Chapter 7: Stepper Motors. (Revision 6.0, 27/10/2014)

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1 Chapter 7 Stepper Motors (Revision 6.0, 7/10/014) 1. Stepping Angle Analysis The following analysis derives the formula for the stepping angle of the stepper motor. It has been reproduced and edited from Clarence de Silva s Mechatronics: An Integrated Approach. Assume that: n s is the number of stator poles n r is the number of rotor poles n is the number of steps per revolution θs is the stator angle θr is the rotor angle Δθ is the stepping angle r is the largest feasible integer p is the number of phases of the stator θs = 360 ns. (1) θr = 360 nr. () The number of steps per revolution is related to the stepping angle as follows: 360 θ = n. (3) Note that when the motor has been given a number of pulses equal to the number of phases of the stator, it advances by the angle of the rotor: θ = θ r p. (4) Case 1: Assume that the number of stator poles is more than the number of rotor poles (i.e., In this case we can find the stepping angle as follows: θ = θ r rθ s. (5) Case : Assume that the number of stator poles is less than the number of rotor poles (i.e., In this case we can find the stepping angle as follows: θ = θ s rθ r. (6) The value of r is selected such that it is the largest possible integer that keeps the stepping angle positive. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 1 of 15

2 . Example on calculation of the stepping angle for various stepper motors. The following two examples illustrate the use of the relevant equations to arrive at the stepping angle of the stepper motor. The stepper motor that is used in the lab has the following parameters: 8 stator poles (thus the stator angle is 45 degrees). 50 rotor poles (thus the rotor angle is 7. degrees). So using the value of r as 6 gives a stepping angle 1.8 degrees. The stepper motor used in the example shown in the notes has the following parameters: 1 stator poles. 3 phases 4 stator poles per phase 8 rotor poles. Note that the number of phases has no effect on the stepping angle for the motor. In the case the rotor angle will be larger than the stator angle. The stator angle is 30 degrees. The rotor angle is 45 degrees. So the largest value of the integer r is 1, giving a stepping angle of 15 degrees. It is worth noting that not all rotor and stator arrangement are physically realisable. For example, some soft iron rotors can rotate but in an undefined direction (i.e., for a specific sequence of excitation pulses, the motor might rotate in a clockwise or anticlockwise direction). If already travelling in one direction, the inertia will force it to carry on moving in that direction. But if stationery it might run in either direction. 3. Stator poles and phases The number of phases of the stator will not necessarily be equal to the number of stator poles. For example a motor might have 3 phases and 4 poles per phase resulting in 1 stator poles. The number of stator phases does not affect the stepping angle. The stepping angle only depends on the number of rotor poles and the number of stator poles. 4. Unipolar and bipolar; unifilar and bifilar Coils can either be unipolar or bipolar. Unipolar windings can have current only flowing in one direction. Unipolar coils usually have the common wire of all the coils connected together to the ground. Usually in a motor with a permanent magnet rotor, there is a requirement to reverse the current in the stator pole in order to produce the required pole in order to attract the nearby permanent magnet rotor pole. This cannot be done with a unipolar stator pole. In such a case, two coils can be wound around the same stator pole and it is called in this case bifilar windings. Where the polarity of the stator pole has to be reversed, one of the two coils is switched off and the other switched on. Bipolar windings can have current flowing in both directions. Each coil has two wires coming out to allow full connection flexibility. In this case, the polarity of the stator pole can be reversed by reversing the polarity of the current in the coil (this can be either done by having two power supplies, or the use a switching Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page of 15

3 arrangement that is referred to as an H bridge, because it comprises a four transistors used as switches in an H shape arrangement). The use of bifilar windings makes the size of the coils smaller for the same size of stator pole compared to the unifilar windings. This has an effect of increasing the torque from the unifilar motor compared to the bifilar motor. However, it takes longer to reverse the direction of current in a unifilar coil and a higher stepping speed can be achieved with a bifilar windings motor compared to a unifilar windings motor. So to summarise, where reversal of magnetic polarity is required in the same stator pole, there are two options to use: Unifilar bipolar windings Bifilar unipolar windings 5. Full stepping, half stepping and micro-stepping By applying full voltage to one phase of the motor, full stepping is achieved. The motor will advance by the stepping angle (derived in the equations discussed earlier using the number of stator poles and the number of rotor poles). By applying equal voltage to adjacent phases, half stepping can be achieved. By applying fractions of voltage to adjacent phases, micro-stepping is achieved. Micro-stepping can be applied in very small steps down to around 15 th of the full step. However, it is important to note that the ratio of voltages or currents applied to adjacent phases is not necessarily linear. 6. Stepper motor sizing example This example is used to illustrate the method of sizing of a stepper motor for a conveyor belt application. It is based on the example 8.9 of Clarence de Silva s book Mechatronics: An integrated approach. 6.1 The Problem A conveyor belt that moves objects is to be powered using a stepper motor. It has a gearing ratio of r g and a diameter of the driving pulley of d s. The stepper motor has an inertia I m, the gearbox has a high speed shaft inertia of I g1 and a low speed shaft inertia I g. The conveyor belt pulleys have inertias of I d and I s. The conveyor belt has a mass of m C and the objects on it have a total mass of m L. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 3 of 15

