--- title: "Reversible HVDC power flow: firing angle relation" subject: "Power Electronics" chapter: 7 key: "semester_7/power_electronics:hvdc-reversible-firing-angle-relation" source: https://www.bctengineer.com/q/semester_7/power_electronics/hvdc-reversible-firing-angle-relation site: BCT Engineering --- # Reversible HVDC power flow: firing angle relation Power Electronics · Chapter 7: HVDC power transmission - Asked in: 2070 Chaitra, 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-reversible-firing-angle-relation - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-7 ## Question Explain the process of reversible power flow on HVDC line and obtain the relation between firing angle of both converters. ## Short answer Power reversal in HVDC uses thyristor converters that cannot reverse current; instead, firing angles are changed to reverse voltage polarity. In normal flow, α1<90° (rectifier) and α2>90° (inverter). For reversal, α1>90° and α2<90°, making Vd negative. Current flows only if α1+α2<180°, and for small line resistance, α2≈180°–α1. ## Answer 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 ```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 ``` #### 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 $\alpha < 90°$. | Mode | $\alpha_1$ | $\alpha_2$ | Line polarity | Power | |---|---|---|---|---| | Normal | $< 90°$ (rectifier) | $> 90°$ (inverter) | + | 1 → 2 | | Reversed | $> 90°$ (inverter) | $< 90°$ (rectifier) | − | 2 → 1 | #### Waveform idea ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is power reversal in an HVDC line done by reversing the voltage rather than the current? A. Voltage cannot be controlled B. Thyristors conduct in one direction only, so current cannot reverse C. The line resistance blocks reverse current D. Reversing current would damage the AC systems Answer: B. Thyristors pass current one way only; so firing angles are used to change the sign of the DC voltage and hence of P = Vd Id. (Key point: Why voltage polarity is reversed) 2. With Vd1 = Vdo cos(a1) and converter 2 seen as -Vdo cos(a2), the line current is: A. Id = Vdo(cos a1 - cos a2)/R B. Id = Vdo(cos a1 + cos a2)/R C. Id = Vdo(cos a1)(cos a2)/R D. Id = R/(Vdo(cos a1 + cos a2)) Answer: B. Id = (Vd1 - Vd2')/R = Vdo(cos a1 + cos a2)/R. (Key point: Current expression) 3. Id >= 0 requires cos a1 + cos a2 > 0, which gives: A. a1 + a2 < 180° B. a1 + a2 > 180° C. a1 = a2 D. a1 - a2 < 90° Answer: A. cos a1 > cos(180° - a2) means a1 < 180° - a2, i.e. a1 + a2 < 180°; for small R, a2 is approximately 180° - a1. (Key point: Firing angle relation) 4. In normal mode (power from system 1 to 2), the converter states and line polarity are: A. Conv 1 inverter, Conv 2 rectifier, polarity negative B. Both converters rectifiers C. Conv 1 rectifier (a1 < 90°), Conv 2 inverter (a2 > 90°), polarity positive D. Both converters inverters Answer: C. The sending end has alpha < 90° (rectifier) and the receiving end alpha > 90° (inverter) with positive line voltage. (Key point: Direction of power) 5. What does P1 - P2 = Id^2 R represent? A. Power lost in the line resistance B. Reactive power of the converters C. Power stored in the smoothing reactor D. Power delivered to system 2 Answer: A. The sending-end power minus the receiving-end power equals the I^2 R loss in the DC line. (Key point: Power balance) ## Related in Power Electronics - [Reversible power flow on a DC line](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Areversible-power-flow-dc-line) — Power Electronics, chapter 7 - [Reverse power flow in bipolar HVDC link](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahvdc-reverse-power-bipolar) — Power Electronics, chapter 7 - [HVDC power loss is 2/3 of HVAC](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahvdc-two-thirds-loss) — Power Electronics, chapter 7, asked 2× - [HVDC vs HVAC line conductor saving](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahvdc-conductor-saving) — Power Electronics, chapter 7, asked 2× - [HVDC advantages and power transfer capacity](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahvdc-power-transfer-capacity) — Power Electronics, chapter 7, asked 2× - [Compare HVAC and HVDC; 2/3 power loss](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Acompare-ac-dc-two-thirds-loss) — Power Electronics, chapter 7 ## Related in other subjects - [3-phase half controlled bridge: no-load speed, firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ahalf-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [3-phase thyristor bridge: no-load speed and firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [Non-salient power angle relation for salient machines](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Anon-salient-power-angle-salient) — Electric Machines II, chapter 1 - [Two semi converters (50 N-m at 1000 rpm)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Asemi-converters-field-armature-50nm-1000rpm) — Utilization of Electrical Energy, chapter 3, asked 3×