A bearingless permanent-magnet sheet motor rotor displacement measurement method in a hall sensor fault state

Abstract

The application discloses a kind of bearingless permanent-magnet slice motor rotor displacement measurement methods under Hall sensor fault state, when one of Hall sensor fails, first deduce the output signal of Hall1-Hall6, the method for constructing fault Hall output signal is used to indirectly obtain the output signal of fault Hall;Three Hall output signals of electric angle difference 120 ° are not eccentric permanent-magnet flux linkage item total offset, and the eccentric permanent-magnet flux linkage item of the output signal of fault Hall is obtained by the remaining non-fault Hall of the electric angle difference 120 ° with fault Hall;Again, the output signal of the remaining non-fault Hall can be obtained by equation S1, and the output signal of the remaining non-fault Hall is multiplied by the vector amplitude perpendicular to armature reaction vector after being added, to obtain the projection of displacement in another time-varying angle, as equation S2;The equation set is associated, and the radial displacement x and y under the starting state of rotor are obtained;The application can accurately estimate rotor displacement information under the condition that motor Hall fails.

Description

Technical Field

[0001] This invention relates to the field of motor control technology, and mainly to a method for measuring the rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions. Background Technology

[0002] Bearingless motors eliminate the limitations of mechanical shafts, making them widely applicable in high-speed, high-cleanliness environments. In clean medical devices, the requirements for motor levitation performance, stability, and size are extremely high. Traditional bearingless control often uses eddy current sensors to sample displacement signals to achieve closed-loop control, but eddy current sensors are large and complex. Therefore, some researchers have used small, low-cost Hall effect sensors as a replacement. However, Hall effect sensors are inherently fragile. Due to vibration, humidity, and electromagnetic interference, Hall effect devices are prone to damage and failure under prolonged high-load operation. Once a Hall effect sensor fails, the motor rotor cannot achieve levitation, generating large abnormal currents that can cause greater damage to the bearingless system and lead to safety accidents. Therefore, research is needed on a Hall effect fault-tolerant displacement-free method for bearingless permanent magnet thin-film motors.

[0003] Since bearingless thin-plate motors inherently require Hall sensors to measure angles, the patent "Rotor Displacement Identification Method in Start-up State of Bearingless Permanent Magnet Thin-plate Motor" (application number: 202111623440.9) proposes to integrate Hall sensors for dual purposes, simultaneously measuring rotor displacement and angle. This solution can be used for displacement identification in the start-up state, but it does not consider the possibility of Hall sensor failure. When one of the Hall sensors fails, the Hall output signal decreases, and the original process of decoupling the angle and displacement from the Hall signal cannot be used, making it impossible to obtain the displacement in the start-up state, and thus the bearingless motor cannot achieve levitation.

[0004] The paper "Wang Xiaolin et al., Analysis of Fault Operation Characteristics of Multiphase Bearingless Permanent Magnet Thin-Layer Motor, Proceedings of the CSEE, 2011, 31(18): 73-78" reconstructed the stator currents of other non-faulty phases considering the influence of the faulty phase on the levitation force and torque. This satisfied the levitation force and torque required for stable operation of the motor under fault conditions, realizing fault-tolerant control of the motor under short-circuit and open-circuit fault states. However, this paper's research was based on the normal function of the sensors, studying the fault-tolerant control of short-circuit and open-circuit faults in the motor windings. However, compared with the motor windings, Hall sensors are more fragile. Due to the interference of static electricity on the internal components of the Hall sensor, the service life of the Hall sensor is greatly shortened, and it is prone to failure.

[0005] like Figure 2The diagram shown is a flowchart of the rotor displacement estimation method proposed in patent 2021116234409. Considering that summing two relative Hall signals can eliminate permanent magnet flux interference, the summation of the two relative Hall signals yields three sets of summed data, which are then summed again to obtain one of the equations. The three summed data are multiplied by the vector amplitude perpendicular to the armature reaction vector and then added together to obtain another equation, from which the displacement information is calculated. Figure 3 The given and estimated displacement results when the Hall sensor is working normally show a very high degree of overlap. However, when the Hall sensor malfunctions, the overlap between the given and estimated displacements is very low under the same algorithm conditions, as shown in the following figure. Figure 4 As shown, such a degree of overlap cannot be used to calculate displacement information. Summary of the Invention

[0006] Purpose of the invention: To address the problems existing in the background technology, the present invention provides a method for measuring the rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions. The method indirectly obtains the output signal of the fault Hall by constructing a fault Hall output signal, and redesigns the displacement and angle decoupling process to bypass the fault Hall sensor and re-estimate the rotor displacement information when the Hall fails. This solves the problem that the motor cannot levitate under Hall failure conditions in existing Hall displacement-free technology.

