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9.P — Worked Problems: Power Electronics and Machines
Problem 1 — Transformer design
Design a 230 V to 24 V, 100 VA mains transformer at 50 Hz. Core area 12 cm², maximum flux density 1.2 T. Find the turns, the wire sizes and the efficiency.
Solution
Primary turns, from the EMF equation:
N_1=\frac{V}{4.44fAB_{max}}=\frac{230}{4.44\times50\times12\times10^{-4}\times1.2}=\frac{230}{0.32}=719
Turns per volt: 719/230 = 3.13.
Secondary turns, adding about 5% to compensate for the voltage drop under load:
N_2=24\times3.13\times1.05=79\ \text{turns}
Currents:
I_1=\frac{100}{230}=0.435\ \text{A}, \qquad I_2=\frac{100}{24}=4.17\ \text{A}
Wire sizes at 3 A/mm² current density, which is the usual figure for a naturally cooled transformer:
A_1=\frac{0.435}{3}=0.145\ \text{mm}^2 \;\Rightarrow\; d=0.43\ \text{mm}
A_2=\frac{4.17}{3}=1.39\ \text{mm}^2 \;\Rightarrow\; d=1.33\ \text{mm}
Copper loss. Assume a mean turn length of 20 cm on each winding:
Primary: \ell = 719\times0.2=144 m, R=\rho\ell/A=1.68\times10^{-8}\times144/1.45\times10^{-7}=16.7\ \Omega
P_{cu1}=0.435^2\times16.7=3.16\ \text{W}
Secondary: \ell=79\times0.2=15.8 m, R=1.68\times10^{-8}\times15.8/1.39\times10^{-6}=0.19\ \Omega
P_{cu2}=4.17^2\times0.19=3.30\ \text{W}
Core loss, taking 1.5 W/kg for silicon steel at 1.2 T and 50 Hz, with a core mass of about 1.5 kg:
P_{core}=2.25\ \text{W}
Efficiency:
\eta=\frac{100}{100+3.16+3.30+2.25}=\frac{100}{108.7}=92\%
Check the efficiency-optimum condition. Copper loss is 6.46 W against 2.25 W of core loss — copper dominates, so maximum efficiency occurs below full load, at
\text{load fraction}=\sqrt{\frac{P_{core}}{P_{cu(full)}}}=\sqrt{\frac{2.25}{6.46}}=0.59
Maximum efficiency at 59% load, which is exactly where a general-purpose transformer should be optimised. ✓
Regulation, referring the secondary resistance to the primary and adding:
R_{eq}=R_2+R_1\left(\frac{N_2}{N_1}\right)^2=0.19+16.7\times(0.11)^2=0.19+0.20=0.39\ \Omega
\text{drop}=4.17\times0.39=1.63\ \text{V} \;\Rightarrow\; \text{regulation}=\frac{1.63}{24}=6.8\%
Which is why the 5% turns compensation was added, and it is still slightly short — 7% would have been better.
Problem 2 — Grid transmission losses
Compare transmitting 200 MW over 150 km at 132 kV and 400 kV. Line resistance is 0.06 Ω/km per phase, power factor 0.95.
Solution
At 132 kV:
I=\frac{200\times10^6}{\sqrt3\times132\times10^3\times0.95}=\frac{200\times10^6}{2.172\times10^5}=921\ \text{A}
R=0.06\times150=9\ \Omega \text{ per phase}
P_{loss}=3\times921^2\times9=3\times8.49\times10^6\times\ldots
More carefully: 921^2 = 848{,}241, times 9 = 7.63 MW per phase, times 3:
P_{loss}=22.9\ \text{MW}=11.4\%
At 400 kV:
I=\frac{200\times10^6}{\sqrt3\times400\times10^3\times0.95}=304\ \text{A}
P_{loss}=3\times304^2\times9=3\times0.832\ \text{MW}=2.49\ \text{MW}=1.25\%
A factor of nine reduction, exactly as (400/132)^2 = 9.18 predicts.
Annual saving at 8000 hours and €50/MWh:
(22.9 − 2.49) MW x 8000 h x 50 euro/MWh = 8.16 million euro per year.
