GTO full converter, extinction angle 30°

In extinction angle control, the switches are turned ON at the voltage zero crossing ($\omega t = 0$) and turned OFF at $\omega t = \pi - \beta$, where $\beta$ is the extinction angle. GTOs are needed because the turn-off occurs while the supply voltage is still positive (forced turn-off). The input current then leads the voltage, which improves the power factor.

Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $I_o = 20$ A, $\beta = 30^\circ = \pi/6$.

              P (+)
        +-------+-------+
        |               |
       S1              S3
        |               |
 a o----+               +----o b
        |               |
       S4              S2
        |               |
        +-------+-------+
              N (-)
 S1..S4: GTOs (or IGBTs) with anti-parallel diodes
 vs = 230 V, 50 Hz between a and b
 Highly inductive load (Io) between P and N

i) Waveforms and operation

During the freewheeling intervals the load is shorted by two switches of one leg, so $v_o = 0$ and no current is drawn from the supply.

 vo  (load voltage, period 180 deg)
     _____          _____
    /     |        /     |
   /      |       /      |
 -/-------+------/-------+-----
  0    180-b   180    360-b  360
 io  ============================  Io (constant)
 is
 +Io +-------+
     |       |
  0 -+       +---+         +---
                 |         |
 -Io             +---------+
  0    180-b 180      360-b 360     (b = beta)

ii) Average and RMS output voltage

$$\begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi-\beta} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\beta) \\ &= \frac{325.27}{\pi}(1 + 0.866) = 103.54 \times 1.866 = 193.20\ \text{V} \end{aligned}$$ $$\begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_0^{\pi-\beta} V_m^2\sin^2\omega t\, d(\omega t)\right]^{1/2} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2} \\ &= 230\left[\frac{1}{\pi}\left(2.618 + 0.433\right)\right]^{1/2} = 230 \times 0.9855 = 226.66\ \text{V} \end{aligned}$$

iii) Fundamental component of input current

Fourier analysis of $i_s$ (amplitude $I_o$, width $\pi-\beta$, starting at $0$):

$$\begin{aligned} a_1 &= \frac{2I_o}{\pi}\sin\beta = \frac{40}{\pi}(0.5) = 6.366\ \text{A} \\ b_1 &= \frac{2I_o}{\pi}(1+\cos\beta) = \frac{40}{\pi}(1.866) = 23.759\ \text{A} \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\beta}{2} = 24.597\ \text{A} \\ I_{s1} &= \frac{24.597}{\sqrt2} = 17.39\ \text{A (rms)} \\ \phi_1 &= \tan^{-1}\frac{a_1}{b_1} = \frac{\beta}{2} = 15^\circ\ \text{(leading)} \end{aligned}$$

So $i_{s1} = 24.60\sin(\omega t + 15^\circ)$ A.

Answer: $V_{dc} = 193.20$ V, $V_{rms} = 226.66$ V, $I_{s1} = 17.39$ A (rms) leading $v_s$ by $15^\circ$. (Displacement factor $= \cos 15^\circ = 0.966$ leading.)