AC voltage controller RL load extinction angle

For an AC voltage controller with R-L load, current continues after the voltage zero until the extinction angle $\beta$, found from the load current equation.

Given: $V_s = 230$ V, $R = 2\ \Omega$, $X_L = 2\ \Omega$, $\alpha = 45°$.

$$\begin{aligned} Z &= \sqrt{2^2 + 2^2} = 2.828\ \Omega \\ \phi &= \tan^{-1}\frac{X_L}{R} = \tan^{-1}1 = 45° \end{aligned}$$

Load current and extinction condition

For $\alpha \le \omega t \le \beta$:

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

At $\omega t = \beta$, $i = 0$:

$$\sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-(\beta - \alpha)/\tan\phi}$$

(i) Extinction angle

Here $\alpha = \phi = 45°$, so $\sin(\alpha - \phi) = 0$ and the transient term vanishes:

$$\sin(\beta - 45°) = 0 \;\Rightarrow\; \beta = 180° + 45° = 225°$$

(ii) Conduction angle

$$\gamma = \beta - \alpha = 225° - 45° = 180°$$

(iii) RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\ &= 230\sqrt{\frac{1}{\pi}\left[\pi + \frac{\sin 90° - \sin 450°}{2}\right]} = 230\sqrt{1} = 230\ \text{V} \end{aligned}$$

Load current $I_o = 230/2.828 = 81.3$ A (pure sinusoid).

Note: $\alpha = \phi$ is the boundary between discontinuous ($\alpha > \phi$) and continuous conduction. Each SCR conducts exactly 180°, the current is a pure sine wave lagging by 45°, and the controller gives full output; any $\alpha < 45°$ also gives full output and loses control.

(iv) Waveforms

 vs,vo  /\        /\
       /  \      /  \
 -----/----\----/----\---
     0  45  \  / 225
  io    __   \/
       /  \        /
 -----/----\------/------
     45    225   405
 T1: 45 -> 225   T2: 225 -> 405

Answer: $\beta = 225°$, $\gamma = 180°$, $V_o = 230$ V.