[0007] Technical solution: To achieve the above objectives, the technical solution adopted by this invention is as follows:

[0008] A method for measuring rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions is disclosed. The method employs a six-tooth, one-pole bearingless permanent magnet thin-film motor, comprising six L-shaped stators. Each L-shaped stator includes an axial stator yoke and radial stator teeth, surrounding a thin-film rotor. The radial stator teeth are flush with the rotor. Each axial stator yoke is wound with a levitation winding and a torque winding, respectively. The torque winding has one pole, and the levitation winding has two poles, simultaneously achieving levitation control and rotation control. The bottom of the L-shaped stators is connected via a core magnetic ring. The outer side of the thin-film rotor is attached with… A pair of permanent magnets; six Hall sensors (Hall1-Hall6) are sequentially installed in the radial stator slots, with Hall1 located at the center of the stator slot directly opposite the N-pole rotor, and Hall2-Hall6 arranged counterclockwise in the other stator slots, with the same spacing angle between adjacent Hall sensors; when one Hall sensor fails and cannot output a signal, the faulty Hall sensor number is identified, and a system of equations is constructed based on the output signals of the five non-faulty Hall sensors to solve for the rotor displacement in the start-up state; specifically, the following steps are included:

[0009] Step S1: Derive the output signal expressions of Hall1-Hall6 under normal working conditions. When a Hall sensor malfunctions, select two Hall sensors that are not faulty and have opposite mechanical angles. Differentiate the output signals to obtain two differential results. Sum the differential results and take the opposite value to obtain the permanent magnet flux term. Sum the output signal of the Hall sensor with the opposite mechanical angle to the identified faulty Hall sensor with the permanent magnet flux term result to eliminate the influence of the permanent magnet flux.

[0010] Step S2: Sum the output signals of two fault-free Hall sensors with opposite mechanical angles. Add the two sums to the summation result of step S1 to obtain equation S1. Multiply the three sums by the vector amplitude perpendicular to the armature reaction vector and add them together to obtain the projection of the offset displacement on another time-varying angle, which is equation S2. Solve equations S1 and S2 simultaneously to obtain the radial displacements x and y in the rotor starting state.

[0011] Further, in step S1, the output signal expressions for Hall1-Hall6 are first established; taking Hall1 as an example, the output signal is related to the rotor displacement along the Hall1 direction and perpendicular to the Hall1 direction, the rotor angle, the levitation current flux linkage, and the permanent magnet flux linkage, and is expressed as follows:

[0012] Hall1 = 0.5 * k1 * l * cos(θ) l -θ0)cos(ωt-θ0)+0.5*k2*l*sin(θ l -θ0)sin(ωt-θ0)

[0013] +0.5*k3*i s *cos(θ s -2θ0)+0.5*k4*cos(ωt-θ0)

[0014] Where θ l The eccentricity is represented by the angle, l by the length of the eccentricity, ωt by the rotor angle, and θ0 by the mechanical angle where the Hall sensor is located. For Hall1, θ0 = 0. s Represents the instantaneous value of the levitation current, θ s K represents the levitation current angle, k1 is the displacement coefficient along the Hall1 direction, k2 is the displacement coefficient perpendicular to the Hall1 direction, k3 is the levitation leakage flux coefficient, and k4 is the permanent magnet flux coefficient when there is no eccentricity; the output signal expressions of Hall2-Hall6 can be obtained similarly.

[0015] Furthermore, taking the failure of Hall1 as an example, the method for solving the permanent magnet flux linkage term in step S1 is as follows:

[0016] Hall1 and Hall4 are mechanically opposite each other. Replacing θ0 with θ0+pi, the output signal of Hall4 is represented as follows:

[0017] Hall4=0.5*k1*l*cos(θ l -θ0)cos(ωt-θ0)+0.5*k2*l*sin(θ l -θ0)sin(ωt-θ0)

[0018] +0.5*k3*i s *cos(θ s -2θ0)-0.5*k4*cos(ωt-θ0)

[0019] For other Hall sensors, the output signal of Hall2 can be represented as follows:

[0020] Hall2=0.5*k1*l*cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3)

[0021] +0.5*k2*l*sin(θ l -θ0-pi / 3)sin(ωt-θ0-pi / 3)

[0022] +0.5*k3*i s *cos(θ s -2θ0-2pi / 3)-0.5*k4*cos(ωt-θ0+2pi / 3)

[0023] Hall5 is mechanically opposite to Hall2, and the output signal is represented as follows:

[0024] Hall5=0.5*k1*l*cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3)