4 Objects (m L ) Ø=d s I d Conveyor (m C ) I s Stepper motor I g1 I g Gearbox I m Figure 1: Overview of the conveyor system. The conveyor has to be advanced by a fixed distance d regularly in a specified period of time T. In order to achieve this in the specified time, the conveyor has to be accelerated at a constant acceleration until it covers half the distance d and then decelerated at a constant deceleration until it comes to rests within half the distance d as well. The values of acceleration and deceleration are equal. Speed Distance d T Time Figure : Speed-time profile for conveyor. High positioning accuracy is required in this application, as the objects need to be positioned ready for the next operation (e.g., a manufactured items that need further processing, either automatic or manual). A stepper motor is ideal for this application, as it allows good positioning accuracy. The following are the parameters of the conveyor: Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 4 of 15

5 Parameter Value Unit Description I m see table below kg m Inertia of the rotor of the stepper motor I g1 50 μkg m Inertia of the high speed shaft of the gearbox I g 00 μkg m Inertia of the low speed shaft of the gearbox I d.0 mkg m Inertia of the driving pulley of the conveyor I s.0 mkg m Inertia of the idler pulley of the conveyor m C 5.0 kg Mass of the conveyor belt m L 5.0 kg Mass of the objects on the conveyor d s 0. m Diameter of the driving pulley of the conveyor d 0.1 m Distance to be advanced r g 1: - Gear reduction ratio T 0. s Time during which to advance η 80% - Overall system efficiency 6. Stepper motor models The selection of the stepper motor is to be based on the table shown below. It shows 4 motor models with each motor s available torque (holding torque) and its motor inertia. Motor rotor inertia (kg m ) Motor Model Available Torque (N m) at ω max 50 SM x SM x SM x SM x Calculation We need to first calculate the value of the required acceleration. Note that in this case there is no out of balance torque as the conveyor is horizontal 1. Referring to Figure the top attained speed will be: (d/)=((t/) v max )/ (0.1/)=(0./) (v max /) v max =1 m s -1 From this we can deduce that the acceleration is: v max =a (T/) a=10 m s - So the stepper motor to be selected must be capable of accelerating the conveyor with its load at this acceleration. We shall refer all loads to the high speed shaft (i.e. the motor shaft). So first we calculate the required value of angular acceleration at the high speed shaft: 1 Had the conveyor been inclined, then there would be an out of balance torque that the motor has to overcome first. The value of the out of balance torque would be m L g (d s /) (sinθ) where θ is the angle of incline to the horizontal. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 5 of 15

6 α=(a/(d s /)) r g = (10/0.1)=00 rad s - We next need to calculate the equivalent inertia of the whole system at the high speed shaft. I eq I = m I = I = m + I m g1 ds I g + Id + I s + + rg rg η η ( m + m ) C ( 10) Im = L This physical significance of this equivalent inertia is that we could have replaced the whole installation at the motor shaft with one flywheel that has the equivalent inertia and the motor would have not seen any difference. The general equation for finding the required torque is: THSS = α HSS I HSS Let us assume we select the motor model number 310 SM, that has a rotor inertia of 187x10-6 kg m. The required torque would be: THSS = Tm = = N But this value is more that what this motor can provide. So we need to move to the next motor up, model 1010 SM, that has an inertia of 805x10-6 kg m. So the torque required in this case would be: THSS = Tm = = N..which is less than what the motor can provide, so this motor is suitable for this application. 6 6 m m Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 6 of 15

7 It is important to ensure that the selected motor has a torque higher than that required to accelerate the load at the required acceleration value. However, this does not mean that when installed and operated it will accelerate the load at the higher value. The control of the speed and acceleration in the case of a stepper motor is set by the speed of the pulses driving the stepper motor by the drive circuit. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 7 of 15

8 Problems 1. For the stepper motor shown below, answer the following questions: What type of stepper motor is this? Derive the stepping angle in full stepping mode. Assume one-phase-on excitation (i.e., only one coil is on at any one point in time).. Draw the stepper motor with the following parameters and comment on the stepping angle, type of motor needed and the realisability of the design: Two phases, stator poles, 5 rotor poles. 3. A stepper motor is to be selected to drive the inclined conveyor system shown in Figure 3 below. The stepper motor has to advance the system from standstill a specified distance, d, in the up direction, during a specific period of time, T. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 8 of 15

