# Industrial Brushless Servomotors--MOTOR AND LOAD DYNAMICS

 Home | Articles | Forum | Glossary | Books
 1. Introduction The aim of this Section is to study the electrical and dynamic characteristics of the loaded motor, and the effects of the interplay between the motor and load masses. The emphasis is on understanding the characteristics of the motor and load as one component of the overall system, rather than on the detailed methods of system control which are well explained in existing text , . The treatment of servomotor control theory is limited here to a very short introduction to the main principles. The electrical time constant of the brushed motor is often ignored, but this is generally not possible in the brushless machine due to its relatively high inductance. The main part of this Section starts by developing the basic equations for the brushless motor in terms of the mechanical and electrical time constants. An introduction is given to the method of Laplace transformation, which is then used to study the effects of the time constants on the electromechanical behavior of the motor. When the brushless motor is used for incremental motion, the motor rating may be affected by the relative values of the motor and load inertias. Here, we look to the optimization of the design of the transmission mechanism between the two masses. A gear train, belt and pulley or ball screw drive can each be designed in a way which keeps the drive motor heating to a minimum. Such transmission mechanisms can be applied in ways which differ widely, and it is unrealistic to describe any numerical example as typical. Some examples are included here in order to illustrate the main principles involved in the efficient connection of the motor to the load. The basic analysis of the performance of a loaded motor usually assumes that the shafts and couplings between the rotor of the motor and the load are completely rigid. In practice, flexibility of the motor-load connection may alter performance and cause problems with the control system. Flexibility in the connection is normally known as compliance. The electromechanical effects of the compliance of the connection between the motor and the load are treated at the end of the Section. Note that throughout this Section, the speed of the motor is denoted by Wm instead of the symbol w used previously. 2. Motor control In Section 3 we studied the inverter, which is one of the two main components of the drive. The drive also has the task of controlling the motor shaft position and speed, using input command signals and signals from the respective sensors. The motor, sensor and drive form a closed circuit normally known as a closed loop. Closed loops At a fixed supply voltage, the voltage drop across the winding resistance of a permanent magnet brushed motor causes the speed to fall as the load torque rises. In the brushless motor, the speed is affected by voltage drops which occur across the winding inductance as well as across the winding resistance. In the ideal case, the requirement of most motion control applications is for the motor speed to be independent of everything except the speed command. There is therefore the need for feedback of information about the motor speed from moment to moment, so that the motor input current may be correctly and continuously adjusted. FIG. 1 Closed-loop speed control The basic blocks of the control unit are shown as a comparator and an amplifier in FIG. 1. The closed loop is made up of the sensor, control unit, inverter output circuit and of course the motor. The difference between the sensor output and the desired reference input is transmitted as an error signal from the comparator to the amplifier. The drive output is then adjusted at the inverter according to the amplifier output so as to reduce the error in speed. Speed regulation FIG. 2 shows the four possible combinations of positive and negative velocity and torque as a four-quadrant diagram. The shaded area of FIG. 2(a) indicates that the control system is designed for only positive velocity and torque, and the unidirectional system is normally known as a speed regulator. FIG. 2 Motor torque and velocity Servo control A servo system controls the direction of rotation as well as the torque direction of the motor, and so the motor is operated over all four quadrants of the velocity-torque plane shown in FIG. 2(b). For many applications there may also be the need for control of the position of the load, and therefore of the motor shaft. FIG. 3 shows the additional feedback loop required for position control. FIG. 3 Closed-loop control of velocity and position Of major importance in the design of the control system is a knowledge of the 'open-loop' properties of the motor and load unit, before its inclusion in the closed loop. Sub-Sections 3 and 4, below, look at the characteristics of the motor and load, and at how they affect the open-loop behavior. 3. Motor equations Equations were developed in Section 1 for the steady-state characteristics of the brushed motor. It was seen in Section 2 that one difference between the brushed and brushless motors is that one motor incorporates a mechanical inverter consisting of brushes and commutator whereas the other has an external, power electronic inverter. Both machines operate from a DC supply. Constant speeds For the brushed motor, the current is subjected to commutation in only a small part of the winding at any one time. The distorting effect of the inductance of the part under commutation is low in comparison to the smoothing effect of the inductance of the remaining winding. The direct current to the brushed motor is therefore largely unaffected by the commutation process, and the speed at steady-state is usually assumed to be independent of motor inductance. Such an assumption cannot be made for the brushless motor, where commutation occurs at the same moment for a complete winding. Any formulation of steady-state equations must take account of the voltage drop across the winding inductance as well as that across the winding resistance. Transient demands of torque and speed are, however, the common requirements for a brushless servomotor, and steady-state equations are unlikely to