22-Elec-B8 Power Electronics and Drives · December 2015
Question 5 of 6: High-Frequency PWM Drawbacks and a Current-Source-Inverter-Fed Induction Motor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-B8 Power Electronics and Drives. Open Book, 3 hours. Six problems, all of equal value (20 marks each); the rubric states that any five constitute a complete paper and only the first five presented in the answer book are marked. All six are solved below, in full and including every sub-part, because this set is a study resource rather than an examination script.
Reference texts. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. (primary EGBC reference for this code); N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed.; C. W. Lander, Power Electronics, 3rd ed.; B. K. Bose, Modern Power Electronics and AC Drives; R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control. Angles are quoted in degrees but every integral is evaluated with the angle in radians, and each conduction integral is taken over the real conduction window rather than over an assumed half cycle.
Question 5: High-Frequency PWM Drawbacks and a Current-Source-Inverter-Fed Induction Motor (20 marks)
Part (a) — Undesirable effects of high switching frequency
Insulation stress from $dv/dt$ and reflected waves. Fast IGBT edges present rise times of the order of 50–100 ns, so the motor cable behaves as a transmission line. When the cable length exceeds roughly half the critical length for the rise time, the wave reflected from the high-impedance motor terminals doubles the voltage seen at the terminals, and the first few turns of each coil absorb most of that step because the voltage distributes non-linearly along the winding. The result is partial discharge in the turn-to-turn insulation and progressive failure, which is why inverter-duty machines specify reinforced magnet-wire enamel and why $dv/dt$ or sine filters are fitted on long runs.
Bearing currents and shaft voltages. The three inverter pole voltages do not sum to zero instant by instant, so a common-mode voltage appears between the stator neutral and earth at the switching frequency. This drives capacitive current through the stator-to-rotor and rotor-to-frame capacitances and across the bearing lubricant film. When the film breaks down, the discharge machines the raceway, producing the pitting and fluting that shortens bearing life dramatically. Insulated or ceramic bearings, shaft grounding rings and symmetrical shielded cable are the standard countermeasures.
Switching loss, EMI and secondary effects. Device loss during turn-on and turn-off is proportional to switching frequency, so raising the carrier trades converter efficiency and heatsink size for waveform quality. The same fast edges radiate and conduct electromagnetic interference across a wide spectrum, interfering with instrumentation and telemetry and causing nuisance operation of earth-leakage and ground-fault protection through the common-mode current path. Additional effects worth listing are the extra iron and eddy-current loss the ripple produces in the machine, high-frequency acoustic and ultrasonic noise from magnetostriction, and the reduced immunity of the drive itself to conducted disturbance.
Parts (b) and (c) — The current-source drive calculation
Given.
Quantity
Symbol
Value
Supply frequency
$f$
50 Hz
Poles / connection
$p$
8, star
Stator resistance
$R_s$
$0.2\ \Omega$
Rotor resistance (referred)
$R_r$
$0.3\ \Omega$
Stator leakage reactance
$X_s$
$1.0\ \Omega$
Rotor leakage reactance (referred)
$X_r$
$1.5\ \Omega$
Magnetising reactance
$X_m$
$10.42\ \Omega$
Inverter output current (constant)
$I_i$
30 A
Developed torque
$T$
120 N·m
Find. The slip and the corresponding rotor speed, then the terminal voltage per phase and the input power factor at that operating point.
Figure 5.1 — The approximate constant-current equivalent circuit of figure (2). The magnetising branch sits directly across the terminals, so the terminal voltage per phase is simply $I_mX_m$.
Approach. Multiplying the constant-current torque expression through by $s^2$ turns it into a quadratic in slip with two roots; the physically admissible root is chosen on air-gap flux, and the terminal quantities then follow from the current divider between the magnetising branch and the rotor branch.
Establish the synchronous speed. With eight poles at 50 Hz,
$$n_s = \frac{120f}{p} = \frac{120\times50}{8} = 750\ \text{rev/min}, \qquad \omega_s = \frac{2\pi\times750}{60} = 78.540\ \text{rad/s}.$$
The total series reactance appearing in the torque expression is $X = X_m + X_s + X_r = 10.42+1.0+1.5 = 12.92\ \Omega$.
Adopt the constant-current torque expression that the given circuit actually yields. With the inverter forcing $I_i$, the current divider between the magnetising branch and the rotor branch gives $I_r = I_iX_m/\sqrt{(R_s+R_r/s)^2+X^2}$, and the developed torque is the air-gap power divided by synchronous speed:
$$T = \frac{3I_r^2(R_r/s)}{\omega_s} = \frac{3(X_mI_i)^2(R_r/s)}{\omega_s\left[\left(R_s+\dfrac{R_r}{s}\right)^2+X^2\right]}.$$
Turn it into a quadratic in slip. Multiplying numerator and denominator by $s^2$ clears the $R_r/s$ terms and yields
$$T\omega_s\big[(R_ss+R_r)^2 + X^2s^2\big] = 3(X_mI_i)^2R_rs,$$
that is $As^2+Bs+C = 0$ with $A = T\omega_s(R_s^2+X^2)$, $B = 2T\omega_sR_sR_r - 3(X_mI_i)^2R_r$ and $C = T\omega_sR_r^2$.
