Reverse power flow in bipolar HVDC link

In a bipolar HVDC link, each end has two fully controlled (thyristor) converters in series with the mid-point earthed; one pole is at $+V_d$ and the other at $-V_d$. Because a full converter can work as a rectifier ($\alpha < 90°$) or as an inverter ($\alpha > 90°$), power flow is reversed simply by changing the firing angles at both ends; the current direction cannot change, so the pole voltages reverse.

Bipolar link

  Station 1                    Station 2
 [Conv 1P] +Vd ======Id=====> [Conv 2P]
     |                            |
   earth                        earth
     |                            |
 [Conv 1N] -Vd <=====Id====== [Conv 2N]

Each pole forms a link like the following (one pole shown):

   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

Mathematical analysis

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 $\alpha < 90°$.

In the bipolar link

• Both poles are controlled together. In normal mode the upper pole is $+V_d$ and the lower $-V_d$; on reversal both polarities swap (upper becomes $-V_d$, lower $+V_d$), while the pole currents keep their directions.
• Total power $P = 2V_dI_d$ reverses with no mechanical switching.
• In practice, the rectifier controls current (constant current) and the inverter controls extinction angle ($\gamma$ control); reversal is done by ramping current to a low value, reversing the voltage by changing $\alpha$, then raising current.