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HVDC transmission

Electrical and Electronics Engineering (B.E./B.Tech) · Power systems · free study notes for exams, placements and interviews

In short

HVDC transmission converts AC to DC at a sending converter station, transmits bulk power over a DC line or cable, and inverts it back to AC at the receiving end. It is used where AC transmission is uneconomic or impossible: very long overhead lines, long submarine or underground cables, and asynchronous interconnections between grids or systems of different frequency. In India, POWERGRID operates several ±500 kV and ±800 kV bipoles (Rihand–Dadri, Talcher–Kolar, Ballia–Bhiwadi, Biswanath Chariali–Agra, Champa–Kurukshetra, Raigarh–Pugalur) plus back-to-back stations at Vindhyachal, Chandrapur, Vizag and Sasaram that tie the regional grids. The topic carries regular GATE marks (converter equations, firing angle, bridge waveforms, control modes) and appears in interviews for POWERGRID, NTPC, state transcos, Siemens Energy, Hitachi Energy, GE Vernova, BHEL and CPRI/ERDA-type research roles.

Explained

Why DC, and where it wins

Link configurations and components

Converter theory: the 6-pulse and 12-pulse bridge

LCC versus VSC

Control of an HVDC link

Harmonics, insulation and Indian practice

Key points

Formulas and examples

No-load (ideal) DC voltage of a 6-pulse bridge: Vd0 = (3√2/π)·VLL ≈ 1.35·VLL
where VLL = line-to-line rms secondary voltage of the converter transformer (V), Vd0 in V
Rectifier DC voltage with overlap: Vd = Vd0·cos α − (3/π)·Xc·Id = 1.35·VLL·cos α − 3Xc·Id/π
α = firing angle (deg), Xc = commutating (leakage) reactance per phase (Ω), Id = DC current (A)
Inverter DC voltage: Vd = Vd0·cos γ − (3/π)·Xc·Id (magnitude form; γ = extinction angle)
Commutation relation: cos γ = −cos(α + μ); γ = 180° − α − μ
DC current and power: Id = (Vdr − Vdi)/Rline ; Pd = Vd·Id ; for a bipole Pd = 2·Vpole·Id
Rline = loop resistance of the DC line (Ω)
Overlap angle: cos α − cos(α + μ) = 2·Xc·Id/(√2·VLL)
Converter power factor (approx.): cos φ ≈ Vd/Vd0 = cos α − 3Xc·Id/(π·Vd0); Q = P·tan φ
AC line current (fundamental, 6-pulse): I1 = (√6/π)·Id ≈ 0.78·Id ; rms of ideal square-wave current Iac = √(2/3)·Id ≈ 0.816·Id
Harmonic orders: AC side h = pk ± 1, DC side h = pk (p = pulse number, k = 1, 2, 3…); Ih ≈ I1/h for an ideal rectifier
12-pulse bridge (two series bridges): Vd = 2[Vd0·cos α − 3Xc·Id/π]

Example 1 — DC voltage, overlap and power factor of a 6-pulse bridge. Converter transformer secondary VLL = 345 kV, Xc = 15 Ω/phase, α = 15°, Id = 2,000 A.

Example 2 — Bipole power and loss. A ±500 kV bipole carries 1,800 A per pole over 800 km; conductor resistance 0.015 Ω/km per pole.

Example 3 — Inverter firing angle for a required γ. Inverter with Vd0 = 500 kV per bridge-group, Xc = 12 Ω, Id = 2,500 A, required γ = 18°.

Example 4 — Link current and control mode. Rectifier Vd0r = 420 kV at α = 10°, Xc = 10 Ω; inverter holds Vdi = 390 kV constant; DC line loop resistance 8 Ω.

Example 5 — Break-even distance. AC option: terminals ₹400 crore, line ₹4.25 crore/km. HVDC option: terminals ₹2,600 crore, line ₹0.75 crore/km.

FeatureAC transmissionHVDC (LCC)HVDC (VSC)
Conductors for same power32 (bipole)2
Line reactance / chargingLimits lengthNoneNone
Needs synchronismYesNoNo
Power control speedSlow (via FACTS/generation)MillisecondsMilliseconds
Reactive power at terminalsSupplies/absorbsConsumes ~50–60% of MWIndependently controllable
Converter losses per station—~0.7–0.8%~1%
Black start into dead gridYesNoYes
Fault level contributionIncreasesNoneNone
Typical max rating—Up to ~8,000 MW (±800 kV)A few thousand MW
ParameterRectifierInverter
Angleα ≈ 12–18° normal, 5° minimumγ ≈ 15–18° minimum, α ≈ 140–150°
Normal controlConstant current (CC)Constant extinction angle / constant Vd
On AC voltage dipHits α-min, loses current controlTakes over current control at I − margin
Vd polarity for power reversalBecomes negativeBecomes positive (roles swap)

Where it is used

Interview questions

Common mistakes

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