Which pays for the substations quickly. A pair of 400 kV substations costs perhaps €40 million, so the payback is under five years.
What sets the practical limit. Higher voltage needs taller towers, longer insulator strings and wider corridors, and corona discharge — ionisation of the air at the conductor surface — begins around 400 kV and causes both loss and audible noise.
The fix is bundled conductors: several conductors per phase held apart by spacers, which increases the effective diameter and reduces the surface field. A 400 kV line typically uses two or four conductors per phase, and that is what those spacers are for.
Problem 3 — Induction motor performance
A 4-pole, 50 Hz, 11 kW induction motor runs at 1440 rpm at full load, drawing 21 A at 400 V with a power factor of 0.85. Find the slip, torque, efficiency and rotor losses.
Solution
Synchronous speed:
n_s=\frac{120\times50}{4}=1500\ \text{rpm}
Slip:
s=\frac{1500-1440}{1500}=0.04=4\%
Rotor frequency:
f_r=0.04\times50=2\ \text{Hz}
Torque:
\omega=\frac{2\pi\times1440}{60}=150.8\ \text{rad/s}
T=\frac{11{,}000}{150.8}=72.9\ \text{N·m}
Input power:
P_{in}=\sqrt3\times400\times21\times0.85=\sqrt3\times7140=12{,}367\ \text{W}
Efficiency:
\eta=\frac{11{,}000}{12{,}367}=89\%
Total loss: 1367 W.
Air-gap power:
P_{gap}=\frac{P_{mech}}{1-s}\ \ldots
more precisely, mechanical output plus friction. Taking friction and windage as 200 W, the developed power is 11,200 W:
P_{gap}=\frac{11{,}200}{1-0.04}=11{,}667\ \text{W}
Rotor copper loss:
P_{rotor}=s\,P_{gap}=0.04\times11{,}667=467\ \text{W}
Remaining loss for stator copper and core:
1367-467-200=700\ \text{W}
The point to extract. Rotor loss is exactly the slip fraction of the air-gap power, and it can only be removed through the rotor surface into the air gap — there is no path for a heatsink. That is why running an induction motor at high slip for long periods destroys it, and why a motor that is repeatedly stalled or started fails from rotor damage.
Starting current is roughly six times full load, so about 126 A. At s=1 the rotor loss equals the entire air-gap power — every watt entering the rotor becomes heat, which is why a motor that fails to start must be disconnected within seconds.
Problem 4 — Motor efficiency economics
Compare an IE2 motor at 91.6% with an IE4 at 94.8%, both 37 kW, running 7000 hours a year at €0.18/kWh. The IE4 costs €1400 more.
Solution
Input power:
P_{IE2}=\frac{37}{0.916}=40.39\ \text{kW}, \qquad P_{IE4}=\frac{37}{0.948}=39.03\ \text{kW}
Difference: 1.36 kW.
Annual energy saving:
1.36\times7000=9520\ \text{kWh}
Annual cost saving:
9520 kWh x 0.18 euro = 1714 euro per year.
Payback:
\frac{1400}{1714}=0.82\ \text{years}=10\ \text{months}
Over a 15-year life:
1714 euro x 15 years − 1400 euro = 24,310 euro saved.
Now the number that reframes the whole decision. Total electricity over 15 years for the IE4:
39.03 kW x 7000 h x 15 years x 0.18 euro = 737,700 euro.
The motor itself costs perhaps €4000. So the purchase price is 0.5% of the lifetime cost.
And that is why efficiency classes had to be made mandatory. The buyer of a motor is often not the payer of the electricity bill — a machine builder specifies the cheapest motor that meets the specification, and the operator pays for fifteen years. The market failure is structural, not a matter of ignorance, and regulation is the standard fix for exactly this shape of problem.
One caution on the arithmetic. The saving is proportional to running hours. A motor running 500 hours a year saves €122 and takes eleven years to pay back — so for intermittent duty the cheaper motor is genuinely the right choice, and blanket rules are not always right.
Problem 5 — Battery pack sizing
Design a battery for an electric bicycle: 500 W motor, 36 V nominal, 60 km range at 25 km/h, using 3.6 V 3400 mAh cells.
