University of Sydney · FACULTY OF ELECTRICAL ENGINEERING

ELEC5206 Chap.11 Single-Phase Grid-Side Conversion

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Chapter 11 of 14 · ELEC5206

Single-Phase Grid-Side Conversion

The grid side converter turns the DC link into a sinusoidal current injected into a single-phase grid. The lecture starts with the four-switch bridge: diagonal conduction produces plus or minus the DC voltage, flat conduction produces zero, and inserting the zero state lets pulse-width modulation control the output.

Most single-phase PV inverters are grid following, so an L-type grid link turns the bridge voltage into a current that tracks the grid voltage, with amplitude 2PDC/Vmag from power balance and the requirement that the DC link exceeds the grid peak. A hysteresis current controller chooses the diagonal pair from the current error.

The second lecture assembles a 6 kW two-channel system and shows how lower power reduces DC-link ripple. A simulation of a 2.2 kW inverter shows the grid current held inside a 1 A band around its sinusoidal reference, switching at about 16 kHz on average, which makes the band-versus-frequency trade-off concrete.

In this chapter

What this chapter covers

  • 01

    The four-switch bridge and the shoot-through rule

  • 02

    Diagonal and flat conduction states

  • 03

    Pulse-width modulation by inserting the zero state

  • 04

    Grid following through an L-type grid link

  • 05

    Current amplitude from power balance

  • 06

    Why the DC link must exceed the grid peak

  • 07

    Hysteresis current control and band selection

  • 08

    A two-channel 6 kW system and its DC and AC sections

Worked example · free

Current command for a 3 kW inverter on a 240 V supply

Q [3 marks]. A single-phase PV inverter delivers 3 kW at unity power factor into a 240 V RMS grid from a 380 V DC link. Find the current command and check that the bridge can control it. The 3-mark allocation is our own practice weighting, not the university's marking scheme.
  • 1Grid peak: Vmag = 240√2 = 339.4 V.
  • 1Current amplitude: Imag = 2 × 3000/339.4 = 17.7 A, so iref = 17.7 sin ωt.
  • 1Check: vdc = 380 V exceeds Vmag = 339.4 V by 40.6 V, so the rising-current state is available even at the grid crest.
Imag = 17.7 A, and the 380 V link leaves about 41 V of headroom above the grid peak.
Sia tip — Convert RMS to peak before using Imag = 2P/Vmag; using 240 V directly overstates the current by 41%.
Glossary

Key terms

Shoot-through
The forbidden state in which both switches of one bridge leg conduct and short the DC link.
Diagonal conduction
Switching the high side of one leg with the low side of the other to place plus or minus the DC voltage across the output.
Flat conduction
Switching both high-side or both low-side switches so that the bridge output is zero.
L-type grid link
A single inductor between bridge and grid that sets how the bridge voltage drives the grid current.
Grid-following inverter
An inverter that synchronises its current to the measured grid voltage rather than forming the voltage itself.
FAQ

Single-Phase Grid-Side Conversion FAQ

Why must the DC-link voltage be higher than the grid peak?

To raise the grid current the bridge applies plus the DC-link voltage, and the inductor current only rises if that exceeds the instantaneous grid voltage. Near the crest of the grid sine wave, a DC link below the peak cannot push current upward, so the current stops tracking its reference.

How does the hysteresis band affect power quality?

A narrower band keeps the grid current closer to its sinusoidal reference, so distortion falls; the lecture shows the improvement going from a 5 A demonstration band to 0.74 A. The cost is a higher switching frequency, which is also shaped by the size of the grid inductor.

Why does the 6 kW system use two MPPT channels?

Each channel's boost converter tracks one string independently, so mismatch between strings no longer costs power and a problem on one string does not drag down the other. The two converter outputs sum into a shared DC link feeding one grid side converter.

What happens to DC-link ripple when irradiance falls?

It falls roughly in proportion to the delivered power, because the ripple equals the DC power divided by the product of grid angular frequency, DC-link voltage and capacitance, with the capacitor and DC voltage unchanged. The lecture's integrated simulation shows smaller ripple after irradiance drops from 800 to 400 W/m².

Why does inserting a zero state matter for a single-phase bridge?

With diagonal conduction alone the bridge produces a square wave of plus and minus the DC voltage, whose RMS value is fixed. Adding flat conduction, with both high or both low switches on, creates a zero level, so the pulse widths can be modulated and the RMS output controlled.

Study strategy

Exam move

Draw the four-switch bridge and fill in the switch-state table until vab for each state is automatic. Then practise three short calculations: the current command from power, the DC-link headroom check and a band as a percentage of the current amplitude. Finish by explaining in two sentences how band width and inductor size together set power quality and switching frequency.

Practise converting between RMS and peak grid voltages until it is automatic.

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