Question 5 of 6: High-Frequency PWM Drives and Constant-Volts-per-Hertz Breakdown Torque

Question 5: High-Frequency PWM Drives and Constant-Volts-per-Hertz Breakdown Torque (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — undesirable effects of high-frequency PWM

Motor insulation stress from fast voltage edges and reflected waves. A modern switching device produces edges of a few tens of nanoseconds. When the cable between drive and motor is longer than about half the rise-time wavelength, the impedance mismatch at the motor terminals reflects the wave and the terminal voltage can reach twice the d.c.-link value. The high $dv/dt$ also distributes non-uniformly across the stator winding, concentrating most of the step across the first few turns of the first coil, where the turn-to-turn insulation was never designed for it. Partial discharge and premature insulation failure follow; inverter-duty motors (NEMA MG-1 Part 31) and output $dv/dt$ filters exist to address exactly this.

Bearing currents and shaft voltages. The three-phase PWM output has a non-zero common-mode component that couples capacitively to the rotor, charging the shaft until it discharges through the bearing lubricant film. The resulting electrical-discharge-machining pits and fluting destroy bearings in months rather than years. High switching frequency makes it worse because the coupling is capacitive. Insulated bearings, shaft grounding rings and common-mode chokes are the standard countermeasures.

Switching loss, EMI and derating. Loss per switching event is essentially fixed by the device and the load, so switching loss rises in direct proportion to frequency, forcing a larger heatsink or a derated inverter output. The same fast edges radiate and conduct electromagnetic interference into control wiring, encoder feedback and nearby instrumentation, which is why drives require screened motor cable, proper 360-degree gland terminations and line filters to meet the emission limits in CAN/CSA-CISPR 11. Long cables additionally load the inverter with capacitive charging current at every edge.

Two further effects are worth naming if marks allow: acoustic and ultrasonic noise moves out of the audible band but into the range where magnetostrictive stress and eddy loss in the core rise; and any ground-fault or residual-current protection sees the common-mode charging current as leakage, causing nuisance tripping.

Parts (b) and (c) — frequency, line current and voltage

Given.

Find. The line frequency and line current at the first operating point, and the line-to-line voltage required at the second.

Approach. Use the supplied breakdown-torque expression in each direction — solve it for frequency in part (b) and for voltage in part (c) — and obtain the line current from the fact that at breakdown the referred rotor resistance equals the total leakage reactance.

1. Convert the given voltage to line-to-line. The formula is written in terms of $V_{LL}$ but the paper supplies a line-to-neutral value, so $V_{LL}=\sqrt{3}\times 200=346.41$ V. Missing this conversion divides the computed $\omega$ by $\sqrt{3}$ and poisons everything downstream.
2. Solve the torque expression for the angular frequency. Rearranging $T_{\max}=V_{LL}^{2}P/(4\omega^{2}L_{T})$:$$\omega=\sqrt{\frac{V_{LL}^{2}P}{4\,T_{\max}L_{T}}}=\sqrt{\frac{346.41^{2}\times 4}{4\times 275\times 1.75\times 10^{-3}}}=499.35\ \text{rad/s}$$$$\boxed{f=\frac{499.35}{2\pi}=79.5\ \text{Hz}}$$
3. Find the line current from the breakdown condition. At the torque peak the referred rotor resistance divided by slip equals the total leakage reactance, $R_{2}^{\prime}/s=\omega L_{T}$. With the stator resistance neglected the per-phase impedance is therefore$$|Z|=\sqrt{(\omega L_{T})^{2}+(\omega L_{T})^{2}}=\sqrt{2}\,\omega L_{T}=\sqrt{2}\times 0.87386=1.2358\ \Omega$$and the machine operates at a power factor of $1/\sqrt{2}=0.707$ — the classic result that an induction motor at breakdown runs at 0.707 lagging when resistance is ignored.
4. Evaluate the current. The winding is fed at 200 V per phase, so$$\boxed{I_{\text{line}}=\frac{V_{ph}}{\sqrt{2}\,\omega L_{T}}=\frac{200}{1.2358}=161.9\ \text{A}}$$Cross-checking through the air-gap power, $T=3I^{2}(R_{2}^{\prime}/s)/\omega_{s,\text{mech}}$ with $\omega_{s,\text{mech}}=2\omega/P=249.7$ rad/s, returns 275.0 N·m — the stated torque, which validates both the current and the impedance argument.
5. Use the quoted rotor speed as a consistency check. The speed of 1800 rev/min is not needed for either answer, but it is not redundant: the synchronous speed at 79.47 Hz is $120f/P=2384$ rev/min, so the breakdown slip is $s=1-1800/2384=0.245$ and the implied referred rotor resistance is $R_{2}^{\prime}=s\,\omega L_{T}=0.214\ \Omega$. That is an entirely plausible value for a machine of this rating, so the data are mutually consistent.
6. Invert the same expression for part (c). At $f=65$ Hz, $\omega=408.41$ rad/s, and$$V_{LL}=\sqrt{\frac{4\,T_{\max}\,\omega^{2}L_{T}}{P}}=\sqrt{240\times 408.41^{2}\times 1.75\times 10^{-3}}$$$$\boxed{V_{LL}=264.7\ \text{V}\quad(152.8\ \text{V line to neutral})}$$
7. Test the two points against the volts-per-hertz rule. The first point runs at $346.41/79.47=4.359$ V/Hz and the second at $264.68/65=4.072$ V/Hz — a 7.0 per cent difference, so the drive is not on a single volts-per-hertz line. That is exactly why the breakdown torque changes: the supplied expression rewrites as $T_{\max}=(V_{LL}/f)^{2}P/(16\pi^{2}L_{T})$, so torque depends only on the ratio, and $(4.072/4.359)^{2}=0.873=240/275$ exactly. In practice the higher ratio at the lower frequency is the low-speed voltage boost that real drives apply to compensate the stator resistance drop.