be of use. Speed variations An equation for the motor speed under transient conditions must take account of all mechanical and electrical factors which affect a change in speed. The rate of change of the motor torque is limited by the rate at which the motor current can be changed, which is in turn limited by the motor inductance. The inductance can be found by applying a sinusoidal AC voltage of angular frequency w rad/s to the stator, after first locking the rotor shaft in a fixed position. The line-to-line impedance of the motor to the flow of alternating current is shown in FIG. 4 to consist of resistance R and the motor reactance wL, where L is the electrical inductance of the motor. Resistance RM accounts for the power losses in the magnetic circuit but as its value is normally high in comparison with ~L, its effect on the overall circuit is usually ignored. The voltage applied across the lines is Vrms = Irms(R + j~L) L is the only unknown and is normally assumed to have the same value over a wide range of frequency. FIG. 5 shows an equivalent of the stator input circuit, which consists of the line-to-line back emf, inductance and resistance. We will study how quickly the motor speed can be changed on the assumption that the input voltage V is applied suddenly, as a step input. FIG. 4 Locked rotor equivalent circuit FIG. 5 An equivalent circuit for the brushless motor The electrical equation In FIG. 5, the volt drop across L acts in the direction shown when the rate of change of current is positive. In other words, VL opposes the change in current. The electrical equation of the motor is seen to be V = L' d/tt i + Ri + ICE O.;m where V is the applied voltage and i is the current at time t. FIG. 6 The dynamic system The dynamic equation The rate at which the motor speed can change is clearly affected by the moment of mechanical inertia of the driven load, and also by the moments of inertia of the rotor and sensor. The unit of the moment of inertia is the kgm 2. In FIG. 6, the rotor of a motor of inertia Jm is connected to a load mass of inertia JL and to a sensor of inertia Js. The torque T produced by the motor is opposed by the torque Tf, due to bearing and beating seal friction, and by the torque TD due to the viscous damping from iron losses and windage. The motor must also react to the torque TA imposed by the mass of the motor, load and sensor during speed changes. The dynamic equation is T_j d d t C~m + Dwm + Tf + TL where J = Jm + JL + Js. At constant speed, the expression reduces to the steady-state form. Electrical and mechanical time constants Following the application to a stationary motor of a constant voltage in the form of a step input, the rotor speed and stator current each rise over time. Suppose that the rotor is locked in position and stationary throughout. With Wm set to zero, the electrical equation becomes d V- L-~i+ Ri Dividing through by R gives the final stator current as: Ld I - -R-~t i + i Solution of this expression shows that the current rises exponentially towards its final value according to i - I(1 - e -Rt/L) FIG. 7(a) shows the form of the current rise following the application of the step input of voltage. When t- L/R, the current reaches 100(1-e-])%, or 63.2% of its final value. The electrical time constant is defined as L re R FIG. 7 (a) Rise of current with rotor locked (b) Rise of speed when L = 0 Suppose now that the rotor is stationary but free to rotate, with no load and no supply voltage. Following the application of the step-input voltage to the stator, current flows into the stator winding and the rotor accelerates. If the opposing torques due to viscous damping and friction are assumed to be insignificant, the accelerating torque on the rotor is found from the dynamic equation to be: d T- Jm ~-~ Wm ... where Jm is from this point taken to include the inertia of the sensor. The rate of rise of stator current, and therefore of torque and rotor speed, is affected by the electrical time constant. In order to study the other factors which affect the rate of speed rise after the sudden application of V, assume for the moment that the motor has no inductance and therefore an electrical time constant of zero. The electrical equation reduces to: V= Ri + KEWm The final no-load speed at voltage V would be: V W~NL = KE Replacing T by KTi and combining the last three equations above gives: RJm d W~L KT KE d t Wm + ~ Om Solving the last expression shows the speed of the unloaded motor with no inductance to rise with time according to: R&. Wm - WNL(1 -- e -t/~') where 1",,, = KTKE In FIG. 7(b), the speed reaches 63.2% of its final value when t equals the mechanical time constant T m. Resistance R places the limit on the current and torque for a motor with no inductance, which accounts for the appearance of the electrical resistance in the mechanical time constant. A hypothetical motor with R = 0 and L = 0 would reach full speed at the instant of application of the supply voltage in response to an infinite current impulse. The rate of rise of speed of a real motor is, of course, subject to the combined effects of the electrical and mechanical time constants. The electrical time constant of a brushed motor is usually low compared to its mechanical constant, and analysis is often eased by ignoring the motor inductance. This simplification cannot be used for the brushless motor, where in many cases 7"m < 7"e. Taking, for example, the trapezoidal motor in Table 1, Table 1 Specification of a four-pole brushless servomotor The electrical time constant of this motor is therefore about five times the mechanical value. When the supply voltage is switched on, the rate of acceleration of the rotor mass would clearly be affected by the opposition of the voltage Ldi/dt to the build-up of motor current, as well as by the moment of inertia of the rotor The electromechanical equation of the unloaded motor Assume now that we have a stationary motor which is free to rotate, and which has inductance, resistance and inertia. A step input of voltage V is applied at t- 0. We already know that the electrical and mechanical equations for the unloaded motor are: d V- L--;-: i + Ri + KEWm (1t and d T- Jm -'~ Wm Viscous damping and friction are