Power Electronics

Everything in this chapter comes back to one idea: three families of semiconductor switch, arranged in different circuits, do all the converting aboard a modern ship. What separates them is how much control you have over when they conduct and when they stop — and that single property decides which circuit topology becomes possible.

The diode is the simplest case: a one-way valve with no control input at all. It conducts as soon as it is forward biased and blocks as soon as it is reverse biased, purely as a function of the voltage across it. There is no gate, no drive circuit and nothing to fail electronically — which is exactly why diode bridges are the default choice for a simple, robust rectifier front end where you don't need to vary the output.

The thyristor (silicon-controlled rectifier) adds a gate: a brief pulse of gate current fires it into conduction at a moment you choose, which is what lets a phase-controlled rectifier vary its DC output by delaying the firing point within each half-cycle. The catch is that the gate only turns it on. Once conducting, a thyristor behaves like a diode and stays on regardless of the gate until the main current is driven to zero by the circuit itself — on an AC supply this happens naturally every half-cycle, which is why thyristors still turn up in older battery chargers and DC drives, but why building an inverter from them needs extra commutation circuitry to force them off on demand.

The IGBT (insulated-gate bipolar transistor) closes that gap: a small gate voltage switches it fully on or fully off at will, thousands of times a second, with the low drive power of a MOSFET gate but the current-carrying capability of a power transistor. That full, on-demand controllability is what makes high-frequency PWM inverters practical, and it is why the IGBT — not the thyristor — is the switch you will find in virtually every modern drive, inverter and active front end on board.

A drive's front end takes whatever AC the ship's generators are producing and turns it into DC. On anything above a small single-phase load this means a six-pulse bridge: six diodes (or thyristors, in an older controlled design) arranged so each phase contributes two conduction periods per cycle, giving six overlapping "humps" of DC per supply cycle rather than the single lumpy hump you'd get from a single-phase bridge. The more conduction periods per cycle, the closer the raw output sits to a flat DC level before any filtering is even applied.

The average value of that output is a fixed multiple of the supply's line-to-line voltage, and it's worth carrying this relationship in your head as a sanity check on any DC link reading you take.

What is left after rectification is not smooth DC but a rippling waveform at six times supply frequency, and this is where the DC link earns its name: a bank of electrolytic capacitors (sometimes with a series inductor ahead of them) that smooths that ripple, stores enough energy to ride through brief supply dips, and — just as importantly — decouples the motor side of the drive from disturbances on the supply side. A voltage sag on the generator bus, for instance, barely troubles a well-designed DC link, because the capacitors simply supply the shortfall for the fraction of a second it takes for the rectifier to recover.

That stored energy has a downside you meet at both ends of a drive's life. At switch-on, an empty capacitor bank looks like a short circuit to the supply, so drives use a pre-charge resistor or a soft-start sequence to limit the inrush current while the capacitors come up to voltage. At switch-off, the same capacitors stay charged at the full DC link voltage — several hundred volts on a typical drive — for some time after the drive has been isolated, and this single fact is behind more near-misses in drive maintenance than almost anything else in this chapter.

The inverter stage takes the DC link and builds a variable-frequency, variable-voltage AC supply for the motor out of six switches — two per phase, one connecting that phase to the positive DC rail and one to the negative rail, switched so they are never both closed together. On its own that only gives a square-edged output, so the inverter doesn't switch each phase on for half a cycle and off for the other half; it switches at a much higher rate and varies how long each pulse stays on, which is the essence of pulse-width modulation.

The usual method compares a sine wave at the wanted output frequency (the reference) against a triangular wave at a much higher frequency (the carrier). Wherever the reference sits above the carrier, the switch is commanded on; where it sits below, the switch is commanded off. The result is a train of pulses whose individual width varies smoothly through the cycle — wide pulses near the peak of the sine wave, narrow ones near the zero crossing — so that the average voltage over each switching period traces out the sine wave, even though the instantaneous voltage is never anything but full DC link voltage or zero.

What actually reaches the motor as current looks far more sinusoidal than that voltage waveform suggests, and this is worth understanding rather than just remembering: the motor's own winding inductance opposes rapid changes in current, so it naturally averages out — smooths — the high-frequency switching content, leaving the low-frequency fundamental to drive the motor while the switching frequency component is left mostly as a small ripple. It is the winding, not any filter in the drive, doing that smoothing.

