# A Novel Rotor Eddy Current Loss Estimation Method for Permanent Magnet Synchronous Machines with Small Inductance and a Conductive Rotor Sleeve

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Problem Statement

_{0}; then, the copper sleeve used to resist eddy currents, whose thickness is d

_{r}; inside the copper sleeve is the permanent magnet layer, whose thickness is d

_{p}; and the innermost layer is the rotor yoke, whose thickness is d

_{z}. These layers are depicted in Figure 2.

_{b}is the base frequency, n is the order of the current harmonics, L

_{s}is the phase inductance, and I

_{n}is the amplitude of current harmonics.

_{sov}and K

_{dpv}are coefficients accounting for the slot opening effect and winding distribution effect, respectively. The direction of the travelling current sheet depends on the space harmonic order as well as the current harmonic order. Then, the ECL caused by each current sheet is calculated individually, then summed together to form the overall eddy current loss J

_{Loss}. To do so, the vector potential of the motor A

_{z}should be calculated. The governing law in the object small inductance PMSM is listed in Equation (3):

_{i}and μ

_{i}have different values, depending on the material. For example, in the airgap, σ

_{1}is 0, and μ

_{1}is the permeability of the vacuum. Boundary conditions are given in Equations (4)~(6):

_{z}is obtained, Poynting’s Theorem is used to calculate the loss:

_{0}and D

_{0}are related to the specific vector potential of the motor. Detailed calculations are shown in Appendix A.

_{0}is the modulation wave frequency, and ω

_{c}is the carrier wave frequency. However, ω

_{c}is chosen so that the switching losses of power devices are within a reasonable range and need not be exact multiples of ω

_{0}. This made it difficult for all harmonic components to find a common base. If the fundamental current frequency continues to act as the base for time-FFT analysis, spectrum leakage is unavoidable, making the eddy current loss calculation inaccurate. Furthermore, while traditional methods only consider the frequencies of harmonic currents, it will be seen in Section 5 that only one part of the harmonic order pair determines the travelling direction of magneto waves. This error is especially serious in high-speed applications because the modulation wave frequency is relatively high compared to conventional machines. To solve these problems, improved methods for the decomposition of PWM voltage and a novel estimation algorithm to improve the ECL estimation accuracy with different travelling wave direction judgement methods are proposed in this article.

## 3. SVPWM Working Principle and Expression Deduction of Modulation Wave Per Phase

## 4. Refined SVPWM Frequency Spectrum Structure

_{o}/V

_{DC}; then, the integral limits of the coefficients are listed as shown in Table 3.

_{n}denotes the n-th order Bessel function.

_{mn}= 0. This implies that the ideal modulation process does not change the phase of the modulation wave. Furthermore, when m ≥ 1, n ≠ 3k, A

_{mn}does not have to be zero; however, when m = 0, n ≠ 3k, A

_{mn}does have to be zero. This implies that an ideal modulation process does introduce new frequency components into the modulated voltage waveform.

_{mn}+jB

_{mn}is the same as phase A.

- When n = 3k, the coefficients in f
_{a}(t), f_{b}(t) and f_{c}(t) are the same. Thus, the harmonic voltage does not produce a respective resulting harmonic current. - Switching harmonics does not exist in phase currents; i.e., the voltage harmonics bearing the form in Equation (20)$$\sum _{m=1}^{\infty}\left[{A}_{m0}\mathrm{cos}({\omega}_{c}t+{\theta}_{c})+{B}_{m0}\mathrm{sin}({\omega}_{c}t+{\theta}_{c})\right]$$
- Voltage harmonics with m ± n being even do not produce current harmonics either. The reasons for this are as follows.Components bearing the form in Equation (21)$$\frac{\pi}{6}\mathrm{sin}\left([m+n]\frac{\pi}{2}\right)\left\{{J}_{n}\left(m\frac{3\pi}{4}M\right)+2\mathrm{cos}n\frac{\pi}{6}\cdot {J}_{n}\left(m\frac{\sqrt{3}\pi}{4}M\right)\right\}$$Components bearing the form in Equation (22)$${\frac{1}{n}\mathrm{sin}m\frac{\pi}{2}\mathrm{cos}n\frac{\pi}{2}\mathrm{sin}n\frac{\pi}{6}\left\{{J}_{0}\left(m\frac{3\pi}{4}M\right)-{J}_{0}\left(m\frac{\sqrt{3}\pi}{4}M\right)\right\}|}_{n\ne 0}$$Components bearing the form in Equation (23)$$\sum _{\begin{array}{l}k=1\\ k\ne -n\end{array}}^{\infty}\left[\begin{array}{l}\frac{1}{n+k}\mathrm{sin}\left([m+k]\frac{\pi}{2}\right)\mathrm{cos}\left([n+k]\frac{\pi}{2}\right)\mathrm{sin}\left([n+k]\frac{\pi}{6}\right)\\ \cdot \left\{{J}_{k}\left(m\frac{3\pi}{4}M\right)+2\mathrm{cos}\left([2n+3k]\frac{\pi}{6}\right){J}_{k}\left(m\frac{\sqrt{3}\pi}{4}M\right)\right\}\end{array}\right]$$Components bearing the form in Equation (24)$$\sum _{\begin{array}{l}k=1\\ k\ne n\end{array}}^{\infty}\left[\begin{array}{l}\frac{1}{n-k}\mathrm{sin}\left([m+k]\frac{\pi}{2}\right)\mathrm{cos}\left([n-k]\frac{\pi}{2}\right)\mathrm{sin}\left([n-k]\frac{\pi}{6}\right)\\ \cdot \left\{{J}_{k}\left(m\frac{3\pi}{4}M\right)+2\mathrm{cos}\left([2n-3k]\frac{\pi}{6}\right){J}_{k}\left(m\frac{\sqrt{3}\pi}{4}M\right)\right\}\end{array}\right]$$