[0025] +0.5*k2*l*sin(θ l -θ0-pi / 3)sin(ωt-θ0-pi / 3)

[0026] +0.5*k3*i s *cos(θ s -2θ0-2pi / 3)+0.5*k4*cos(ωt-θ0+2pi / 3)

[0027] Hall2 and Hall5 are mechanically opposite each other, differing by 180°, and also differing by 180° electrically. When not eccentric, the effects of the permanent magnet flux linkage are opposite. Subtracting the output signals of Hall5 and Hall2 yields the following results for the permanent magnet flux linkage:

[0028] Hall5-Hall2=k4*cos(ωt-θ0+2pi / 3)

[0029] Similarly, by subtracting the output signals of Hall3 and Hall6, the difference between Hall3 and Hall6 is obtained as follows:

[0030] Hall3-Hall6=k4*cos(ωt-θ0-2pi / 3)

[0031] Since the output signal differences between the first permanent magnet flux linkage term, the second permanent magnet flux linkage term, and the constructed third permanent magnet flux linkage term (i.e., Hall1 and Hall4) differ by 120 electrical degrees, and the sum of the permanent magnet flux linkage terms is 0, we can conclude that:

[0032] k4*cos(ωt-θ0+2pi / 3)+k4*cos(ωt-θ0-2pi / 3)+

[0033] k4*cos(ωt-θ0)=0

[0034] but:

[0035] k4*cos(ωt-θ0)=-k4*cos(ωt-θ0+2pi / 3)

[0036] -k4*cos(ωt-θ0-2pi / 3)

[0037] =Hall2-Hall5+Hall6-Hall3

[0038] Since Hall1 is faulty, the sum of twice the output signal of Hall4 and the result of the permanent magnet flux linkage term above can be obtained as follows:

[0039] A=k1*l*cos(θ l -θ0)cos(ωt-θ0)+

[0040] k2*l*sin(θ l -θ0)sin(ωt-θ0)+k3*i s *cos(θ s -2θ0).

[0041] =2Hall4+Hall2-Hall5+Hall6-Hall3

[0042] Furthermore, the method for obtaining equation S1 in step S2 is as follows:

[0043] Adding the output signals of Hall2 and Hall5, and Hall3 and Hall6, which are at opposite mechanical angles, yields the following data:

[0044] Hall2 + Hall5 = k1 * l * cos(θ) l -θ0+2pi / 3)cos(ωt-θ0+2pi / 3)+

[0045] k2*l*sin(θ l -θ0+2pi / 3)sin(ωt-θ0+2pi / 3)+k3*i s *cos(θ s -2θ0-2pi / 3)

[0046] Hall3 + Hall6 = k1 * l * cos(θ) l -θ0-2pi / 3)cos(ωt-θ0-2pi / 3)+

[0047] k2*l*sin(θ l -θ0-2pi / 3)sin(ωt-θ0-2pi / 3)+k3*i s *cos(θ s -2θ0+2pi / 3)

[0048] Adding the above summation data to the permanent magnet flux linkage term A obtained in step S1, the last term, the levitation leakage magnetic field effect, is canceled out due to the 120-degree electrical angle difference. The summation result is only related to the rotor displacement, and is used as equation S1 to obtain equation S1 as follows:

[0049] S1=2Hall2+2Hall4+2Hall6=1.5*(k1+k2)*l*cos(θ l -ωt)

[0050] The method for obtaining equation S2 is as follows:

[0051] Sampling the floating current i along the dq axis sd and i sq , where i sq Let i be the sine function value of the levitation angle. sd It is the cosine function value of the suspension angle; the suspension angle of rotation dq is obtained through arctangent operation, and the suspension angle θ in the three-phase stationary coordinate system is obtained by adding the synchronous electric angle. s The specific expression is as follows:

[0052]

[0053] The result of the arctangent in the formula is between -pi / 2 and pi / 2. By adding a compensation term, the output suspension angle θ is guaranteed. s Between 0 and 2π; since the rotor has one pair of poles and the motor's suspension winding has two pairs of poles, the initial electrical angle position of Hall1 is 2θ0, and the angle induced by Hall1 is as follows:

[0054] θ s '=θ s -2θ0

[0055] Multiply each of the three sums by the vector magnitude perpendicular to the armature reaction vector and then sum them to obtain the projection of the offset displacement onto another time-varying angle, which is used as equation S2, as follows:

[0056] S2=cos(θ s -2θ0+pi / 2)*(Hall1+Hall4)+cos(θ s -2θ0+pi / 2+2pi / 3)*

[0057] (Hall3+Hall6)+cos(θ s -2θ0+pi / 2-2pi / 3)*(Hall5+Hall2)

