Question 2 of 6: Problem 1 — half-wave controlled rectifier with an R-L load

Question 2: Problem 1 — half-wave controlled rectifier with an R-L load (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.

Find. For each case, the delay angle $\alpha$, and then the missing one of the pair (mean output current, load resistance).

Approach. The load angle follows from the power factor; the extinction angle condition for a half-wave rectifier into R-L turns the stated conduction angle into the delay angle; the mean output voltage over the full period then gives the missing current or resistance, because the mean voltage across an inductance over a complete cycle is zero.

1. Fix the load angle from the power factor. The load is R in series with L, so the impedance angle is $$\phi=\arccos(0.707)=45.01^{\circ},\qquad \tan\phi=\omega L/R=1.0003$$ The stated power factor of 0.707 is the familiar rounding of $1/\sqrt2$, so this is a load with $\omega L=R$ to within three parts in ten thousand. Both cases share this angle; only the conduction angle differs.
2. Write the current waveform and the extinction condition. With the thyristor fired at $\alpha$, the loop equation $L\,\mathrm{d}i/\mathrm{d}t+Ri=V_m\sin\omega t$ with $i(\alpha)=0$ gives the standard result $$i(\omega t)=\frac{V_m}{Z}\Bigl[\sin(\omega t-\phi)-\sin(\alpha-\phi)\, e^{(\alpha-\omega t)/\tan\phi}\Bigr],\qquad \alpha\le\omega t\le\beta$$ where the first term is the steady-state response and the second the decaying transient that enforces zero initial current. Conduction ends at the extinction angle $\beta$, where the current returns to zero, so setting $i(\beta)=0$ gives the condition that must be solved for $\alpha$: $$\boxed{\;\sin(\beta-\phi)=\sin(\alpha-\phi)\,e^{(\alpha-\beta)/\tan\phi},\qquad \beta=\alpha+\gamma\;}$$ Note what this says physically: the load, not the gate, decides where conduction ends. Specifying $\gamma$ is therefore an indirect way of specifying $\alpha$.
3. Solve for the Case A delay angle. With $\gamma_A=145^{\circ}$ the exponential factor is $e^{-\gamma/\tan\phi}=e^{-2.5307}=0.07966$, and a numerical solve (Brent bracketing on 46° to 89.9°) returns $$\boxed{\;\alpha_A=77.55^{\circ}\;}\qquad \beta_A=\alpha_A+\gamma_A=222.55^{\circ}$$ As a check, the closed-form estimate $\alpha\approx 180^{\circ}+\phi-\gamma$, which is exact only when the exponential term vanishes, gives 80.01°. Its error of 2.5° tracks the size of that exponential factor (0.0797 here), so at this modest power factor the closed form is a sanity check and not the answer.
4. Find the mean output voltage in Case A. The load terminals follow the supply from $\alpha$ to $\beta$ and are at zero for the rest of the period, so averaging over the full $2\pi$ of a half-wave circuit, $$V_{dc}=\frac{1}{2\pi}\int_{\alpha}^{\beta}V_m\sin\theta\,\mathrm{d}\theta =\frac{V_m}{2\pi}\bigl(\cos\alpha-\cos\beta\bigr)$$ Substituting $\cos 77.55^{\circ}=0.21554$ and $\cos 222.55^{\circ}=-0.73666$, $$V_{dc,A}=\frac{169.706}{2\pi}\,(0.95220)=25.718\ \text{V}$$ The divisor is $2\pi$, not $\pi$: this is a half-wave circuit with one conduction interval per supply cycle. Using $\pi$ here would halve the deduced resistance.
5. Deduce the Case A load resistance. Over a complete cycle the mean voltage across an inductance is zero, because the current is periodic and $\bar v_L=(L/2\pi)\oint \mathrm{d}i=0$. The whole mean output voltage therefore appears across R, and $$\begin{aligned} I_{dc}&=\frac{V_{dc}}{R}\ R_A&=\frac{V_{dc,A}}{I_A}=\frac{25.718}{25}=\boxed{\;1.0287\ \Omega\;} \end{aligned}$$ This is why no integration of the current waveform is needed: one stated average current measures R directly.
6. Repeat for Case B. With $\gamma_B=150^{\circ}$ the exponential factor is $e^{-2.6180}=0.07301$ and the same solve gives $$\boxed{\;\alpha_B=73.04^{\circ}\;}\qquad \beta_B=223.04^{\circ}$$ Then, with $\cos 73.04^{\circ}=0.29167$ and $\cos 223.04^{\circ}=-0.73085$, $$V_{dc,B}=\frac{169.706}{2\pi}\,(1.02252)=27.618\ \text{V}$$
7. Deduce the Case B mean current. With the resistance given as 1.2 ohm, $$I_B=\frac{V_{dc,B}}{R_B}=\frac{27.618}{1.2}=\boxed{\;23.01\ \text{A}\;}$$ The two cases are mutually consistent as a sanity check: the longer conduction angle of Case B admits more voltage (27.62 V against 25.72 V), and the resistances deduced or supplied, 1.029 and 1.200 ohm, are of the same order, as one would expect of two operating points on one machine.
8. Confirm the results are physically sensible. Retarding the firing angle from 73.04° to 77.55°, a change of 4.51°, moves the extinction angle by only 0.49° (223.04° to 222.55°). That insensitivity is the point the question is built around: with $\omega L=R$ the current runs about 43° past the supply zero, and where it stops is set by the load time constant, not by the gate.