current are in phase with each other, which produces a unity power factor of 1. The true power in this circuit will be computed using the formula: Example #2. In the next example, a delta connected alternator is connected to a wye connected resistive load, as shown in ill. 19. The alternator produces a line voltage of 240 volts and the resistors have a value of 6 ohm each. The following values will be found:
E_L(LOAD) - Line voltage of the load EP(LOAD) - Phase voltage of the load IP(LOAD) - Phase current of the load I_L(LOAD) - Line current to the load IL(ALT) - Line current delivered by the alternator EP(ALT) - Phase voltage of the alternator P - True power As in the first example, the load is connected directly to the output of the alternator. The line voltage of the load must, therefore, be the same as the line voltage of the alternator.
P_ALT () 277.13 volts =
1.732 LALT ()
× LALT ()
× × =
1.732 480 × 103.92 × 1 × =
86,394.93 watts =
L LOAD () 240 volts =
The phase voltage of a wye connection is less than the line voltage by a factor of 1.732.
Each of the three 6 ohm resistors comprises one phase of the wye connected load. Since the phase voltage is 138.57 volts, this voltage is applied to each of the three resistors. The amount of phase current can now be determined using Ohm's law.
The amount of line current needed to supply a wye connected load is the same as the phase current of the load.
In this example there is only one load connected to the alternator. The line current supplied to the load is the same as the line current of the alternator.
The phase windings of the alternator are connected in delta. In a delta connection the phase current is less than the line current by a factor of 1.732.
L LOAD () 23.1 amps =
LALT () 23.1 amps =
P LOAD () 23.1 amps =
The phase voltage of a delta is the same as the line voltage.
Since the load in this example is pure resistive, the power factor has a value of unity or 1.
Power will be computed by using the line values of voltage and current.
PALT () 240 volts =
1.732 L × L × × =
1.732 240 × 23.1 × 1 × =
9,602.21 watts =
Date _ Materials Required
2 AC voltmeters
AC ammeter, in-line or clamp-on. (If a clamp-on type is used, it's recommended to use a 10:1 scale divider.) 6 100-watt lamps
In this experiment six 100 watt lamps will be connected to form different three-phase loads.
Two lamps will be connected in series to form three separate loads. These loads will be connected to form wye or delta connections.
1. Connect the two 100 watt lamps in series to form three separated load banks. Connect the load banks in wye by connecting one end of each bank together to form a center point, as shown in ill. 20. It is assumed that this load is to be connected to a 208 VAC three-phase line. Connect an AC ammeter in series with the line supplying power to the load.
2. Turn on the power and measure the line voltage supplied to the load.
3. Calculate the value of phase voltage for a wye connected load.
E(PHASE) __ volts
4. Measure the phase voltage and compare this value with the computed value.
5. Measure the line current.
6. Turn off the power supply.
7. In a wye connected system, the line current and phase current are the same. Reconnect the circuit as shown in ill. 21.
8. Turn on the power and measure the phase current.
9. Turn off the power supply.
10. Reconnect the three banks of lamps to form a delta connected load, as shown in ill. 22.
11. Turn on the power and measure the line voltage supplied to the load.
ill. 20 Measuring the line current in a wye connected load.
ill. 21 Measuring the phase current in a wye connected load.
12. Measure the phase value of voltage.
E(PHASE) volts 13. Are the line and phase voltage values the same or different?
14. Measure the line current.
I(LINE) amp(s) 15. Turn off the power supply.
16. In a delta connected system, the phase current will be less than the line current by a factor of 1.732. Calculate the phase current value for this connection.
ill. 22 Measuring the voltage and line current values of a delta connected load.
I(PHASE) amp(s) 17. Reconnect the circuit as shown in ill. 23.
18. Turn on the power supply and measure the phase current. Compare this value with the computed value.
I(PHASE) amp(s) 19. Turn off the power supply.
20. Disconnect the circuit and return the components to their proper place.
1. How many degrees out of phase with each other are the voltages of a three-phase system?
2. What are the two main types of three-phase connections?
3. A wye connected load has a voltage of 480 volts applied to it. What is the voltage dropped across each phase?
4. A wye connected load has a phase current of 25 amps. How much current is flowing through the lines supplying the load?
5. A delta connection has a voltage of 560 volts connected to it. How much voltage is dropped across each phase?
6. A delta connection has 30 amps of current flowing through each phase winding. How much current is flowing through each of the lines supplying power to the load?
7. A three-phase load has a phase voltage of 240 volts and a phase current of 18 amperes.
What is the apparent power of this load?
ill. 23 Measuring the voltage and phase current values of a delta connected load.
