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.

Resonance

Resonance

Note that the expression for the series impedance goes to infinity at high frequency because of the presence of the inductor, which produces a large emf if the current varies rapidly. Similarly it is large at very low frequencies because of the capacitor, which has a long time in each half cycle in which to charge up. As we saw in the plot of Z_seriesω above, there is a minimum value of the series impedance, when the voltages across capacitor and inductor are equal and opposite, ie v_L(t) = - v_C(t) so V_L(t) = V_C, so

ωL = 1/ωC        so the frequency at which this occurs is

where ω_o and f_o are the angular and cyclic frequencies of resonance, respectively. At resonance, series impedance is a minimum, so the voltage for a given current is a minimum (or the current for a given voltage is a maximum).

This phenomenon gives the answer to our teaser question at the beginning. In an RLC series circuit in which the inductor has relatively low internal resistance r, it is possible to have a large voltage across the the inductor, an almost equally large voltage across capacitor but, as the two are nearly 180° degrees out of phase, their voltages almost cancel, giving a total series voltage that is quite small. This is one way to produce a large voltage oscillation with only a small voltage source. In the circuit diagram at right, the coil corresponds to both the inducance L and the resistance r, which is why they are drawn inside a box representing the physical component, the coil. Why are they in series? Because the current flows through the coil and thus passes through both the inductance of the coil and its resistance.You get a big voltage in the circuit for only a small voltage input from the power source. You are not, of course, getting something for nothing. The energy stored in the large oscillations is gradually supplied by the AC source when you turn on, and it is then exchanged between capacitor and inductor in each cycle. For more details about this phenomenon, and a discussion of the energies involved.

RL Series combinations

RL Series combinations

In an RL series circuit, the voltage across the inductor is aheadof the current by 90°, and the inductive reactance, as we saw before, is X_L = ωL. The resulting v(t) plots and phasor diagram look like this.

It is straightforward to use Pythagoras' law to obtain the series impedance and trigonometry to obtain the phase. We shall not, however, spend much time on RL circuits, for three reasons. First, it makes a good exercise for you to do it yourself. Second, RL circuits are used much less than RC circuits. This is because inductors are always* too big, too expensive and the wrong value, a proposition you can check by looking at an electronics catalogue. If you can use a circuit involving any number of Rs, Cs, transistors, integrated circuits etc to replace an inductor, one usually does. The third reason why we don't look closely at RL circuits on this site is that you can simply look at RLC circuits (below) and omit the phasors and terms for the capacitance.* Exceptions occur at high frequencies (~GHz) where only small value Ls are required to get substantial ωL. In such circuits, one makes an inductor by twisting copper wire around a pencil and adjusts its value by squeezing it with the fingers.

Impedance of components

Impedance of components

Let's recap what we now know about voltage and curent in linear components. The impedance is the general term for the ratio of voltage to current. Resistance is the special case of impedance when φ = 0, reactance the special case when φ = ± 90°. The table below summarises the impedance of the different components. It is easy to remember that the voltage on the capacitor is behind the current, because the charge doesn't build up until after the current has been flowing for a while.

The same information is given graphically below. It is easy to remember the frequency dependence by thinking of the DC (zero frequency) behaviour: at DC, an inductance is a short circuit (a piece of wire) so its impedance is zero. At DC, a capacitor is an open circuit, as its circuit diagram shows, so its impedance goes to infinity.