Carrier (switching) frequency is a trade-off, typically chosen somewhere between about 2 kHz and 16 kHz: push it higher and the current gets smoother and quieter (above roughly 15–16 kHz it leaves the audible range entirely), but every switching event costs a small amount of energy in the IGBTs, so switching losses and heating rise with carrier frequency too, regardless of how fast the motor itself is turning.

Every UPS on board does the same basic job — keep a load running through a mains interruption — but the three common architectures differ in exactly how much of a break the load feels, and that difference is what should drive the selection for a given piece of equipment, not habit or cost alone.

• Off-line (standby). The load runs directly off the mains supply during normal operation; only when that supply fails does the UPS switch the load onto its battery-fed inverter. The transfer takes a few milliseconds — fine for most general electronic loads, which ride through it on their own internal capacitance, but not something you'd accept for equipment that cannot tolerate any interruption at all.
• Line-interactive. Adds a voltage regulator (often a tap-changing autotransformer) in the normal mains path, so it corrects sags and surges without resorting to the battery at all; it still transfers to battery, with a similarly short break, on an actual outage.
• On-line (double-conversion). The load is fed continuously from the inverter, which is itself fed from a rectifier/charger with the battery floating across the DC link. Mains failure simply removes the rectifier's contribution — the battery picks up the DC link seamlessly, with no break at the output at all. The cost is that every watt to the load is converted twice, so on-line units run warmer and less efficiently than the alternatives.

Match the type to what the load can tolerate: GMDSS radio equipment, ECDIS and similar navigation-critical loads generally justify an on-line UPS; routine office or general electronic loads are usually well served by off-line or line-interactive units at lower cost and complexity.

Behind almost every shipboard UPS sits a lead-acid battery bank, and its condition is only as good as its charging regime and upkeep. A fully charged cell reads a specific gravity of around 1.28, falling as it discharges; charging normally runs through a bulk stage, then a lower float or trickle stage to hold the bank topped up without overcharging it. Vented (flooded) cells give off hydrogen while charging, so the compartment must be ventilated and kept free of ignition sources; whichever cell type is fitted, routine checks — electrolyte level, specific gravity or cell voltage, terminal cleanliness, and temperature during charge — are what catch a failing bank before it lets the load down.

A rectifier doesn't draw a smooth sine-wave current from the supply — it only pulls current while a diode is actually conducting, which happens in short, peaky bursts near the top of each half-cycle. Any current waveform that isn't a pure sine wave can be shown, mathematically, to be built from the fundamental plus a set of higher-frequency sine waves at whole-number multiples of it: the harmonics. A six-pulse bridge is a textbook case, producing significant 5th, 7th, 11th and 13th harmonic current on top of the fundamental it draws.

Those harmonic currents don't stay politely inside the drive — they flow back through the ship's cabling and, wherever the supply has some source impedance (it always does), they produce a matching harmonic distortion in the supply voltage itself, which every other load on that board then sees too.

The damage this does is mostly thermal, and it's easy to miss because it doesn't show up as overcurrent on an ordinary ammeter. Transformers suffer extra eddy-current and stray losses that rise sharply with harmonic order, so a transformer running within its rated current can still run hot and age prematurely if that current is badly distorted. Cables see increased effective resistance at higher frequencies (skin effect), and motors pick up extra rotor losses and torque pulsation. Worst of all in a three-phase, four-wire system, the triplen harmonics — the 3rd, 9th, 15th and so on — do not cancel in the neutral the way the fundamental does between balanced phases; they add arithmetically, so a neutral conductor sized only for the phase current can overheat even though every phase current looks perfectly normal.

Because harmonic content is exactly what an averaging-type meter is bad at measuring — it reads a scaled average and assumes the waveform is a clean sine — any measurement on the supply or load side of power electronics needs a true-RMS instrument, or the figure you record can be badly wrong in either direction.

Mitigation follows the same three routes wherever you meet this problem: passive filters tuned to remove specific troublesome harmonics, multi-pulse rectification (a twelve-pulse arrangement cancels the 5th and 7th harmonics by combining two six-pulse bridges 30° apart), or phase-shifting transformers feeding multiple drives so that their harmonic currents cancel each other out upstream rather than adding together.

