Question 1 of 6: Snubbers, and a Full-Wave AC Voltage Controller Feeding a Motor

Question 1: Snubbers, and a Full-Wave AC Voltage Controller Feeding a 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.

Part (a) — What a snubber is, and why it is fitted

A snubber is a small auxiliary network — in its commonest form a resistor and a capacitor, often with a bypass diode — connected directly across a power semiconductor (or across a group of them) for the sole purpose of controlling the rate of change of voltage and current that the device experiences during switching. It carries no load current in the steady state; it exists to shape the switching trajectory. Because the capacitor cannot change its voltage instantaneously, placing one across a thyristor or transistor forces the device voltage to rise gradually after turn-off instead of stepping up at the rate the circuit stray inductance would otherwise impose. The series resistor limits the capacitor discharge current into the device at the next turn-on and damps the resonance formed by the snubber capacitance with the circuit inductance.

Snubbers are used for four distinct reasons, and a designer normally needs all four. First, dv/dt protection: a thyristor that sees a re-applied forward voltage rising faster than its rated $\mathrm{d}v/\mathrm{d}t$ will turn on spuriously, because the displacement current through its internal junction capacitance acts exactly like gate current. An RC snubber across the device holds the re-applied rate below the rating. Second, di/dt protection: at turn-on, conduction begins in a small area next to the gate and spreads outward at a finite velocity, so an excessive current slope concentrates the whole current in that small area and destroys the device thermally. A series reactor (the di/dt inductor, which is the dual of the RC snubber and is usually considered part of the same protection package) limits the slope until the conducting area has spread. Third, transient voltage suppression: the energy trapped in the stray inductance of the commutation loop, and the reverse-recovery charge of the freewheel or bridge diodes, would otherwise appear as a high-voltage spike across the device; the snubber capacitor absorbs it and the resistor dissipates it. Fourth, keeping the device inside its safe operating area: a turn-off snubber diverts current away from the device while its voltage is rising, so the instantaneous product of device voltage and device current — the switching loss — stays inside the SOA envelope and the switching energy is transferred to a resistor that is easy to cool.

The cost is real. Every joule stored in the snubber capacitor at turn-off is dissipated in the snubber resistor at the next turn-on, so snubber loss rises linearly with switching frequency and sets a practical ceiling on it. This is why hard-switched converters with modern IGBTs use the smallest snubber that satisfies the SOA and dv/dt limits, and why resonant (soft-switching) topologies, which shape the trajectory using the main circuit itself, have displaced dissipative snubbers in high-frequency designs. In line-commutated equipment of the kind in this question — 2300 V thyristor controllers switching at 60 Hz — snubber loss is negligible and generous RC networks are standard practice.

Parts (b) and (c) — Delay angle and the equivalent motor impedance

Given. A single-phase full-wave (back-to-back thyristor) ac voltage controller supplies a motor from a 2300 V, 60 Hz source; the measured conduction angle is 162.5°, the motor operates at 0.85 power factor lagging, and each thyristor carries an average current of 575 A.

Find. The firing (delay) angle $\alpha$ that produces the stated conduction angle, and then the series resistance $R$ and inductive reactance $X$ that represent the motor.

Approach. The extinction angle of an inductive load is set by the load, not by the gate: write the exact conduction-interval current, force it to zero at $\beta=\alpha+\gamma$ to get $\alpha$, then integrate the same waveform over one full cycle to relate the measured mean thyristor current to the unknown impedance magnitude, which the power factor resolves into $R$ and $X$.

