DC motors

DC motors

A simple DC motor has a coil of wire that can rotate in a magnetic field. The current in the coil is supplied via two brushes that make moving contact with a split ring. The coil lies in a steady magnetic field. The forces exerted on the current-carrying wires create a torque on the coil.

The force F on a wire of length L carrying a current i in a magnetic field B is iLB times the sine of the angle between B and i, which would be 90° if the field were uniformly vertical. The direction of F comes from the right hand rule*, as shown here. The two forces shown here are equal and opposite, but they are displaced vertically, so they exert a torque. (The forces on the other two sides of the coil act along the same line and so exert no torque.)

* A number of different nmemonics are used to remember the direction of the force. Some use the right hand, some the left. For students who know vector multiplication, it is easy to use the Lorentz force directly:  F  =  q v X B ,  whence F  =  i dL X B . That is the origin of the diagram shown here.

The coil can also be considered as a magnetic dipole, or a little electromagnet, as indicated by the arrow SN: curl the fingers of your right hand in the direction of the current, and your thumb is the North pole. In the sketch at right, the electromagnet formed by the coil of the rotor is represented as a permanent magnet, and the same torque (North attracts South) is seen to be that acting to align the central magnet.Note the effect of the brushes on the split ring. When the plane of the rotating coil reaches horizontal, the brushes will break contact (not much is lost, because this is the point of zero torque anyway – the forces act inwards). The angular momentum of the coil carries it past this break point and the current then flows in the opposite direction, which reverses the magnetic dipole. So, after passing the break point, the rotor continues to turn anticlockwise and starts to align in the opposite direction. In the following text, I shall largely use the 'torque on a magnet' picture, but be aware that the use of brushes or of AC current can cause the poles of the electromagnet in question to swap position when the current changes direction.
The torque generated over a cycle varies with the vertical separation of the two forces. It therefore depends on the sine of the angle between the axis of the coil and field. However, because of the split ring, it is always in the same sense. The animation below shows its variation in time, and you can stop it at any stage and check the direction by applying the right hand rule.

Some notes about AC and DC motors for high power applications

Some notes about AC and DC motors for high power applications

This site was originally written to help high school students and teachers in New South Wales, Australia, where a new syllabus concentrating on the history and applications of physics, at the expense of physics itself, has been introduced. The new syllabus, in one of the dot points, has this puzzling requirement: "explain that AC motors usually produce low power and relate this to their use in power tools".

AC motors are used for high power applications whenever it is possible. Three phase AC induction motors are widely used for high power applications, including heavy industry. However, such motors are unsuitable if multiphase is unavailable, or difficult to deliver. Electric trains are an example: it is easier to build power lines and pantographs if one only needs one active conductor, so this usually carries DC, and many train motors are DC. However, because of the disadvantages of DC for high power, more modern trains convert the DC into AC and then run three phase motors.Single phase induction motors have problems for applications combining high power and flexible load conditions. The problem lies in producing the rotating field. A capacitor could be used to put the current in one set of coils ahead, but high value, high voltage capacitors are expensive. Shaded poles are used instead, but the torque is small at some angles. If one cannot produce a smoothly rotating field, and if the load 'slips' well behind the field, then the torque falls or even reverses.
Power tools and some appliances use brushed AC motors. Brushes introduce losses (plus arcing and ozone production). The stator polarities are reversed 100 times a second. Even if the core material is chosen to minimise hysteresis losses ('iron losses'), this contributes to inefficiency, and to the possibility of overheating. These motors may be called 'universal' because they can operate on DC. This solution is cheap, but crude and inefficient. For relatively low power applications like power tools, the inefficiency is usually not economically important.
If only single phase AC is available, one may rectify the AC and use a DC motor. High current rectifiers used to be expensive, but are becoming less expensive and more widely used. If you are confident you understand the principles, it's time to go to How real electric motors work by John Storey. Or else continue here to find out about loudspeakers and transformers.

