Linear analysis assumes that the inductance is unaffected by the current: that is, there is no magnetic saturation. For simplicity we also ignore the effect of fringing flux around the pole corners, and assume that all the flux crosses the airgap in the radial direction. Mutual coupling between phases is normally zero or very small, and is ignored. The voltage equation for one phase is
where v is the terminal voltage, i is the current, R is the flux-linkage in volt-seconds, R is the phase resistance, L is the phase inductance, 2 is the rotor position, and Tm is the angular velocity in rad/s. The last term is sometimes interpreted as a “back-EMF” e:
It is helpful to visualize the supply voltage as being dropped across the three terms in (4.1): namely, the resistance voltage drop, the L di/dt term, and the back-EMF e. The instantaneous electrical power vi is
The rate of change of magnetic stored energy at any instant is given by
According to the law of conservation of energy, the mechanical power conversion p = TmTe is what is
left after the resistive loss Ri2 and the rate of change of magnetic stored energy are subtracted from the power input vi , Te being the instantaneous electromagnetic torque. Thus from eqns. (4.2) and (4.3),
writing Te = p/Tm = vi ! Ri2 ! d(½Li2)/dt, we get
Note that dL/d2 is the slope of the inductance graph in Fig. 4.5.
d dt 1 2Li 2 ' 1 2i 2dL dt % Li di dt ' 1 2i 2T m dL d2 % Li di dt (4.4) Te ' 1 2i 2dL d2. (4.5)
Fig. 4.6 Single phaseleg circuit
Fig. 4.7 Alternative single-phaseleg circuits
Drive circuit — unidirectional current, bidirectional voltage
Eqn. (4.5) says that the torque does not depend on the direction of the current, since i2 is always positive. However, the voltage must be reversed at the end of each stroke, to return the flux- linkage to zero. By Faraday’s law, this requires a negative voltage applied to the coil, to ensure that dR/dt < 0. Fig. 4.6 shows the half-bridge phaseleg circuit that accomplishes this. When Q1 and Q2 are both on, the voltage across the motor windings is v = Vs, the supply voltage. When Q1 and Q2 are both
off, v = !Vs while the current freewheels through D1 and D2.
To reduce R and i to zero as quickly as possible, as in Fig. 4.5, the reverse voltage must be much larger than the forward voltage, otherwise the flux-linkage will persist beyond the aligned position, producing an unwanted negative pulse of torque. At low speeds this is achievable by chopping the forward voltage, reducing its effective value compared to the reverse voltage. The circuit of Fig. 4.6 can operate the machine as a motor or as a generator, since the electrical power vi can be positive or
negative. If the average power is negative (i.e., generating), the energy supplied during transistor conduction in one stroke must be less than the energy recovered during freewheeling. The transistor conduction period (with positive applied voltage) is still necessary to establish the flux, which is built up from zero and returned to zero each stroke. The voltage-time integrals during transistor conduction and freewheeling must be approximately equal (apart from resistive volt-drop), regardless of whether the machine is motoring or generating. As regards control, the main difference between motoring and generating is the phasing of the conduction pulse relative to the rotor position. From Fig. 4.4 it appears that generating current pulses must coincide with AK, just as motoring pulses coincide with JA.
Other circuits use only one transistor per phase, and employ various means to produce the "suppression voltage" (i.e., reverse voltage) needed to de-flux the windings at the end of each stroke. The circuit in Fig. 4.7a uses separate voltage sources for “fluxing” and “de-fluxing”. The circuit in Fig. 4.7b goes one stage further by having two isolated windings with a common magnetic circuit. The two parts of the winding could be bifilar-wound or they could have completely different numbers of turns and wire sizes. Unfortunately the leakage inductances of the two parts of the winding are usually quite large, even with a bifilar winding, and this leads to problems with transistor overvoltage. Consequently the circuit of Fig. 4.7b is rarely used, although it is common in stepper-motor drives.
Fig. 4.8 Linear energy conversion diagram W ' OJA ' Te)2 ' 1 2im 2dL d2)2 ' 1 2im 2)L. (4.6) ei)t ' ABCJ ' im2T m dL d2 × )2 Tm ' im 2)L (4.7) Q ' 8 ! 1 28 ! 1. (4.10) S ' im2)L % 1 2im 2L u. (4.8) Q ' energy converted energy supplied ' W S % W W % R (4.9)
Additional phases simply use additional drive circuits of the same form as the first phase, usually with a common voltage source. With four-phase motors it is possible to use a common chopping transistor between two phases, so that only six power transistors are required; but the control range of conduction angles must be limited limited to allow de-fluxing of the “complementary” phase in each pair.
Limitations of the ideal linear model
At the aligned position A in Fig. 4.5, the current must be switched off quickly to avoid the production of negative torque after the poles have passed the aligned position. The magnetic stored energy ½Lai2 must be returned to the supply. In a
nonsaturating reluctance machine of this type, the magnetic energy stored at A is large, because both
L and i are at their maximum values. We can get
further insight from an energy audit taken over one stroke as the rotor moves from J to A. The process is shown in the energy conversion diagram, Fig. 4.8, which plots flux-linkage against current. The slope of OU is the inductance at the unaligned position,
Lu, and the slope of OA is the inductance at the
aligned position, La. At intermediate positions the
inductance is represented by a line of intermediate
slope between Lu and La. At the J position (start of overlap), if fringing is neglected (as in the idealized
inductance variation in Fig. 4.4), LJ = Lu. The complete stroke is represented by the locus OJAO. In motoring operation it is traversed in the counterclockwise direction, and in generating operation in the clockwise direction.