[0058] =0.75*(k1-k2)*l*cos(θ) l +ωt-θ s -pi / 2)

[0059] Finally, by solving equations S1 and S2 simultaneously, the radial displacements x and y of the rotor under starting conditions can be obtained using the elimination method. The system of equations is as follows:

[0060]

[0061] Solving for:

[0062]

[0063] Beneficial effects:

[0064] This invention provides a method for measuring rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions. Based on the fact that the permanent magnet flux linkage term accounts for the largest proportion of the Hall output signal, this invention proposes the concepts of signal proportion and Hall signal reconstruction. It indirectly obtains the output signal of the faulty Hall sensor by constructing a faulty Hall output signal. New constraint equations are constructed using the remaining non-faulty Hall output signals to achieve displacement and angle decoupling, enabling the re-estimation of rotor displacement information under Hall sensor failure. The algorithm proposed in this invention can accurately estimate rotor displacement information even when the motor Hall sensor fails and the Hall output signal decreases, solving the problem of existing Hall sensor displacement-free technologies failing to levitate the motor under Hall sensor failure conditions. In constructing the faulty Hall output signal, permanent magnet flux linkage is selected as the main component, resulting in a high signal-to-noise ratio. Attached Figure Description

[0065] Figure 1 This is an axial cross-sectional view of the six-tooth, one-pole, bearingless permanent magnet thin-film motor provided by the present invention;

[0066] Figure 2 It is a flowchart of the rotor displacement calculation method proposed in patent 2021116234409;

[0067] Figure 3 It is a simulation diagram of the given displacement and estimated displacement under normal working conditions, as shown in patent 2021116234409;

[0068] Figure 4 It is a simulation diagram of given displacement and estimated displacement under Hall sensor failure, as shown in patent 2021116234409;

[0069] Figure 5 This is a schematic diagram of the mechanical position of the bearingless permanent magnet thin-film motor Hall sensor used in this invention;

[0070] Figure 6 This is a schematic diagram of the rotor eccentricity of the bearingless permanent magnet thin-film motor provided by the present invention;

[0071] Figure 7 This is a flowchart of the method for measuring the rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions provided by the present invention;

[0072] Figure 8 These are simulation diagrams of the given displacement and estimated displacement of the motor rotor displacement measurement method provided by this invention;

[0073] Figure 9 This is an overall block diagram of a bearingless permanent magnet thin-film motor system using a displacement-free algorithm. Detailed Implementation

[0074] The present invention will be further described below with reference to the accompanying drawings. Obviously, the described embodiments are only some, not all, of the embodiments of the present invention. All other embodiments obtained by those skilled in the art based on the embodiments of the present invention without inventive effort are within the scope of protection of the present invention.

[0075] The present invention provides a method for measuring rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions, employing, as follows: Figure 1 The diagram shows a six-tooth, one-pole, bearingless permanent magnet motor. The motor system comprises six L-shaped stators. Each L-shaped stator includes an axial stator yoke and radial stator teeth, surrounding a thin-plate rotor. The radial stator teeth are flush with the rotor. Each axial stator yoke has a levitation winding and a torque winding wound on each side. The torque winding has one pair of poles, and the levitation winding has two pairs of poles, simultaneously achieving levitation and rotation control. The bottom of the L-shaped stators is connected via a core magnetic ring. A pair of permanent magnets is attached to the outer side of the thin-plate rotor. Figure 9This is an overall block diagram of the bearingless permanent magnet thin-film motor system employing a displacement-free algorithm in this invention. The torque control section uses vector control, with an outer speed loop using the speed calculated from the Hall effect sensor as feedback, and an inner current loop. The displacement control section has an outer displacement loop, using the displacement obtained from the Hall effect sensor signal through a displacement-free algorithm as feedback, and the inner loop is also a current loop.

[0076] The present invention provides a method for measuring rotor displacement of a bearingless permanent magnet thin-film motor under Hall sensor failure conditions. Six Hall sensors (Hall1-Hall6) are sequentially installed in the radial stator slots. Hall1 is located at the center of the stator slot directly opposite the N-pole rotor. Hall2-Hall6 are arranged counter-clockwise in the other stator slots. The spacing angle between adjacent Hall sensors is the same. Specific positions are as follows: Figure 5 As shown, the output signal of the fault Hall sensor is indirectly obtained by constructing a fault Hall output signal. This embodiment assumes that Hall1 has failed and cannot output a signal. It should be noted that in practical applications, regardless of which Hall sensor fails, the method described in this embodiment can be used for rotor displacement estimation. The specific steps are as follows:

[0077] Step S1: Derive the output signal expressions for Hall1-Hall6 under normal operating conditions. When a Hall sensor malfunctions, select two fault-free Hall sensors with opposite mechanical angles. Differentiate the output signals to obtain two differential results. Summate the differential results and take the opposite value to obtain the permanent magnet flux linkage term. Summate the output signal of the Hall sensor with the opposite mechanical angle to the identified faulty Hall sensor with the permanent magnet flux linkage term result to eliminate the influence of the permanent magnet flux linkage.