Solution
Journey time:
t=\frac{60}{25}=2.4\ \text{hours}
Energy needed, assuming an average draw of 60% of peak (a realistic duty for cycling with pedal assistance):
E=500\times0.6\times2.4=720\ \text{Wh}
Add margin. Never discharge below 20% for cycle life (Chapter 9.3), and allow 10% for drivetrain and controller losses:
E_{pack}=\frac{720\times1.1}{0.8}=990\ \text{Wh}
Cell configuration.
Series count for 36 V:
\frac{36}{3.6}=10 \;\Rightarrow\; \textbf{10S}
Pack voltage range: 42.0 V full (10\times4.2) to 30 V empty (10\times3.0).
Energy per cell:
3.4\times3.6=12.24\ \text{Wh}
Cells needed:
\frac{990}{12.24}=81 \;\Rightarrow\; \text{round to } 80 = \textbf{10S8P}
Pack capacity:
8\times3.4=27.2\ \text{Ah}, \qquad 27.2\times36=979\ \text{Wh}
Current check. Peak draw:
I=\frac{500}{36}=13.9\ \text{A}
Per cell:
\frac{13.9}{8}=1.74\ \text{A}=0.51\text{C}
Comfortable — most 18650 cells handle 1C continuously and many handle 2C.
Mass:
80\times48\ \text{g}=3.84\ \text{kg of cells}
Plus BMS, wiring, casing: about 5 kg total.
Energy density:
\frac{979}{5}=196\ \text{Wh/kg at pack level}
against 255 Wh/kg at cell level — a 23% penalty from packaging, which is typical and worth knowing when comparing cell specifications with pack specifications.
Charging. At 0.5C the pack takes 13.6 A, so a charger of
42\times13.6=571\ \text{W}
That is a substantial charger. At 0.25C it is 285 W and takes four hours, which is the usual compromise and is also better for cycle life.
Cycle life estimate. At 80% depth of discharge, roughly 1000 cycles. At 60 km per cycle:
1000\times60=60{,}000\ \text{km}
Which is beyond most bicycles' lifetimes, so calendar ageing will end the pack's life before cycling does — meaning storage conditions matter more than usage.
Problem 6 — Buck converter design
Design a buck converter: 24 V in, 3.3 V out, 5 A, 400 kHz. Choose the inductor and capacitor, and estimate efficiency with and without synchronous rectification.
Solution
D=\frac{3.3}{24}=0.1375
Inductor for 30% ripple:
\Delta I_L=0.3\times5=1.5\ \text{A}
L=\frac{(24-3.3)\times0.1375}{1.5\times400{,}000}=\frac{2.846}{600{,}000}=4.74\ \mu\text{H} \;\to\; 4.7\ \mu\text{H}
Saturation current rating must exceed the peak:
I_{peak}=5+\frac{1.5}{2}=5.75\ \text{A} \;\to\; \text{specify } 8\ \text{A}
Capacitor for 30 mV of ripple, with 3 mΩ ESR:
\Delta V_{ESR}=1.5\times0.003=4.5\ \text{mV}
leaving 25.5 mV:
C=\frac{1.5}{8\times400{,}000\times0.0255}=18.4\ \mu\text{F} \;\to\; 2\times22\ \mu\text{F ceramic}
Two in parallel halves the ESR and shares the substantial ripple current.
Efficiency with a Schottky diode (V_F = 0.4 V, MOSFET R_{DS(on)}=15 mΩ):
- High-side conduction: 5^2\times0.015\times0.1375=51 mW
- Diode conduction: 0.4\times5\times0.8625=1725 mW
- Switching: \tfrac12\times24\times5\times30\ \text{ns}\times400\ \text{kHz}=720 mW
- Inductor DCR at 15 mΩ: 5^2\times0.015=375 mW
- Gate drive and control: 100 mW
P_{loss}=2.97\ \text{W}, \qquad P_{out}=16.5\ \text{W}
\eta=\frac{16.5}{19.47}=85\%
The diode is 58% of the loss, and the low duty cycle is why — it conducts for 86% of every cycle.
