Figure 1.a shows a single phase a.c. voltage controller with RL load.The waveforms for source voltage $E_{s},$ gate currents $i_{g_{1}}$ and $i_{g_{2}},$ load and source currents $i_{0}$ and $i_{s},$ load voltage $e_{0},$ and thyristor voltages are shown in Fig.1.b.

During the interval zero to $\pi$ , thyristor $T_{1}$ is forward biased. At $\omega t=\alpha, T_{1}$ is triggered and $i_{0}=i_{T_{1}}$ starts building up through the load. At $\pi$ load and source voltages are zero but the current is not zero because of the presence of inductance in the load circuit.

Thyristor $T_{1}$ will continue to conduct until its current falls to zero at $\omega t=\beta$ . Angle $\beta$ is called as the extinction angle. The load is subjected to the source votage from $\alpha$ to $\beta$ . At $\beta,$ when $i_{0}$ is zero, $T_{1}$ is turned-off as it is already reversed biased. After the commutation of $T_{1}$ at $\beta,$ a voltage of magnitude $E_{m} \sin \beta$ at once appears as a reverse bias across $T_{1}$ and as a forward bias across $T_{2},$ as shown in Fig.1.b.

From $\beta$ to $\pi+\alpha,$ no current exists in the power circuit. Thyristor $T_{2}$ is turned-on at $(\pi+\alpha) \gt\beta .$ Current $i_{0}=i_{T_{2}}$ starts building up in the reversed direction through the load. At $2 \pi, e_{s}$ and $e_{0}$ are zero but $i_{T_{2}}=i_{0}$ is not zero. At $(\pi+\alpha+\gamma), i_{T_{2}}=0$ and $T_{2}$ is turned-off because it is already reverse biased. At $(\pi+\alpha+\gamma), E_{m} \sin (\pi+\alpha+\gamma)$ appears as a forward bias across $T_{1}$ and as a reverse bias across $T_{2},$ as shown in Fig.1.b . From $(\pi+\alpha+\gamma )$ to $(2 \pi+\alpha)$ no current exists in the power circuit. At $(2 \pi+\alpha), T_{1}$ is turned-on and current starts building up as before.

The expression for load current $i_{s}$ and $\beta$ can be obtained as follows:

For $\alpha \leq \omega t \leq \beta,$ the $\mathrm{KVL}$ for the circuit of Fig.1.a gives,

$$e_{s}=E_{m} \sin \omega t=R \cdot i_{0}+L \frac{\mathrm{d} i_{0}}{\mathrm{d} t}-----(1)$$

The solution of this equation is of the form

$$i_{0}=\frac{E_{m}}{Z} \sin (\omega t-\phi)+A \cdot e^{-(R / L) t}-----(2)$$

$$\begin{array}{ll}{\text { where }} & {Z=\left(R^{2}+\omega^{2} L^{2}\right)^{1 / 2}}-----(3) \\ {\text { and }} & {\phi=\tan ^{-1} \frac{\omega L}{R}}-----(4)\end{array}$$

Constant $A$ can be obtained from the initial condition according to which $i_{0}=0$ at $\omega t=\alpha,$ i.e. at $t=\alpha / \omega .$ Therefore, $$0=\frac{E_{m}}{Z} \sin (\alpha-\phi)+A \cdot e^{-R \alpha / \omega L} \ or \ A=\frac{-E_{m}}{Z} \sin (\alpha-\phi) e^{R \alpha / \omega L}-----(5)$$

Substitution of A from Eq. $(5)$ in Eq. $(2)$ yields

$$i_{0}=\frac{E_{m}}{Z}\left[\sin (\omega t-\phi)-\sin (\alpha-\phi) \exp \left\{\frac{R}{L}\left(\frac{\alpha}{\omega}-t\right)\right\}\right]-----(6)$$

From Fig.1.b, it becomes clear that the load current $i_{0}$ again falls to 0 at angle $\omega t=\beta .$ Substituting this condition in Eq. $(6)$ yields, $$\sin (\beta-\phi)=\sin (\alpha-\phi) \exp \left[\frac{R}{L}\left(\frac{\alpha-\beta}{\omega}\right)\right]-----(7)$$

Extinction angle $\beta$ can be determined by the solution of this transcendental equation. Once $\beta$ is known, the conduction angle $\gamma$ during which current flows from angle $\alpha$ to angle $\beta$ can be found,

$$\gamma=\beta-\alpha-----(8)$$,

From Eqs $(7)$ and $(8),$ one can obtain a relationship between $\gamma$ and $\alpha$ for a giyen value of $\phi$.

