HVDC power transmission

HVDC transmission sends bulk power over long distances at high DC voltage, with a rectifier station at the sending end and an inverter station at the receiving end. It is preferred over HVAC for large power over long distances because the saving in line cost and losses exceeds the extra cost of the converter stations.

Why HVDC for large power over long distance

• Lower line cost: a bipolar DC line needs only 2 conductors instead of 3, smaller towers and narrower right of way. Terminal (converter) cost is high, but beyond the break-even distance (about 500–800 km for overhead lines, 40–80 km for cables) the total cost of DC is lower.
• No stability limit: AC power $P = \dfrac{V_1V_2}{X}\sin\delta$ falls as line reactance grows with length; DC has no reactance or load angle, so power is limited only by thermal rating.
• No reactive power or charging current: no need for line compensation; voltage stays nearly constant along the line.
• Lower losses: no skin effect, less corona, fewer conductors.
• Asynchronous link and fast power control by firing angle.

 Cost
  |                      / AC
  |                    /
  |                  /    _____ DC
  |            _____X-----
  |     _____-    / :
  | DC -        /   :
  |           /     :
  |  AC     /       :
  |       /         :
  |     /           :
  +-----------------:------> km
             break-even distance

Differences between HVAC and HVDC lines

Mathematical comparison

(a) Power transfer. Assumptions: same conductor size and current rating (same rms current $I$ per conductor), same insulation, i.e. the same peak voltage to ground $V_m$; three-phase AC line with 3 conductors; bipolar DC line with 2 conductors at $\pm V_d$, $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_m I\cos\phi \\ P_{dc} &= 2V_d I_d = 2V_m I \end{aligned}$$ $$\frac{P_{dc}}{P_{ac}} = \frac{2V_m I}{2.121\,V_m I\cos\phi} = \frac{0.943}{\cos\phi}$$

Per conductor:

$$\frac{P_{dc}/2}{P_{ac}/3} = \frac{V_m I}{0.707\,V_m I\cos\phi} = \frac{1.414}{\cos\phi}$$

With $\cos\phi \approx 0.943$, a 2-conductor DC line carries the same power as a 3-conductor AC line, and each DC conductor carries about 1.41 times more power at unity pf.

(b) Losses. For the same power and same loss per conductor, total loss is

$$\frac{\text{Loss}_{dc}}{\text{Loss}_{ac}} = \frac{2p}{3p} = \frac{2}{3}$$

(c) Conductor material. For the same power, voltage and loss, the DC line needs only $0.5\cos^2\phi$ times the conductor volume of the AC line (50% at unity pf).

(d) Insulation. AC insulation must withstand the peak $\sqrt{2}V_{rms}$; DC insulation is set by $V_d$, so for the same working level a DC line needs less insulation.

HVDC transmission uses a rectifier at the sending end, a DC line, and an inverter at the receiving end. For large power over long distances it is cheaper and more efficient than HVAC because the line is cheaper and has no stability or reactive-power limits; the high converter cost is recovered beyond the break-even distance.

Why HVDC for large power over long distance

• Lower line cost: a bipolar DC line needs only 2 conductors instead of 3, smaller towers and narrower right of way. Terminal (converter) cost is high, but beyond the break-even distance (about 500–800 km for overhead lines, 40–80 km for cables) the total cost of DC is lower.
• No stability limit: AC power $P = \dfrac{V_1V_2}{X}\sin\delta$ falls as line reactance grows with length; DC has no reactance or load angle, so power is limited only by thermal rating.
• No reactive power or charging current: no need for line compensation; voltage stays nearly constant along the line.
• Lower losses: no skin effect, less corona, fewer conductors.
• Asynchronous link and fast power control by firing angle.

Advantages of HVDC over HVAC lines

1. Only 2 conductors (bipolar) or 1 conductor with earth return (monopolar); lighter towers, narrower corridor.
2. No charging current, so no limit on length of underground or submarine cables.
3. No skin effect; full conductor area is used.
4. Less corona loss and radio interference for the same voltage.
5. Fast and accurate control of power flow by firing angle; can damp oscillations in connected AC systems.
6. Can link AC systems of different frequencies or not in synchronism (e.g. 50 Hz and 60 Hz).
7. Does not raise the fault level of the connected AC systems; DC line faults are cleared quickly by converter control.
8. One pole can continue to work if the other fails (bipolar link) with earth return.

