22-Elec-B8 Power Electronics and Drives · May 2013

Question 5 of 6: Effects of High-Frequency PWM, and Constant Volts-per-Hertz Control of an Induction Motor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Effects of High-Frequency PWM, and Constant Volts-per-Hertz Control of an Induction Motor (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.

(a) Three undesirable effects of high-frequency PWM drives

Switching losses and thermal penalty. Every transition of an IGBT or MOSFET dissipates a finite energy as the device traverses the active region, and diode reverse-recovery adds to it. Since these losses are incurred once per switching event, total switching loss is directly proportional to carrier frequency. Raising the carrier therefore demands larger heatsinks or a derated converter, and it erodes the efficiency advantage that the drive was installed to obtain. High-frequency flux ripple adds eddy-current and hysteresis loss in the motor iron on top of this.

Insulation stress, reflected waves and bearing currents. Fast switching means high $dv/dt$ at the inverter terminals. Three distinct problems follow. The steep edges distribute themselves unevenly across the first few turns of the stator winding, so those turns and the phase insulation see far more than their share of the voltage and suffer partial discharge and premature failure. On a long motor cable the pulse behaves as a travelling wave on a transmission line and reflects from the high-impedance motor terminals, so the terminal voltage can approach twice the dc-link voltage. And the $dv/dt$ drives common-mode capacitive current through the stator-to-frame capacitance, part of which returns through the bearing lubricant film as a discharge, producing the fluting and pitting characteristic of inverter-fed bearing damage.

Electromagnetic interference and leakage current. High-frequency switching is a broadband source of both conducted and radiated interference. Conducted emissions travel back along the supply and can corrupt nearby instrumentation, encoder feedback and communication circuits; common-mode leakage current flowing to earth through cable and motor capacitance can cause nuisance tripping of ground-fault protection and makes proper screening, bonding and filter design mandatory. A carrier chosen in the audible range also produces tonal acoustic noise from the motor laminations, which is a real objection in occupied buildings.

(b), (c) Constant volts-per-hertz operation at breakdown torque

Given. A three-phase, four-pole induction motor on a constant volts-per-hertz drive, with negligible resistance, is described by the data below.

Find. For case (b), the maximum output torque and the line current at that operating point. For case (c), the leakage inductance that makes the line current 85 A at 420 Hz, together with the line voltage the constant V/Hz law then demands and the resulting maximum torque.

Approach. Use the supplied torque formula for the peak torque. For the current, recognise that at breakdown torque the referred rotor resistance equals the leakage reactance, so the magnitude of the per-phase impedance is $\sqrt{2}\,\omega L_T$; inverting that relation in case (c) yields the required inductance.

1. Maximum torque at 460 V, 410 Hz — part (b). The electrical angular frequency is $\omega_i = 2\pi(410) = 2576.1$ rad/s, so the leakage reactance is $\omega_i L_T = 2576.1 \times 1.25\times 10^{-3} = 3.220\ \Omega$. Substituting into the formula supplied on the paper,$$T_{max} = \frac{[V_{LL}]^2 P}{4[\omega_i]^2 L_T} = \frac{(460)^2 (4)}{4(2576.1)^2(1.25\times 10^{-3})} = \frac{846\,400}{33\,182}$$which gives$$\boxed{T_{max} = 25.51\ \text{N}\cdot\text{m}}$$
2. Line current at that operating point. Because the resistances are negligible, the per-phase circuit is just the referred rotor resistance $R_2'/s$ in series with the leakage reactance $\omega_i L_T$. Maximum torque occurs when these two are equal, so at that slip the impedance magnitude is$$|Z| = \sqrt{(R_2'/s)^2 + (\omega_i L_T)^2} = \sqrt{2}\,\omega_i L_T = \sqrt{2}(3.220) = 4.554\ \Omega$$With a phase voltage of $V_{ph} = 460/\sqrt{3} = 265.58$ V the current is$$I = \frac{V_{ph}}{\sqrt{2}\,\omega_i L_T} = \frac{265.58}{4.554}$$so$$\boxed{I = 58.31\ \text{A (line)}}$$This is worth verifying independently through the air-gap power, since the two routes are logically distinct: $T = 3I^2(R_2'/s)(P/2)/\omega_i = 3(58.31)^2(3.220)(2)/2576.1 = 25.51$ N·m, which reproduces part (b) exactly.
3. The constant V/Hz law fixes the new voltage — part (c). A constant volts-per-hertz drive holds the ratio $V_{LL}/f$ fixed so as to keep the air-gap flux constant:$$\frac{V_{LL}}{f} = \frac{460}{410} = 1.1220\ \text{V/Hz} \quad\Longrightarrow\quad V_{LL} = 1.1220 \times 420$$so at 420 Hz$$\boxed{V_{LL} = 471.22\ \text{V}}$$and the phase voltage becomes $471.22/\sqrt{3} = 272.06$ V.
4. Leakage inductance for 85 A. Inverting the current relation of step 2 to solve for the reactance instead,$$\omega_i L_T = \frac{V_{ph}}{\sqrt{2}\,I} = \frac{272.06}{\sqrt{2}(85)} = \frac{272.06}{120.21} = 2.2632\ \Omega$$Dividing by $\omega_i = 2\pi(420) = 2638.9$ rad/s,$$\boxed{L_T = 0.858\ \text{mH}}$$The required inductance is smaller than the original 1.25 mH, which is consistent: a lower leakage reactance lets the machine draw more current (85 A against 58.3 A) at breakdown.
5. Maximum torque with the new inductance. Applying the torque formula again with the new voltage, frequency and inductance,$$T_{max} = \frac{(471.22)^2(4)}{4(2638.9)^2(0.858\times 10^{-3})} = \frac{888\,193}{23\,882}$$giving$$\boxed{T_{max} = 37.19\ \text{N}\cdot\text{m}}$$There is an instructive shortcut hiding in this result. Since $\omega_i = 2\pi f$, the torque formula can be rewritten as $T_{max} = (V_{LL}/f)^2 P/(16\pi^2 L_T)$, in which frequency does not appear at all. Under a constant V/Hz law the breakdown torque therefore depends only on the volts-per-hertz setting and on $1/L_T$, so the torque ratio must equal the inverse inductance ratio: $37.19/25.51 = 1.458$ and $1.25/0.858 = 1.457$. The agreement confirms both answers.