A drive fault-finds far faster if you resist the urge to start prodding the first accessible terminal and instead work the problem stage by stage: incoming supply, rectifier, DC link, inverter, motor and cable — in that order. Each stage has a distinct symptom set, and localising the fault to one of them before you open anything up turns a vague "the drive has tripped" into a short list of real possibilities.

Start upstream of the drive itself. Check the incoming supply is present and balanced on all three phases — a single blown fuse or a lost phase upstream will produce drive faults that look, at first glance, like an internal failure. Only once the supply is confirmed sound is it worth opening the drive's own diagnostics.

The DC link voltage is the single most useful diagnostic point in the whole drive, because it sits between the two stages you're trying to tell apart. A link voltage that is low or absent points back at the rectifier or its pre-charge circuit; a link voltage that is present but excessively rippled or unstable points at a failed rectifying device or a capacitor that has lost capacitance with age (electrolytic capacitors dry out over years of service — their effective series resistance rises, ripple grows, and they run hotter still, which accelerates the same ageing). A link that reads correctly but with a drive still faulting shifts the search downstream to the inverter and motor side.

On the inverter side, look for symmetry: an inverter with one failed switching leg tends to show up as a motor that runs rough, vibrates, or draws unbalanced current on one phase, rather than failing outright, because the remaining legs often keep something turning. Blocked or failed cooling fans and clogged filters are a mundane but very common root cause behind drives that trip on over temperature under load that used to be well within their rating — worth checking before assuming a semiconductor fault.

Whatever the symptom, treat the DC link capacitors as charged and dangerous until you have proved otherwise with a meter rated for the link voltage — many drives include a discharge resistor and a "safe to work" indicator, but neither is a substitute for measuring zero volts yourself before you touch anything inside the enclosure.

The three examples below carry the chapter's ideas through complete numerical problems, each built from more than one reasoning step — the kind of layered question you should expect on the exam rather than a single formula substitution.

Worked example 1

Six-pulse bridge output voltage and ripple, and the case for twelve-pulse

A cargo pump drive is fed from the ship's 400 V, 50 Hz three-phase supply through a standard six-pulse diode bridge rectifier feeding the drive's DC link. (a) Find the average DC link voltage. (b) Find the ripple frequency and the approximate ripple voltage. (c) The yard also offers a twelve-pulse option for a large HVAC drive on the same board — state, with a reason, how the ripple frequency and amplitude would change.

Given

Line-to-line supply voltage V_LL = 400 V (RMS) Supply frequency f = 50 Hz Six-pulse bridge ripple ≈ 4% of Vdc (typical figure for this topology) Twelve-pulse rectification approximately doubles the ripple frequency and halves its amplitude

Required

(a) Find the average DC link voltage
(b) Find the ripple frequency and the approximate ripple voltage

1. Average DC link voltage.

   Vdc=1.35 × V_LL =1.35 × 400 =540 V

   A three-phase, six-pulse full-wave bridge produces a DC output whose average value is a fixed multiple of the supply line voltage — this comes from integrating the six overlapping sine segments that make up one bridge output cycle.

2. Ripple frequency.

   f_ripple=6 × f =6 × 50 =300 Hz

   Each of the six diodes conducts for one-sixth of a cycle, so the output ripples six times per supply cycle, not once.

3. Ripple amplitude.

   V_ripple≈0.04 × Vdc =0.04 × 540 =21.6 V

   Applying the typical 4% figure to the average DC voltage gives the approximate peak-to-peak ripple that the DC link capacitor has to smooth.

4. Twelve-pulse comparison.

   f_ripple(12-pulse)=12 × 50 = 600 Hz V_ripple(12-pulse)≈21.6 / 2 = 10.8 V

   A twelve-pulse rectifier uses two six-pulse bridges fed from windings phase-shifted by 30°, so the two sets of ripple partly cancel: the ripple frequency doubles and its amplitude roughly halves.

AnswerVdc ≈ 540 V; six-pulse ripple ≈ 300 Hz at about 21.6 V; a twelve-pulse arrangement would raise the ripple frequency to 600 Hz while roughly halving its amplitude to about 10.8 V — smoother DC for the same filtering, at the cost of an extra phase-shifting transformer.