The RMS output voltage can be found from,

$$E_{0}=\left[\frac{2}{2 \pi} \int_{\alpha}^{B} 2 \cdot E_{s}^{2} \sin ^{2} \omega t \cdot \mathrm{d}(\omega t)\right]^{1 / 2}$$

$$=\left[\frac{4 E_{s}^{2}}{4 \pi} \int_{\alpha}^{\beta}(1-\cos 2 \omega t) \cdot \mathrm{d}(\omega t)\right]^{1 / 2}-----(9)$$

$$=E_{s}\left[\frac{1}{\pi} \int_{\alpha}^{\beta}\left(\beta-\alpha+\frac{\sin 2 \alpha}{2}-\frac{\sin 2 \beta}{2}\right]^{1 / 2}\right.-----(10)$$

The RMS thyristor current can be found from Eq. $(6)$ as

$$I_{\mathrm{rms}}=\left[\frac{1}{2 \pi} \int_{\alpha}^{\beta} i_{0}^{2} \cdot \mathrm{d}(\omega t)\right]^{1 / 2}$$

$$=\frac{E_{s}}{Z}\left[\frac{1}{\pi} \int_{\alpha}^{\beta}\left\{\sin (\omega t-\phi)-\sin (\alpha-\phi) e^{R / L\left(\frac{\alpha}{\omega}-t\right)}\right\}^{2} \mathrm{d}(\omega t)\right]^{1 / 2}-----(11)$$

and the RMS output current can then be determined by combining the RMS current of each thyristor as

$$I_{0}=\sqrt{I_{\mathrm{rms}}^{2}+I_{\mathrm{rms}}^{2}}=\sqrt{2} I_{\mathrm{rms}}-----(12)$$

The average value of thyristor current can also be found from Eq. $(6)$ as

$$I_{\mathrm{av}}=\frac{1}{2 \pi} \int_{\alpha}^{\beta} i_{0} \cdot \mathrm{d}(\omega t)$$

$$=\frac{\sqrt{2} E_{s}}{2 \pi Z} \int_{\alpha}^{\beta}\left[\sin (\omega t-\phi)-\sin (\alpha-\phi) e^{R / L\left(\frac{\alpha}{\omega}-t\right)} \mathrm{d}(\omega t)\right]$$

Operation with $\alpha \leq \phi :$ Assume that a.c. voltage controller is working under steady-state with $\alpha=\phi$ . From zero to $\phi, T_{2}$ conducts and from $\phi$ to $(\pi+\phi), T_{1}$ conducts; from $(\pi+\phi)$ to $(2 \pi+\phi), T_{2}$ conducts, and so on.

Now, let $\alpha$ be decreased below $\phi$ . When $t_{1}$ is triggered at $\alpha \lt \phi, T_{1}$ will not get turned-on because it is reverse biased by voltage drop in $T_{2}$ which is conducting current $i_{T_{2}} \cdot T_{1}$ will get turned-on only at $\phi$ when $i_{T_{2}}=0$ and reverse bias due to voltage drop in $T_{2}$ vanishes. Now $T_{1}$ will conduct from $\phi$ to $(\pi+\phi) . T_{2}$ will be triggered at an angle $(\pi+\alpha) \lt (\pi+\phi)$ . As $T_{1}$ is conducting, a voltage drop in $T_{1}$ will apply a reverse bias across $T_{2},$ as a result $T_{2}$ will not get turned-on at $(\pi+\alpha)$ but only at $(\pi+\phi),$ when $i_{T_{1}}=0 .$ Now $T_{2}$ will conduct from $(\pi+\phi)$ to $(2 \pi+\phi)$ and so on.

This shows that load voltage and current waveforms will not change from what they are at $\alpha=\phi$ . Thus, the reduction of $\alpha$ below $\phi$ is not able to control the load voltage and load current. The a.c. output power can be controlled only for $a \gt\phi$ Note that for $\alpha \leq \phi, \gamma$ remains equal to $\pi$ . Thus, the control range of firing angle is $\phi \lt \alpha \lt 180^{\circ} .$