[0078] First, establish the output signal expressions for Hall1-Hall6. Taking Hall1 as an example, the output signal is related to the rotor displacement along and perpendicular to Hall1, the rotor angle, the levitation current flux linkage, and the permanent magnet flux linkage, expressed as follows:

[0079] Hall1 = 0.5 * k1 * l * cos(θ) l -θ0)cos(ωt-θ0)+0.5*k2*l*sin(θ l -θ0)sin(ωt-θ0)

[0080] +0.5*k3*i s *cos(θ s -2θ0)+0.5*k4*cos(ωt-θ0)

[0081] Where θ lωt represents the eccentricity angle, l represents the eccentricity length, ωt represents the rotor angle, and θ0 represents the mechanical angle where the Hall sensor is located. For Hall1, θ0 = 0. s Represents the instantaneous value of the levitation current, θ s K represents the levitation current angle, k1 is the displacement coefficient along the Hall1 direction, k2 is the displacement coefficient perpendicular to the Hall1 direction, k3 is the levitation leakage flux coefficient, and k4 is the permanent magnet flux coefficient when there is no eccentricity. The expression for the output signal of Hall2-Hall6 can be obtained similarly.

[0082] Then, the permanent magnet flux linkage term is calculated. Hall1 and Hall4 are mechanically opposite each other. Replacing θ0 with θ0+pi, the output signal of Hall4 is represented as follows:

[0083] Hall4=0.5*k1*l*cos(θ l -θ0)cos(ωt-θ0)+0.5*k2*l*sin(θ l -θ0)sin(ωt-θ0)

[0084] +0.5*k3*i s *cos(θ s -2θ0)-0.5*k4*cos(ωt-θ0)

[0085] For other Hall sensors, the output signal of Hall2 can be represented as follows:

[0086] Hall2=0.5*k1*l*cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3)

[0087] +0.5*k2*l*sin(θ l -θ0-pi / 3)sin(ωt-θ0-pi / 3)

[0088] +0.5*k3*i s *cos(θ s -2θ0-2pi / 3)-0.5*k4*cos(ωt-θ0+2pi / 3)

[0089] Hall5 is mechanically opposite to Hall2, and the output signal is represented as follows:

[0090] Hall5=0.5*k1*l*cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3)

[0091] +0.5*k2*l*sin(θ l-θ0-pi / 3)sin(ωt-θ0-pi / 3)

[0092] +0.5*k3*i s *cos(θ s -2θ0-2pi / 3)+0.5*k4*cos(ωt-θ0+2pi / 3)

[0093] Hall2 and Hall5 are mechanically opposite each other, differing by 180°, and also by 180° electrically. When not eccentric, their effects on the permanent magnet flux linkage are opposite. Subtracting the output signals of Hall5 and Hall2 yields the following results for the permanent magnet flux linkage:

[0094] Hall5-Hall2=k4*cos(ωt-θ0+2pi / 3)

[0095] Similarly, by subtracting the output signals of Hall3 and Hall6, the difference between Hall3 and Hall6 is obtained as follows:

[0096] Hall3-Hall6=k4*cos(ωt-θ0-2pi / 3)

[0097] Since the sum of the three mutually differentiated electrical angle signals is zero, that is, the sum of the first permanent magnet flux linkage term, the second permanent magnet flux linkage term, and the constructed third permanent magnet flux linkage term (the difference between the output signals of Hall1 and Hall4) can be obtained as follows:

[0098] k4*cos(ωt-θ0+2pi / 3)+k4*cos(ωt-θ0-2pi / 3)+

[0099] k4*cos(ωt-θ0)=0

[0100] but:

[0101] k4*cos(ωt-θ0)=-k4*cos(ωt-θ0+2pi / 3)

[0102] -k4*cos(ωt-θ0-2pi / 3)

[0103] =Hall2-Hall5+Hall6-Hall3

[0104] Since Hall1 is faulty, the sum of twice the output signal of Hall4 and the result of the permanent magnet flux linkage term above can be obtained as follows:

[0105] A=k1*l*cos(θ l -θ0)cos(ωt-θ0)+

[0106] k2*l*sin(θ l -θ0)sin(ωt-θ0)+k3*is *cos(θ s -2θ0).

