Chapter 1 of 30
- What power electronics is, and the four roles inside every converter: source, power circuit, load, and control.
- Why a switching converter reaches 95–99% efficiency where a linear regulator wastes the difference as heat.
- The ideal switch — why \( p(t) = v\,i = 0 \) at every instant, and how real devices fall short of it.
- The four conversion families — AC–DC, DC–DC, DC–AC, AC–AC — and the converter that performs each.
- The two workhorse relations of the whole subject: the duty ratio \( D = t_{on}/T \) and volt-second balance on an inductor.
- How converters are classified: by device controllability, commutation, switching type, and quadrants of operation.
What Is Power Electronics?
Power electronics is the use of solid-state switching devices to convert and control electrical power.
It sits where three subjects meet:
- Power engineering — supplies the grid, the machines and the loads.
- Electronics — supplies the switching devices themselves.
- Control systems — supplies the feedback that holds the output where we want it.
What makes it a distinct subject is not the size of the signal but its purpose. In a communications amplifier the signal carries information and we may spend energy freely to keep it faithful. Here the signal is the energy, so every watt we spend processing it is a watt the customer paid for and did not receive.
The four roles inside every converter
No matter how complicated the circuit looks, it decomposes into the same four blocks:
- Source — where the energy comes from A utility line, a battery, a PV array, a DC link fed by another converter.
- Power circuit — where the conversion happens The arrangement of switches, inductors and capacitors that chops the input and reshapes it. This is what most of the book analyses.
- Load — where the energy goes A motor, a heater, a data-centre rack, or the grid itself. The load's character decides which waveform quantity matters — a point we return to constantly.
- Control circuit — the part that decides It senses the output, compares it with a reference, and issues gate pulses telling each switch when to turn on and off.
The engineering goal never changes: deliver the demanded voltage and current efficiently, reliably, and with acceptable waveform quality. Those three words — efficiency, reliability, harmonics — generate almost every design constraint in this book.
Why Not a Resistor? Linear versus Switching
Suppose we need 5 V from a 12 V supply. The obvious answer is to drop the extra 7 V across a series element — a resistor, or a transistor held in its active region. This is linear regulation. It works, it is quiet, and it is simple.
It is also a heater. Every ampere delivered to the load drags 7 W of loss into the pass device, because that device is standing in the current path with 7 V across it.
The switching alternative never sits in between. The switch is either closed (full current, almost no voltage) or open (full voltage, no current); the inductor and capacitor smooth the chopped waveform into a steady output. Nothing is deliberately dissipated, so the efficiency of a well-designed converter is set only by its parasitics — typically 90–99%. For a linear regulator the ceiling is fixed by physics alone:
Read that result carefully, because it is stronger than it looks. Stepping 12 V down to 5 V caps efficiency at 41.7% — and the transistor's quality does not appear anywhere in the formula. A perfect, zero-resistance, infinitely fast pass device would still throw away 58.3% of the input.
The penalty then compounds:
- Wasted energy costs money every hour the equipment runs.
- The waste appears as heat inside the enclosure, which must be removed.
- Removing it needs heatsinks, fans and air gaps — so the box grows.
- A bigger, hotter box is heavier, costlier and less reliable (every 10 °C roughly halves semiconductor life).
That chain is why almost every power supply built since the 1980s is a switching one.
| Property | Linear regulator | Switching converter |
|---|---|---|
| Device operating region | Active (partially on) | Saturation / cut-off only |
| Efficiency | \(V_o/V_{in}\); often 30–60% | Typically 90–99% |
| Heat & cooling | Large heatsink required | Small; often convection only |
| Size & weight | Bulky (50 Hz magnetics) | Compact (high-frequency magnetics) |
| Conversion direction | Step-down only | Step-up, step-down, invert, reverse |
| Output noise | Very low | Switching ripple and EMI |
| Typical use | Low-noise analogue rails | Chargers, drives, inverters, EVs |
A linear regulator produces 3.3 V from a 5 V input, and another produces 3.3 V from a 12 V input. Both deliver 0.5 A. Which runs hotter, and by how much?