## 5. ECL Estimation Algorithm with Different Travelling Current Sheet Direction Judgement Methods

_{0}denotes the width of one slot opening;

_{a}denotes the current in a phase winding.

- (v+n) can be divided by three, while (v-n) cannot:$$c{s}_{tot}(i,x)={\displaystyle \sum _{v}{K}_{sov}{K}_{dpv}\frac{2}{\tau}\mathrm{sin}(-\frac{v\pi}{2})N\cdot}{\displaystyle \sum _{m}{\displaystyle \sum _{n}\frac{3}{2}{I}_{mn}\cdot \mathrm{sin}(m{\omega}_{c}t+n{\omega}_{o}t+v\frac{\pi}{\tau}x+\phi )}}$$In this situation, the travelling wave travels backwards with regard to the rotor. Suppose the velocity of the outer rotor of the permanent magnet synchronous machine is v
_{r}; then, the travel velocity v_{tr}of the travelling current sheet is expressed as in Equation (31):$${v}_{tr}=\frac{m\cdot {\omega}_{c}+n\cdot {\omega}_{o}}{v\cdot \pi}\tau +{v}_{r}$$ - (v-n) can be divided by three, while (v+n) cannot:$$c{s}_{tot}(i,x)={\displaystyle \sum _{v}{K}_{sov}{K}_{dpv}\frac{2}{\tau}\mathrm{sin}(-\frac{v\pi}{2})N\cdot}{\displaystyle \sum _{m}{\displaystyle \sum _{n}\frac{3}{2}{I}_{mn}\cdot \mathrm{sin}(m{\omega}_{c}t+n{\omega}_{o}t-v\frac{\pi}{\tau}x+\phi )}}$$

_{r}; then, the travel velocity v

_{tr}of the travelling current sheet is expressed as in Equation (33):

## 6. Simulation Study and Experiments

## 7. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## Appendix A. Details of the Analytical Model for the Eddy Current Loss Estimation

_{0}is the slot opening width, τ is the pole pitch, and α is the winding distribution angle.

_{z}= 0 at the rotor shaft, the following can be obtained:

_{z}is found, the eddy current loss of the rotor J

_{Loss}is expressed as Equation (A11):

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**Figure 1.**Structure of small inductance permanent magnet synchronous machine (PMSM) (

**a**) and simplified model for the calculation of eddy current losses (

**b**).

**Figure 10.**Phase voltage spectrum at a base frequency of 200 Hz (left) and 240 Hz (right) under SVPWM modulation. The DC bus voltage is 30 V.

**Figure 11.**Phase harmonic current prediction of 200 Hz (left) and 240 Hz (right) under SVPWM modulation. The DC bus voltage is 30 V and the phase inductance is 48 μH.

**Figure 12.**Experimental current waveform under sensorless vector control mode. The base current frequency is 200 Hz in (

**a**) and 240 Hz in (

**b**).

**Figure 13.**Performance comparison of current spectrum prediction by newly proposed method (double Fourier analysis (DFA)) and traditional fast Fourier transform (FFT). The base current frequency is 200 Hz in (

**a**) and 24 0Hz in (

**b**). The experimental current spectrum served as the standard.

**Figure 14.**Eddy current losses estimated with different methods. The base current frequency is 200 Hz. The FEM software calculation is shown in (

**a**), and data are listed in (

**b**).

**Figure 15.**Eddy current losses estimated with different methods. The base current frequency is 240 Hz. The FEM software calculation is shown in (

**a**), and data are listed in (

**b**).

θ_{o} = ω_{o}t | Active Space Vectors | Effect Time Span |
---|---|---|

0 < θ_{o} ≤ π/3 | $\begin{array}{c}\overline{S{V}_{1}}=A\overline{B}\overline{C}\\ \overline{S{V}_{2}}=AB\overline{C}\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{2})\end{array}$ |

