--- title: "AC voltage controller rms and ELC use" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-rms-elc-mhp" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rms-elc-mhp site: BCT Engineering --- # AC voltage controller rms and ELC use Power Electronics · Chapter 6: AC voltage controller - Asked in: 2080 Baishakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rms-elc-mhp - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question Explain the operation single phase AC voltage controller. Derive the expression for RMS value of the output voltage. How it can be used on ELC of MHP scheme? ## Short answer RMS output voltage is V_s sqrt((1/pi)(pi - alpha + sin(2 alpha)/2)). In micro hydro ELC, the controller senses frequency and adjusts alpha to vary dump load power, keeping total load constant. ## Answer A single-phase AC voltage controller varies the rms value of the AC voltage applied to a load, at the supply frequency, by delaying the firing of two anti-parallel thyristors (phase angle control). #### Circuit and operation ```text T1 --->|--- +---| |----+ | ---|<--- | | | T2 | ( ~ ) vs R vo | | +-------------------+ ``` 1. $0 < \omega t < \alpha$: T1 not yet fired, $v_o = 0$. 2. At $\alpha$: T1 fired, $v_o = v_s$ until $\pi$, where the current falls to zero and T1 turns off (natural commutation). 3. $\pi < \omega t < \pi + \alpha$: $v_o = 0$. 4. At $\pi + \alpha$: T2 fired, $v_o = v_s$ until $2\pi$. ```text vo . | \ -------|--\--------.------ 0 a pi | / 2pi | / |/ pi+a ``` #### RMS output voltage $$ \begin{aligned} V_o &= \left[\frac{2}{2\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\left\{\omega t - \frac{\sin 2\omega t}{2}\right\}_{\alpha}^{\pi}\right]^{1/2} \\ &= \frac{V_m}{\sqrt{2}}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)\right]^{1/2} \end{aligned} $$ $$ V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} $$ As $\alpha$ goes from 0 to $\pi$, $V_o$ goes from $V_s$ to 0, and load power $V_o^2/R$ from full to zero. #### Use in the ELC of a micro hydro (MHP) scheme Micro hydro plants usually run with a fixed water input and no mechanical governor. The **electronic load controller (ELC)** keeps the total electrical load constant: $$ P_{gen} = P_{consumer} + P_{dump} = \text{constant} $$ ```text Gen ---+-------> Consumers | +--[ AC voltage controller ]--> Dump heater | ^ alpha +-> sense f/V -> controller ``` - The controller senses generator frequency (or voltage). - Consumer load decreases → frequency tends to rise → $\alpha$ is reduced → more power to the dump (ballast) heater. - Consumer load increases → $\alpha$ is increased → less power to the dump load. - So speed, frequency and voltage stay nearly constant without a governor. - The dump load is usually water or air heaters, and the heat can be used. Drawbacks of phase control here are harmonics and poorer power factor at mid-range $\alpha$; binary-weighted dump loads or integral cycle control reduce this. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the AC voltage controller, what happens at wt = alpha? A. T2 is fired B. T1 is fired and vo = vs until pi C. T1 turns off D. The supply is disconnected Answer: B. T1 is fired at alpha, then conducts until current falls to zero at pi. (Key point: Operation sequence) 2. Why does the rms integral use limits alpha to pi with a factor 2/(2pi)? A. Only the positive half cycle exists B. The positive and negative half cycles give identical contributions, so they are combined C. T1 conducts for the whole cycle D. The frequency is doubled Answer: B. By symmetry the negative half gives the same result, so the integral is doubled over a full period 2pi. (Key point: Derivation step) 3. How do Vo and the load power change as alpha goes from 0 to pi? A. Both rise from 0 to maximum B. Vo goes from Vs to zero and power from full to zero C. Vo stays constant and power falls D. Frequency changes from 50 Hz to 0 Answer: B. A larger delay shortens conduction, reducing rms voltage and power V^2/R. (Key point: Effect of alpha) 4. When consumer load increases and frequency tends to fall, what does the ELC do? A. Reduces alpha to send more power to the dump heater B. Disconnects the generator C. Increases alpha so less power goes to the dump load D. Increases the water flow Answer: C. The total of consumer and dump load is kept constant, so the dump power is reduced as consumer load rises. 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