Mathematical comparison of power transfer capacity

Assumptions: same conductor size and current rating (same rms current $I$ per conductor), same insulation, i.e. the same peak voltage to ground $V_m$; three-phase AC line with 3 conductors; bipolar DC line with 2 conductors at $\pm V_d$, $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_m I\cos\phi \\ P_{dc} &= 2V_d I_d = 2V_m I \end{aligned}$$ $$\frac{P_{dc}}{P_{ac}} = \frac{2V_m I}{2.121\,V_m I\cos\phi} = \frac{0.943}{\cos\phi}$$

Per conductor:

$$\frac{P_{dc}/2}{P_{ac}/3} = \frac{V_m I}{0.707\,V_m I\cos\phi} = \frac{1.414}{\cos\phi}$$

Interpretation:

• At $\cos\phi = 0.943$, $P_{dc} = P_{ac}$: a DC line with two conductors carries the same power as an AC line with three conductors of the same size and insulation.
• At unity power factor, $P_{dc}/P_{ac} = 0.943$, but per conductor the DC line carries $\sqrt{2} = 1.414$ times more power, because the DC conductor works continuously at the peak voltage $V_m$ while the AC conductor's rms voltage is only $V_m/\sqrt{2}$.
• So for the same power, DC needs one conductor less and its loss is about 2/3 of that of the AC line.

An HVDC line is preferred for large power over long distance because, after the break-even distance, the cheaper DC line (fewer conductors, less insulation, lower losses, no reactive power) outweighs the extra cost of converter stations.

Reasons

• Lower line cost: a bipolar DC line needs only 2 conductors instead of 3, smaller towers and narrower right of way. Terminal (converter) cost is high, but beyond the break-even distance (about 500–800 km for overhead lines, 40–80 km for cables) the total cost of DC is lower.
• No stability limit: AC power $P = \dfrac{V_1V_2}{X}\sin\delta$ falls as line reactance grows with length; DC has no reactance or load angle, so power is limited only by thermal rating.
• No reactive power or charging current: no need for line compensation; voltage stays nearly constant along the line.
• Lower losses: no skin effect, less corona, fewer conductors.
• Asynchronous link and fast power control by firing angle.

Mathematical comparison: conductor saving

Assumptions: same power $P$, same length $l$, same peak voltage to ground $V_m$ (same insulation), same total loss, same conductor material (resistivity $\rho$).

AC (3 conductors, area $A_{ac}$) and DC (2 conductors, area $A_{dc}$):

$$\begin{aligned} I_{ac} &= \frac{P}{3(V_m/\sqrt{2})\cos\phi}, \qquad I_{dc} = \frac{P}{2V_m} \\ \text{Loss}_{ac} &= 3I_{ac}^2\frac{\rho l}{A_{ac}} = \frac{2P^2\rho l}{3V_m^2\cos^2\phi\,A_{ac}} \\ \text{Loss}_{dc} &= 2I_{dc}^2\frac{\rho l}{A_{dc}} = \frac{P^2\rho l}{2V_m^2A_{dc}} \end{aligned}$$

Equating the losses:

$$\frac{A_{dc}}{A_{ac}} = \frac{1/2}{2/(3\cos^2\phi)} = 0.75\cos^2\phi$$

Ratio of conductor volume (material):

$$\frac{\text{Vol}_{dc}}{\text{Vol}_{ac}} = \frac{2A_{dc}l}{3A_{ac}l} = \frac{2}{3}\times 0.75\cos^2\phi = 0.5\cos^2\phi$$

Result:

At unity power factor, the DC line needs only half the conductor material of an equivalent AC line for the same power, voltage stress and losses; at lower AC power factor the saving is even greater. Fewer and lighter conductors also mean lighter towers and a narrower right of way, which is why HVDC lines are cheaper per km.

