--- title: "Reverse power flow in bipolar HVDC link" subject: "Power Electronics" chapter: 7 key: "semester_7/power_electronics:hvdc-reverse-power-bipolar" source: https://www.bctengineer.com/q/semester_7/power_electronics/hvdc-reverse-power-bipolar site: BCT Engineering --- # Reverse power flow in bipolar HVDC link Power Electronics · Chapter 7: HVDC power transmission - Asked in: 2075 Asoj, 8 marks - Video: [Problem & Power reversal of HVDC links](https://www.youtube.com/watch?v=OceNNLeO610) — Birendra Krishna Ghosh - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/hvdc-reverse-power-bipolar - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-7 ## Question With mathematical aid, explain how full converters can be used for reverse power flow in bipolar link of HVDC transmission. ## Short answer Both converters are fully controlled, with V_d = V_do cos α. In normal mode, α1<90° (rectifier) and α2>90° (inverter); reversal makes α1>90° and α2<90°, reversing voltage polarity while current direction stays same. For current flow, α1+α2<180°, and for low resistance, α2≈180°–α1. ## Answer 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 ```text 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): ```text 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 $\alpha < 90°$. | Mode | $\alpha_1$ | $\alpha_2$ | Line polarity | Power | |---|---|---|---|---| | Normal | $< 90°$ (rectifier) | $> 90°$ (inverter) | + | 1 → 2 | | Reversed | $> 90°$ (inverter) | $< 90°$ (rectifier) | − | 2 → 1 | #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a thyristor HVDC link, how is power flow reversed? A. By reversing the DC current direction B. By reversing the DC voltage polarity through firing angles; current direction stays the same C. By switching off one converter D. By changing the frequency Answer: B. Thyristors conduct in one direction only, so the current cannot reverse; the voltage polarity reverses so P = Vd Id changes sign. (Key point: Direction of current) 2. What firing angle makes a full converter act as a rectifier and as an inverter? A. alpha < 90° rectifier; alpha > 90° inverter B. alpha > 90° rectifier; alpha < 90° inverter C. alpha = 0 for both D. alpha = 180° for rectifier Answer: A. Vd = Vdo cos(alpha) is positive for alpha < 90° (rectifier) and negative for alpha > 90° (inverter). (Key point: Rectifier and inverter modes) 3. What is the line current in the two-converter analysis? A. Id = Vdo(cos a1 - cos a2)/R B. Id = Vdo(cos a1 + cos a2)/R C. Id = Vdo(sin a1 + sin a2)/R D. Id = Vdo/R Answer: B. Converter 2 appears in opposition, V_d2' = -Vdo cos a2, so Id = (Vd1 - V_d2')/R = Vdo(cos a1 + cos a2)/R. (Key point: Line current expression) 4. What condition must the firing angles satisfy for positive current? A. a1 + a2 > 180° B. a1 + a2 < 180° C. a1 = a2 D. a1 - a2 = 90° Answer: B. cos a1 + cos a2 > 0 requires a1 + a2 < 180°; for small R, a2 is approximately 180° - a1. 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