Linear motors

Linear motors

A set of coils can be used to create a magnetic field that translates, rather than rotates. The pair of coils in the animation below are pulsed on, from left to right, so the region of magnetic field moves from left to right. A permanent or electromagnet will tend to follow the field. So would a simple slab of conducting material, because the eddy currents induced in it (not shown) comprise an electromagnet. Alternatively, we could say that, from Faraday's law, an emf in the metal slab is always induced so as to oppose any change in magnetic flux, and the forces on the currents driven by this emf keep the flux in the slab nearly constant. (Eddy currents not shown in this animation.)


Alternatively, we could have sets of powered coils in the moving part, and induce eddy currents in the rail. Either case gives us a linear motor, which would be useful for say maglev trains. (In the animation, the geometry is, as usual on this site, highly idealised, and only one eddy current is shown.)

Build a simple motor

Build a simple motor

To build this simple but strange motor, you need two fairly strong magnets (rare earth magnets about 10 mm diameter would be fine, as would larger bar magnets), some stiff copper wire (at least 50 cm), two wires with crocodile clips on either end, a six volt lantern battery, two soft drink cans, two blocks of wood, some sticky tape and a sharp nail.



Make the coil out of stiff copper wire, so it doesn't need any external support. Wind 5 to 20 turns in a circle about 20 mm in diameter, and have the two ends point radially outwards in opposite directions. These ends will be both the axle and the contacts. If the wire has lacquer or plastic insulation, strip it off at the ends

The supports for the axle can be made of aluminium, so that they make electrical contact. For example poke holes in a soft drink cans with a nail as shown. Position the two magnets, north to south, so that the magnetic field passes through the coil at right angles to the axles. Tape or glue the magnets onto the wooden blocks (not shown in the diagram) to keep them at the right height, then move the blocks to put them in position, rather close to the coil. Rotate the coil initially so that the magnetic flux through the coil is zero, as shown in the diagram.Now get a battery, and two wires with crocodile clips. Connect the two terminals of the battery to the two metal supports for the coil and it should turn. Note that this motor has at least one 'dead spot': It often stops at the position where there is no torque on the coil. Don't leave it on too long: it will flatten the battery quickly.The optimum number of turns in the coil depends on the internal resistance of the battery, the quality of the support contacts and the type of wire, so you should experiment with different values. As mentioned above, this is also a generator, but it is a very inefficient one. To make a larger emf, use more turns (you may need to use finer wire and a frame upon which to wind it.) You could use eg an electric drill to turn it quickly, as shown in the sketch above. Use an oscilloscope to look at the emf generated. Is it AC or DC? This motor has no split ring, so why does it work on DC? Simply put, if it were exactly symmetrical, it wouldn't work. However, if the current is slightly less in one half cycle than the other, then the average torque will not be zero and, because it spins reasonably rapidly, the angular momentum acquired during the half cycle with greater current carries it through the half cycle when the torque is in the opposite direction. At least two effects can cause an asymmetry. Even if the wires are perfectly stripped and the wires clean, the contact resistance is unlikely to be exactly equal, even at rest. Also, the rotation itself causes the contact to be intermittent so, if there are longer bounces during one phase, this asymmetry is sufficient. In principle, you could partially strip the wires in such a way that the current would be zero in one half cycle.

An alternative relisation of the simple motor, by James Taylor.

An even simpler motor (one that is also much simpler to understand!) is the homopolar motor.

Power in resistive and reactive AC circuits

Consider a circuit for a single-phase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a resistive load: (Figure below)

Ac source drives a purely resistive load.

In this example, the current to the load would be 2 amps, RMS. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. If we were to plot the voltage, current, and power waveforms for this circuit, it would look like Figure below.

Current is in phase with voltage in a resistive circuit.

Note that the waveform for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads. If the source were a mechanical generator, it would take 240 watts worth of mechanical energy (about 1/3 horsepower) to turn the shaft.

Also note that the waveform for power is not at the same frequency as the voltage or current! Rather, its frequency is double that of either the voltage or current waveforms. This different frequency prohibits our expression of power in an AC circuit using the same complex (rectangular or polar) notation as used for voltage, current, and impedance, because this form of mathematical symbolism implies unchanging phase relationships. When frequencies are not the same, phase relationships constantly change.

As strange as it may seem, the best way to proceed with AC power calculations is to use scalar notation, and to handle any relevant phase relationships with trigonometry.