The claim compares a three-phase HVAC line (3 conductors) with a bipolar HVDC line (2 conductors) of the same power capacity, assuming each conductor has the same loss.

Step 1: Show that both lines carry the same power

Let both lines use the same conductors (same rms current $I$, resistance $R$ per conductor) and the same insulation (same peak voltage to earth $V_m$). For DC: $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi \\ P_{dc} &= 2V_dI_d = 2V_mI \end{aligned}$$

Equating, $P_{dc} = P_{ac}$ when $\cos\phi = \dfrac{2\sqrt{2}}{3} = 0.943$, which is a typical operating power factor. So a 2-conductor DC line has practically the same capacity as a 3-conductor AC line.

Step 2: Loss per conductor

Since both carry the same current $I$ (rms for AC, DC value for DC) through the same resistance $R$:

$$p_{ac} = I^2R = p_{dc} = p$$

(This is the given assumption.)

Step 3: Total losses

$$\begin{aligned} \text{Loss}_{ac} &= 3I^2R = 3p \\ \text{Loss}_{dc} &= 2I^2R = 2p \end{aligned}$$ $$\frac{\text{Loss}_{dc}}{\text{Loss}_{ac}} = \frac{2p}{3p} = \frac{2}{3}$$

Hence the total power loss of the HVDC line is only 2/3 of that of the HVAC line of the same capacity.

 HVAC:  ===== a  (loss p)      HVDC: ===== +Vd (loss p)
        ===== b  (loss p)            ===== -Vd (loss p)
        ===== c  (loss p)
 Total: 3p                    Total: 2p

Further points

• Efficiency of DC is also higher because there is no skin effect (AC resistance > DC resistance) and corona loss is lower.
• No reactive (charging) current flows in a DC line, so the full current rating carries real power.
• Converter station losses (about 0.5–1% per station) must be added, which is why HVDC pays off only for long distances.

HVDC has advantages over HVAC when large power is to be sent over a long distance (beyond the break-even distance), and in some special cases.

Circumstances favouring HVDC

1. Bulk power over long overhead lines (beyond about 500–800 km).
2. Submarine or underground cables longer than about 40–80 km, where AC charging current would use up the cable rating.
3. Linking two AC systems of different frequency, or not synchronised (asynchronous tie).
4. Where fast power-flow control or stability support is needed, and to avoid raising fault levels.
5. Where right of way is limited (fewer conductors, smaller towers).

Proof: 2 DC conductors carry the same power as 3 AC conductors

Assumptions: all conductors same size (same current $I$), same insulation level (peak voltage to earth $V_m$).

AC three-phase line, phase voltage rms $V_m/\sqrt{2}$:

$$P_{ac} = 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi$$

Bipolar DC line, $\pm V_d$ with $V_d = V_m$, $I_d = I$:

$$P_{dc} = 2V_dI_d = 2V_mI$$

Ratio:

$$\frac{P_{dc}}{P_{ac}} = \frac{2}{2.121\cos\phi} = \frac{0.943}{\cos\phi}$$

For the usual power factor $\cos\phi = 0.943$ (or close to it), $P_{dc} = P_{ac}$. So a 2-conductor HVDC line transmits the same power as a 3-conductor HVAC line with the same conductor size and insulation. In addition, its line loss is only $2I^2R$ against $3I^2R$, i.e. 2/3.

Converter used in HVDC and why

• Line-commutated, fully controlled three-phase thyristor bridge converters, usually connected as 12-pulse units (two 6-pulse bridges fed by Y-Y and Y-Δ transformers).
• The same bridge works as a rectifier ($\alpha < 90°$) at the sending end and an inverter ($\alpha > 90°$) at the receiving end, so power can be reversed by changing firing angles.
• Thyristors are available in very high voltage and current ratings and can be series-connected into valves; they use natural commutation (no extra commutation circuit).
• 12-pulse operation cancels the 5th and 7th harmonics, reducing filter size.
• Newer schemes also use voltage source converters (VSC) with IGBTs, which allow independent reactive power control and connection to weak networks.