assumed to be negligible. Replacing T by KTi and combining the last two equations above gives: v=LJm { d2 R d KTKE } KT -~ ~m + "~ -~t ~m Jr L Jm O.)m Using the electrical and mechanical time constants already developed in terms of R, L, Jm, KT and KE, the last equation above may be written as: { d ) V- KE 7"eYm ~-'/~m --b Tin ~-~m q" aJm This equation can be used to study the response over a commutation period. For the brushed motor, the equation reduces at steady state to the familier form V- KE~m Frequency response Up to this point, we have dealt with the motor speed response to a step input of voltage as a transient disturbance in terms of the function of time am(t). We have seen that the speed would rise exponentially for a hypothetical motor with no inductance. However, it is possible for a sinusoidal element to be introduced into the transient response of the motor speed, in the practical case when both inductance and inertia are present. The actual shape of the waveform of motor speed against time then depends on the frequency of the sinusoid, as well as on the exponential change in the magnitude of the response. Both features can be displayed on a four-quadrant diagram known as the s-plane. The s-plane The s-plane is shown in FIG. 8. For linear systems the upper and lower quadrants form mirror images, and the points of such a plane are defined by: s-cr+j.~ FIG. 8 The s-plane of f(t) A purely sinusoidal part of the speed response has a frequency of ~ and appears on the jw axis, and also at the conjugate position-ja~. The point has a co-ordinate along the negative horizontal axis which is determined by the rate of the exponential decay of the response. Any exponentially rising and therefore unstable component appears on the positive horizontal axis. The steady-state component appears at the origin, where s-0. In FIG. 9, the response with a noticeable oscillatory component is positioned relatively close to the j~o axis. Only a small overshoot of speed occurs at the 45 degree position, and no overshoot occurs at the lowest position shown where the frequency of the sinusoid is relatively low and the exponential decay of its magnitude is relatively large. The question which now arises is how do we find the position on the s-plane of a response which is given as a function of time. FIG. 9 Speed responses associated with various points on the s-plane The Laplace transformation The Laplace approach allows us to represent time-varying functions on the s-plane. The Laplace transform of f(t) is given by O(3 f(s)- f f(t) -Tdt 0 Note that f(s) is not defined for t < 0. In the integral, e st is dimensionless and so s has the dimension of t -1, that is of frequency. The Laplace integration is normally done from a list of standard transforms. Five examples are shown in Table 4.2. The first four are known as functional transforms and are found from the Laplace integral given above, using the normal method of integration. The first line of the table shows the transform of a step function. The step function has a value of zero for t < 0 and has a constant value of V for t > 0. Table 2 Laplace transforms Function of time Laplace transform The last line of the table shows an operational transform, developed using integration by parts. It states that the transform of the rate of change of a time-varying function is simply equal to s multiplied by the transform of the function. For example, the Laplace transform of the rate of change with time of motor speed is f (s) - S~3m(S) where ~dm (S) is the Laplace transform of the motor speed ~m (l). It follows that the transform of the second order of the rate of speed change is s2a;m(S), and so on. The poles of the frequency response Let us now return to the case of the unloaded motor which has inductance and inertia, and which is supplied with step voltage V at t = 0. The electromechanical equation has been shown to be { d } V -- KE TeTm ~ Wm -'b Tmdt Wm -Jr- Wm Applying Laplace transforms gives V -- TeTmS20.;m(S) + TmSOdm(S) -Jr-Calm(S) sKE Rearranging, we obtain V ~m(S) = KES(S2TeTm + STm -Jr l) The last expression above gives the speed response of the motor to a step input. Values of s which make the denominator equal to zero are known as the system poles, defining the composition of the response on the s-plane. The pole at s = 0 represents a steady-state part of the transform. Transfer functions A transfer function is written for a system output relative to a system input. There is a separate transfer function for each combination of input and output. In FIG. 10 the system input is the voltage V(s) and the output is the motor speed w(s). The figure shows the transfer function of the unloaded motor as &m(S) V(S) KE(S2TeTm + STm -I- 1) The denominator of the transfer function may be used to find the poles of the transient speed response to the step voltage input. The steady-state pole at s- 0 does not appear in the transfer function. The trapezoidal motor specified in Table 1 has already been shown to have a ratio of electrical to mechanical time constant of approximately 5:1, and so we would expect its speed response to a step input to be dominated by the effect of inductance. The numerical values of the motor constants are KE(-- KT) - 0.84, n - 0.0069, Tin -- 0.0014 Using these values in the above transfer function gives: ~m(S) l V(s) 8.1 • 10-6s 2 + 11.8 • 10-as + 0.84 The poles are defined when the denominator of the transfer function equals zero. Solving the quadratic in the normal way gives: sl--73+j314 and sz=-73-j314 FIG. 10 Transfer function of motor speed relative to voltage input FIG. 11 shows the two poles on the s-plane. The poles are relatively close to the jw axis which indicates that the response is oscillatory, i.e. with speed overshoot. The time constant of the decay in the oscillatory waveform is 1/73 or 13.7 ms. Viscous damping and friction have been ignored and so the decay is the result of i2R loss in the stator winding. FIG. 11 Poles of an unloaded motor with T e ~ 5Tm 4. Variation of load inertia Inclusion of a load inertia JL increases the overall mechanical time constant to 7"M - (1 + JL/Jm)Ym The ratio of electrical to mechanical time constant falls from Te/Ym to 7"e/YM. It is of interest to plot