With synchronous rectification (R_{DS(on)}=8 mΩ on the low side):
- Low-side conduction: 5^2\times0.008\times0.8625=173 mW
- Body diode during 30 ns of dead time, twice per cycle: 0.7\times5\times60\ \text{ns}\times400\ \text{kHz}=84 mW
P_{loss}=51+173+84+720+375+150=1.55\ \text{W}
\eta=\frac{16.5}{18.05}=91.4\%
From 85% to 91.4% — and the low duty cycle is exactly why the gain is so large. A converter with a high duty cycle would gain much less, since the diode would conduct for only a small fraction of each cycle.
Where the remaining loss is: switching, at 720 mW. Halving the switching frequency to 200 kHz would halve it, at the cost of doubling the inductor to 9.4 µH. That is the frequency-versus-size trade, and 400 kHz was a reasonable choice.
Problem 7 — Boost converter and its limit
Design a boost from 3.7 V (a lithium cell) to 12 V at 0.5 A. Then examine what happens as the cell discharges to 3.0 V.
Solution
D=1-\frac{V_{in}}{V_{out}}=1-\frac{3.7}{12}=0.692
Input current, from power balance at an assumed 90% efficiency:
I_{in}=\frac{12\times0.5}{3.7\times0.9}=1.80\ \text{A}
Inductor for 30% ripple on the input current, at 500 kHz:
\Delta I=0.54\ \text{A}
L=\frac{V_{in}D}{\Delta I f}=\frac{3.7\times0.692}{0.54\times500{,}000}=\frac{2.56}{270{,}000}=9.5\ \mu\text{H} \;\to\; 10\ \mu\text{H}
Peak inductor current:
1.80+0.27=2.07\ \text{A}
Now at 3.0 V:
D=1-\frac{3.0}{12}=0.75, \qquad I_{in}=\frac{6}{3.0\times0.88}=2.27\ \text{A}
Higher duty and higher current — the converter works harder as the battery empties, which is the opposite of intuition and is why boost converters are hardest at end of discharge.
Now push further. At 2.5 V:
D=0.792, \qquad I_{in}=\frac{6}{2.5\times0.85}=2.82\ \text{A}
Conduction loss in the switch rises as I^2:
\left(\frac{2.82}{1.80}\right)^2=2.45\ \text{times the loss at 3.7 V}
And there is a hard ceiling. With parasitic resistance R in the inductor and switch, the real transfer function is
\frac{V_{out}}{V_{in}}=\frac{1}{1-D}\cdot\frac{1}{1+\frac{R}{R_{load}(1-D)^2}}
As D\to1 the second factor collapses towards zero. With R = 100 mΩ and R_{load}=24\ \Omega:
At D=0.9: \frac{1}{1+\frac{0.1}{24\times0.01}}=\frac{1}{1.417}=0.706 — so the output is 30% below the ideal.
At D=0.95: \frac{1}{1+\frac{0.1}{24\times0.0025}}=\frac{1}{2.67}=0.375 — 62% below.
The output voltage actually falls as duty is increased beyond about 0.9, which means the control loop can chase itself into a collapse.
Practical conclusion: keep D below about 0.8, giving a maximum boost ratio of 5:1. For more, use a transformer-based topology, where the turns ratio supplies the bulk of the step-up and the duty cycle only trims it.
Problem 8 — Flyback charger design
Design a phone charger: 85–265 V AC input, 5 V 3 A output, flyback topology at 65 kHz. Find the turns ratio, and check the switch voltage stress.
Solution
DC bus after rectification (Chapter 2.2):
V_{DC(min)}=85\times\sqrt2-\text{ripple}\approx100\ \text{V}
V_{DC(max)}=265\times\sqrt2=375\ \text{V}
Choose the maximum duty at minimum input as 0.45, leaving margin below the 0.5 that a flyback should not exceed without complications.