The 12 V one, by a factor of about 4.
Loss is \((V_{in}-V_o)I_o\), so the 5 V version dissipates \((5-3.3)(0.5) = 0.85\) W and the 12 V version dissipates \((12-3.3)(0.5) = 4.35\) W.
Notice the load is getting exactly the same 1.65 W in both cases. Nothing about the load changed — only the input voltage — and yet the heat went up five-fold. This is why a linear regulator's input voltage should always be chosen as close to its output as the circuit allows.
True or false: making the pass transistor bigger and better-cooled will improve a linear regulator's efficiency.
False. It will improve the regulator's reliability and let it handle more current without overheating — but efficiency is fixed at \(\eta = V_o/V_{in}\) by the physics of the arrangement.
The transistor is not "losing" power because it is imperfect; it is losing power because it is required to stand off the voltage difference while passing the load current. A better transistor still has to do exactly that job.
The Ideal Switch and the Real Device
The entire subject rests on one idealisation. An ideal switch has zero voltage across it when closed, carries zero current when open, and changes state instantaneously. The instantaneous power it dissipates is therefore
When the switch is ON, \(v = 0\) and the product vanishes. When it is OFF, \(i = 0\) and the product vanishes again. Only during the transition, when \(v\) and \(i\) are simultaneously non-zero, is energy lost — which is why switching loss scales with frequency and why fast devices matter.
A real device departs from that ideal in exactly four measurable ways. Each departure has a name, a datasheet symbol, and a loss it is responsible for:
| Departure from ideal | Typical symbol | Consequence | Studied in |
|---|---|---|---|
| Finite on-state voltage drop | \(V_{CE(sat)}\), \(R_{DS(on)}\) | Conduction loss while carrying current | Ch. 2, 5 |
| Small leakage current when off | \(I_{leak}\) | Blocking loss — usually negligible, but grows fast with temperature | Ch. 2, 3 |
| Finite rise and fall times | \(t_r,\ t_f\) | Switching loss — voltage and current overlap during the transition | Ch. 5 |
| Limited voltage and current ratings | \(V_{DRM}\), \(I_{T(AV)}\) | Sets the safe operating area; exceeded means destruction, not degradation | Ch. 2–5 |
Chapters 2–5 study how each device family trades these four against one another. For now, hold on to the pairing: rows 1 and 2 produce conduction loss; row 3 produces switching loss.
The Four Conversion Families
Electrical power comes in only two forms, AC and DC, so there are exactly four conversions to perform — and every converter you will ever meet is one of them, or a cascade of them. Learn this map and the structure of the whole book follows from it.
| Conversion | Converter name | Typical application | Covered in |
|---|---|---|---|
| AC → DC | Rectifier (uncontrolled or phase-controlled) | DC drives, battery chargers, front end of every SMPS | Part 2 |
| DC → DC | Chopper / switched-mode converter | EV traction, MPPT stages, point-of-load supplies | Part 3 |
| DC → AC | Inverter | Solar and battery inverters, VFDs, UPS | Part 4 |
| AC → AC | AC voltage controller, cycloconverter, matrix converter | Fan and light dimming, soft starters, cement mill drives | Part 5 |
Why is there no such thing as a "DC transformer"? What has to happen before a transformer can be used to change a DC voltage?
A transformer works by changing flux: \(v = N\,d\phi/dt\). A steady DC voltage produces a steadily rising flux until the core saturates, after which the winding is just a piece of wire and the current runs away.
So to use a transformer on DC you must first manufacture an alternating waveform — that is, do a DC–AC conversion, pass it through the transformer, then rectify back to DC (AC–DC). That three-stage chain, run at tens of kilohertz so the transformer can be tiny, is exactly what an isolated DC–DC converter is (Chapter 14).
This is also the answer to "why does a phone charger have no heavy iron transformer?" — it has a transformer, but it operates at 100 kHz instead of 50 Hz, so the core is roughly two thousand times smaller.