For comparison, let's consider a simple AC circuit with a purely reactive load in Figure below.

AC circuit with a purely reactive (inductive) load.

Power is not dissipated in a purely reactive load. Though it is alternately absorbed from and returned to the source.

Note that the power alternates equally between cycles of positive and negative. (Figure above) This means that power is being alternately absorbed from and returned to the source. If the source were a mechanical generator, it would take (practically) no net mechanical energy to turn the shaft, because no power would be used by the load. The generator shaft would be easy to spin, and the inductor would not become warm as a resistor would.

Now, let's consider an AC circuit with a load consisting of both inductance and resistance in Figure below.

AC circuit with both reactance and resistance.

At a frequency of 60 Hz, the 160 millihenrys of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152^o. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.

We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works in Figure below.

A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.

As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.

Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.

Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance that we've worked with so far.

• REVIEW:
• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90^o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0^o and 90^o.

AC motors

AC motors

With AC currents, we can reverse field directions without having to use brushes. This is good news, because we can avoid the arcing, the ozone production and the ohmic loss of energy that brushes can entail. Further, because brushes make contact between moving surfaces, they wear out.The first thing to do in an AC motor is to create a rotating field. 'Ordinary' AC from a 2 or 3 pin socket is single phase AC--it has a single sinusoidal potential difference generated between only two wires--the active and neutral. (Note that the Earth wire doesn't carry a current except in the event of electrical faults.) With single phase AC, one can produce a rotating field by generating two currents that are out of phase using for example a capacitor. In the example shown, the two currents are 90° out of phase, so the vertical component of the magnetic field is sinusoidal, while the horizontal is cosusoidal, as shown. This gives a field rotating counterclockwise.
(* I've been asked to explain this: from simple AC theory, neither coils nor capacitors have the voltage in phase with the current. In a capacitor, the voltage is a maximum when the charge has finished flowing onto the capacitor, and is about to start flowing off. Thus the voltage is behind the current. In a purely inductive coil, the voltage drop is greatest when the current is changing most rapidly, which is also when the current is zero. The voltage (drop) is ahead of the current. In motor coils, the phase angle is rather less than 90¡, because electrical energy is being converted to mechanical energy.)


In this animation, the graphs show the variation in time of the currents in the vertical and horizontal coils. The plot of the field components B_x and B_y shows that the vector sum of these two fields is a rotating field. The main picture shows the rotating field. It also shows the polarity of the magnets: as above, blue represents a North pole and red a South pole.
If we put a permanent magnet in this area of rotating field, or if we put in a coil whose current always runs in the same direction, then this becomes a synchronous motor. Under a wide range of conditions, the motor will turn at the speed of the magnetic field. If we have a lot of stators, instead of just the two pairs shown here, then we could consider it as a stepper motor: each pulse moves the rotor on to the next pair of actuated poles. Please remember my warning about the idealised geometry: real stepper motors have dozens of poles and quite complicated geometries!

Bandwidth and Q factor

Bandwidth and Q factor At resonance, the voltages across the capacitor and the pure inductance cancel out, so the series impedance takes its minimum value: Zo = R. Thus, if we keep the voltage constant, the current is a maximum at resonance. The current goes to zero at low frequency, because XC becomes infinite (the capacitor is open circuit for DC). The current also goes to zero at high frequency because XL increases with ω (the inductor opposes rapid changes in the current). The graph shows I(ω) for circuit with a large resistor (lower curve) and for one with a small resistor (upper curve). A circuit with low R, for a given L and C, has a sharp resonance. Increasing the resistance makes the resonance less sharp. The former circuit is more selective: it produces high currents only for a narrow bandwidth, ie a small range of ω or f. The circuit with higher R responds to a wider range of frequencies and so has a larger bandwidth. The bandwidth Δω (indicated by the horiztontal bars on the curves) is defined as the difference between the two frequencies ω+ and ω- at which the circuit converts power at half the maximum rate.Now the electrical power converted to heat in this circuit is I2R, so the maximum power is converted at resonance, ω = ωo. The circuit converts power at half this rate when the current is Io/√2. The Q value is defined as the ratio Q  =  ωo/Δω.