Extinction angle control: current leads by β/2

In extinction angle control, the forced-commutated switch of a single-phase converter is turned ON at $\omega t = 0$ and turned OFF at $\omega t = \pi-\beta$ ($\beta$ = extinction angle); the load current freewheels from $\pi-\beta$ to $\pi$. With a highly inductive load ($i_o = I_a$ constant) the input current is:

$$i_s(\omega t) = \begin{cases} +I_a, & 0 < \omega t < \pi-\beta \\ 0, & \pi-\beta < \omega t < \pi \\ -I_a, & \pi < \omega t < 2\pi-\beta \\ 0, & 2\pi-\beta < \omega t < 2\pi \end{cases}$$

 vs   ___
     /   \             /
 ---/-----\-----------/----
           \___/
 is
 +Ia +------+
     |      |
 ----+      +--+       +---
               |       |
 -Ia           +-------+
     0  180-b 180  360-b 360
 (pulse centre at (180-b)/2, i.e. b/2
  before the voltage peak at 90)

Fourier analysis

Let $i_s = \frac{a_0}{2} + \sum (a_n\cos n\omega t + b_n\sin n\omega t)$. The waveform has half-wave symmetry, so $a_0 = 0$.

$$\begin{aligned} a_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_a\cos\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_a\cos\omega t\,d(\omega t)\right] \\ &= \frac{I_a}{\pi}\left[\sin(\pi-\beta) - \{\sin(2\pi-\beta) - \sin\pi\}\right] = \frac{I_a}{\pi}\left[\sin\beta + \sin\beta\right] = \frac{2I_a}{\pi}\sin\beta \end{aligned}$$ $$\begin{aligned} b_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_a\sin\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_a\sin\omega t\,d(\omega t)\right] \\ &= \frac{I_a}{\pi}\left[\{1 - \cos(\pi-\beta)\} + \{\cos(2\pi-\beta) - \cos\pi\}\right] \\ &= \frac{I_a}{\pi}\left[(1+\cos\beta) + (\cos\beta + 1)\right] = \frac{2I_a}{\pi}(1+\cos\beta) \end{aligned}$$

Magnitude and phase of the fundamental

$$i_{s1} = a_1\cos\omega t + b_1\sin\omega t = I_{s1,m}\sin(\omega t + \phi_1)$$

with $I_{s1,m}\sin\phi_1 = a_1$ and $I_{s1,m}\cos\phi_1 = b_1$. Therefore

$$\begin{aligned} \tan\phi_1 &= \frac{a_1}{b_1} = \frac{\sin\beta}{1+\cos\beta} = \frac{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}{2\cos^2\frac{\beta}{2}} = \tan\frac{\beta}{2} \\ \phi_1 &= +\frac{\beta}{2} \end{aligned}$$ $$\begin{aligned} I_{s1,m} &= \sqrt{a_1^2 + b_1^2} = \frac{2I_a}{\pi}\sqrt{\sin^2\beta + (1+\cos\beta)^2} = \frac{2I_a}{\pi}\sqrt{2(1+\cos\beta)} \\ &= \frac{4I_a}{\pi}\cos\frac{\beta}{2} \end{aligned}$$

So

$$i_{s1} = \frac{4I_a}{\pi}\cos\frac{\beta}{2}\,\sin\left(\omega t + \frac{\beta}{2}\right)$$

Since the supply voltage is $v_s = V_m\sin\omega t$, the positive phase angle $+\beta/2$ shows that the fundamental input current leads the input voltage by $\beta/2$. Hence the displacement factor is $\cos(\beta/2)$ leading, and the converter draws leading (capacitive) reactive power, which is the basis of power factor improvement by this method.

Example: $\beta = 30^\circ$, $I_a = 20$ A gives $I_{s1} = \frac{2\sqrt2 \times 20}{\pi}\cos15^\circ = 17.39$ A rms, leading by $15^\circ$.