8. If the load in question 7 is connected in a wye, what would be the line voltage and line current supplying the load?
9. An alternator with a line voltage of 2,400 volts supplies a delta connected load. The line current supplied to the load is 40 amperes. Assuming the load is a balanced three-phase load, what is the impedance of each phase?
10. What is the apparent power of the circuit in question 9?
Nearly all motors exploit the force which is exerted on a current-carrying conductor placed in a magnetic field . The force can be demonstrated by placing a bar magnet near a wire carrying current, but anyone trying the experiment will probably be disappointed to discover how feeble the force is, and will doubtless be left wondering how such an unpromising effect can be used to make effective motors.
We will see that in order to make the most of the mechanism, we need to arrange a very strong magnetic field , and make it interact with many conductors, each carrying as much current as possible. We will also see later that although the magnetic field (or working of the motor, it acts only as a catalyst, and all of the mechanical output power comes from the electrical supply to the conductors on which the force is developed. It will emerge later that in some motors the parts of the machine responsible for the excitation and for the energy converting functions are distinct and self-evident. In the DC motor, for example, the excitation is provided either by permanent magnets or by field coils wrapped around clearly defined projecting field poles on the stationary part, while the conductors on which force is developed are on the rotor and supplied with current via sliding brushes. In many motors, however, there is no such clear-cut physical distinction between the ‘excitation’ single stationary winding serves both purposes. Nevertheless, we will find that identifying and separating the excitation and energy-converting functions is always helpful in understanding how motors of all types operate.
ill. 1 Mechanical force produced on a current-carrying wire in a magnetic field.
Returning to the matter of force on a single conductor, we will first look at what determines the magnitude and direction of the force, before turning to ways in which the mechanism is exploited to produce rotation. The concept of the magnetic circuit will have to be explored, since this is central to understanding why motors have the shapes they do. A brief introduction to magnetic field , magnetic flux, and flux density is included before that for those who are not familiar with the ideas involved.
Magnetic field and magnetic flux
When a current-carrying conductor is placed in a magnetic field , it experiences a force. Experiment shows that the magnitude of the force depends directly on the current in the wire, and the strength of the magnetic field , and that the force is greatest when the magnetic field is perpendicular to the conductor.
In the set-up shown in ill. 1, the source of the magnetic field is a bar magnet, which produces a magnetic field.
The notion of a ‘magnetic field’ idea that helps us to come to grips with the mysterious phenomenon of magnetism: it not only provides us with a convenient pictorial way of picturing the directional effects, but it also allows us to quantify the ‘strength ’of the magnetism and hence permits us to predict the various effects produced by it.
The dotted lines are referred to as magnetic flux lines, or simply flux lines. They indicate the direction along which iron filings (or small steel pins) would align themselves when placed in the field of the bar magnet. Steel pins have no initial magnetic field of their own, so there is no reason why one end or the other of the pins should point to a particular pole of the bar magnet.
However, when we put a compass needle (which is itself a permanent magnet)in the field we find that it aligns itself. In the upper half of the figure, the S end of the diamond-shaped compass settles closest to the N pole of the magnet, while in the lower half of the figure, the N end of the compass seeks the S of the magnet. This immediately suggests that there is a direction associated with the lines of flux, as shown by the arrows on the flux lines, which conventionally are taken as positively directed from the N to the S pole of the bar magnet.
The sketch in ill. 2 might suggest that there is a ‘source ’near the top of the bar magnet, from which flux lines emanate before making their way to a corresponding ‘sink ’at the bottom. However, if we were to look at the flux lines inside the magnet, we would find that they were continuous, with no ‘start ’or ‘finish ’.(In ill. 2 the internal flux lines have been omitted for the sake of clarity, but a very similar field pattern is produced by a circular coil of wire carrying a DC. See ill. 6 where the continuity of the flux lines is clear.).Magnetic flux lines always form closed paths, as we will see when we look at the ‘magnetic circuit ’,and draw a parallel with the electric circuit, in which the current is also a continuous quantity.(There must be a ‘cause ’of the magnetic flux, of course, and in a permanent magnet this is usually pictured in terms of atomic-level circulating currents within the magnet material.
Fortunately, discussion at this physical level isn't necessary for our purpose.)
Magnetic flux density
Along with showing direction, the flux plots also convey information about the intensity of the magnetic field. To achieve this, we introduce the idea that between every pair of flux lines (and for a given depth into the paper) there is a same ‘quantity ’of magnetic flux. Some people have no difficulty with such a concept, while others find that the notion of quantifying something so abstract represents a serious intellectual challenge. But whether the approach seems obvious or not, there is no denying of the practical utility of quantifying the mysterious stuff we call magnetic flux, and it leads us next to the very important idea of magnetic flux density (B ).