Turns ratio. For a flyback:
V_{out}+V_F=\frac{N_2}{N_1}\cdot\frac{D}{1-D}V_{in}
5.4=\frac{N_2}{N_1}\times\frac{0.45}{0.55}\times100=\frac{N_2}{N_1}\times81.8
\frac{N_2}{N_1}=0.066 \;\Rightarrow\; \frac{N_1}{N_2}=15.2 \;\to\; \text{use } 15:1
Duty at maximum input:
5.4=\frac{1}{15}\times\frac{D}{1-D}\times375 \;\Rightarrow\; \frac{D}{1-D}=0.216 \;\Rightarrow\; D=0.178
Switch voltage stress. The MOSFET sees the input voltage plus the reflected output voltage plus the leakage spike:
V_{reflected}=15\times5.4=81\ \text{V}
V_{DS}=375+81+V_{spike}
The leakage inductance spike is the dangerous part. When the switch turns off, the energy in the leakage inductance has nowhere to go and produces a voltage spike that can reach 100 V or more.
A snubber is compulsory — an RCD network or a Zener clamp — limiting the spike to about 80 V:
V_{DS(max)}=375+81+80=536\ \text{V}
Specify a 650 V MOSFET, allowing margin for mains surges.
Primary inductance, for discontinuous conduction at minimum input:
L_p=\frac{(V_{in}D)^2}{2P_{in}f}=\frac{(100\times0.45)^2}{2\times18\times65{,}000}=\frac{2025}{2.34\times10^6}=865\ \mu\text{H}
Peak primary current:
I_{pk}=\frac{V_{in}D}{L_pf}=\frac{45}{865\times10^{-6}\times65{,}000}=0.80\ \text{A}
Secondary peak: 0.80\times15=12 A. The output capacitor and the rectifier must handle that pulse, which is why a flyback needs low-ESR output capacitors and a substantial Schottky.
Why a flyback for this job. It provides isolation from the mains, which is a safety requirement, and it needs only one magnetic component and one switch — which is why a 15 W charger costs a few pounds. Above about 75 W the flyback's high peak currents become inefficient and a forward or LLC topology takes over.
Problem 9 — Inverter design
Design a 3 kW single-phase grid-tied inverter for a solar array. The array produces 300–450 V DC and the grid is 230 V, 50 Hz. Find the modulation index and check the DC bus is adequate.
Solution
Required output:
V_{rms}=230\ \text{V} \;\Rightarrow\; V_{peak}=230\sqrt2=325\ \text{V}
For a full bridge with sinusoidal PWM:
V_{peak}=m\times V_{dc}
At V_{dc}=450 V:
m=\frac{325}{450}=0.72 \ ✓
At V_{dc}=300 V:
m=\frac{325}{300}=1.08 \ ✗
Over-modulation — the inverter cannot produce the required peak, so the output clips and distorts badly.
Two solutions:
A boost stage ahead of the inverter, raising the array voltage to a fixed 400 V. This is what almost all string inverters do, and it doubles as the maximum power point tracker.
More panels in series so the minimum array voltage exceeds about 360 V. Cheaper, and it constrains the installation.
Take the boost approach, with a 400 V bus:
m=\frac{325}{400}=0.81
Comfortable, with room for grid voltage variation of ±10%: at 253 V the peak is 358 V and m=0.89 ✓.
Current:
I_{rms}=\frac{3000}{230}=13.0\ \text{A}, \qquad I_{peak}=18.4\ \text{A}
Switch rating: 400 V bus plus transients, so 600 V devices, rated at 30 A for margin.
Switching frequency: 16 kHz. Above the audible range, and low enough that switching losses stay manageable.
Output filter. An LCL filter, with the corner about a tenth of the switching frequency:
f_c=1.6\ \text{kHz}
For L giving 10% current ripple:
L=\frac{V_{dc}}{8f\Delta I}=\frac{400}{8\times16{,}000\times1.84}=1.7\ \text{mH}
Efficiency estimate. IGBTs with 1.8 V saturation voltage:
- Conduction: 1.8\times13\times2 devices conducting =46.8 W
- Switching, at 1 mJ per event: 2\times16{,}000\times0.001=32 W... too high, so use lower-loss devices.
With 600 V MOSFETs at 50 mΩ:
- Conduction: 13^2\times0.05\times2=169 W. Worse — this is the crossover Chapter 2.P analysed, and at 13 A the IGBT wins on conduction.