The trap: reading "six-pulse" and assuming the ripple repeats at the supply frequency — it repeats at six times the supply frequency, which changes what a scope trace or a ripple specification should actually show.

Worked example 2

PWM pulse ratio at rated speed and at reduced speed

A thruster motor drive uses sinusoidal PWM with a fixed carrier (switching) frequency of 4 kHz. (a) At the motor's rated output frequency of 50 Hz, how many switching pulses occur per fundamental cycle? (b) The thruster is throttled back so the inverter's output frequency falls to 10 Hz — find the new pulse ratio and explain, in terms of switching losses and current smoothness, what changes and what does not.

Given

Carrier (switching) frequency f_sw = 4000 Hz (fixed by the drive) Rated output frequency f_1 = 50 Hz Reduced output frequency f_2 = 10 Hz

1. Pulse ratio at rated speed.

   N1=f_sw / f_1 =4000 / 50 =80 pulses per cycle

   The pulse (or carrier) ratio is simply how many times the inverter switches for every one cycle of the output waveform it is building.

2. Pulse ratio at reduced speed.

   N2=f_sw / f_2 =4000 / 10 =400 pulses per cycle

   The carrier frequency is fixed by the drive's electronics, but the output (fundamental) frequency has dropped, so the same 4 kHz of switching now has to cover a much longer fundamental period.

3. What this means in practice.

   Switching losses depend on how often the devices switch per second, which is set by the carrier frequency alone — so they stay essentially unchanged as speed is reduced. But the motor current, filtered by the same winding inductance, is now built from five times as many pulses per cycle, so it becomes smoother, not rougher, at low speed.

AnswerN1 = 80 pulses/cycle at 50 Hz; N2 = 400 pulses/cycle at 10 Hz. Switching losses are essentially unchanged, since they track the carrier frequency rather than the output frequency; current smoothness actually improves at the lower speed because more pulses now shape each fundamental cycle.

The trap: assuming a lower output frequency means less switching and therefore cooler power devices — it is the carrier frequency, not the output frequency, that sets the switching loss.

Worked example 3

Sizing a UPS battery bank and estimating recharge time

An on-line UPS supplies a critical navigation and communications load of 2400 W (AC side) from a 120 V DC battery bank, with the inverter section operating at 80% efficiency. The bank must be able to carry the full load for 30 minutes, and the designer adds a 20% margin for battery ageing. The shipboard charger available for recharging after a discharge can deliver 5 A. Find (a) the battery capacity to specify, and (b) the approximate time to recharge the bank from empty.

Given

AC load power = 2400 W Inverter efficiency η = 80% (0.8) Battery bank nominal voltage = 120 V DC Required backup time = 30 min = 0.5 h Ageing margin = 20% Charger output = 5 A

Required

Find (a) the battery capacity to specify, and (b) the approximate time to recharge the bank from empty

1. DC-side power.

   P_dc=P_ac / η =2400 / 0.8 =3000 W

   The battery has to supply not just the AC load but also the loss in the inverter that converts DC to AC, so the DC draw is larger than the AC load figure.

2. Battery discharge current.

   I=P_dc / V =3000 / 120 =25 A

   Dividing the DC power by the nominal bank voltage gives the current the battery must supply for the duration of the backup.

3. Capacity with margin.

   Ah_bare=I × t =25 × 0.5 =12.5 Ah Ah_specified=12.5 × 1.2 =15 Ah

   The bare ampere-hour figure for a 30-minute discharge is the current multiplied by the time in hours; the 20% ageing margin is then added so the bank still meets the duty once it has lost some capacity with age.

4. Recharge time.

   t_recharge≈Ah_specified / I_charge =15 / 5 =3 h

   Once discharged, the bank recovers roughly its rated ampere-hours divided by the charger's output current — this ignores the tapering of current near full charge, so treat it as a lower-bound estimate.

AnswerSpecify a battery bank of at least 15 Ah; recharging it from empty at 5 A takes roughly 3 hours as a first estimate, longer in practice once the charger tapers off near full charge.

The trap: sizing the battery from the AC load figure directly and forgetting the inverter's own conversion loss, which understates the true DC-side draw and leaves the bank undersized.