A regenerative-braking system on a metro train sends energy from the motor back to the overhead line. Which conversion family is operating, and in which direction?
The same hardware that ran DC–AC while motoring now runs AC–DC while braking. The motor becomes a generator, its three-phase AC is rectified by the inverter's own switches, and the energy flows back to the DC link.
The important lesson: the conversion family is a statement about the direction of power flow at a given instant, not a permanent property of the circuit. A converter that can do this is called bidirectional, and it needs devices and control that support current in both directions — which is why Section 1-7 classifies converters by quadrants of operation.
Duty Ratio, Averaging, and Volt-Second Balance
If a switch only ever produces a chopped square wave, how do we get a controlled output? By controlling the fraction of each cycle for which the switch conducts. Let the switching period be \(T = 1/f_s\) and let the switch be closed for \(t_{on}\). The duty ratio is
Watch the two dashed lines. They meet only at \(D = 0\) and \(D = 1\); everywhere in between the RMS line sits above the average, and the gap is widest around \(D \approx 0.25\). A motor armature or a battery responds to the red line. A resistor heats according to the green one.
Why does controlling time control the output at all? Because of a property of the load, not of the switch:
- A motor winding, an LC filter or a battery has far too much inertia to follow a 20 kHz square wave.
- It therefore ignores the individual pulses and responds only to what they add up to — the average.
- Chop harder (smaller \(D\)) and the average drops. Chop less and it rises. Nothing is dissipated to make that happen.
Volt-second balance, in four steps
The second workhorse relation follows from one observation about steady state.
- Start from the inductor law The voltage across an inductor is \(v_L = L\,\dfrac{di_L}{dt}\), so the change in its current over any interval is \(\Delta i_L = \frac{1}{L}\int v_L\,dt\).
- Impose steady state "Steady state" means every cycle looks like the last one. So the inductor current at the end of a period must equal the current at the start: \(\Delta i_L = 0\) over one full period \(T\).
- Conclude
If \(\Delta i_L = 0\) and \(L \ne 0\), the integral itself must vanish:
In words: whatever volt-seconds the inductor absorbs while the switch is on, it must give back while the switch is off.Volt-second balance\[ \int_0^{T} v_L(t)\,dt = 0 \]
- Use it Write down \(v_L\) in each sub-interval, multiply each by how long it lasts, set the sum to zero, and solve. Two lines of algebra replace a differential equation — see Worked Example 5.
The first says the output is controlled by time, not by resistance — this is why switching is both efficient and adjustable. The second says an inductor in steady state must gain and lose exactly the same volt-seconds each cycle; applying it to the buck converter gives \(V_o = D V_s\) directly, as Worked Example 5 shows.
| Symbol | Name | Meaning |
|---|---|---|
| \(V_s\) | Source voltage | Input to the converter |
| \(V_o,\ I_o\) | Output voltage, current | Average quantities delivered to the load |
| \(D\) | Duty ratio | \(t_{on}/T\); the control variable |
| \(T,\ f_s\) | Switching period, frequency | \(T = 1/f_s\) |
| \(v_L,\ \Delta i_L\) | Inductor voltage, current ripple | Sets filter size and ripple current |
| \(\eta\) | Efficiency | \(P_o/P_{in}\) |
| \(\alpha\) | Firing (delay) angle | Control variable of a phase-controlled rectifier |
A chopper runs at \(D = 0.25\) from 400 V. A colleague measures the output with a cheap multimeter set to DC and reads 100 V; with another meter set to AC+DC true-RMS they read 200 V. Is one of the meters broken?
No — both are correct, and they are measuring different things.
The DC setting reports the average: \(V_o = DV_s = 0.25 \times 400 = 100\) V. The true-RMS setting reports \(V_{rms} = V_s\sqrt{D} = 400\sqrt{0.25} = 200\) V.
The ratio \(V_{rms}/V_o = 1/\sqrt{D} = 2\) here. A chopped waveform genuinely has two different "sizes" and you must know which one your load cares about. This is also why a cheap average-responding AC meter gives nonsense on chopper and inverter outputs — it assumes a sine wave, and this is not one.