the locus of the poles on the s-plane as the ratio changes. The poles appear at the values of s which make the denominator of the transfer function for the motor-load unit equal to zero, that is when 1 1 S 2 --I- S--}-~ = 0 Ye Te T M The roots of the quadratic are S "- - 2-'~e -+ 77.2 TeT M When rM > 4re, there is no sinusoidal component and the two poles lie along the horizontal axis of the s-plane. The physical interpretation of the two rates of exponential decay is that the transient response dies away relatively quickly at first, and then more slowly as time goes on. When rM < 4re the response is partly exponential and partly sinusoidal. The poles appear at s = rr 4-jw = 2re + j e~'M 4r 2 For example, when 7"M = "re, s = (--0.5 4-jv/3/2)/r~, i.e. at the 60 degree position. When rM = 4re both poles lie at 1 S -- O" -- 2r~ FIG. 12 shows how the poles move as the ratio JL/Jm rises to make the value of the overall mechanical time constant change from one-fifth to more than four times the value of the electrical time constant. For the example of the unloaded motor with the poles shown in FIG. 11, this would mean increasing the load inertia from zero to greater than 19Jm. Overshoot FIG. 13 shows an exponential and oscillatory variation with time of the speed of an initially stationary motor-load unit, following a step input of voltage. We know that 1 cr = and 2re TeT M 41-2 FIG. 12 Movement of poles as load inertia is increased The response is shown as per-unit of a final steady-state speed and has a maximum value during the first overshoot. The per- unit speed at T/2 is 1 + e -~r/2, or 0.,/m = 1 + e -Tr/2wre toss ,/3 When rM --Te W--2~re and therefore a3m = 1 + e -r/v/3 = 1.163 (.Uss 1 When TM - 2re w =~--- and ZTe = 1 +e -~- 1.043 ~SS This means that the speed overshoot shown in FIG. 13 increases from about 4% to 16% of a final steady, state speed as the pole position shown in FIG. 12 is changed from 45 degree to 60 degree As the pole angle increases, the frequency a; of the oscillatory component also increases and so the time in which ~m reaches the zone of the first overshoot shortens. Reducing the response time in this way is, however, at the expense of increasing both the overshoot and the risk of system instability. FIG. 13 Overshoot of motor speed System control We have studied the transient responses which are taken into account when the motor and load are incorporated into the system as a whole. In practice, the control system is designed to eliminate the voltage step-input poles of the motor-load unit. The motor current is controlled to give the motor and load a relatively rapid change of speed, with less than 5% overshoot. The appearance of a larger than designed overshoot and associated instability is often the result of a load inertia which differs from the value used in the design of the control system. The best results are achieved when the amplifier tuning parameters include the range of inertias of the load masses likely to be driven by a particular motor . 5. Optimization Understanding the dynamics of the motor and load is particularly important when an application involves incremental motion, where the load is required to move in discrete steps with a specific velocity profile. The steps can be in the form of an angle of rotation of a load driven through a direct or geared shaft coupling with the motor, or in the form of linear translation where the load is moved, for example, by a belt and pulley mechanism. FIG. 14 shows a handling system used in the manufacture of filters for the automobile industry which allows three-axis translation and also rotation of the loads. The need for rapid rotation or translation often means that a load must be accelerated and decelerated back to rest over a relatively short time. Two main factors are involved in the minimization of the stator i2R loss, and therefore of the required size and cost of a motor. We start by looking at the effect of the profile of the waveform of load velocity against time, and then go on to include the effect of the ratio between the moments of inertia of the motor and load. FIG. 14 Four-axis handling system. (Hauser division of Parker Hannifin) Load velocity profiles FIG. 15 shows a motor connected to a rotating load through a geared reducer. The inertias Jm and JL are assumed to include the inertias of the shaft and gear on the respective sides of the reducer. FIG. 16 shows the general trapezoidal load velocity profile, with unequal periods of acceleration, constant speed and deceleration. Here, we are using the trapezoidal term to describe a four-sided figure with two parallel sides. By adding together the angles of rotation during the three periods and equating to the total angle 0p, the constant speed is found to be Op tp[1 -- 0.5(pl +P2)] FIG. 15 Geared drive: ~m -- GWL For the symmetrical profile of FIG. 17(a), Pl - P2 - P, and the constant speed of the load is 0p ~0 c --- tp(1 -p) The rate of acceleration and deceleration of the load in the case of the symmetrical profile is d ~c -dt O'; L -- p t p Combining the last two equations above and writing WL as a;m/G gives the rate of acceleration and deceleration of the motor as d GOp dt wm- t~p(1 - p) FIG. 16 General trapezoidal, velocity profile Stator fiR loss In FIG. 15, the motor drives a rotating load through a reducer of ratio G. Torque TL is assumed to be constant. If the load velocity is to follow the symmetrical profile of FIG. 17, the motor torque required during the acceleration and deceleration periods is: T- KTi- jdwm + TLG The motor current during the same periods is therefore: J d TL i--~ ~~m -~-~ KT dt GKT or i-I~ +I2 ...where Ii is the component of motor current required for the constant acceleration or deceleration of the load and 12 is the component which provides a constant output torque. FIG. 17(b) shows the current waveform (rms in the case of the sinusoidal motor). Note that /1 is negative during deceleration. The total energy in joules produced in the form of heat by the iZR loss is: e = [p(I2 + II )2 _+. (1 - 2p)I 2 + p(I2 - I1 )2]Rtp (J) = (2pI 2 + I2)Rtp giving e_Rtp d / 2 2] Replacing d wm with the form already found in terms of G, 0p, lit tp and p gives the stator heating energy as P 2 where