1. Convert the power factor into a load angle. For a series R–L load the fundamental displacement angle is $$\varphi=\cos^{-1}(\mathrm{pf})=\cos^{-1}(0.85)=31.79^\circ,\qquad \tan\varphi=0.6197 .$$ This single number governs both the shape of the current pulse and the eventual split of the impedance into resistance and reactance.
2. Write the current during conduction. With the thyristor fired at $\alpha$ the load current is the sum of a steady-state sinusoid and the natural decay that enforces $i(\alpha)=0$: $$i(\omega t)=\frac{V_m}{Z}\left[\sin(\omega t-\varphi)-\sin(\alpha-\varphi)\, e^{(\alpha-\omega t)/\tan\varphi}\right],\qquad \alpha\le\omega t\le\beta .$$ Both the amplitude factor $V_m/Z$ and the shape factor in brackets matter, but only the bracket depends on $\alpha$. Call the bracket $u(\omega t)$; it is the normalised current, independent of the still-unknown $Z$.
3. Impose the extinction condition. Conduction ends when the current returns to zero, at $\omega t=\beta=\alpha+\gamma$. Setting $u(\beta)=0$, $$\sin(\alpha+\gamma-\varphi)=\sin(\alpha-\varphi)\,e^{-\gamma/\tan\varphi}.$$ This is transcendental in $\alpha$; it is the equation the question asks to be read backwards, since $\gamma$ is given and $\alpha$ is wanted. Solving it numerically over the physically sensible bracket 20°–120° gives $$\boxed{\alpha=49.11^\circ},\qquad \beta=\alpha+\gamma=211.61^\circ .$$
4. Confirm the root with the closed-form estimate. Because $e^{-\gamma/\tan\varphi}=e^{-4.576}=0.0103$ is small, the right-hand side nearly vanishes and $\alpha+\gamma-\varphi\approx180^\circ$, giving $\alpha\approx180^\circ+\varphi-\gamma=49.29^\circ$. The estimate is 0.18° high, consistent with the size of the exponential; it validates the numerical root but is not itself the answer.
5. Relate the mean thyristor current to the impedance. Each thyristor conducts once per supply cycle, so its mean current is the integral of the pulse divided by the full $2\pi$: $$I_{T,\text{avg}}=\frac{1}{2\pi}\int_{\alpha}^{\beta} i\,\mathrm{d}(\omega t) =\frac{V_m}{2\pi Z}\int_{\alpha}^{\beta} u\,\mathrm{d}(\omega t).$$ Numerical integration of the normalised waveform over 49.11° to 211.61° gives $\int u\,\mathrm{d}(\omega t)=1.7720$, so the mean of the normalised current is 0.28202.
6. Solve for the impedance magnitude. Rearranging and substituting the measured 575 A, $$Z=\frac{V_m}{I_{T,\text{avg}}}\cdot\frac{1}{2\pi}\int_{\alpha}^{\beta}u\,\mathrm{d}(\omega t) =\frac{3252.69}{575}\times 0.28202=1.5953\ \Omega .$$ The load impedance never had to be assumed: the shape of the pulse is fixed by $\alpha$ and $\varphi$ alone, so $Z$ falls out of a single measured average.
7. Split the impedance using the power factor. With $\varphi=31.79^\circ$, $$R=Z\cos\varphi=1.5953\times0.85=1.356\ \Omega,\qquad X=Z\sin\varphi=1.5953\times0.5268=0.840\ \Omega,$$ so $$\boxed{R=1.356\ \Omega,\qquad X_L=0.840\ \Omega}$$ which at 60 Hz corresponds to a motor inductance of $L=X/(2\pi f)=2.23\ \text{mH}$.
8. Cross-check on the rms quantities. The rms load current follows from the same normalised waveform, $I_{\text{rms}}=(V_m/Z)\sqrt{(1/\pi)\int u^2}=1336.7$ A, and the rms output voltage over the real conduction window is $$V_o=V_s\sqrt{\frac{\gamma-\tfrac12\left(\sin2\beta-\sin2\alpha\right)}{\pi}}=2203.9\ \text{V}.$$ The device-selection shortcut that treats the pulse as a half sine gives $\sqrt2 I_{\text{rms}}/\pi=601.7$ A against the stated 575 A — 4.6 per cent high, and conservative, which is exactly the behaviour expected of that approximation.

One subtlety is worth naming because it looks like an error and is not. Dividing the rms output voltage by the rms current gives 1.649 Ω, not the 1.595 Ω found above. The two disagree because the chopped waveform is far from sinusoidal: the harmonics in $v_o$ see reactance $nX$ rather than $X$, so the load draws proportionally less harmonic current than harmonic voltage, and the ratio of total rms quantities is not the fundamental impedance. The impedance asked for is the fundamental-frequency model of the motor, and that is what the mean-current route delivers.