Single-phase half-wave thyristor rectifier, R load

A single-phase half-wave controlled rectifier uses one thyristor in series with the load. The thyristor conducts only in the positive half cycle, from the firing angle $\alpha$ to $\pi$; the output is controlled by changing $\alpha$.

        T (thyristor)
 +----->|---------+
 |   ig ->        |
(~) vs =         R    vo
 |  Vm sin wt     |
 +----------------+

Operation (resistive load)

• $0 < \omega t < \alpha$: thyristor is forward biased but not fired, so it blocks; $v_o = 0$, $v_T = v_s$ (positive).
• At $\omega t = \alpha$ a gate pulse turns it ON; $v_o = v_s$, $i_o = v_s/R$, $v_T \approx 0$.
• At $\omega t = \pi$, $v_s$ and hence $i_o$ fall to zero; the thyristor turns off naturally.
• $\pi < \omega t < 2\pi$: thyristor is reverse biased; $v_o = 0$, $v_T = v_s$ (negative). The thyristor must withstand peak reverse voltage $V_m$ and peak forward voltage $V_m$ (if $\alpha \ge 90^\circ$).

Waveforms

 vs  ___
    /   \
 --/-----\---------/--
          \_______/
 vo     _
       | \
 ------+--\----------+--
 vT  _
    / |              
 --/--+---+---------/--
           \_______/
     0 a  180       360   (deg)
 vo: sine from a to 180; vT: vs from 0 to a
 and from 180 to 360 (reverse), zero while ON

Average output voltage

$$V_{dc} = \frac{1}{2\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{2\pi}(1+\cos\alpha)$$

Derivation of RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{1}{2\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\, d(\omega t) \\ &= \frac{V_m^2}{4\pi}\int_{\alpha}^{\pi}(1-\cos2\omega t)\, d(\omega t) \\ &= \frac{V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\alpha}^{\pi} \\ &= \frac{V_m^2}{4\pi}\left[\pi - \alpha + \frac{\sin2\alpha}{2}\right] \end{aligned}$$ $$V_{rms} = \frac{V_m}{2}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin2\alpha}{2}\right)\right]^{1/2}$$

Check: at $\alpha = 0$, $V_{rms} = V_m/2$, the value for an uncontrolled half-wave rectifier. RMS load current is $I_{rms} = V_{rms}/R$.