AC voltage controller rms at α = 90°

A single-phase AC voltage controller controls the rms voltage across a load by phase-delayed firing of two anti-parallel thyristors, T1 for the positive half and T2 for the negative half.

Operation with resistive load

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2          |
 ( ~ ) 220 V, 50 Hz    R   vo
   |                   |
   +-------------------+

• $0$ to $\alpha$: no thyristor on, $v_o = 0$.
• At $\alpha$, T1 is fired; $v_o = v_s$ until $\pi$, where current becomes zero and T1 turns off naturally.
• $\pi$ to $\pi+\alpha$: $v_o = 0$.
• At $\pi + \alpha$, T2 is fired; $v_o = v_s$ until $2\pi$.

 vo (alpha = 90 deg)
        |\
        | \
 -------+--\-------+-------
 0     90  180     |   /360
                   |  /
                   | /
                  270

Each half cycle the load gets only the second quarter of the sine wave. The output frequency stays 50 Hz.

RMS output voltage formula

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

Calculation

Given $V_s = 220$ V, $\alpha = 90° = \pi/2$, so $\sin 2\alpha = \sin 180° = 0$:

$$\begin{aligned} V_o &= 220\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{2} + 0\right)} \\ &= 220\sqrt{0.5} = 220 \times 0.7071 \\ &= 155.56\ \text{V} \end{aligned}$$

Answer: $V_{o,rms} = 155.56$ V (half the power of full conduction, since $V_o^2/V_s^2 = 0.5$).