[0107] =2Hall4+Hall2-Hall5+Hall6-Hall3

[0108] Step S2: Sum the output signals of two fault-free Hall sensors with opposite mechanical angles. Add the two sums to the sum from step S1 to obtain equation S1. Multiply each of the three sums by the vector amplitude perpendicular to the armature reaction vector and then sum them to obtain the projection of the offset displacement onto another time-varying angle, which is equation S2. Solve equations S1 and S2 simultaneously to obtain the radial displacements x and y under rotor starting conditions. Specifically,

[0109] The method for obtaining equation S1 is as follows:

[0110] Adding the output signals of Hall2 and Hall5, and Hall3 and Hall6, which are at opposite mechanical angles, yields the following data:

[0111] Hall2 + Hall5 = k1 * l * cos(θ) l -θ0+2pi / 3)cos(ωt-θ0+2pi / 3)+

[0112] k2*l*sin(θ l -θ0+2pi / 3)sin(ωt-θ0+2pi / 3)+k3*i s *cos(θ s -2θ0-2pi / 3)

[0113] Hall3 + Hall6 = k1 * l * cos(θ) l -θ0-2pi / 3)cos(ωt-θ0-2pi / 3)+

[0114] k2*l*sin(θ l -θ0-2pi / 3)sin(ωt-θ0-2pi / 3)+k3*i s *cos(θ s -2θ0+2pi / 3)

[0115] Adding the above summation data to the permanent magnet flux linkage term A obtained in step S1, the last term, the levitation leakage magnetic field effect, is canceled out due to the 120-degree electrical angle difference. The summation result is only related to the rotor displacement, and is used as equation S1 to obtain equation S1 as follows:

[0116] S1=2Hall2+2Hall4+2Hall6=1.5*(k1+k2)*l*cos(θ l -ωt)

[0117] The method for obtaining equation S2 is as follows:

[0118] Sampling the floating current i along the dq axis sd and i sq , where i sq Let i be the sine function value of the levitation angle. sd It is the cosine function value of the suspension angle. The suspension angle of rotation dq is obtained through arctangent calculation, and then the synchronous electric angle is added to obtain the suspension angle θ in the three-phase stationary coordinate system. s The specific expression is as follows:

[0119]

[0120] The result of the arctangent in the formula is between -pi / 2 and pi / 2. By adding a compensation term, the output suspension angle θ is guaranteed. s Between 0 and 2π. Since the rotor has one pair of poles and the motor's suspension winding has two pairs of poles, the initial electrical angle position of Hall1 is 2θ0, and the angle induced by Hall1 is as follows:

[0121] θ s '=θ s -2θ0

[0122] Multiply each of the three sums by the vector magnitude perpendicular to the armature reaction vector and then sum them to obtain the projection of the offset displacement onto another time-varying angle, which is used as equation S2, as follows:

[0123] S2=cos(θ s -2θ0+pi / 2)*(Hall1+Hall4)+cos(θ s -2θ0+pi / 2+2pi / 3)*

[0124] (Hall3+Hall6)+cos(θ s -2θ0+pi / 2-2pi / 3)*(Hall5+Hall2)

[0125] =0.75*(k1-k2)*l*cos(θ) l +ωt-θ s -pi / 2)

[0126] Finally, by solving equations S1 and S2 simultaneously, the radial displacements x and y of the rotor under starting conditions can be obtained using the elimination method. The system of equations is as follows:

[0127]

[0128] Solving for:

[0129]

[0130] Figure 7 This is a simplified flowchart of the rotor displacement estimation method for the motor under Hall sensor failure conditions described above. The equations are reconstructed using non-faulty Hall signals, and the displacement information is obtained by solving them. Figure 8 The simulation diagrams of the given displacement and the estimated displacement clearly show a high degree of overlap, indicating that the displacement information can be correctly calculated, thus verifying the feasibility of the scheme.

[0131] The above are merely preferred embodiments of the present invention. It should be noted that those skilled in the art can make various improvements and modifications without departing from the principle of the present invention, and these improvements and modifications should also be considered within the scope of protection of the present invention.