With SiC MOSFETs at 25 mΩ:
- Conduction: 13^2\times0.025\times2=84.5 W
- Switching, at 0.15 mJ: 2\times16{,}000\times0.00015=4.8 W
P_{loss}=89\ \text{W}+\text{filter and control}\approx110\ \text{W}
\eta=\frac{3000}{3110}=96.5\%
And SiC allows the frequency to rise to 50 kHz, shrinking the filter inductor by a factor of three, which is why modern solar inverters are so much smaller than those of a decade ago.
Anti-islanding is compulsory: the inverter must detect grid loss and disconnect within 200 ms, because an inverter continuing to energise a supposedly dead line endangers anyone working on it.
Problem 10 — Variable frequency drive
A 15 kW, 400 V, 50 Hz induction motor must run from 5 to 50 Hz. Explain the V/f control and find the voltage at each frequency.
Solution
Constant volts per hertz:
\frac{V}{f}=\frac{400}{50}=8\ \text{V/Hz}
| Frequency | Voltage | Speed (4-pole) |
|---|---|---|
| 50 Hz | 400 V | 1500 rpm |
| 40 Hz | 320 V | 1200 rpm |
| 25 Hz | 200 V | 750 rpm |
| 10 Hz | 80 V | 300 rpm |
| 5 Hz | 40 V | 150 rpm |
Why the ratio must be held. From Chapter 9.1, B\propto V/f. Reduce the frequency without reducing the voltage and the core saturates, drawing enormous magnetising current and overheating.
But at low frequency the rule breaks down, and this is the practical wrinkle. The stator resistance drop I R_s becomes a significant fraction of the applied voltage.
At 5 Hz with 40 V applied, and a stator resistance of 0.5 Ω carrying 25 A:
V_{Rs}=25\times0.5=12.5\ \text{V}
V_{available\ for\ flux}=40-12.5=27.5\ \text{V}
Only 69% of the intended flux, so the torque falls by the same factor.
The fix is voltage boost: add a fixed offset at low frequency.
V=V_{boost}+\frac{V_{rated}-V_{boost}}{f_{rated}}f
With a 15 V boost:
V(5\ \text{Hz})=15+\frac{385}{50}\times5=15+38.5=53.5\ \text{V}
Now the flux is correct at 5 Hz ✓.
And the boost must not be excessive, or the motor saturates and overheats at low speed with no load — which is a common misconfiguration on drives set up by trial and error.
Above 50 Hz the voltage cannot rise further, so the ratio falls and the flux weakens:
T\propto\frac{1}{f} \ \text{above base speed}
This is the constant power region — the motor can run to 100 Hz at half the torque, giving the same power. Useful for spindles and fans, and it requires the mechanical design to tolerate double speed.
Cooling is the other low-speed problem. A self-cooled motor's fan is on its own shaft, so at 10% speed it delivers essentially no cooling while the motor may still be at full torque and full current. A separately powered blower is required for continuous low-speed operation, and its absence is a common cause of failure in retrofitted drives.
Problem 11 — Wireless charging efficiency
A Qi pad has coils with $L=10\ \mu$H and R=0.2\ \Omega at 150 kHz, with k=0.35. Find the efficiency, and the effect of a 5 mm misalignment reducing k to 0.2.
Solution
Q=\frac{\omega L}{R}=\frac{2\pi\times150{,}000\times10\times10^{-6}}{0.2}=\frac{9.42}{0.2}=47
Figure of merit:
k\sqrt{Q_1Q_2}=0.35\times47=16.5
\eta_{max}=\frac{k^2Q_1Q_2}{\left(1+\sqrt{1+k^2Q_1Q_2}\right)^2}=\frac{272}{\left(1+\sqrt{273}\right)^2}=\frac{272}{(1+16.52)^2}=\frac{272}{307}=88.6\%
With k=0.2:
k\sqrt{Q_1Q_2}=9.4, \qquad \eta_{max}=\frac{88.4}{(1+9.45)^2}=\frac{88.4}{109}=81\%
From 88.6% to 81% for 5 mm of misalignment.
In practical terms, delivering 10 W to the phone:
P_{in}=\frac{10}{0.886}=11.3\ \text{W}, \qquad \text{loss}=1.3\ \text{W}
versus
P_{in}=\frac{10}{0.81}=12.3\ \text{W}, \qquad \text{loss}=2.3\ \text{W}
Nearly double the heat, and that heat is generated in the coils immediately behind the phone's battery.
