AC voltage controller RL load at α = 90°

With an R-L load, current lags; since $\alpha$ is greater than the load angle $\phi$, conduction is discontinuous and each SCR conducts from $\alpha$ to the extinction angle $\beta$.

Given: $V_s = 230$ V, 50 Hz, $R = 20\ \Omega$, $L = 50$ mH, $\alpha = 90°$.

Load parameters

$$\begin{aligned} X_L &= 2\pi f L = 2\pi(50)(0.05) = 15.708\ \Omega \\ Z &= \sqrt{20^2 + 15.708^2} = 25.431\ \Omega \\ \phi &= \tan^{-1}\frac{15.708}{20} = 38.15°, \quad \tan\phi = 0.7854 \end{aligned}$$

Since $\alpha = 90° > \phi = 38.15°$, conduction is discontinuous.

Extinction angle

$$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right]$$

Setting $i(\beta) = 0$:

$$\sin(\beta - 38.15°) = \sin(51.85°)\,e^{-(\beta - 90°)\pi/(180°\times 0.7854)}$$

Solving numerically (trial and error):

$$\beta = 215.37°, \qquad \gamma = \beta - \alpha = 125.37°$$

Check: $\beta < 180° + \alpha = 270°$, so T1 stops before T2 is fired (discontinuous), as expected.

RMS load voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\ &= 230\sqrt{\frac{1}{\pi}\left[2.1880 + \frac{0 - 0.9440}{2}\right]} = 230\sqrt{\frac{1.7160}{\pi}} \\ &= 230 \times 0.7391 = 169.99\ \text{V} \end{aligned}$$

(Here $\beta - \alpha = 125.37° = 2.1880$ rad, $\sin 2\alpha = \sin 180° = 0$ and $\sin 2\beta = \sin 430.73° = 0.9440$.)

RMS load current

$$I_o = \left[\frac{1}{\pi}\int_{\alpha}^{\beta} i^2(\omega t)\, d(\omega t)\right]^{1/2}$$

Evaluating with $V_m/Z = 325.27/25.431 = 12.79$ A gives

$$I_o = 5.575\ \text{A}$$

Check by power balance: $P = I_o^2 R = 5.575^2 \times 20 = 621.6$ W, equal to the average of $v_o i_o$.

Answer: $\beta = 215.37°$, $V_{o,rms} = 169.99$ V, $I_{o,rms} = 5.575$ A.