the profile constant is cp- p(1 -p)2" For a given motor, load, and reducer ratio G, the stator heating is at a minimum when d d 2 e-0, i.e. when -- dp dp p(1 _p)2 giving =0 1 P-~ The most efficient symmetrical profile is therefore equally distributed, as shown in FIG. 18. When p- 1/3, the profile constant is Cp - 13.5 The last expression for c above can be shown to apply when the load velocity follows the general trapezoidal profile of FIG. 16. The profile constant becomes: -1 p] + p~-i Cp -- [1 - 0.5(pl "[-p2)] 2 The constant has a value greater than 13.5 at all relevant values of p] and p2 other than p~ = P2 -1. Stator heating is therefore at a minimum when the trapezoidal profile of load velocity is equally distributed. FIG. 17 Motor current for the symmetrical, load velocity profile The second term of the last expression above for c does not depend on Cp, and profile optimization is unnecessary when the load torque requires most of the KTI product. Optimization of the velocity profile can be of benefit if the acceleration and deceleration of the load mass is responsible for a high proportion of the total stator i2R loss. Such cases are likely to arise when a substantial load mass is moved rapidly and repeatedly. FIG. 18 Optimum trapezoidal profile for incremental motion The inertia match for the geared drive When the speed of a load is changed many times per second, backlash in the transmission components can obviously cause problems. Gear reducers used for incremental motion must be of very high quality and are relatively expensive. A planetary reducer with a maximum backlash of 2 to 3 arc-minutes may have the same order of price as the drive motor itself. We have already seen how the load velocity profile affects the motor losses, and now follow on by including the effect of the ratio between the inertias of the motor and load masses. The inertial load In FIG. 15 the load is purely inertial when load torque TL is zero. Ignoring reducer losses, motor friction and viscous damping, the motor torque is given by: d T- KTi= J~t m where the sum of the motor inertia and the equivalent load inertia reflected at the motor side of the reducer is JL J=Jm+~ G 2 Writing a;m as GU;L, the motor current is therefore: JG d KT dt cOL The energy dissipated in the form of heat in the motor stator over the time dt is: de = flRdt Suppose that the load is to move a complete step in a time interval of tp over any load velocity profile. Using the last three equations above and integrating gives the energy dissipated in the stator winding as tp R,2o2 / ,. ]2 e -- z-------F-- WL dt T 0 Varying the reducer ratio minimizes the stator heating energy when d d(G) 2 ~e--0 The integration term in the above expression for e depends on the velocity profile of the load during time tp, but the profile itself does not depend on the gear ratio. Stator heating is therefore at a minimum for any particular profile when d G2(j m + GJ_~) d(G) 2 2_0, i.e. when Go- ~jJ~L m Go is the gear ratio which minimizes the motor stator heating during incremental rotation of a purely inertial load, for any velocity profile. Note that although Go is independent of the velocity profile, the same is not true of the heating energy e. However, for any particular combination of motor, inertial load and velocity profile, the energy is a function of the gear ratio alone. For reducer ratios G and Go, the energies are JL JL~2 e -- 7GZ(Jm + ~_~)2 and co - 7G02(Jm + where 7 is a constant. Dividing e by e0 and rearranging gives the extra heating factor for a non-optimum gear ratio as: FIG. 19 Increase in stator heating when G # Go(7L = 0) Ref.  develops the above expression and plots the energy ratio as the gear ratio falls in comparison to the optimum value. The curve has the same shape when it is plotted for rising values of gear ratio in comparison to the optimum. FIG. 19 shows how the fiR loss increases as the gear ratio deviates from the ideal. As a rule of thumb: G --<1.14 when 0.7<~-<1.4 E0 ~,0 Note that the stator heating doubles when the gear ratio is more than twice or less than half the optimum value. Effect of load torque on the optimum gear ratio Let us now return to the case where the load is due partly to the inertias of the motor rotor and the load, and partly to a constant torque which is delivered over a specific velocity profile. The stator heating energy when the load velocity follows the general trapezoidal profile of FIG. 16 has been shown to be + Writing J in terms of JL, Jm and G as before and rearranging gives the heating energy as / P L G 2 Jm s -- r K2 t3 -~L X P +~-5 + where AI--- r OpJL Differentiating the last expression for e above with respect to G 2 and equating to zero gives the gear ratio for minimum stator heating as G~ - 6;o(1 + Z~) ~ Gz is the gear ratio which minimizes stator heating when the load consists of an inertial mass and an output torque, for any trapezoidal velocity profile. We have optimized the ratio of a reducer for any trapezoidal velocity profile, and we found earlier that the most efficient shape is equally distributed. The trapezoidal profile has the advantage of ease of control by the drive. It is not, however, the most efficient velocity profile in general. The lowest losses of all occur when the profile of load velocity against time is parabolic . Example 1 A sinusoidal motor is connected to a load through a geared reducer The load is to be rotated incrementally, using an equally distributed, trapezoidal velocity profile. The motor details are given in Table 4.1. The system constants are as follows: Jm = 0.00022 kgm 2 Op -" 2 rad tp = O.06 s Cp = 13.5 TL= lONm JL = 0.0022 kgm 2 If there were no opposing load torque, A1 would equal zero and Go = ~j~ 3.2 For the constant opposing torque of 10 Nm, giving , A1 ---- -- 5.0 Cp OpJL GA -- 3.2(1 + 5.0) ~ -- 5.0 The total stator heating energy is formed from two components. One arises from the effort of rotating the load mass from one stationary position to another, increasing as already shown in FIG. 19 as G is changed from Go. The second component is generated when the load mass is subjected to the opposing torque TL. The motor output torque required (for a constant TL) falls as G rises, together with the associated