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

Question 5 of 6: Side Effects of High-Frequency PWM Drives, and Constant Volts-per-Hertz Operation

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

Question 5: Side Effects of High-Frequency PWM Drives, and Constant Volts-per-Hertz Operation (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.

Introductory part — undesirable effects of high-frequency PWM drives

Raising the carrier frequency of a PWM inverter improves the current waveform and moves the acoustic noise out of the audible band, but it does so at a cost. At least five effects are worth naming.

First, insulation stress from high $dv/dt$. A fast IGBT edge presents the motor with a step of several thousand volts per microsecond. That step does not distribute itself evenly around the stator winding: the first few turns of the first coil see a disproportionate share of it, because at those rise times the winding behaves as a transmission line rather than a lumped inductance. Repeated at the carrier rate, the resulting partial discharge erodes the interturn insulation and shortens motor life, which is why inverter-duty machines specify reinforced magnet wire.

Second, and closely related, reflected-wave over-voltage on long motor cables. The cable surge impedance and the motor impedance are badly mismatched, so each edge reflects and can nearly double the terminal voltage. The critical cable length falls as the rise time shortens, so the faster the switching the shorter the cable that will cause trouble; $dv/dt$ filters or terminating networks are the usual remedy.

Third, shaft voltages and bearing currents. The common-mode component of the PWM output couples through the parasitic capacitances between stator winding, rotor and frame, impressing a voltage on the shaft. When it exceeds the dielectric strength of the lubricant film it discharges through the bearing, and the resulting electrical-discharge-machining pits progress to the characteristic fluting pattern and premature bearing failure. Insulated bearings, shaft grounding brushes and common-mode chokes are the standard countermeasures.

Fourth, electromagnetic interference. The same fast edges radiate and conduct broadband noise that couples into instrumentation, encoder cables and communication buses, and appears as conducted emission on the supply. Shielded cable, careful grounding and line filters are needed, and compliance with the relevant CSA/IEC emission standards has to be demonstrated.

Fifth, switching losses and derating. Loss in the semiconductors is very nearly proportional to switching frequency, so a higher carrier means more heat, larger heat sinks, and either a derated inverter or a more expensive one. Against this must be set the reduced harmonic loss in the machine — there is an optimum, and it is why general-purpose drives sit at a few kilohertz rather than tens of kilohertz.

Parts (a) and (b) — constant volts-per-hertz operation at breakdown torque

Given.

Find. (a) the line-to-line supply voltage and the line current at the 60 Hz operating point, and (b) the line-to-line supply voltage and the maximum output torque at 62.5 Hz.

Approach. Invert the supplied breakdown-torque approximation for the voltage, then use the fact that with negligible resistance the breakdown condition is $R_2'/s = \omega L_T$, so the stator sees $|Z| = \sqrt{2}\,\omega L_T$ and the current follows directly.

1. Part (a) — confirm the operating point. A four-pole machine at 60 Hz has a synchronous speed of $n_s = 120f/P = 120(60)/4 = 1800$ rpm, so the stated 1500 rpm at maximum torque corresponds to a breakdown slip of $$s_{\max} = \frac{1800 - 1500}{1800} = 0.1667 .$$ The leakage reactance at this frequency is $X_T = \omega_i L_T = 2\pi(60)(1.25\times10^{-3}) = 0.4712\ \Omega$.
2. Invert the torque approximation for the voltage. Rearranging $T_{\max} = V_{LL}^{2}P/(4\omega_i^{2}L_T)$, $$V_{LL} = \sqrt{\frac{4T_{\max}\omega_i^{2}L_T}{P}} = \sqrt{\frac{4(245)(376.99)^{2}(1.25\times10^{-3})}{4}} = \sqrt{43525} ,$$ $$\boxed{\;V_{LL} = 208.6 \ \text{V (line to line)}\;}$$ so the phase voltage is $V_{ph} = 208.6/\sqrt{3} = 120.45$ V.
3. Establish the impedance at breakdown. With the resistances neglected, the per-phase equivalent circuit reduces to $R_2'/s$ in series with $jX_T$, and the torque is maximised when the two are equal in magnitude, $R_2'/s = X_T$. The impedance seen by the supply at that point is therefore $$|Z| = \sqrt{\left(\frac{R_2'}{s}\right)^{2} + X_T^{2}} = \sqrt{2}\,X_T = 1.41421 \times 0.47124 = 0.66642\ \Omega .$$
4. Compute the line current. For a star-referred equivalent circuit the line current equals the phase current: $$\boxed{\;I_L = \frac{V_{ph}}{\sqrt{2}\,X_T} = \frac{120.45}{0.66642} = 180.7 \ \text{A}\;}$$ As a check, the air-gap power route must return the given torque. With $\omega_{sm} = \omega_i/(P/2) = 376.99/2 = 188.50$ rad/s, $$T = \frac{3I^{2}(R_2'/s)}{\omega_{sm}} = \frac{3(180.74)^{2}(0.47124)}{188.50} = 245.0 \ \text{N}\cdot\text{m},$$ which reproduces the given maximum torque exactly and confirms both the voltage and the current. Incidentally the referred rotor resistance follows as $R_2' = s_{\max}X_T = 0.0785\ \Omega$.
5. Part (b) — apply the constant volts-per-hertz law. The drive holds the voltage-to-frequency ratio at the value established in part (a), $$\frac{V_{LL}}{f} = \frac{208.63}{60} = 3.4771 \ \text{V/Hz} \quad\Rightarrow\quad \boxed{\;V_{LL2} = 3.4771 \times 62.5 = 217.3 \ \text{V}\;}$$ giving $V_{ph2} = 125.47$ V, and the new synchronous speed is $120(62.5)/4 = 1875$ rpm.
6. Recognise what the stated current implies. Under constant $V/f$ with an unchanged $L_T$, the breakdown current $V_{ph}/(\sqrt{2}\,\omega L_T)$ is independent of frequency, because $V_{ph}$ and $\omega$ rise together — it would still be 180.7 A at 62.5 Hz. The question specifies 185 A, so the total leakage inductance must have been changed. Solving the same current expression for $L_T$, $$L_{T2} = \frac{V_{ph2}}{\sqrt{2}\,\omega_2 I_L} = \frac{125.47}{1.41421(392.70)(185)} = 1.221 \ \text{mH}.$$
7. Compute the new maximum torque. Substituting into the supplied approximation with $\omega_2 = 2\pi(62.5) = 392.70$ rad/s, $$T_{\max 2} = \frac{V_{LL2}^{2}P}{4\omega_2^{2}L_{T2}} = \frac{(217.32)^{2}(4)}{4(392.70)^{2}(1.2212\times10^{-3})} = \frac{47228}{188.33},$$ $$\boxed{\;T_{\max 2} = 250.8 \ \text{N}\cdot\text{m}\;}$$
8. Cross-check by the invariance rule. Rewriting the supplied formula as $T_{\max} = (V_{LL}/f)^{2}P/(16\pi^{2}L_T)$ shows that under constant $V/f$ the breakdown torque depends on nothing but $L_T$. Therefore $T_{\max2}/T_{\max1}$ must equal $L_{T1}/L_{T2} = 1.250/1.221 = 1.0236$, giving $245 \times 1.0236 = 250.8$ N·m — identical to the direct calculation. The air-gap-power route agrees as well: $3(185)^{2}(392.70 \times 1.2212\times10^{-3})/196.35 = 250.8$ N·m.