Volt-second balance says \(\int_0^T v_L\,dt = 0\). Does that mean the inductor voltage is zero?
No. It means the average inductor voltage over a full switching period is zero. The instantaneous voltage is large and alternating — that is the whole point, because that is what ramps the current up and down.
An analogy: your bank balance can be the same on the 1st of January two years running while millions have flowed through the account. Zero net change does not mean zero activity.
A useful corollary: because the average voltage across an ideal inductor is zero in steady state, the average voltage at one end of an inductor equals the average voltage at the other end. That single sentence solves a surprising number of DC–DC problems in Part 3 by inspection.
Losses in Real Devices
Since the ideal switch is lossless, every watt of loss in a converter can be traced to a specific departure from the ideal. Two dominate. Conduction loss arises from the on-state voltage drop while the device carries current, and is proportional to the fraction of the period spent conducting. Switching loss arises during the finite transitions, when voltage and current overlap, and is proportional to how often those transitions happen.
Look at what each expression does not contain — that is where the diagnostic power lives:
| Property | Conduction loss | Switching loss |
|---|---|---|
| Grows with | Current squared, and duty ratio | Frequency, and transition time |
| Independent of | Frequency | Duty ratio |
| Worst at | Heavy load | High frequency, high voltage |
| Reduce it by | Lower \(R_{DS(on)}\), paralleling devices, better cooling | Lower \(f_s\), faster gate drive, soft switching (Ch. 20) |
The remaining contributors are smaller but rarely negligible:
- Gate-drive power — charging and discharging the gate capacitance every cycle; grows with \(f_s\).
- Magnetic core and copper loss — hysteresis and eddy currents in the core, \(I^2R\) in the winding (Chapter 23).
- Capacitor ESR loss — ripple current heating the equivalent series resistance.
- Diode reverse-recovery loss — the subject of Chapter 2, and often the surprise in a hard-switched bridge.
Chapter 5 develops all of these into a loss budget you can actually calculate.
Efficiency is not just an economic figure — the numerator is what the customer paid for and the denominator's extra terms are heat that must leave the enclosure. In a 100 kW converter, moving from 96% to 98% efficiency halves the heat load from 4 kW to 2 kW, and that single change can decide whether the design needs liquid cooling.
Classification of Power Converters
Beyond the four conversion families, converters and their devices are sorted along several independent axes. Knowing where a circuit sits tells you which analysis applies — and which chapters you will need.
Uncontrolled (diode — conducts when the circuit says so), semi-controlled (SCR — turn-on commanded, turn-off not), and fully controlled (MOSFET, IGBT, GTO — both commanded).
Natural or line commutation lets the AC supply reverse and turn the device off; forced or self-commutation uses an auxiliary circuit or a gate command. Chapter 4 treats both.
Hard switching transitions with voltage and current overlapping; soft switching (ZVS, ZCS) times the transition to a natural zero crossing, cutting switching loss (Chapter 20).
Which signs of \(V\) and \(I\) the converter supports: one quadrant (motoring only), two (adds braking or reverse), four (full regeneration, as in EV traction and lifts).
Non-isolated converters share a ground; isolated ones insert a high-frequency transformer for safety, voltage matching, or multiple outputs.
From milliwatt energy-harvesting circuits at megahertz, to gigawatt HVDC valves switching at line frequency — the same principles, six orders of magnitude apart.
Classifying one circuit on every axis
The axes are only useful if you can apply them. Here are two familiar circuits classified completely — try covering the right-hand columns and filling them in yourself first.
| Axis | Buck converter in a laptop | Phase-controlled rectifier on a DC drive |
|---|---|---|
| Conversion family | DC–DC | AC–DC |
| Device controllability | Fully controlled (MOSFET) | Semi-controlled (SCR) — turn-on commanded, turn-off is not |
| Commutation | Forced / self-commutated by the gate | Natural (line) commutation — the supply reverses and does it for you |
| Switching type | Hard switched (unless a resonant topology) | Effectively soft — the device turns off at a natural current zero |
| Quadrants | One (\(V\) and \(I\) both positive) | Two — voltage can reverse for regenerative braking, current cannot |
| Isolation | Non-isolated (shares a ground) | Isolated only if a supply transformer is fitted |
| Power / frequency | Tens of watts, 300 kHz–2 MHz | Tens of kilowatts, 50 Hz line frequency |
Worked Examples
Problem. A 12 V supply must deliver 5 V at 1 A. Compare a linear regulator with a switching converter of 92% efficiency: find the loss and the current drawn from the source in each case.