Advantages of HVDC over HVAC

• Only 2 conductors (bipolar) or 1 (monopolar with earth return): cheaper towers and line.
• No stability limit, so power capacity does not fall with line length.
• No reactive power and charging current; any cable length is possible.
• No skin effect; lower corona and line losses (about 2/3 of AC for the same power).
• Can connect asynchronous AC systems or systems of different frequencies.
• Fast power-flow control by firing angle; does not add to AC fault level.

Why negative polarity is used in homopolar HVDC

A homopolar link has two or more conductors of the same polarity with earth return. Negative polarity is chosen because:

• Corona loss is lower with a negative conductor than with a positive one at the same voltage.
• Radio and audible interference produced by corona is much less for negative polarity.
• Since all conductors have the same polarity, choosing negative gives these benefits on every conductor.

  Rect ===(-Vd)=== Inv
  Rect ===(-Vd)=== Inv
   |               |
  earth ......... earth (return)

HVAC transmission sends power as three-phase AC through transformers; HVDC converts AC to DC at the sending end (rectifier), transmits DC, and converts back to AC (inverter). Each is best in a different range.

Comparison of AC and DC transmission

AC is cheaper for short and medium distances because terminals are simple; DC is cheaper for long distances (beyond the break-even distance) and for long cables.

Justification of the 2/3 loss

Assumptions: same power capacity, same conductor size and current $I$, same loss per conductor $p = I^2R$, same insulation (peak voltage to earth $V_m$, with $V_d = V_m$).

Same capacity:

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi \\ P_{dc} &= 2V_mI \end{aligned}$$

These are equal for $\cos\phi = 0.943$, so a 2-conductor bipolar DC line has the same capacity as a 3-conductor AC line.

Total losses:

$$\frac{\text{Loss}_{dc}}{\text{Loss}_{ac}} = \frac{2I^2R}{3I^2R} = \frac{2p}{3p} = \frac{2}{3}$$

Hence the HVDC line loss is only 2/3 of the HVAC line loss for the same power. In practice the saving is a little more, because the AC resistance is higher due to skin effect and AC lines have more corona loss.

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.

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.

HVDC needs costly converter stations (thyristor valves, converter transformers, filters, reactive compensation) at both ends, but its line is cheaper per km and has lower losses. So HVDC becomes economical only when the line saving is large enough to pay for the terminals, i.e. for bulk power over a long distance.

Justification of the statement

• Terminal cost: HVDC terminal cost is much higher than AC substations. This is a fixed cost, independent of distance.
• Line cost: DC line needs 2 conductors instead of 3, lighter towers, less insulation and narrower right of way, so cost per km is lower. Its losses are about 2/3 of AC.
• Total cost = terminal cost + (cost per km × distance). The AC curve starts low but rises steeply; the DC curve starts high but rises slowly. They cross at the break-even distance (about 500–800 km for overhead lines, 40–80 km for cables).

 Cost
  |                      / AC
  |                    /
  |                  /    _____ DC
  |            _____X-----
  |     _____-    / :
  | DC -        /   :
  |           /     :
  |  AC     /       :
  |       /         :
  |     /           :
  +-----------------:------> km
             break-even distance

• Power level: converter cost per MW falls as rating rises, and the 2-conductor saving grows with power; for small power the fixed terminal cost cannot be recovered.
• Also, long AC lines need series and shunt compensation and are stability-limited ($P = \frac{V_1V_2}{X}\sin\delta$), adding cost; DC has no such limit.

So HVDC is cheaper than HVAC only when large power is sent beyond the break-even distance (or for long cables / asynchronous links).

Power transmission capacity of HVDC and HVAC lines are equal

Assumptions: same conductor size and current $I$, same insulation (peak voltage to earth $V_m$), DC voltage $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi \\ P_{dc} &= 2V_dI_d = 2V_mI \\ \frac{P_{dc}}{P_{ac}} &= \frac{0.943}{\cos\phi} \end{aligned}$$

For a normal power factor of $\cos\phi = 0.943$, $P_{dc} = P_{ac}$. Thus a bipolar DC line with 2 conductors has the same capacity as a three-phase AC line with 3 conductors, while its losses are $2I^2R$ against $3I^2R$ (2/3) and it saves one conductor and its insulators.

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.