π/3 < θ_{o} ≤ 2π/3 | $\begin{array}{c}\overline{S{V}_{2}}=AB\overline{C}\\ \overline{S{V}_{3}}=\overline{A}B\overline{C}\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{5\pi}{6})\end{array}$ |

2π/3 < θ_{o} ≤ π | $\begin{array}{c}\overline{S{V}_{3}}=\overline{A}B\overline{C}\\ \overline{S{V}_{4}}=\overline{A}BC\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{2})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{7\pi}{6})\end{array}$ |

π < θ_{o} ≤ 4π/3 | $\begin{array}{c}\overline{S{V}_{4}}=\overline{A}BC\\ \overline{S{V}_{5}}=\overline{A}\overline{B}C\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{5\pi}{6})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{3\pi}{2})\end{array}$ |

4π/3 < θ_{o} ≤ 5π/3 | $\begin{array}{c}\overline{S{V}_{5}}=\overline{A}\overline{B}C\\ \overline{S{V}_{6}}=A\overline{B}C\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{7\pi}{6})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{11\pi}{6})\end{array}$ |

5π/3 < θ_{o} ≤ 2π | $\begin{array}{c}\overline{S{V}_{6}}=A\overline{B}C\\ \overline{S{V}_{1}}=A\overline{B}\overline{C}\end{array}$ | $\begin{array}{c}\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{3\pi}{2})\\ \frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\end{array}$ |

θ_{o} = ω_{o}t | f_{A}(θ_{o}) |
---|---|

0 < θ_{o} ≤ π/3 | $\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})$ |

π/3 < θ_{o} ≤ 2π/3 | $\frac{3}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o})$ |

2π/3 < θ_{o} ≤ π | $\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})$ |

π < θ_{o} ≤ 4π/3 | $\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})$ |

4π/3 < θ_{o} ≤ 5π/3 | $\frac{3}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o})$ |

5π/3 < θ_{o} ≤ 2π | $\frac{\sqrt{3}}{2}\frac{{U}_{o}\cdot \Delta T}{{V}_{DC}}\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})$ |

i | y_{s}(i) | y_{e}(i) | x_{r}(i) | x_{f}(i) |
---|---|---|---|---|

1 | 0 | π/3 | $-\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\right]$ | $\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\right]$ |

2 | π/3 | 2π/3 | $-\frac{\pi}{2}\left[1+\frac{3}{2}M\mathrm{cos}({\theta}_{o})\right]$ | $\frac{\pi}{2}\left[1+\frac{3}{2}M\mathrm{cos}({\theta}_{o})\right]$ |

3 | 2π/3 | π | $-\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})\right]$ | $\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})\right]$ |

4 | π | 4π/3 | $-\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\right]$ | $\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}-\frac{\pi}{6})\right]$ |

5 | 4π/3 | 5π/3 | $-\frac{\pi}{2}\left[1+\frac{3}{2}M\mathrm{cos}({\theta}_{o})\right]$ | $\frac{\pi}{2}\left[1+\frac{3}{2}M\mathrm{cos}({\theta}_{o})\right]$ |

6 | 5π/3 | 2π | $-\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})\right]$ | $\frac{\pi}{2}\left[1+\frac{\sqrt{3}}{2}M\mathrm{cos}({\theta}_{o}+\frac{\pi}{6})\right]$ |

Parameter | Value |
---|---|

Poles/slots | 2/12 |

Stator inner diameter | 36 mm |

Rotor outer diameter | 30.2 mm |

Air gap flux density | 0.4 T |

Effective length | 50 mm |

Magnet material | N40EH |

Phase inductance | 48 μH |

Parameter | Value |
---|---|

Width of slot opening | 0.8 mm |

Winding turns of a single phase | 12 |

Electrical angle between adjacent slot opening | 30° |

Pole pitch | 56.6 mm |

© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

## Share and Cite

**MDPI and ACS Style**

Pei, L.; Li, L.; Guo, Q.; Yang, R.; Du, P.
A Novel Rotor Eddy Current Loss Estimation Method for Permanent Magnet Synchronous Machines with Small Inductance and a Conductive Rotor Sleeve. *Energies* **2019**, *12*, 3760.
https://doi.org/10.3390/en12193760

**AMA Style**

Pei L, Li L, Guo Q, Yang R, Du P.
A Novel Rotor Eddy Current Loss Estimation Method for Permanent Magnet Synchronous Machines with Small Inductance and a Conductive Rotor Sleeve. *Energies*. 2019; 12(19):3760.
https://doi.org/10.3390/en12193760

**Chicago/Turabian Style**

Pei, Le, Liyi Li, Qingbo Guo, Rui Yang, and Pengcheng Du.
2019. "A Novel Rotor Eddy Current Loss Estimation Method for Permanent Magnet Synchronous Machines with Small Inductance and a Conductive Rotor Sleeve" *Energies* 12, no. 19: 3760.
https://doi.org/10.3390/en12193760