Reversible HVDC power flow: firing angle relation

In an HVDC line, the converters at both ends are fully controlled thyristor bridges. Thyristors pass current in one direction only, so power reversal is done by reversing the DC voltage polarity using the firing angles, not by reversing the current.

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

Process and firing angle relation

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 $\alpha < 90°$.

Waveform idea

 Vd
 +|----____        normal (1 -> 2)
  |        \
 0+---------\----------- time
  |          \____     reversed (2 -> 1)
 -|
 Id  constant direction throughout

Result: current flows only if $\alpha_1 + \alpha_2 < 180°$; for a low-resistance line $\alpha_2 \approx 180° - \alpha_1$, and the converter with $\alpha < 90°$ is the sending end.