When the flux lines are close together, the ‘tube ’of flux is squashed into a smaller space, whereas when the lines are further apart the same tube of flux has more breathing space. The flux density (B) is simply the flux in the ‘tube ’(phi )divided by the cross sectional area (A )of the tube, i.e. B = Phi / A [1.1]
The flux density is a vector quantity, and is therefore often written in bold type: its magnitude is given by equation, and its direction is that of the prevailing flux lines at each point. Near the top of the magnet in ill. 2, for example, the flux density will be large (because the flux is squashed into a small area),and pointing upwards, whereas on the equator and far out from the body of the magnet the flux density will be small and directed downwards.
It will be seen later that in order to create high flux densities in motors, the flux spends most of its life inside well-de fined ‘magnetic circuits ’made of iron or steel, within which the flux lines spread out uniformly to take full advantage of the available area. In the case shown in ill. 3, for example, the cross-sectional area at bb ’is twice that at aa ’,but the flux is constant so the flux density at bb ’is half that at aa ’.
ill. 3 Magnetic flux lines inside part of an iron magnetic circuit.
It remains to specify units for quantity of flux, and flux density. In the SI system, the unit of magnetic flux is the weber (Wb).If one weber of flux is distributed uniformly across an area of 1m^2 perpendicular to the flux, the flux density is clearly one weber per square meter (Wb=m2). This was the unit of magnetic flux density until about 40 years ago, when it was decided that one weber per square meter would henceforth be known as one tesla (T), in honor of Nikola Tesla who is generally credited with inventing the induction motor. The widespread use of B (measured in tesla) in the design stage of all types of electromagnetic apparatus means that we are constantly reminded of the importance of tesla; but at the same time one has to acknowledge that the outdated unit did have the advantage of conveying directly what flux density is, i.e. flux divided by area.
In the motor world we are unlikely to encounter more than a few milliwebers of flux, and a small bar magnet would probably only produce a few microwebers. On the other hand, values of flux density are typically around 1T in most motors, which is a reflection of the fact that although the quantity of flux is small, it's also spread over a small area.
Force on a conductor
We now return to the production of force on a current-carrying wire placed in a magnetic field , as revealed by the setup.
The direction of the force is shown in ill. 1: it's at right angles to both the current and the magnetic flux density. With the flux density horizontal and to the right, and the current flowing out of the paper, the force is vertically upward. If either the field or the current is reversed, the force acts downwards, and if both are reversed, the force will remain upward.
We find by experiment that if we double either the current or the flux density, we double the force, while doubling both causes the force to increase by a factor of four. But how about quantifying the force? We need to express the force in terms of the product of the current and the magnetic flux density, and this turns out to be very straightforward when we work in SI units.
The force on a wire of length l, carrying a current I and exposed to a uniform magnetic flux density B throughout its length is given by the simple expression
F = BIl [1.2]
… where F is in newtons when B is in tesla, I in amperes, and l in meters.
This is a simple formula, and it may come as a surprise to some readers that there are no constants of proportionality involved in equation 2. The simplicity isn't a coincidence, but stems from the fact that the unit of current (the ampere) is actually defined in terms of force.
Strictly, equation 2 only applies when the current is perpendicular to the field . If this condition isn't met, the force on the conductor will be less; and in the extreme case where the current was in the same direction as the field , the force would fall to zero. However, every sensible motor designer knows that to get the best out of the magnetic field it has to be perpendicular to the conductors, and so it's safe to assume in the subsequent discussion that B and I are always perpendicular. In the remainder of this book, it will be assumed that the flux density and current are mutually perpendicular, and this is why, although B is a vector quantity (and would usually be denoted by bold type), we can drop the bold notation because the direction is implicit and we are only interested in the magnitude.
The reason for the very low force detected in the experiment with the bar magnet is revealed by equation 2. To obtain a high force, we must have a high flux density, and a lot of current. The flux density at the ends of a bar magnet is low, perhaps 0.1 tesla, so a wire carrying 1 amp will experience a force of only 0.1 N/m (approximately 100 gm wt). Since the flux density will be confined to perhaps 1 cm across the end face of the magnet, the total force on the wire will be only 1 gm. This would be barely detectable, and is too low to be of any use in a decent motor. So how is more force obtained?