stator loss. When G is changed from 3.2 to 5.0 in the present example, the extra loss due to the mismatch of motor to load inertia is compensated by the fall in the loss associated with the supply of TL. One way to study the relative importance of the two components of the stator heating energy is to calculate the values as the step time tp is varied, all other parameters (including G) remaining constant. FIG. 20 shows the variation in the two energies in the present example when G-5.0. The motor torque required for load translation rises as tp falls, generating rapidly rising i2R losses below tp -- 100 ms. On the other hand, the stator heating due to the supply of TL becomes dominant as tp is increased. A relatively large increase in G would then be needed if the overall losses are to be minimized at the higher values of tp. FIG. 20 Relative stator heating due to (a) load translation and (b) load output torque. Limitations to minimization of the i2R loss Minimization of the i2R loss is desirable when a significant part of the stator heating is due to the effect of load and motor inertia on the motor torque requirement. Minimization may still be worthwhile when the output load torque forms a relatively high proportion of the total torque, but the optimum gear ratio (and therefore the motor speed) rises with the output torque. A non-optimum solution is often imposed by practical limitations on the speed of the motor and reducer input, and by reducer losses at the higher speeds. The belt and pulley drive Belt drives are often used in pick-and-place robotics, for example in the loading of electronic components onto printed circuit boards. FIG. 21 shows a load attached to a belt and pulley which may be driven directly by the motor, or through a reducer of ratio G. The reducer may be in the form of a gear train, or another belt and pulley system. Mass m is the total mass of the load and the load conveyor belt. If present, force F is assumed constant over the step of length x. FIG. 21 Belt and pulley drive The system can be optimized at the reducer or at the drive pulley. If optimization is to take place at the reducer, the optimum ratio is: 1 (TLt2) 2 G6 - JL (1 + A 1)0"25 where A 1 -- -- r OpJL From the diagram, 0p = x/r TL =Fr JL = mr 2 The optimum ratio of a reducer for the belt and pulley drive is therefore: i GA -- r (1%- A2) where A 2 ---- r k xm ) When the use of an existing reducer or a direct drive is convenient, the belt and pulley mechanism may be optimized by selecting the correct radius of the drive pulley. Take first the case where there is little or no opposing force F, making A 2 = 0. Writing r 0 as the pulley radius which makes the ratio G' of the existing reducer equal to the optimum value gives: G' = r 0 ~m The optimum pulley radius when a reducer of nominal ratio G' is already in place is therefore: r o - ....where G' = 1 when the motor is coupled directly to the pulley. r 0 is the optimum radius of the belt drive pulley when there is no opposing load force, for any velocity profile. Working in the same way for the case where F :/: 0 gives rzx - ff v/1 + A2 rA is the optimum pulley radius when the load includes an opposing force, for any trapezoidal velocity profile. The inertias of the drive pulley and reducer have so far been ignored. Where the motor shaft is connected directly to the drive pulley and optimization takes place according to the drive pulley radius, the inertia of the drive pulley should be added to the motor inertia and a correction made to the optimum pulley radius. Where optimization takes place according to the ratio GA of a reducer, Ga should be corrected by adding to mass m the mass which would have the same inertia as that of the reducer output plus drive pulley, if distributed at the drive pulley circumference. The inertia of the reducer input should be added to Jm. Example 2 A load is to be belt driven and moved incrementally The force opposing the load is negligible. The following are already available: (a) The motor of the previous example. (b) A reducer of ratio G'= 3.5. (c) A belt and pulley drive. The system constants are: Jm = 0.00022 kgm 2 Mass of belt and load m = 1.0 kg Existing drive pulley radius r = 0.03 m The load force is negligible in this case, and so the load velocity profile, the step time tp and the step distance x have no effect on the optimization. The optimum pulley radius is r o - G'V~ Two or three mechanical configurations are possible: 1. No reducer and a drive pulley radius of: _ ~/0.00022 r~ V i~ -0.015m This radius is close to the minimum for toothed drive belts. 2. Use the existing reducer and a new drive pulley with a radius of ~/0.00022 r 0 - 3.5 1.0 = 0.052 m 3. Use the existing drive pulley and change the ratio of the reducer to ~/0 1"0 Go - 0.03 .00022 = 2.0 Finally, a correction should be made by taking into account the inertias of the prospective pulley or reducer. The ball screw and lead screw drives FIG. 22 shows the principle of the ball screw and lead screw mechanisms, the screw may be driven by the motor, or through a belt and pulley or geared reducer. Screw mechanisms can provide high forces at the load, and also allow accurate load positioning. FIG. 23 shows a screw drive being used to control the vertical position of a two-axis handling machine used in the internal coating of glass medical bottles. In this machine, the horizontal position is controlled by a belt and pulley drive. The primary and secondary threads of a ball screw drive are connected through ball bearings. This results in low backlash and a resolution at the load which can be better than 1 micron. The transmission of energy is very efficient, the losses being low enough to allow the device to be 'backdriven' from the load side. The disadvantage of this is that the ball screw may be back-driven by the load if the motor torque is lost, say through a power failure. Consequently, the ball screw driving motor must be fitted with a brake. In the case of the lead screw, the threads are in direct contact. This results in a transmission which is inefficient in comparison to the ball screw, but which does not backdrive following the loss of motor torque and is relatively inexpensive. The lead screw inevitably has