--- title: "AC voltage controller with R and RL loads" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-r-and-rl-load" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r-and-rl-load site: BCT Engineering --- # AC voltage controller with R and RL loads Power Electronics · Chapter 6: AC voltage controller - Asked in: 2069 Chaitra, 8 marks - Video: [Single phase full wave AC voltage controller with R-load](https://www.youtube.com/watch?v=h3UAP09Hk1Q) — Anet Jose - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r-and-rl-load - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question Explain the operation of ac voltage controller with purely resistive load. Derive the expression for calculating rms value of the output voltage. If the load is inductive (R-L load), derive the expression for time domain equation for load current. ## Short answer An AC voltage controller phase-fires anti-parallel thyristors T1 and T2. For resistive load, Vo = Vs sqrt[(1/pi)(pi - alpha + sin2alpha/2)]. For R-L load, i(omega t) = (Vm/Z)[sin(omega t - phi) - sin(alpha - phi) exp(-(omega t - alpha)/tan phi)], valid alpha to beta, with extinction angle beta given by sin(beta-phi)=sin(alpha-phi)exp(-(beta-alpha)/tan phi). ## Answer 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). ```text 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 $v_o$. ```text 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 $V_o = V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a resistive load, when does T1 turn off? A. At wt = alpha B. When it is gated off C. Naturally at wt = pi when the current falls to zero D. At wt = pi + alpha Answer: C. For R load the current follows the voltage, so it falls to zero at pi and T1 turns off naturally. (Key point: R-load operation) 2. What are the two parts of the R-L load current after T1 fires? A. Steady-state (Vm/Z) sin(wt - phi) plus an exponentially decaying transient B. Two sinusoids of different frequency C. A DC term plus a ramp D. Only a steady-state sinusoid Answer: A. Solving L di/dt + Ri = Vm sin wt gives i = i_ss + A e^(-Rt/L), with the constant A found from i(alpha) = 0. (Key point: R-L current solution) 3. What determines the extinction angle beta in an R-L load? A. i = 0 again: sin(beta - phi) = sin(alpha - phi) e^(-(beta - alpha)/tan phi) B. beta = pi always C. beta = alpha + 180° always D. beta = phi Answer: A. Setting the current expression to zero for wt > alpha gives the transcendental equation for beta. (Key point: Extinction angle) 4. For R-L load, what is the useful range of firing angle for controlling output voltage? A. 0 to phi B. phi to pi C. 0 to pi/2 regardless of phi D. pi to 2pi Answer: B. If alpha <= phi the current is continuous and Vo = Vs, so control is lost; control exists for phi <= alpha <= pi. 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