The first step is to obtain the highest possible flux density. This is achieved by designing a Secondly, as many conductors as possible must be packed in the space where the magnetic field exists, and each conductor must carry as much current as it can without heating up to a dangerous temperature. In this way, impressive forces can be obtained from modestly sized devices, as anyone who has tried to stop an electric drill by grasping the chuck will testify.
So far we have assumed that the source of the magnetic field is a permanent magnet. This is a convenient starting point as all of us are familiar with magnets, even if only of the fridge-door variety. But in the majority of motors, the working magnetic field is produced by coils of wire carrying current, so it's appropriate that we spend some time looking at how we arrange the coils and their associated iron ‘magnetic circuit’ Welds which then interact with so as other current-carrying conductors to produce force, and hence rotation.
ill. 4 Magnetic flux lines produced by a straight, current-carrying wire.
First, we look at the simplest possible case of the magnetic field surrounding an isolated long straight wire carrying a steady current. (In the figure, the + sign indicates that current is flowing into the paper, while a dot is used to signify current out of the paper: these symbols can perhaps be remembered by picturing an arrow or dart, with the cross being the rear view of the fletch, and the dot being the approaching point.) The flux lines form circles concentric with the wire, the field strength being greatest close to the wire. As might be expected, the field strength at any point is directly proportional to the current. The convention for determining the direction of the field is that the positive direction is taken to be the direction that a right-handed corkscrew must be rotated to move in the direction of the current.
ill. 4 is somewhat artificial as current can only flow in a complete circuit, so there must always be a return path. If we imagine a parallel ‘go’ field can be obtained by and superimposing the field produced by the positive current in the go side with the field produced by the negative current in the return side, as shown in ill. 5.
ill. 5 Magnetic flux lines produced by current in a parallel go and return circuit.
ill. 5 Magnetic flux lines produced by current in a parallel go and return circuit
ill. 6 Multi-turn cylindrical coil and pattern of magnetic flux produced by current in the coil. (For the sake of clarity, only the outline of the coil is shown on the right.)
We note how the field is increased in the region between the conductors, and reduced in the regions outside. Although ill. 5 strictly only applies to an infinitely long pair of straight conductors, it will probably not come as a surprise to learn that the field produced by a single turn of wire of rectangular, square or round form is very much the same as that shown in ill. 5. This enables us to build up a picture of the field that would be produced in air, by the sort of coils used in motors, which typically have many turns.
The coil itself is shown on the left in ill. 6 while the flux pattern produced is shown on the right. Each turn in the coil produces a field pattern, and when all the individual field components are superimposed we see that the field inside the coil is substantially increased and that the closed flux paths closely resemble those of the bar magnet we looked at earlier. The air surrounding the sources of the field offers a homogeneous path for the flux, so once the tubes of flux escape from the concentrating influence of the source, they are free to spread out into the whole of the surrounding space. Recalling that between each pair of flux lines there is an equal amount of flux, e see that because the flux lines spread out as they leave the con fines of the coil, the flux density is much lower outside than inside: for example, if the distance ‘b ’is say four times ‘a ’the flux density B b is a quarter of B a.
Although the flux density inside the coil is higher than outside, we would find that the flux densities which we could achieve are still too low to be of use in a motor. What is needed firstly is a way of increasing the flux density, and secondly a means for concentrating the flux and preventing it from spreading out into the surrounding space.
Magnetomotive force (MMF)
One obvious way to increase the flux density is to increase the current in the coil, or to add more turns. We find that if we double the current, or the number of turns, we double the total flux, thereby doubling the flux density everywhere.
We quantify the ability of the coil to produce flux in terms of its magnetomotive force (MMF).The MMF of the coil is simply the product of the number of turns (N )and the current (I ),and is thus expressed in ampere-turns. A given MMF can be obtained with a large number of turns of thin wire carrying a low current, or a few turns of thick wire carrying a high current: as long as the product NI is constant, the MMF is the same.
Electric circuit analogy
We have seen that the magnetic flux which is set up is proportional to the MMF driving it. This points to a parallel with the electric circuit, where the current (amps)that flows is proportional to the EMF (volts) driving it.
In the electric circuit, current and EMF are related by Ohm's Law, which is
Current = EMF / Resistance i.e., :I = V/R
For a given source EMF (volts), the current depends on the resistance of the circuit, so to obtain more current, the resistance of the circuit has to be reduced.