the backlash associated with screw threads. FIG. 22 The screw drive The load mass m travels distance x in time tp, for both the belt and pulley drive of FIG. 21 and the screw drive of FIG. 22. The load inertia is mr 2 for both (in the case of the screw, imagine that the load is travelling around a stationary screw). The screw driven load travels distance d for one revolution of the screw, and so the screw drive has a 'gear ratio' of 27rr/d. The inertia of the load mass reflected on the motor side of the screw threads is therefore: d )2 md 2 J1 - mr2 ~r 471-2 The reflected inertia of the load mass should be matched with the inertia which already exists on the motor side of the screw threads, and so J~ = Jm + Jsw where Jsw is the inertia of the screw. Combining the last two equations and including the effect of force F gives the optimum screw pitch as / 1m +]~w where Az--u n Cp xm dzx is the optimum screw pitch when the load follows any trapezoidal velocity profile and is subject to an opposing force. FIG. 23 Two-axis, pick-and-place handling machine. (Hauser division of Parker Hannifin) Example 3 The sinusoidal motor in Table 1 is to be used to drive a ball screw. The load velocity is to follow the profile shown in FIG. 24. The system constants are Jm = 0.00022 kgm 2 Jsw - 0.00003 kgm 2 x - 0.025 m t? = 0.120 s F- 1000N m=5kg pql+p2-1 The profile constant is Cp = [1 - 0.5(p, +p2)] 2" In this case Pl - 20/120 and p2 -60/120, giving Cp load force factor is ~8 (1000x0"1221 2 A2 "- 0.025 x 5 - 737 - 18. The The optimum screw pitch is 2zr /0.00022 + 0.00003 dA -- V 5q-1 + 737 = 8.5 mm In this example, the optimum pitch is dictated mainly by the effect of the load force. In the hypothetical case of zero load force, the optimum screw pitch given by the calculation would be do - 44 mm! FIG. 24 Load velocity profile for example 4.3 6. Torsional resonance In Sub-Sections 4 and 5 the mechanical connection between the motor and load is assumed to be inelastic, leaving the increase in system inertia as the only mechanical effect of an added load. In practice some flexibility in the connection is unavoidable, and an error may develop in the position of the load relative to that of the hub of the rotor of the motor as torsional forces come into play. When the error becomes oscillatory, the condition is known as torsional resonance. The problem can also arise in the section of shaft between the hub and the sensor, but we will assume throughout that any such effects have been eliminated through the design of the motor and sensor. Under these circumstances, the error between the sensor and the load can be assumed to be the same as the error between the hub and the load. FIG. 25 shows a rotor of inertia Jm connected to a rotating load through a shaft which is subject to twist. Following the approach used in Section 4.4 for the totally rigid shaft shows that the poles of the transfer function of motor speed response are given by JL[\$4TeTm -~- S3Tm -]- S 2] -1- C -1 [S2TeTM Jr- STM -q- 1] -- 0 (1) where TM--Tm(Jm-~-JL)/Jm. If JL = 0, TM = Tm ...and the equation reduces to the form already derived for an unloaded motor: S 2 TeTm -1"- STm -[-- 1 - 0 C is the compliance of the mechanical connection, or error factor for the angle between the motor position sensor and the driven load, normally expressed in microradians/Nm. The compliance varies widely according to the length and diameter of the drive shaft and the types of transmission between the motor and the load. Typical values are in the range 10-100 #rad/Nm. Solution of expression (1) gives the theoretical locations of the poles for the case where the damping effects of eddy currents and friction are ignored. There are two pairs of poles, one pair at low frequency and the other at the frequency of potential torsional resonance. FIG. 25 Shaft compliance Effect of compliance at a fixed load inertia The resonant frequency is affected by the ratio of the motor to load inertias, and also by the value of compliance. We start by looking at the way the resonant frequency varies as the compliance is changed, for a case where the inertias are approximately equal. Example 4 A rotating load is connected to the shaft of a brushless servomotor. The motor and load inertias are approximately equal. Torsional resonance frequency values are required for a wide range of shaft compliance. The system constants are Jm = 0.00215 kgm 2 JL = 0.00200 kgm 2 Te = 5.0 ms Tm = 2.6 ms TM-- 5.0ms Inserting the numerical values in equation (1) above and dividing through by JLTeTm gives S 4 + 200s 3 + (962C -1 + 77 • 103)s 2 + 192 • 103C-Is + 38 • 106C -1 = 0 FIG. 26 shows how the poles move as C is varied from infinity to 10 #rad/Nm. The physical interpretation of infinite compliance is of course that the load is disconnected from the motor, at which point the last expression is reduced to: S 4 + 200s 3 + 77 • 103S 2 = 0 The motor and disconnected load therefore have four poles, two showing the normal response of an unloaded motor to a step voltage input. The other two poles remain at the origin as long as the load stays unconnected. The arrows show the shift in position of the four poles as the compliance is reduced from infinity, or in other words as the stiffness of the transmission is increased from zero. As the compliance is reduced, the motor-load poles move from the position (at the origin) for a disconnected transmission towards the positions P1, P2 for a totally rigid connection. The poles at positions P3, P4 for the normal response of the unloaded motor rise in frequency but become increasingly oscillatory as the compliance falls, taking up relatively undamped positions close to the boundary between stable and unstable operation of the system. In practice it is found that the lower the frequency of such lightly damped responses, the more likely it becomes for the frequency to be excited by the system in general and for system instability to occur. FIG. 26 Pole loci as compliance is reduced Resonant frequency predictions and tests The low frequency poles in FIG. 26 are relatively well damped and are in any case eliminated through the design of the control system. The other pair travel towards infinity as the compliance falls towards zero. In the