Claims

1. A method for measuring the displacement of the rotor of a bearingless permanent-magnet thin-slice motor in a Hall sensor fault state, using a bearingless permanent-magnet thin-slice motor with six teeth and a pair of poles, comprising six L-shaped stators; each L-shaped stator comprises an axial stator yoke and a radial stator tooth, which surrounds a thin-slice rotor, and the radial stator tooth is flush with the rotor; each axial stator yoke is wound with a suspension winding and a torque winding, the torque winding is a pair of poles, and the suspension winding is two pairs of poles, thereby achieving suspension control and rotation control; the bottom of the L-shaped stator is connected through a core magnetic conductive ring; the outer side of the thin-slice rotor is attached with a pair of pole permanent magnets; characterized in that, The six Hall sensors Hall1-Hall6 are sequentially installed in the radial stator slots, wherein Hall1 is located at the center of the stator slot opposite to the N-pole rotor, and Hall2-Hall6 are sequentially arranged in the other stator slots in anticlockwise direction, and the interval angle between adjacent Hall sensors is the same; when one of the Hall sensors fails to output a signal, the number of the faulty Hall sensor is identified, and an equation group is constructed based on the output signals of the non-faulty five Hall sensors to solve the rotor displacement in the starting state; specifically comprising the following steps: Step 1, the output signal expression of Hall1-Hall6 under normal working condition is derived, when a Hall sensor fails, two Hall sensors without faults and with opposite mechanical angles are selected, the output signals are differentiated to obtain two differential results, the differential results are summed and the negative is taken to obtain a permanent magnet flux linkage term; the output signals of the Hall sensors with opposite mechanical angles to the identified faulty Hall sensor are summed with the permanent magnet flux linkage term result to obtain sum 1, and the influence of the permanent magnet flux linkage is eliminated; Step 2, the output signals of the two Hall sensors without faults and with opposite mechanical angles are summed to obtain sum 2 and sum 3, sum 1, sum 2 and sum 3 are added to obtain equation S1; the three data of sum 1, sum 2 and sum 3 are multiplied by the vector amplitude perpendicular to the armature reaction vector and then added to obtain the projection of the offset displacement on another time-varying angle as equation S2; equations S1 and S2 are solved to obtain the radial displacements x and y of the rotor in the starting state.

2. A method of rotor displacement measurement for a bearingless permanent magnet sheet motor in a fault condition of a Hall sensor according to claim 1, characterized in that, In step 1, the output signal expression of Hall1-Hall6 is established first; the output signal of Hall1 is related to the displacements of the rotor in the direction along Hall1 and perpendicular to Hall1, the rotor angle, the suspension current flux linkage and the permanent magnet flux linkage, and the expression form is as follows: Hall1 = 0.5 * k1 * l * cos(θ - θ0) l + 0.5 * k2 * l * sin(θ - θ0) l Hall2 = 0.5 * k1 * l * sin(θ - θ0) + 0.5 * k2 * l * cos(θ - θ0) +0.5*k3*i s *cos(θ s -2θ0)+0.5*k4*cos(ωt-θ0) where θ l represents the angle of eccentricity, l represents the length of eccentricity, ωt represents the rotor angle, θ0 represents the mechanical angle where the Hall sensor is located, θ0 = 0 for Hall1; i s represents the instantaneous value of the suspension current, θ s represents the angle of suspension current, k1 is the displacement coefficient along the direction of Hall1, k2 is the displacement coefficient perpendicular to the direction of Hall1, k3 is the suspension leakage flux coefficient, k4 is the permanent magnet flux coefficient when there is no eccentricity; the output signal expressions of Hall2-Hall6 can be similarly obtained.