We can make use of an equivalent 'magnetic Ohm's law 'by introducing the idea of reluctance (R).The reluctance gives a measure of how difficult it's for the magnetic flux to complete its circuit, in the same way that resistance indicates how much opposition the current encounters in the electric circuit. The magnetic Ohm 's law is then
Flux = MMF / Reluctance i.e., Φ = NI/R
We see from equation 1.4 that to increase the flux for a given MMF, we need to reduce the reluctance of the magnetic circuit. In the case of the example (ill. 6), this means we must replace as much as possible of the air path (which is a 'poor 'magnetic material, and therefore constitutes a high reluctance) with a 'good 'magnetic material, thereby reducing the reluctance and resulting in a higher flux for a given MMF.
The material which we choose is good quality magnetic steel, which for historical reasons is usually referred to as 'iron '.This brings several very dramatic and desirable benefits, as shown.
ill. 7 Flux lines inside low-reluctance magnetic circuit with air-gap Firstly, the reluctance of the iron paths is very much less than the air paths which they have replaced, so the total flux produced for a given MMF is very much greater. (Strictly speaking therefore, if the MMFs and cross-sections of the coils in ills 1.6 and 1.7 are the same, many more flux lines should be shown in ill. 7 than in ill. 6,but for the sake of clarity similar numbers are indicated.) Secondly, almost all the flux is confined within the iron, rather than spreading out into the surrounding air. We can therefore shape the iron parts of the magnetic circuit as shown in ill. 7 in order to guide the flux to wherever it's needed. and finally, we see that inside the iron, the flux density remains uniform over the whole cross-section, there being so little reluctance that there is no noticeable tendency for the flux to crowd to one side or another.
Before moving on to the matter of the air-gap, we should note that a question which is often asked is whether it's important for the coils to be wound tightly onto the magnetic circuit, and whether, if there is a multi-layer winding, the outer turns are as effective as the inner ones.
The answer, happily, is that the total MMF is determined solely by the number of turns and the current, and therefore every complete turn makes the same contribution to the total MMF, regardless of whether it happens to be tightly or loosely wound. Of course it does make sense for the coils to be wound as tightly as is practicable, since thisn't only minimizes the resistance of the coil (and thereby reduces the heat loss) but also makes it easier for the heat generated to be conducted away to the frame of the machine.
In motors, we intend to use the high flux density to develop force on current-carrying conductors. We have now seen how to create a high flux density inside the iron parts of a magnetic circuit, but, of course, it's physically impossible to put current-carrying conductors inside the iron.
We therefore arrange for an air-gap in the magnetic circuit. We will see shortly that the conductors on which the force is to be produced will be placed in this air-gap region.
If the air-gap is relatively small, as in motors, we find that the flux jumps across the air-gap as shown in ill. 7, with very little tendency to balloon out into the surrounding air. With most of the flux lines going straight across the air-gap, the flux density in the gap region has the same high value as it does inside the iron.
In the majority of magnetic circuits consisting of iron parts and one or more air-gaps, the reluctance of the iron parts is very much less than the reluctance of the gaps. At first sight this can seem surprising, since the distance across the gap is so much less than the rest of the path through the iron. The fact that the air-gap dominates the reluctance is simply a reflection of how poor air is as a magnetic medium, compared to iron.
To put the comparison in perspective, if we calculate the reluctances of two paths of equal length and cross-sectional area, one being in iron and the other in air, the reluctance of the air path will typically be 1000 times greater than the reluctance of the iron path.(The calculation of reluctance will be discussed in later.) Returning to the analogy with the electric circuit, the role of the iron parts of the magnetic circuit can be likened to that of the copper wires in the electric circuit. Both offer little opposition to flow (so that a negligible fraction of the driving force (MMF or EMF) is wasted in conveying the flow to where it's usefully exploited) and both can be shaped to guide the flow to its destination. There is one important difference, however. In the electric circuit, no current will flow until the circuit's completed, after which all the current is confined inside the wires. With an iron magnetic circuit, some flux can flow (in the surrounding air) even before the iron is installed. and although most of the flux will subsequently take the easy route through the iron, some will still leak into the air.
We won't pursue leakage flux here, though it's sometimes important, as will be seen later.
ill. 8 Air-gap region, with MMF acting across opposing pole-faces
Reluctance and air-gap flux densities
If we neglect the reluctance of the iron parts of a magnetic circuit, it's easy to estimate the flux density in the air-gap. Since the iron parts are then in effect 'perfect conductors 'of flux, none of the source MMF (NI ) is used in driving the flux through the iron parts, and all of it's available to push the flux across the air-gap. The situation depicted in ill. 7 therefore reduces to that shown in ill. 8, where an MMF of NI is applied directly across an air-gap of length g .