Substitute the numbers. With $T\omega_s = 120\times78.540 = 9424.8$ W and $3(X_mI_i)^2R_r = 3\times(10.42\times30)^2\times0.3 = 87\,947$,
$$A = 9424.8\times166.966 = 1\,573\,678, \quad B = 9424.8\times0.12 - 87\,947 = -86\,816, \quad C = 9424.8\times0.09 = 848.2 .$$
Solving,
$$s_1 = 0.012689, \qquad s_2 = 0.042481 .$$
Locate the breakdown slip to confirm the two roots straddle it.
$$s_{\max} = \frac{R_r}{\sqrt{R_s^2+X^2}} = \frac{0.3}{\sqrt{0.04+166.926}} = 0.023217, \qquad T_{\max} = 142.23\ \text{N}\cdot\text{m}.$$
The demanded 120 N·m is 84 % of the breakdown torque, so a solution exists and the two roots lie one on each side of $s_{\max}$, exactly as expected.
Select the root on air-gap flux, not on static stability. Evaluating the current divider at each root, the low-slip root draws a magnetising current of 26.52 A out of the 30 A injected, which is 88 % of the inverter output and implies a terminal voltage of 276.35 V per phase — roughly 1.7 times the flux of the other root, deep into saturation and far above any plausible rating for this machine. The high-slip root draws 15.55 A of magnetising current, a normal 52 % for an eight-pole machine. The physically realizable point is therefore
$$\boxed{s = 0.04248}$$
and the rotor speed follows as
$$\boxed{N_r = n_s(1-s) = 750(1-0.04248) = 718.1\ \text{rev/min}}$$
a slip speed of 31.9 rev/min.
Evaluate the branch impedance and split the inverter current. At $s = 0.042481$, $R_r/s = 7.0621\ \Omega$, so the rotor branch is
$$Z_{br} = (R_s + R_r/s) + j(X_s+X_r) = 7.2621 + j2.500 = 7.680\angle 19.00^\circ\ \Omega .$$
The divider against the magnetising branch gives
$$I_r = I_i\frac{jX_m}{jX_m+Z_{br}} = 21.09\ \text{A}, \qquad I_m = I_i\frac{Z_{br}}{jX_m+Z_{br}} = 15.55\ \text{A}.$$
Read the terminal voltage off the magnetising branch. In the approximate circuit $X_m$ is connected directly across the machine terminals, so
$$\boxed{V_s = I_mX_m = 15.55\times10.42 = 162.0\ \text{V per phase}}$$
equivalent to 280.6 V line-to-line.
Obtain the power factor from the input impedance angle. The impedance the inverter sees is $Z_{in} = jX_m\,\|\,Z_{br}$, whose angle is $48.35^\circ$, hence
$$\boxed{\cos\varphi = 0.665\ \text{lagging}}$$
Close the power balance as an independent check. The input power computed from the terminal quantities must equal the developed mechanical power plus the stator copper loss:
$$3V_sI_i\cos\varphi = 3\times161.99\times30\times0.6648 = 9691.7\ \text{W},$$
$$T\omega_s + 3I_r^2R_s = 9424.8 + 3\times21.09^2\times0.2 = 9424.8 + 266.9 = 9691.7\ \text{W}.$$
The two agree to five figures, which simultaneously validates the chosen root, the current divider and the parallel combination.
Figure 5.2 — Constant-current torque–slip locus. The 120 N·m demand line cuts the curve twice; the low-slip intersection starves the rotor branch and over-fluxes the machine, so the high-slip intersection is the operating point.
Check
Two points are worth recording. First, the torque formula printed on the question paper carries an additional factor of $s$ in its denominator. Taken literally it returns $s = 0.235$, a slip at which the equivalent circuit of figure (2) develops only 28.2 N·m rather than the stated 120 N·m, so the printed form is not self-consistent with the circuit it accompanies. The expression used above is the one the given circuit yields, $T = 3I_r^2(R_r/s)/\omega_s$, and it is the standard constant-current result; a candidate should state the discrepancy and proceed on the consistent version, as the rubric on stating assumptions invites. Second, the selected operating point lies beyond the breakdown slip, so it is statically unstable under open-loop current control. That is not an error: it is precisely why current-source inverter drives are always closed on slip frequency or on flux, and the instability is a property of the drive topology rather than of this calculation.
