AC voltage controller with R and RL loads

An AC voltage controller varies the rms AC voltage applied to a load at the supply frequency by phase-delayed firing of two anti-parallel thyristors, T1 (positive half) and T2 (negative half).

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2         R
 ( ~ ) vs = Vm sin wt  |
   |                 (L)  vo
   +-------------------+

Operation with purely resistive load

• $0$ to $\alpha$: no SCR conducts, $v_o = 0$.
• At $\alpha$, T1 is fired and $v_o = v_s$ till $\pi$, where current falls to zero and T1 turns off naturally.
• At $\pi + \alpha$, T2 is fired and $v_o = v_s$ till $2\pi$.
• Load current $i_o = v_o/R$ has the same shape as $v_o$.

 vo     .
        | \
 -------|--\--------.------
 0      a   pi      |    / 2pi
                    |  /
                    |/
                   pi+a

RMS output voltage (R load)

$$\begin{aligned} V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\; d(\omega t)\right]^{1/2} = \left[\frac{V_m^2}{2\pi}\left(\omega t - \frac{\sin 2\omega t}{2}\right)\Big|_{\alpha}^{\pi}\right]^{1/2} \\ &= \frac{V_m}{\sqrt{2}}\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \end{aligned}$$

Load current with R-L load

When T1 is fired at $\omega t = \alpha$, the circuit equation is

$$L\frac{di}{dt} + Ri = V_m\sin\omega t, \qquad i(\alpha) = 0$$

The solution has a steady-state part and a transient part:

$$i(t) = i_{ss} + i_{tr} = \frac{V_m}{Z}\sin(\omega t - \phi) + A\,e^{-\frac{R}{L}t}$$

where $Z = \sqrt{R^2 + (\omega L)^2}$ and $\phi = \tan^{-1}(\omega L/R)$.

Applying $i = 0$ at $\omega t = \alpha$ (i.e. $t = \alpha/\omega$):

$$A = -\frac{V_m}{Z}\sin(\alpha - \phi)\,e^{\frac{R\alpha}{\omega L}}$$

Since $\dfrac{R}{\omega L} = \dfrac{1}{\tan\phi}$:

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

Extinction angle $\beta$ (where $i = 0$ again):

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

• If $\alpha > \phi$: $\beta < \pi + \alpha$, current is discontinuous, and output voltage is controlled.
• If $\alpha \le \phi$: current becomes continuous and sinusoidal, $V_o = V_s$, and control is lost. So the control range is $\phi \le \alpha \le \pi$.

The rms output voltage for R-L load is then $V_o = V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]}$.