To determine how much flux will cross the gap, we need to know its reluctance. As might be expected, the reluctance of any part of the magnetic circuit depends on its dimensions, and on its magnetic properties, and the reluctance of a rectangular 'prism 'of air, of cross-sectional area A and length g as in ill. 8 is given by
Rg = g / A μ0 [1.5]
…where μ0 is the so-called 'primary magnetic constant 'or 'permeability of free space '.Strictly, as its name implies, m 0 quantifies the magnetic properties of a vacuum, but for all engineering purposes the permeability of air is also μ0 .The value of the primary magnetic constant (μ0) in the SI system is 4 10 7 H/m; rather surprisingly, there is no name for the unit of reluctance.
In passing, we should note that if we want to include the reluctance of the iron part of the magnetic circuit in our calculation, its reluctance would be given by…
Rfe = lfe / A μfe
and we would have to add this to the reluctance of the air-gap to obtain the total reluctance. However, because the permeability of iron (μfe )is so much higher than 0,the iron reluctance will be very much less than the gap reluctance, despite the path length l being considerably longer than the path length (g )in the air.
Equation 1.5 reveals the expected result that doubling the air-gap would double the reluctance (because the flux has twice as far to go), while doubling the area would halve the reluctance (because the flux has two equally appealing paths in parallel).To calculate the flux, F ,we use the magnetic Ohm's law (equation 1.4),which gives
Phi = MMF/R = NI A μ0 / g [1.6]
We are usually interested in the flux density in the gap, rather than the total flux, so we use equation 1.1 to yield
B = phi / A = μ0 NI / g [1.7]
Equation 1.7 is delightfully simple, and from it we can calculate the air gap flux density once we know the MMF of the coil (NI )and the length of the gap (g ).We don't need to know the details of the coil-winding as long as we know the product of the turns and the current, nor do we need to know the cross-sectional area of the magnetic circuit in order to obtain the flux density (though we do if we want to know the total flux, see equation 1.6).
E.g., suppose the magnetizing coil has 250 turns, the current is 2A, and the gap is 1 mm. The flux density is then given by
B = (4 pi x 10^-7 x 250 x 2) / (1 x 10^-3) = 0.63 tesla
(We could of course obtain the same result using an exciting coil of 50 turns carrying a current of 10A, or any other combination of turns and current giving an MMF of 500 ampere-turns.) If the cross-sectional area of the iron was constant at all points, the flux density would be 0.63 T everywhere. Sometimes ,as has already been mentioned, the cross-section of the iron reduces at points away from the air-gap, as shown for example in ill. 3. Because the flux is com pressed in the narrower sections, the flux density is higher, and in ill. 3 if the flux density at the air-gap and in the adjacent pole-faces is once again taken to be 0.63 T, then at the section aa'(where the area is only half that at the air-gap)the flux density will be 2 x 0.63 = 1.26T.
ill. 9 Sketch showing how the effective reluctance of iron increases rapidly as the flux density approaches saturation
It would be reasonable to ask whether there is any limit to the flux density at which the iron can be operated. We can anticipate that there must be a limit, or else it would be possible to squash the flux into a vanishingly small cross-section, which we know from experience isn't the case. In fact there is a limit, though not a very sharply defined one.
Earlier we noted that the iron has almost no reluctance, at least not in comparison with air. Unfortunately this happy state of a V airs is only true as long as the flux density remains below about 1.6 -1.8 T, depending on the particular steel in question. If we try to work the iron at higher flux densities, it begins to exhibit significant reluctance, and no longer behaves like an ideal conductor of flux. At these higher flux densities, a significant proportion of the source MMF is used in driving the flux through the iron. This situation is obviously undesirable, since less MMF remains to drive the flux across the air-gap. So just as we wouldn't recommend the use of high-resistance supply leads to the load in an electric circuit, we must avoid overloading the iron parts of the magnetic circuit.
The emergence of significant reluctance as the flux density is raised is illustrated qualitatively in ill. 9.
When the reluctance begins to be appreciable, the iron is said to be beginning to 'saturate '.The term is apt, because if we continue increasing the MMF, or reducing the area of the iron, we will eventually reach an almost constant flux density, typically around 2T. To avoid the undesirable effects of saturation, the size of the iron parts of the magnetic circuit are usually chosen so that the flux density does not exceed about 1.5T. At this level of flux density, the reluctance of the iron parts will be small in comparison with the air-gap.
ill. 10 Evolution of d.c. motor magnetic circuit from gapped C-core
Magnetic circuits in motors
The reader may be wondering why so much attention has been focused on the gapped C-core magnetic circuit, when it appears to bear little resemblance to the magnetic circuits found in motors. We will now see that it's actually a short step from the C-core to a magnetic motor circuit, and that no fundamentally new ideas are involved.