Solution. The load takes \(P_o = 5 \times 1 = 5\) W either way. The linear regulator passes the same 1 A through the source, so \(P_{in} = 12 \times 1 = 12\) W and \(P_{loss} = 7\) W, giving \(\eta = 5/12 = 41.7\%\). The switching converter draws only the power it needs:
Loss falls from 7 W to 0.43 W — a factor of sixteen — and source current falls from 1 A to 0.45 A. A switching converter is not a current-in-equals-current-out device; it trades voltage for current the way a transformer does.
Problem. A chopper fed from \(V_s = 220\) V must deliver 150 V average to a DC motor. It switches at 1 kHz. Find \(D\), \(t_{on}\) and \(t_{off}\).
Solution. From \(V_o = D V_s\):
The output is set purely by timing. No series resistance, no dissipation — which is exactly why chopper-fed drives replaced rheostatic speed control.
Problem. The same chopper (\(V_s = 220\) V, \(D = 0.682\)) now feeds a purely resistive load of 10 Ω. Find the RMS output voltage and the load power. Compare with \(V_o^2/R\).
Solution. For a rectangular wave of height \(V_s\) present for a fraction \(D\) of each cycle,
The two differ by nearly 50%. Power always follows the RMS value; only a load that responds to the average (a well-filtered motor armature, a battery) sees \(D V_s\). Confusing the two is the single most common error in early power-electronics work.
Problem. A MOSFET with \(R_{DS(on)} = 0.1\ \Omega\) carries 20 A at \(D = 0.5\) while blocking 300 V. Its transitions take \(t_r = t_f = 50\) ns. Find both losses at \(f_s = 20\) kHz, then at \(f_s = 200\) kHz.
Solution.
Total 26 W at 20 kHz. Raising the frequency tenfold leaves \(P_{cond}\) at 20 W but drives \(P_{sw}\) to 60 W, tripling the total. The magnetics would shrink tenfold — and the heatsink would have to grow. This is the size-versus-efficiency trade of Section 1-3 in numbers.
Problem. In a buck converter the inductor sees \((V_s - V_o)\) while the switch is on and \((-V_o)\) while it is off (the diode freewheels). Derive \(V_o\).
Solution. In steady state the inductor current returns to its starting value each cycle, so the net volt-seconds must be zero:
Two lines of algebra, no differential equations. This is the standard method for every DC–DC topology in Part 3 — write the inductor voltage in each sub-interval, set the weighted sum to zero, solve.
Problem. A rooftop solar plant feeds a battery and the grid. Identify the conversion stages and classify each.
Solution. The PV array produces DC at a voltage that varies with irradiance, so an MPPT stage performs DC–DC conversion (non-isolated, one quadrant, hard-switched, fully controlled devices). The battery is charged and discharged through a bidirectional DC–DC converter (two quadrants — current must flow both ways). The grid interface performs DC–AC conversion in an inverter (fully controlled, PWM, four quadrants if it must both export and absorb reactive power). Naming the stages this way immediately tells you which chapters govern each block — 11–15 for the DC–DC stages, 16–20 for the inverter, and 28 for the system as a whole.
Summary & Formula Sheet
Six ideas carry the rest of the book. If you can state each one in a sentence without looking, you are ready for Chapter 2.
Every converter has a source, a power circuit, a load, and a control circuit issuing gate pulses.
Linear regulation caps efficiency at \(V_o/V_{in}\); switching reaches 90–99% because nothing is deliberately dissipated.
