Reversible power flow on a DC line

In a DC (HVDC) line with thyristor converters at both ends, the direction of current cannot reverse because thyristors conduct only one way. Power flow is reversed by reversing the polarity of the DC voltage, which is done by changing the firing angles: the rectifier becomes an inverter and the inverter becomes a rectifier.

Circuit

   AC sys 1                         AC sys 2
     |   Conv 1      R, Ld      Conv 2   |
     +--[3-ph full]--+--/\/\--+--[3-ph full]--+
        bridge  Vd1 (+)  Id -> (+) Vd2'  bridge
                 |               |
                 +---------------+
 Vd1 = Vdo cos a1     Vd2' = -Vdo cos a2

Both converters are three-phase fully controlled bridges whose average output is $V_{do}\cos\alpha$: positive for $0 < \alpha < 90°$ (rectifier) and negative for $90° < \alpha < 180°$ (inverter).

Operation

Let both converters have the same no-load voltage $V_{do} = \dfrac{3\sqrt{2}V_L}{\pi}$ and the line resistance be $R$. Converter 1 gives $V_{d1} = V_{do}\cos\alpha_1$. Converter 2 is connected in opposition, so its terminal voltage seen by the line is $V_{d2}' = -V_{do}\cos\alpha_2$. Thyristors allow current $I_d$ in one direction only:

$$I_d = \frac{V_{d1} - V_{d2}'}{R} = \frac{V_{do}(\cos\alpha_1 + \cos\alpha_2)}{R} \ge 0$$

Power sent by converter 1 and received by converter 2:

$$P_1 = V_{do}\cos\alpha_1\,I_d, \qquad P_2 = -V_{do}\cos\alpha_2\,I_d, \qquad P_1 - P_2 = I_d^2R$$

Case 1: power from system 1 to system 2.

• $\alpha_1 < 90°$: converter 1 is a rectifier ($V_{d1} > 0$, $P_1 > 0$).
• $\alpha_2 > 90°$: converter 2 is an inverter ($\cos\alpha_2 < 0$, $P_2 > 0$ delivered to system 2).
• Line voltage is positive at the top conductor.

Case 2: power from system 2 to system 1 (reversal).

• Make $\alpha_1 > 90°$: converter 1 becomes an inverter ($V_{d1} < 0$).
• Make $\alpha_2 < 90°$: converter 2 becomes a rectifier.
• The current direction stays the same; only the polarity of the DC line voltage reverses, so $P = V_dI_d$ changes sign.

Condition for current flow in both cases:

$$\cos\alpha_1 + \cos\alpha_2 > 0 \;\Rightarrow\; \cos\alpha_1 > \cos(180° - \alpha_2) \;\Rightarrow\; \alpha_1 + \alpha_2 < 180°$$

For a line of small resistance, $I_dR$ is small, so

$$\alpha_1 + \alpha_2 \approx 180°, \quad \text{i.e. } \alpha_2 \approx 180° - \alpha_1$$

The magnitude of $I_d$ (and power) is set by how much $\alpha_1 + \alpha_2$ is less than $180°$; the direction of power is set by which converter has $\alpha < 90°$.

Practical sequence for reversal

1. Reduce the current order to a minimum value.
2. Increase $\alpha_1$ beyond $90°$ and decrease $\alpha_2$ below $90°$ gradually; the line voltage passes through zero and reverses.
3. Raise the current order again; power now flows from system 2 to system 1.

No switching of conductors is needed and reversal takes only a fraction of a second, which is an important advantage of HVDC links.