GTO full converter, extinction angle: RMS and input current

In extinction angle control the forced-commutated switches (GTOs) are turned ON at the zero crossing of the supply voltage and turned OFF $\beta$ before the next zero crossing ($\omega t = \pi-\beta$). The input current pulse then sits before the voltage peak, so its fundamental leads the voltage: the converter can supply reactive power and the power factor improves.

Data: $V_s = 230$ V, $V_m = 325.27$ V, $I_o = 20$ A, $\beta = 30^\circ$.

              P (+)
        +-------+-------+
        |               |
       S1              S3
        |               |
 a o----+               +----o b
        |               |
       S4              S2
        |               |
        +-------+-------+
              N (-)
 S1..S4: reverse-blocking GTOs (or IGBTs with
 series diodes); no anti-parallel diodes
 vs = 230 V, 50 Hz between a and b
 Highly inductive load (Io) between P and N

Operation

• From $\pi-\beta$ to $\pi$ the load current freewheels through the two switches of one leg, so $v_o = 0$ and $i_s = 0$.
• The load current stays at 20 A because of the large inductance.

Waveforms of $v_o$, $i_o$, $i_s$

 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)

RMS value of output voltage

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

(Average value for reference: $V_{dc} = \frac{V_m}{\pi}(1+\cos\beta) = 193.20$ V.)

Fundamental component of input current

$$\begin{aligned} a_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_o\cos\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_o\cos\omega t\,d(\omega t)\right] = \frac{2I_o}{\pi}\sin\beta = 6.366\ \text{A} \\ b_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_o\sin\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_o\sin\omega t\,d(\omega t)\right] = \frac{2I_o}{\pi}(1+\cos\beta) = 23.759\ \text{A} \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\beta}{2} = \frac{80}{\pi}(0.9659) = 24.60\ \text{A} \\ I_{s1} &= 24.60/\sqrt2 = 17.39\ \text{A (rms)} \\ \phi_1 &= \tan^{-1}\frac{a_1}{b_1} = \tan^{-1}\frac{\sin\beta}{1+\cos\beta} = \frac{\beta}{2} = 15^\circ\ \text{leading} \end{aligned}$$

Answer: $V_{rms} = 226.66$ V; $I_{s1} = 17.39$ A rms (24.60 A peak), leading $v_s$ by $15^\circ$.