The evolution from C-core to motor geometry is shown in ill. 10, which should be largely self-explanatory, and relates to the field system of a d.c. motor.
We note that the first stage of evolution (ill. 10,left)results in the original single gap of length g being split into two gaps of length g /2, reflecting the requirement for the rotor to be able to turn. At the same time the single magnetizing coil is split into two to preserve symmetry.
(Relocating the magnetizing coil at a different position around the magnetic circuit's of course in order, just as a battery can be placed anywhere in an electric circuit.)Next,(ill. 10,centre)the single magnetic path is split into two parallel paths of half the original cross section, each of which carries half of the flux; and finally (ill. 10, right),the flux paths and pole-faces are curved to match the rotor. The coil now has several layers in order to fit the available space, but as discussed earlier this has no adverse effect on the MMF. The air-gap is still small, so the flux crosses radially to the rotor.
Having designed the magnetic circuit to give a high flux density under the poles, we must obtain maximum benefit from it. We therefore need to arrange a set of conductors, fixed to the rotor, as shown in ill. 11, and to ensure that conductors under a N-pole (at the top of ill. 11) carry positive current (into the paper),while those under the S-pole carry negative current. The tangential electromagnetic ('BIl ') force (see equation 1.2)on all the positive conductors will be to the left, while the force on the negative ones will be to the right. A nett couple, or torque will therefore be exerted on the rotor, which will be caused to rotate.
(The observant reader spotting that some of the conductors appear to have no current in them will find the explanation later, in section 3.)
ill. 11 Current-carrying conductors on rotor, positioned to maximize torque.(The source of the magnetic flux lines (arrowed) isn't shown.)
At this point we should pause and address three questions that often crop up when these ideas are being developed. The first is to ask why we have made no reference to the magnetic field produced by the current carrying conductors on the rotor. Surely they too will produce a magnetic field, which will presumably interfere with the original field in the air-gap, in which case perhaps the expression used to calculate the force on the conductor will no longer be valid.
The answer to this very perceptive question is that the field produced by the current-carrying conductors on the rotor certainly will modify the original field (i.e. the field that was present when there was no current in the rotor conductors.) But in the majority of motors, the force on the conductor can be calculated correctly from the product of the current and the 'original 'field. This is very fortunate from the point of view of calculating the force, but also has a logical feel to it. E.g. in ill. 1, we wouldn't expect any force on the current-carrying conductor if there was no externally applied field, even though the current in the conductor will produce its own field (upwards on one side of the conductor and downwards on the other).So it seems right that since we only obtain a force when there is an external field, all of the force must be due to that field alone.
The second question arises when we think about the action and reaction principle. When there is a torque on the rotor, there is presumably an equal and opposite torque on the stator; and therefore we might wonder if the mechanism of torque production could be pictured using the same ideas as we used for obtaining the rotor torque. The answer is yes; there is always an equal and opposite torque on the stator, which is why it's usually important to bolt a motor down securely. In some machines (e.g. the induction motor)it's easy to see that torque is produced on the stator by the interaction of the air-gap flux density and the stator currents, in exactly the same way that the flux density interacts with the rotor currents to produce torque on the rotor. In other motors, (e.g. the d.c. motor we have been looking at), there is no simple physical argument which can be advanced to derive the torque on the stator, but nevertheless it's equal and opposite to the torque on the rotor.
The final question relates to the similarity between the set-up shown in ill. 10 and the field patterns produced for example by the electro magnets used to lift car bodies in a scrap yard. From what we know of the large force of attraction that lifting magnets can produce, might not we expect a large radial force between the stator pole and the iron body of the rotor? and if there is, what is to prevent the rotor from being pulled across to the stator? Again the affirmative answer is that there is indeed a radial force due to magnetic attraction, exactly as in a lifting magnet or relay, although the mechanism whereby the magnetic field exerts a pull as it enters iron or steel is entirely different from the 'BIl 'force we have been looking at so far.
It turns out that the force of attraction per unit area of pole-face is proportional to the square of the radial flux density, and with typical air gap flux densities of up to 1 T in motors, the force per unit area of rotor surface works out to be about 40N/cm^2
.This indicates that the total radial force can be very large: for example the force of attraction on a small pole-face of only 5 x10 cm is 2000N, or about 200 Kg. This force contributes nothing to the torque of the motor, and is merely an unwelcome by-product of the 'BIl 'mechanism we employ to produce tangential force on the rotor conductors.
