Single-phase full converter: rectification and inversion

A single-phase full converter is a fully controlled bridge of four thyristors. With a highly inductive load the load current $I_o$ is continuous and flows in one direction only, but the average output voltage can be positive or negative. It is therefore a two-quadrant converter: it works as a rectifier for $\alpha < 90^\circ$ and as a line-commutated inverter for $\alpha > 90^\circ$.

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load (R-L, Io) between P and N

Operation with highly inductive load

• Positive half cycle: T1 and T2 are forward biased and are fired at $\omega t = \alpha$. Load is connected to the supply, $v_o = v_s$, $i_s = +I_o$.
• At $\omega t = \pi$, $v_s$ reverses, but the large inductance keeps $I_o$ flowing, so T1, T2 continue to conduct and $v_o$ becomes negative from $\pi$ to $\pi+\alpha$ (energy returns from the inductance to the supply).
• At $\omega t = \pi+\alpha$, T3 and T4 are fired. Since $v_s < 0$, they apply reverse voltage across T1, T2, which turn off by natural (line) commutation. Now $v_o = -v_s$, $i_s = -I_o$.
• At $2\pi + \alpha$ T1, T2 are fired again and the cycle repeats.
• Each pair conducts for $180^\circ$; the input current is a square wave of amplitude $I_o$ lagging the voltage by $\alpha$.

Output voltage

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

Rectification mode ($0 < \alpha < 90^\circ$)

• $\cos\alpha > 0$, so $V_{dc}$ is positive; $I_o$ is positive.
• Power $P = V_{dc}I_o > 0$: energy flows from the ac supply to the dc load (first quadrant).
• Example: motoring of a separately excited dc motor.

Inversion mode ($90^\circ < \alpha < 180^\circ$)

• $\cos\alpha < 0$, so $V_{dc}$ is negative while $I_o$ is still positive (thyristors conduct in one direction only).
• $P = V_{dc}I_o < 0$: energy flows from the dc side to the ac supply (fourth quadrant).
• Conditions needed:
  1. A dc source in the load (e.g. back emf of a dc machine or a battery) with its polarity reversed relative to rectification, so that it can drive current through the thyristors against negative $V_{dc}$: $E > |V_{dc}|$.
  2. Continuous current (large inductance).
  3. A firing-angle limit: $\alpha_{max} \approx 180^\circ - (\mu + \gamma)$ where $\mu$ is the overlap angle and $\gamma$ the turn-off margin; otherwise the outgoing thyristor cannot regain blocking ability and commutation failure occurs.
• Example: regenerative braking of a dc motor, HVDC inverter station.

At $\alpha = 90^\circ$, $V_{dc} = 0$ and the average power is zero.