Question 3 of 6: Problem 2 — single-phase ac voltage controller feeding a motor

Question 3: Problem 2 — single-phase ac voltage controller feeding a motor (15 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.

Given. A pair of inverse-parallel thyristors chops a 120 V, 60 Hz supply into an inductive motor load, and the controller is held at a fixed conduction angle while the load power factor changes between starting and full load.

Find. (a) the delay angle $\alpha$ that produces a 135° conduction angle at each power factor; (b) the corresponding rms output-to-input voltage ratio $V_{o}/V_{s}$.

Approach. The current in an inductive load does not stop at the supply zero crossing but at the angle $\beta$ where the natural and forced components of the transient cancel; that extinction condition, read backwards, gives $\alpha$ for a specified $\gamma$, and the rms of the surviving window then gives the voltage ratio.

1. Write the load current from firing to extinction. For an $R$–$L$ load fired at $\alpha$, $$i(\omega t)=\frac{V_{m}}{|Z|}\left[\sin(\omega t-\varphi)-\sin(\alpha-\varphi)\,e^{(\alpha-\omega t)/\tan\varphi}\right]$$ where $\varphi=\arctan(\omega L/R)$ is the load angle.
2. Set the current to zero at the end of the stated window. Putting $\omega t=\beta=\alpha+\gamma$ and cancelling $V_{m}/|Z|$ leaves a relation that contains neither the impedance magnitude nor the supply voltage: $$\sin(\alpha+\gamma-\varphi)=\sin(\alpha-\varphi)\,e^{-\gamma/\tan\varphi}$$
3. Solve it numerically at the starting power factor. With $\cos\varphi=0.4$, $\varphi=66.42^\circ$ and $e^{-\gamma/\tan\varphi}=0.3576$; a bracketed root search on $[\varphi,180^\circ]$ returns $$\boxed{\alpha_{1}=100.01^\circ}\qquad \beta_{1}=235.01^\circ$$
4. Repeat at full load. With $\cos\varphi=0.8$, $\varphi=36.87^\circ$ and $e^{-\gamma/\tan\varphi}=0.04321$, giving $$\boxed{\alpha_{2}=80.17^\circ}\qquad \beta_{2}=215.17^\circ$$

Both roots deserve a sanity check against the familiar closed form $\alpha\approx 180^\circ+\varphi-\gamma$, which follows from neglecting the decaying exponential. At $\cos\varphi=0.8$ that estimate gives 81.87°, only 1.70° high, because the exponential factor is a mere 0.043. At $\cos\varphi=0.4$ it gives 111.42°, a full 11.41° high, because the factor has grown to 0.358. The approximation tracks $e^{-\gamma/\tan\varphi}$ and is a check, never the answer, at poor power factors.

5. Write the rms output over the real conduction window. The output equals the supply between $\alpha$ and $\beta$ in each half cycle and is zero elsewhere, so $$\frac{V_{o}}{V_{s}}=\sqrt{\frac{1}{\pi}\left[\gamma-\frac{\sin 2\beta-\sin 2\alpha}{2}\right]}$$ with $\gamma$ in radians.
6. Evaluate at the starting condition. With $\alpha_{1}=100.01^\circ$, $\beta_{1}=235.01^\circ$: $\sin 2\beta_{1}=0.9396$, $\sin 2\alpha_{1}=-0.3424$, so $$\frac{V_{o}}{V_{s}}=\sqrt{\frac{2.35619-0.64097}{\pi}}=\boxed{0.739}\quad(88.7\ \text{V})$$
7. Evaluate at full load. With $\alpha_{2}=80.17^\circ$, $\beta_{2}=215.17^\circ$: $\sin 2\beta_{2}=0.9417$, $\sin 2\alpha_{2}=0.3364$, so $$\frac{V_{o}}{V_{s}}=\sqrt{\frac{2.35619-0.30267}{\pi}}=\boxed{0.808}\quad(97.0\ \text{V})$$

The pair moves the way physical reasoning demands and that is worth stating as the check on both halves of the problem: at a fixed conduction angle a better power factor needs an earlier gate pulse, and an earlier pulse slides the 135° window towards the crest of the sine, so the rms output rises. Here 0.739 at 100.01° against 0.808 at 80.17° — smaller angle, larger ratio. A solution in which the ratios moved the other way would contain an arithmetic error somewhere.