\(p = vi = 0\) both ON and OFF. All loss comes from on-state drop, leakage, and finite transition time.
AC–DC (rectifier), DC–DC (chopper), DC–AC (inverter), AC–AC (controller, cycloconverter). Real systems cascade them.
\(D = t_{on}/T\) sets the average, \(V_o = DV_s\); volt-second balance \(\int v_L\,dt = 0\) gives every conversion ratio.
Conduction loss follows current and duty; switching loss follows frequency. Together they fix \(\eta\) and the cooling design.
Key terms
- Power converter
- A circuit of switches, inductors and capacitors that changes electrical power from one form to another under control.
- Linear regulation
- Holding a device in its active region so it drops the surplus voltage. Simple and quiet; efficiency capped at \(V_o/V_{in}\).
- Ideal switch
- Zero voltage when closed, zero current when open, instantaneous transition — therefore \(p = vi = 0\) at all times.
- Duty ratio, \(D\)
- Fraction of the switching period for which the switch conducts. The control variable of every chopper and PWM converter.
- Volt-second balance
- In steady state an inductor's voltage integrates to zero over one period, because its current must return to the same value.
- Conduction loss
- Loss from the on-state voltage drop while the device carries current. Scales with \(I^2\) (MOSFET) and duty ratio; independent of frequency.
- Switching loss
- Loss during the finite transition, when voltage and current overlap. Scales with switching frequency; independent of duty ratio.
- Commutation
- The process of turning a device off. Natural when the supply reverses and does it; forced when a gate command or auxiliary circuit does it.
- Quadrants of operation
- Which combinations of output voltage and current sign a converter supports — one, two or four. Four-quadrant means full regeneration.
Test Yourself
Answer each before revealing it. These are conceptual — if you can talk your way through all six, the numerical problems that follow will be arithmetic rather than thinking.
In one sentence: why is a switching converter efficient?
Because its switch is never asked to hold voltage and pass current at the same time, so it has nothing to dissipate — the surplus energy is stored in an inductor or capacitor and handed on rather than converted to heat.
A weaker answer that still earns credit: "because the switch is only ever fully on or fully off." That is true, but the sentence above says why that matters.
A converter steps 400 V down to 100 V at 20 A output, with 95% efficiency. What current does it draw from the source?
\(P_o = 100 \times 20 = 2000\) W, so \(P_{in} = 2000/0.95 = 2105\) W, and \(I_{in} = 2105/400 = \) 5.26 A.
Sanity check on the answer: the voltage went down by a factor of 4, so the current should go up by roughly 4 — and \(20/5.26 = 3.8\), which is 4 discounted by the efficiency. If you answered 20 A, revisit the "current-in-equals-current-out" misconception in Section 1-2.
Two converters deliver the same output power. Converter A runs at 20 kHz, converter B at 200 kHz. Which has the smaller inductor, and which is more likely to need a bigger heatsink?
B has the smaller inductor — inductor size scales roughly as \(1/f_s\), so ten times the frequency needs roughly a tenth the inductance for the same ripple.
B is also the one more likely to need the bigger heatsink, because \(P_{sw} \propto f_s\) has gone up ten-fold while \(P_{cond}\) is unchanged.
That tension — smaller magnetics, larger losses — is the size-versus-efficiency trade, and choosing \(f_s\) is where a converter design usually starts.
A chopper feeds a DC motor. You are asked for (a) the speed the motor will settle at and (b) the rating of the fuse in series with it. Which value do you use for each?
(a) Speed → the average. The motor's speed follows its back-EMF, which follows the average armature voltage \(V_o = DV_s\). The armature inductance filters the switching ripple, so the machine never "sees" the pulses.
(b) Fuse → the RMS. A fuse is a thermal device; it melts on \(I^2R\) heating, which follows the RMS current.
Same circuit, same instant, two different numbers — because the two questions are about two different physical effects. Always ask what the quantity you need is physically caused by.
Classify a domestic ceiling-fan speed regulator (the electronic kind) on the conversion family and the controllability axes. Why can it not use a plain diode?