9 Objects (m L ) Ø=d s I d θ=30º I s Stepper motor I g1 I g Gearbox Conveyor (m C ) I m Figure 3: Overview of the conveyor system. The parameters of the conveyor system are shown in the table below. Parameter Value Unit Description I m see table below kg m Inertia of the rotor of the stepper motor I g1 50 μkg m Inertia of the high speed shaft of the gearbox I g 00 μkg m Inertia of the low speed shaft of the gearbox I d.0 mkg m Inertia of the driving pulley of the conveyor I s.0 mkg m Inertia of the idler pulley of the conveyor m C 5.0 kg Mass of the conveyor belt m L 5.0 kg Mass of the objects on the conveyor d s 0. m Diameter of the driving pulley of the conveyor d 0.1 m Distance to be advanced r g 1: - Gear reduction ratio T 0. s Time during which to advance η 80% - Overall system efficiency θ 30 degree Angle of incline of the conveyor belt The specifications of the stepper motors from Sanyo Denki are shown in the table below. Sanyo Denki: PMM-BA Dimensions (mm) ID Model # Rated current Holding torque Rotor inertia Mass (kg) (A/phase) (N m) (μkg m ) A 103H B 103H C 103H D 103H E 103H F 103H G 103H H 103H Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 9 of 15

10 4. Stepper Motor Selection for a Rotating Plate You are required to select a suitable stepper motor from the list of four stepper motors shown at the end of this exam paper, for the following system. The stepper motor will be used to rotate a steel board that has the following dimensions: 600 mm long. 30 mm wide 10 mm thick The stepper motor will be directly coupled to the axis to which the board is fixed. The stepper motor has to rotate the board at a constant speed of 600 rpm. Show all your calculations and assumptions in detail. Assume an overall system efficiency of 97%. Use the following equation of drag force if needed: F d = 1 C d ρ A v Where: F d is the drag force in N v is the speed in m s -1 of the mid-point of the board (mid-point: halfway between the axis and the edge of the board) C d is the drag coefficient (0.3 in this case) ρ is the density of air in kg m -3 A is the area of the board in m Use the following parameters if needed: g=9.81 m s - Density of steel is 7800 kg m -3 Density of air is 1. kg m -3 Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 10 of 15

11 600 mm 30 mm 10 mm Ø: Ignore the diameter of this shaft Axis to which the stepper motor will be directly coupled Figure 4: Perspective view of the rotating board system. Ignore the diameter of this shaft 600 mm Axis to which the stepper motor will be directly coupled 10 mm Figure 5: Side view of the rotating board system. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 11 of 15

12 Stepper Motor Datasheets Motor number 1 (use the pull-out torque curve not the starting torque curve) Motor Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 1 of 15

13 Motor 3 Solution We first need to identify the worst case position for the motor. As the plate is being moved at constant speed, there is no acceleration torque. The main two torques of interest are: The out of balance torque. The drag torque on the plate by the air resistance. The maximum value of the out of balance torque occurs at the horizontal position where the plate is being lifted. To calculate this out of balance torque, we first find the mass of the plate. m = = kg The out of balance torque can be calculated as follows: l o = m g = = T / b 4.13 N Next we need to calculate the drag force. The rotational speed of the plate is 600 rpm. This equates to a rotational speed of 6.83 rad s -1. The formula for calculating the drag force is: F d = 1 C d ρ A v Applying this to a small strip of the plate of thickness dx gives: Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 13 of 15

14 dt d 1 = 0.3 ρ w dx v x = ( 6.83 x) x dx = 1.3 x dx Integrating along the length of the plate gives: 4 l 3 x Td = 1.3 x dx = 1.3 = N m Note that assuming that the drag force acts at the centre of the plate would lead to a lower value of the drag torque (0.345 N m). The total torque is taking into consideration the overall system efficiency: 0.6 T tot = ( ) 0.97 = 4.97 N m So we need to select a motor that can deliver this torque at a speed of 600 rpm. The speed torque curve for the first motor is shown below. There is no clear mark against the speed of 600 rpm. However, if we carefully examine the figure we can see that the speed of 3000 rpm corresponds to a pulse rate of 10 kpulses per second. Thus a speed of 3000 rpm would correspond to a pulse rate of kpulses per second. These have been marked with a yellow line. From the curve it can be seen that the torque from the motor would be around 7 N m which is sufficient for the required application. A quick examination of the other two curves for the other two motors shows that they cannot produce enough torque. Motor number 1 (use the pull-out torque curve not the starting torque curve) Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 14 of 15

15 5. Additional problem Following on from the last problem (4) repeat the motor selection calculations under the following requirements: The plate needs to be raised from the horizontal position to the fully vertical position within 0.5 seconds (90 degrees). The wind speed could be as high as 150 km per hour in any direction. Copyright held by the author 01: Dr. Lutfi R. Al-Sharif Page 15 of 15

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