present case the resonant frequency is predicted to lie between approximately 1550 and 1100 Hz for compliance values from 10 to 20 #rad/Nm. The motor and load of Example 4 were connected together. The compliance of the length of shaft between the front end of the hub of the rotor and the load was calculated to be C- 14.4 #rad/Nm. The resonant frequency was excited by striking the load, and measured by recording the stator emf produced by the resulting oscillations of the rotor. The result is shown in FIG. 27(a). The low frequency envelope is due to the slow rotation of the rotor after the shaft has been struck. The resonant frequency is close to 1305 Hz, and this compares well with the value of 1300 Hz predicted by equation (1) at the compliance of 14.4 #rad/Nm. As the resonant frequency rises, it becomes less likely to be excited through a well-designed drive system. The resonant frequency rises as the compliance falls, and so the main conclusion is that the compliance should be as low as possible for maximum system stability. Damping Expression (1) automatically includes the damping effects of the i2R loss generated in the stator as the rotor oscillates, but these are negligible. The pole loci in FIG. 26 do not take account of damping due to frictional and eddy current losses. When expression (1) is modified to include the viscous damping due to eddy currents, the effect is predicted to be insignificant in the test motor. The test results do not include the effects of any i2R loss in the stator as measurements must be made with the winding on open-circuit. Damping of the motor under such test conditions is therefore the result of eddy current loss and losses at the bearings, with the bearing loss likely to be the greater part. The time constant of the decay in FIG. 27(a) is approximately 33 ms. The time constant affecting the rate of decay of the oscillations is seen to be high, when the load is mainly inertial. The rate of decay is of course increased in practice when the driven load is subject to friction, and also when damping appears in a transmission mechanism such as a belt and pulley drive. As [....] The effect of load inertia In Example 4, the variation of the motor-load pole positions over a range of shaft compliance was plotted for the case where the motor and load inertias are approximately equal. In many applications a close match of inertias for the purpose of minimization of the i2R loss may be unnecessary or impracticable, and so in practice the load inertia may be several times that of the motor. In order to study the effect of load inertia on the resonant frequency, equation (1) may be rearranged as: JrJm[S4TeTm + S3Tm + \$2] "+" C-I [S2TeTm(1 "+" Jr) -+-STm(1 -t- Jr) + 1] = 0 (2) where Jr--JL/Jm. We can now forecast the resonant frequencies as Jr varies, for a fixed value of compliance. Example 5 Values of the resonant frequency of a motor-load combination are required for load to motor inertia ratios from 1.0 to 10. The motor constants are Jm = 0.000315 kgm 2 T e --- 2.3 ms T m -- 2.8 ms C = 67.5 #rad/Nm FIG. 28 shows the predicted variation in the resonant frequency when the above values are used in expression (2). As Jr varies from 1 to 10, the resonant frequency falls from 1540 to 1145 Hz. Note that the position on the curve for matched inertias is not in any way a special point, and that the matched inertia case has no significance as far as resonance is concerned. In this particular example, the frequency is predicted to fall by 25% as the inertia ratio rises from 1 to 10. The fall is rapid at first and then levels off, and most occurs for a ratio of only 4:1. FIG. 28 Fall in resonant frequency with increasing load inertia Predictions and tests A rotating load with an inertia of 0.002 kgm 2 was fitted to the shaft of the motor of Example 4.5. This gave a ratio of load to motor inertia of Jr = 6.35 The resonant frequency and decay were measured by the method used for the previous example. The results are shown in FIG. 27(b). The resonant frequency is 1167 Hz, which again compares well with the predicted value of 1175 Hz. Damping of the oscillations is again low, the time constant of decay being approximately 20 ms. Compliance and inertia in practice We have looked at the effects of the compliance of the motor shaft. In practice there may be several other points in the transmission mechanism which either add to the compliance problem, or help by providing damping. In the example of the ball screw drive, additional compliance occurs at the coupling between the motor shaft and the screw input shaft, and also along the complete length of the input shaft and screw. The screw itself makes a further contribution to the difficulties by having a compliance which increases as the load moves away from the motor. Damping is added by losses at the screw input bearing and the point of screw contact with the load. Such systems are difficult to analyze by the method used above for the simple case where the load is connected directly to the end of the motor shaft, and modeling using electrical circuit analogues can be a better approach . In general the resonant frequency falls as the compliance increases, and also as the moment of inertia of the load increases in relation to that of the motor. Assuming as much as possible has been done to reduce the compliance, the next step is to reduce the inertia ratio. This can be done by lowering the effective value of the load inertia by means of a reducer, but this method may be prevented by economic and practical considerations. Where this is the case, the best way forward may be to reduce the inertia ratio by increasing the inertia at the motor end of the mechanical drive link. An increase in motor inertia can be made in two ways. One method has been to introduce the required extra inertia by fitting an oversize motor, but this is normally an expensive solution. The other way is to fit a motor of the required size, torque rating and price which has been designed around an increased rotor inertia. Such motors are available over a wide range of servomotor ratings, and offer higher stability for systems with a relatively high load inertia which cannot be reduced.
Top of Page

 PREV NEXT Article Index HOME