3. A method of rotor displacement measurement for a bearingless permanent magnet sheet motor in a fault condition of a Hall sensor according to claim 2, characterized in that, When Hall1 fails, the permanent magnet flux linkage term in step 1 is solved as follows: Hall1 and Hall4 are opposite in mechanical direction, and θ0+pi is used instead of θ0, and the output signal of Hall4 is expressed as follows: Hall4 = 0.5 * k1 * l * cos(θ l -θ0) cos(ωt - θ0) + 0.5 * k2 * l * sin(θ l -θ0) sin(ωt - θ0) +0.5*k3*i s *cos(θ s -2θ0)-0.5*k4*cos(ωt-θ0) For other Hall sensors, the output signal of Hall2 can be expressed as follows: Hall2 = 0.5 * k1 * l * cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3) + 0.5 * k2 * 1 * sin(θ l - θ0- pi / 3) sin(ωt - θ0- pi / 3) +0.5*k3*i s *cos(θ s -2θ0-2pi / 3)-0.5*k4*cos(ωt-θ0+2pi / 3) Hall5 and Hall2 are opposite in mechanical direction, and the output signal is expressed as follows: Hall5 = 0.5 * k1 * l * cos(θ l -θ0-pi / 3)cos(ωt-θ0-pi / 3) + 0.5 * k2 * 1 * sin(θ l - θ0- pi / 3) sin(ωt - θ0- pi / 3) + 0.5 * k3 * i s + 0.5 * k3 * i s - 2θ0- 2pi / 3) + 0.5 * k4 * cos(ωt- θ0+ 2pi / 3) Hall2 and Hall5 are opposite in mechanical angle and differ by 180°, Hall3 and Hall6 also differ by 180 electrical angles, and the influence of the permanent magnet flux linkage is opposite when there is no eccentricity; the output signals of Hall5 and Hall2 are differentiated to obtain the first permanent magnet flux linkage term as follows: Hall5-Hall2=k4*cos(ωt-θ0+2pi / 3) Similarly, the output signals of Hall3 and Hall6 are differentiated to obtain the second permanent magnet flux linkage term as follows: Hall3-Hall6=k4*cos(ωt-θ0-2pi / 3) Since the first permanent magnet flux linkage term, the second permanent magnet flux linkage term and the third permanent magnet flux linkage term constructed, that is, the difference between the output signals of Hall1 and Hall4, differ by 120 electrical angles, and the sum of the permanent magnet flux linkage terms is 0, it can be obtained that: k4*cos(ωt-θ0+2pi / 3)+k4*cos(ωt-θ0-2pi / 3)+ k4*cos(ωt-θ0)=0 Then: k4*cos(ωt-θ0)=-k4*cos(ωt-θ0+2pi / 3) -k4*cos(ωt-θ0-2pi / 3) =Hall2-Hall5+Hall6-Hall3 Due to Hall1 fault, then use twice Hall4 output signal and the above permanent magnet flux linkage term result summation can be obtained: A = k1 * l * cos(θ l - θ0) cos(ωt - θ0) + k2 * l * sin(θ l -θ0)sin(ωt-θ0)+k3*i s *cos(θ s -2θ0) =2Hall4+Hall2-Hall5+Hall6-Hall3.

4. A method of rotor displacement measurement for a bearingless permanent magnet sheet motor in a fault condition of a Hall sensor according to claim 3, characterized in that, The equation S1 in step 2 is obtained as follows: Add the output signals of Hall2 and Hall5, and Hall3 and Hall6 which are opposite in mechanical angle, to obtain the following sum data respectively: Hall2 + Hall5 = k1 * l * cos(θ l - θ0+ 2pi / 3) cos(ωt - θ0+ 2pi / 3) + k2 * l * sin(θ l - θ0+ 2pi / 3) sin(ωt - θ0+ 2pi / 3) + k3 * i s * cos(θ s - 2θ0- 2pi / 3) Hall3 + Hall6 = k1 * l * cos(θ l - θ0- 2pi / 3) cos(ωt - θ0- 2pi / 3) + k2 * l * sin(θ l - θ0- 2pi / 3) sin(ωt - θ0- 2pi / 3) + k3 * i s * cos(θ s - 2θ0+ 2pi / 3) Adding the above sum data to the permanent magnet flux term A obtained in step 1, the last term of the floating leakage flux influence part is cancelled due to the mutual difference of 120 degrees of electrical angle, and the sum result is only related to the rotor displacement, and is obtained as equation S1. Equation S1 is as follows: S1 = 2Hall2 + 2Hall4 + 2Hall6 = 1.5*(k1+k2)*l*cos(θ l - ωt) The equation S2 is obtained as follows: Sampling the floating current i along the dq axis sd and i sq , where i sq Let i be the sine function value of the levitation angle. sd It is the cosine function value of the suspension angle; the suspension angle of rotation dq is obtained through arctangent operation, and the suspension angle θ in the three-phase stationary coordinate system is obtained by adding the synchronous electric angle. s The specific expression is as follows: where the result of the arctangent is between -pi / 2 and pi / 2, the output levitation angle theta is guaranteed to be between -pi and pi by adding the compensation term s between 0 and 2pi; since the rotor is a pair of poles, and the motor levitation winding is two pairs of poles, the initial electrical angle position of Hall 1 is 2theta0, and the angle sensed by Hall 1 is given by the following equation: θ s ' = θ s - 2θ0 Multiply the three sum data by the vector amplitude perpendicular to the armature reaction vector respectively, and then add them to obtain the projection of the offset displacement in another time-varying angle as the equation S2, which is as follows: S2 = cos(θ s -2θ0+pi / 2)*A+cos(θ s -2θ0+pi / 2+2pi / 3)* (Hall3 + Hall6) + cos(θ s -2θ0+pi / 2-2pi / 3)*(Hall5+Hall2) = 0.75 * (k1 - k2) * 1 * cos(θ l + ωt - θ s - pi / 2) Finally, the equations S1 and S2 are combined and solved, and the radial displacements x and y in the starting state of the rotor can be obtained by the elimination method; the combined equation group is as follows: The solution is:

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