In most machines the radial magnetic force under each pole is actually a good deal bigger than the tangential electromagnetic force on the rotor conductors, and as the question implies, it tends to pull the rotor onto the pole. However, the majority of motors are constructed with an even number of poles equally spaced around the rotor, and the flux density in each pole is the same, so that in theory at least the resultant force on the complete rotor is zero. In practice, even a small eccentricity will cause the field to be stronger under the poles where the air-gap is smaller, and this will give rise to an unbalanced pull, resulting in noisy running and rapid bearing wear.
Magnitude of torque
Returning to our original discussion, the force on each conductor is given by equation 1.2, and it follows that the total tangential force F depends on the flux density produced by the field winding, the number of conductors on the rotor, the current in each, and the length of the rotor. The resultant torque or couple (T) depends on the radius of the rotor (r), and is given by:
T = Fr [1.8]
We will develop this further in Section 1.5, after we examine the remark able benefits gained by putting the conductors into slots.
The advantages of slotting
If the conductors were mounted on the surface of the rotor iron, as in ill. 11, the air-gap would have to be at least equal to the wire diameter, and the conductors would have to be secured to the rotor in order to transmit their turning force to it. The earliest motors were made like this, with string or tape to bind the conductors to the rotor.
Unfortunately, a large air-gap results in an unwelcome high-reluctance in the magnetic circuit, and the field winding therefore needs many turns and a high current to produce the desired flux density in the air-gap. This means that the field winding becomes very bulky and consumes a lot of power. The early (Nineteenth-century) pioneers soon hit upon the idea of partially sinking the conductors on the rotor into grooves machined parallel to the shaft, the intention being to allow the air-gap to be reduced so that the exciting windings could be smaller. This worked extremely well as it also provided a more positive location for the rotor conductors, and thus allowed the force on them to be transmitted to the body of the rotor.
Before long the conductors began to be recessed into ever deeper slots until finally (see ill. 12) they no longer stood proud of the rotor surface and the air-gap could be made as small as was consistent with the need for mechanical clearances between the rotor and the stator. The new 'slotted 'machines worked very well, and their pragmatic makers were unconcerned by rumblings of discontent from skeptical theorists.
(1. Older readers will probably have learned the terms Couple and Moment (of a force) long before realizing that they mean the same as torque.)
ill. 12 Influence on flux paths when the rotor is slotted to accommodate conductors
The theorists of the time accepted that sinking conductors into slots allowed the air-gap to be made small, but argued that, as can be seen from ill. 12, almost all the flux would now pass down the attractive low-reluctance path through the teeth, leaving the conductors exposed to the very low leakage flux density in the slots. Surely, they argued, little or no 'BIl' force would be developed on the conductors, since they would only be exposed to a very low flux density.
The skeptics were right in that the flux does indeed flow down the teeth; but there was no denying that motors with slotted rotors produced the same torque as those with the conductors in the air-gap, provided that the average flux densities at the rotor surface were the same. So what could explain this seemingly too good to be true situation? The search for an explanation preoccupied some of the leading thinkers long after slotting became the norm, but finally it became possible to verify theoretically that the total force remains the same as it would have been if the conductors were actually in the flux, but almost all of the tangential force now acts on the rotor teeth, rather than on the conductors themselves.
This is remarkably good news. By putting the conductors in slots, we simultaneously enable the reluctance of the magnetic circuit to be reduced, and transfer the force from the conductors themselves to the sides of the iron teeth, which are robust and well able to transmit the resulting torque to the shaft. A further benefit's that the insulation around the conductors no longer has to transmit the tangential forces to the rotor, and its mechanical properties are thus less critical. Seldom can tentative experiments with one aim have yielded rewarding outcomes in almost every other relevant direction.
There are some snags, however. To maximize the torque, we will want as much current as possible in the rotor conductors. Naturally we will work the copper at the highest practicable current density (typically between 2 and 8A =mm^2), but we will also want to maximize the cross sectional area of the slots to accommodate as much copper as possible.
This will push us in the direction of wide slots, and hence narrow teeth.
But we recall that the flux has to pass radially down the teeth, so if we make the teeth too narrow, the iron in the teeth will saturate, and lead to a poor magnetic circuit. There is also the possibility of increasing the depth of the slots, but this cannot be taken too far or the centre region of the rotor iron which has to carry the flux from one pole to another will become so depleted that it too will saturate.
In the next section we look at what determines the torque that can be obtained from a rotor of a given size, and see how speed plays a key role in determining the power output.
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