Question 1 of 6: Snubbers, and an a.c. voltage controller driving an inductive motor

Question 1: Snubbers, and an a.c. voltage controller driving an inductive motor (20 points)

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) — Snubbers

A snubber is a small auxiliary network fitted around a power semiconductor whose only purpose is to shape the voltage-and-current trajectory the device follows while it is switching. It carries no useful load power; it exists so that the switching locus stays inside the device's safe operating area. Two forms dominate. The turn-off or dv/dt snubber is a capacitor, usually with a series resistor and a bypass diode across that resistor, connected directly across the device. As the device turns off, the capacitor accepts the load current while the device voltage is still low, so voltage and current never reach their peaks simultaneously and the turn-off energy is transferred out of the silicon. The turn-on or di/dt snubber is a small series inductor, often an air-cored or saturable reactor, in the anode lead. It limits the rate of rise of anode current while the gate-triggered conducting plasma is still spreading laterally across the junction, so the initially small conduction area is not asked to carry full current and does not fail by local hot-spotting.

Snubbers are used for four reasons. First, they enforce the manufacturer's $\mathrm{d}v/\mathrm{d}t$ and $\mathrm{d}i/\mathrm{d}t$ ratings, and for a thyristor this is not merely a stress question: a reapplied forward $\mathrm{d}v/\mathrm{d}t$ above the rating injects displacement current into the gate-cathode junction and turns the SCR on spuriously, without any gate signal. Second, they damp the ringing that circuit stray inductance forms with device capacitance during reverse recovery, which otherwise produces overvoltage and radiated interference. Third, by moving the switching locus away from the high-voltage-and-high-current corner they reduce the instantaneous power dissipated in the die, allowing higher switching frequency for a given junction temperature. Fourth, in series-connected strings the same RC network equalises the transient voltage sharing between devices with unequal recovery charge. The cost is real: the energy $\tfrac{1}{2}CV^{2}$ stored in the snubber capacitor is dissipated in the snubber resistor once every switching cycle, so an oversized snubber trades device stress for converter efficiency.

Parts (b) and (c) — The a.c. voltage controller

Given.

Find. The delay angle $\alpha$ that produces the stated conduction angle, and then the equivalent series resistance and inductive reactance of the motor.

Approach. Fix the load angle from the power factor, read the extinction condition backwards to obtain $\alpha$ from the stated $\gamma$, then use the fact that the current pulse shape depends only on $\alpha$ and $\phi$ to convert the one measured mean device current into the load impedance.

1. Part (b) — Fix the load angle from the stated power factor. The motor is represented by a series $R + jX$, so the current lags the voltage by $\phi = \arccos(0.875) = 28.955^\circ$, and $\tan\phi = X/R = 0.55328$. This angle is a property of the load alone and is settled before any converter algebra begins.
2. Write the conduction-interval current and impose extinction. Over $\alpha \le \omega t \le \beta$ the thyristor connects the supply straight to the load, so $$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{(\alpha - \omega t)/\tan\phi}\right],$$ the second term being the natural response that forces $i(\alpha) = 0$. The device turns off when the current returns to zero, and with $\beta = \alpha + \gamma$ that condition reads $$\sin(\alpha + \gamma - \phi) = \sin(\alpha - \phi)\,e^{-\gamma/\tan\phi}.$$
3. Evaluate the damping factor and solve for the delay angle. With $\gamma = 162.5^\circ = 2.83616$ rad, $e^{-\gamma/\tan\phi} = e^{-5.1261} = 5.932\times10^{-3}$. The equation is transcendental in $\alpha$; solving it numerically over the physical range $\phi < \alpha < 180^\circ$ gives $$\boxed{\alpha = 46.353^\circ}\qquad \beta = \alpha + \gamma = 208.853^\circ .$$
4. Confirm with the closed-form estimate. Because the exponential is only $5.9\times10^{-3}$, the right-hand side is nearly zero and the condition collapses to $\alpha + \gamma - \phi \approx 180^\circ$, i.e. $\alpha \approx 180^\circ + \phi - \gamma = 46.455^\circ$. That is 0.10° above the exact root, so the closed form is a sound sanity check here — but only because $\gamma$ is large and the power factor good; at a poorer power factor the same estimate can be several degrees out.
5. Part (c) — Normalise the current pulse. The shape of the conduction pulse depends only on $\alpha$ and $\phi$, not on the size of the load, so define the dimensionless waveform $$u(\omega t) = \frac{Z\,i(\omega t)}{V_m} = \sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{(\alpha-\omega t)/\tan\phi}.$$ Numerical integration over the conduction window gives $\int_{\alpha}^{\beta} u \,\mathrm{d}(\omega t) = 1.78979$.
6. Convert the measured mean device current into the load impedance. Each thyristor of the back-to-back pair conducts once per supply cycle, so its mean current is averaged over $2\pi$, not over $\pi$: $$I_{T,\mathrm{avg}} = \frac{V_m}{Z}\cdot\frac{1}{2\pi}\int_{\alpha}^{\beta} u\,\mathrm{d}(\omega t).$$ Rearranging, $$Z = \frac{V_m}{I_{T,\mathrm{avg}}}\cdot\frac{1.78979}{2\pi} = \frac{3252.69}{625}\times 0.284847 = \boxed{1.4825\ \Omega}.$$ No power rating, efficiency or line current was needed — one average device current is sufficient because $Z$ is the only free scale factor in the waveform.
7. Split the impedance into its resistive and reactive parts. The load angle already fixes the split, so $$R = Z\cos\phi = 1.4825 \times 0.875 = \boxed{1.2972\ \Omega},\qquad X = Z\sin\phi = 1.4825 \times 0.48412 = \boxed{0.7177\ \Omega},$$ which at 60 Hz corresponds to a motor inductance $L = X/(2\pi f) = 1.904$ mH.
8. Cross-check the answer three ways. The rms output voltage over the real conduction window is $$V_{o} = V_s\sqrt{\frac{\gamma - \tfrac{1}{2}\left(\sin 2\beta - \sin 2\alpha\right)}{\pi}} = 2214.7\ \text{V},$$ and the rms load current follows from the same normalised waveform as $I_{rms} = 1452.1$ A. First check: the half-sine device shortcut $\sqrt{2}\,I_{rms}/\pi = 653.7$ A sits 4.6% above the stated 625 A — high and therefore conservative, which is exactly what that approximation should do. Second check: $V_o/I_{rms} = 1.5252\ \Omega$ is 2.9% above $|Z| = 1.4825\ \Omega$, and that gap is physical, not an error: the chopped wave carries harmonics which see $nX$, so the load draws proportionally less harmonic current than harmonic voltage and the ratio of total rms quantities is not the fundamental impedance. Third check: $I_{rms}^{2}R = 2.735$ MW, consistent with a large lagging load drawing 625 A mean per device.

Final results.