--- title: "Single-phase AC voltage controller with R load" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-r-load-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r-load-rms site: BCT Engineering --- # Single-phase AC voltage controller with R load Power Electronics · Chapter 6: AC voltage controller - Asked in: 2081 Baishakh, 8 marks; 2079 Bhadra, 8 marks; 2078 Bhadra, 8 marks; 2076 Asoj, 8 marks; 2072 Chaitra, 8 marks; 2071 Chaitra, 8 marks (asked 6 times) - 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-load-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question With the help of suitable circuit diagram and waveforms, explain the operation of single phase AC voltage controller with resistive load. Derive the expression for rms value of output voltage as a function of firing angle. ## Short answer Two anti-parallel thyristors (T1 positive half, T2 negative half) are fired at angle α. Output is zero until α then equals supply until π, where current naturally extinguishes. The rms output voltage is V_o = V_s √[(1/π)(π - α + sin2α/2)]. ## Answer A single-phase AC voltage controller changes the rms voltage applied to a load by delaying the firing of two anti-parallel thyristors (or a TRIAC) by an angle $\alpha$ in each half cycle (phase control). The output frequency is the same as the input frequency. #### Circuit ```text T1 --->|--- +---| |----+ | ---|<--- | | | T2 | ( ~ ) vs = Vm sin wt R vo | | +-------------------+ ``` T1 handles the positive half cycle and T2 the negative half cycle. #### Operation 1. **$0 < \omega t < \alpha$:** T1 is forward biased but not fired; no current; $v_o = 0$. 2. **$\omega t = \alpha$:** T1 is fired; it conducts and $v_o = v_s$. 3. **$\pi$:** supply and load current fall to zero together (resistive load), so T1 turns off naturally (line commutation). 4. **$\pi < \omega t < \pi + \alpha$:** T2 is forward biased but not yet fired; $v_o = 0$. 5. **$\omega t = \pi + \alpha$:** T2 is fired; $v_o = v_s$ (negative) until $2\pi$. The current has the same shape as the voltage: $i_o = v_o/R$. #### Waveforms ```text vs _ _ / \ / ---/-----\---------/---- \ / \_/ vo . | \ -------|--\--------.------ 0 a pi | / 2pi | / |/ pi+a gates: T1 at a, T2 at pi+a ``` The output is an AC waveform with chopped beginnings of each half cycle. #### RMS output voltage With $v_s = V_m \sin\omega t$ and by symmetry of the two halves: $$ \begin{aligned} V_{o} &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2 \sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\int_{\alpha}^{\pi} (1-\cos 2\omega t)\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)\right]^{1/2} \end{aligned} $$ Since $V_s = V_m/\sqrt{2}$: $$ V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} $$ - $\alpha = 0$: $V_o = V_s$ (full voltage). - $\alpha = \pi$: $V_o = 0$. So the rms output is controlled smoothly from $V_s$ to 0 as $\alpha$ goes from 0 to $\pi$. The input power factor is $V_o/V_s$, which falls as $\alpha$ increases. **Applications:** lamp dimmers, heater control, speed control of fans and small induction motors, soft starters. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does a single-phase ac voltage controller vary the rms output voltage? A. By changing the supply frequency B. By delaying the firing of the anti-parallel thyristors by angle alpha in each half cycle C. By changing the load resistance D. By rectifying the supply Answer: B. Phase control: the load sees the supply only after alpha in each half cycle, at the same frequency. (Key point: Phase control) 2. How does T1 turn off with a resistive load? A. By forced commutation at pi B. By a negative gate pulse C. By firing T2 D. Naturally at wt = pi, when the current falls to zero (line commutation) Answer: D. With resistive load the current falls to zero with the supply at pi. (Key point: Commutation with R load) 3. Which is the rms output voltage as a function of alpha? A. Vs sqrt((1/pi)(pi - alpha + sin(2 alpha)/2)) B. Vs sqrt((1/pi)(pi - alpha - sin(2 alpha)/2)) C. Vs (pi - alpha)/pi D. Vs sqrt(1 - alpha/pi) Answer: A. Integrating Vm^2 sin^2 from alpha to pi gives Vs sqrt((1/pi)(pi - alpha + sin2alpha/2)). (Key point: RMS expression) 4. What happens as alpha increases from 0 to pi? A. Vo rises from 0 to Vs and the power factor rises B. Vo stays Vs and power factor falls C. Vo falls from Vs to 0 and the input power factor (Vo/Vs) falls D. Vo falls from Vs to 0 and the power factor stays 1 Answer: C. At alpha = 0 Vo = Vs, at alpha = pi Vo = 0, and the power factor Vo/Vs decreases. 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