Although the current in Fig. 4.5 has a step from 0 to a maximum value im at the position J, in Fig. 4.8
this step is along OJ and energy OJC = ½Luim2 must be supplied to the magnetic circuit as the current
increases from 0 to im. The step cannot be accomplished in zero time, since that would require infinite voltage, but if the angular velocity is low, the angle of rotation along OJ is small. Along JA the electromechanical energy conversion is W or OJA, given by
Along JA there is a back-EMF e which absorbs energy equal to the area ABCJ:
where )t = )2/Tm is the time taken to rotate through the interval )2 = JA, and )L = La ! Lu is the change in L. The total energy supplied is the sum of areas OCJ and ABCJ, i.e., area S = W + R = OJAB, and
We can now define the energy ratio as
tf
tr ' d. (4.11)
In this type of nonsaturating motor, less than half the energy S supplied by the drive is converted into mechanical work in each stroke (even neglecting losses). During the “working” part of the stroke JA, the energy is partitioned equally between mechanical work and stored field energy; this is evident from the ratio of eqns. (4.6) and (4.7), or OJA/ABCJ. The energy ratio would have a value of 0.5 but for the “overhead” of stored field energy OJC which must be built up before the torque zone JA. This makes
Q < 0.5. The inverse of the energy ratio is the converter volt-ampere-seconds per joule of energy
conversion C = S/W = 1 + R/W, an important quantity in understanding the basic requirement for “silicon” in the converter.
The stored field energy reaches a maximum value R at A, and must be returned to the supply at the end of the stroke by commutating the current into the diodes, so that the voltage reverses and forces the flux-linkage to fall to zero. In the ideal locus this fall is along AO. However this is not possible with finite supply voltage. If the current is chopped with a duty-cycle d along JA, the average forward applied voltage along JA is d × Vs. In a circuit of the form of Fig. 4.6, the reverse voltage after
commutation is !Vs. By integrating Faraday’s law the rise and fall periods of flux-linkage can be shown
to be in the ratio
This shows that “instant” suppression of the flux is effectively achieved at very low speed when d is small. But at higher speed there is no chopping: the forward and reverse voltages are equal in magnitude, and d =, so tf = tr and the flux-linkage waveform is triangular. The time taken along OA is
the same as the time taken along OJA, The waveforms of flux-linkage and current corresponding to Fig. 4.8 are shown in Fig. 4.9, and show a tail in the current waveform extending past the aligned position so that some negative torque must be produced.
Figs. 4.8 and 4.9 represent an important operating condition where the back-EMF is just sufficient to maintain a flat-topped current waveform. With full voltage applied, the speed at which this occurs is called the base speed. The current im could be called the base current. (It is not necessarily the rated
current, because that depends on the cooling arrangements).
Wf ' m idR;
Wc ' m Rdi. (4.13)
Fig. 4.10 Equivalent circuit
Fig. 4.11 A magnetization curve at one rotor position
R ' W × 1 ! Q
Q . (4.12)
In practical terms the nonsaturating switched reluctance motor makes poor use of the power semiconductors, because they have to supply more than 2J of energy in order to get 1 J of mechanical work, and they must also provide a means to recover the unconverted energy at the end of the stroke. The DC link filter capacitance is directly related to the value of R, which is a large fraction of the energy conversion W: in fact
For example, if 8 = 8, Q = 0.467, C = 2.14, and R = 1.14W. This implies that for zero ripple voltage at the DC link, the filter capacitance must be large enough to absorb R joules with negligible change of voltage. This requirement may be reduced by overlapping charge/discharge requirements of adjacent phases, but still it is a serious consideration.
Eqns. (4.1) and (4.2) imply the existence of an equivalent circuit of the form shown in Fig. 4.10, in which there is a back-EMF e = TmidL/d2. Unfortunately e is not an independent parameter, but depends on the current. In a normal equivalent circuit we interpret the product e i as the electromechanical power conversion TmTe, implying that Te = i2dL/d2. However, eqn. (4.5) states that the
torque is only ½i2dL/d2. Of the power ei, only half is
converted into mechanical power during the “working” part of the stroke JA. The other half is being stored as magnetic field energy in the increasing inductance. With
L also varying, the equivalent circuit is misleading and cannot be interpreted in the same way as it can,
for example, with permanent-magnet motors. This means that the simulation of switched reluctance machines and their drives requires the direct solution of eqns. (4.1) and (4.5), even when saturation is ignored. An elegant and thorough solution for the nonsaturating motor was presented by Ray and Davis [1979]. Usually saturation cannot be ignored and the full nonlinear equations must be solved.
4.3 NONLINEAR ANALYSIS OF TORQUE PRODUCTION