Conversion family: AC–AC. Mains AC goes in, chopped AC comes out at a reduced RMS value — it is an AC voltage controller (Chapter 21).
Controllability: semi-controlled — it is built from a TRIAC (or a pair of back-to-back SCRs), where turn-on is commanded by the gate and turn-off happens naturally at the current zero.
Why not a diode: a diode is uncontrolled. It conducts whenever the circuit forward-biases it, so there is no way to command when conduction starts — and therefore no way to vary the output. Control requires, at minimum, command over the turn-on instant.
Historically fans used a series resistor or tapped inductor, which is the linear approach — and exactly why old regulators got hot while modern electronic ones do not.
Your converter delivers 5 kW and measures 94% efficient. Management asks for 97%. How much heat does that remove from the enclosure, and why is the change harder than the numbers suggest?
At 94%: \(P_{in} = 5000/0.94 = 5319\) W, so loss is 319 W.
At 97%: \(P_{in} = 5000/0.97 = 5155\) W, so loss is 155 W.
You must remove 164 W of loss — slightly more than half of it.
Why it is hard: the first few percent of loss are usually the easy, obvious ones (an oversized device, a poor gate drive). What remains at 94% is already the irreducible-looking part — conduction drop, core loss, reverse recovery — spread across many small contributors. Halving a total made of ten roughly equal pieces means attacking nearly all ten. This is why efficiency improvements get exponentially more expensive as you approach the top, and why a "two-point" improvement is a serious engineering programme rather than a component swap.
Problems
Work these with a method rather than by pattern-matching:
- Name the conversion family and sketch the power circuit.
- Decide whether the question is about an average quantity or an RMS one.
- Apply \(V_o = DV_s\), \(V_{rms} = V_s\sqrt{D}\), or the loss expressions.
- Sanity-check the magnitude before writing the answer down.
Problems 1–5 are direct application; 6–9 need a small amount of judgement; 10–12 are open-ended and worth discussing in a tutorial.
- Name the conversion family for each: (a) a mobile-phone charger, (b) a ceiling-fan regulator, (c) a solar grid-tied inverter, (d) an EV traction drive between battery and motor.
- Identify the source, power circuit, load, and control circuit in a variable-frequency drive feeding a 15 kW induction motor.
- A linear regulator produces 3.3 V at 0.8 A from a 9 V input. Find the dissipation in the pass device and the efficiency.
- A chopper operates from 400 V with \(t_{on} = 300\ \mu\text{s}\) and \(f_s = 2\) kHz. Find \(D\), \(V_o\), and \(t_{off}\).
- The same chopper feeds a 5 Ω resistor. Find \(V_{rms}\), the load power, and the ratio of true power to the value \(V_o^2/R\) would predict.
- An IGBT with \(V_{CE(sat)} = 1.8\) V carries 50 A at \(D = 0.6\). Find its conduction loss. If \(t_r + t_f = 400\) ns at 600 V and \(f_s = 5\) kHz, find the switching loss and the total.
- Repeat Problem 6 at \(f_s = 50\) kHz. State which loss now dominates and what design change you would make.
- A boost converter's inductor sees \(V_s\) for \(DT\) and \((V_s - V_o)\) for \((1-D)T\). Apply volt-second balance to show \(V_o = V_s/(1-D)\), and find \(D\) for 48 V from a 24 V battery.
- A converter delivers 20 kW at 97.5% efficiency. Find the heat that must be removed. Repeat at 94% and comment on the cooling implication.
- Classify a phase-controlled thyristor rectifier on every axis of Section 1-7 — controllability, commutation, switching type, quadrants, isolation — and justify each call.
- Explain why raising \(f_s\) shrinks the inductor but not necessarily the heatsink, using the loss expressions to support your answer.
- A 250 kW battery energy-storage system must charge from and discharge to an 11 kV grid. Sketch the full conversion chain, name each stage, state the quadrants required at each, and explain why the DC–DC stage must be bidirectional while the MPPT stage of a PV plant need not be.