Recall Chapter 9.3: high state of charge combined with elevated temperature is the dominant cause of calendar ageing. So poor alignment does not merely waste a watt — it measurably shortens the battery's life, which is a far more consequential outcome than the wasted energy.
Two ways to recover it:
Raise Q. Better Litz wire and lower-loss capacitors could reach Q=100:
0.2\times100=20 \;\Rightarrow\; \eta=90\%
Even misaligned, better than the well-aligned original.
Fix the alignment. Magnets, as Qi2 does, hold k at its best value every time.
And the second is the better engineering. Raising Q costs money in every unit and gives a partial recovery; a ring of magnets costs pennies and eliminates the problem entirely. Removing a variable is almost always better than designing to tolerate it.
Problem 12 — Thermal design
A converter dissipates 8 W in a TO-220 package with a 2 °C/W junction-to-case resistance. Ambient is 50 °C and the maximum junction temperature is 150 °C. Design the cooling.
Solution
Allowable total thermal resistance:
R_{th(j-a)}=\frac{150-50}{8}=12.5\ \text{°C/W}
Subtract the fixed parts. Junction to case is 2 °C/W; case to heatsink with thermal paste is about 0.5 °C/W:
R_{th(s-a)}=12.5-2-0.5=10\ \text{°C/W}
A 10 °C/W heatsink is about 40 mm × 40 mm × 25 mm in natural convection.
Now derate for reliability. Running at the maximum junction temperature gives no margin and shortens life — semiconductor failure rates roughly double for every 10 °C. Design for 110 °C instead:
R_{th(j-a)}=\frac{110-50}{8}=7.5 \;\Rightarrow\; R_{th(s-a)}=5\ \text{°C/W}
Which needs roughly twice the heatsink, about 60 × 60 × 40 mm.
With forced air at 2 m/s, a heatsink's resistance falls by about half, so the smaller 40 mm heatsink would reach 5 °C/W. A fan is often cheaper and smaller than doubling the heatsink — at the cost of a moving part that will eventually fail, and noise.
Check the case temperature:
T_c=T_a+P(R_{cs}+R_{sa})=50+8\times5.5=94\ \text{°C}
Hot enough to burn, so it must not be touchable.
Insulating washer. If the tab is at a live potential and must be isolated from the heatsink, a mica washer adds about 1 °C/W and a silicone pad about 2 °C/W. That directly consumes 8 to 16 °C of the budget, and it is the item most often forgotten in a first thermal calculation.
With a silicone pad:
R_{th(j-a)}=2+2+5=9\ \text{°C/W} \;\Rightarrow\; T_j=50+72=122\ \text{°C}
Over the 110 °C target. Either use a better insulator, a bigger heatsink, or reduce the dissipation — and the last of those is usually the best answer, because it improves efficiency at the same time.
Problem 13 — Power factor correction
A factory draws 400 kW at 0.72 power factor from a 11 kV supply. Find the current, the capacitor bank needed for 0.95, and the annual saving if the tariff penalises reactive power at €8/kvar/month.
Solution
Present state:
S=\frac{400}{0.72}=555.6\ \text{kVA}, \qquad I=\frac{555{,}600}{\sqrt3\times11{,}000}=29.2\ \text{A}
\theta_1=\arccos(0.72)=43.95°, \qquad \tan\theta_1=0.964
Q_1=400\times0.964=385.6\ \text{kvar}
Target:
\theta_2=\arccos(0.95)=18.19°, \qquad \tan\theta_2=0.329
Q_2=400\times0.329=131.6\ \text{kvar}
Capacitor bank:
Q_C=385.6-131.6=254\ \text{kvar}
Capacitance per phase, star-connected at 11 kV line (6.35 kV phase):
C=\frac{Q_C/3}{\omega V_{ph}^2}=\frac{84{,}670}{314\times(6350)^2}=\frac{84{,}670}{1.266\times10^{10}}=6.7\ \mu\text{F}
New current:
S_2=\frac{400}{0.95}=421\ \text{kVA}, \qquad I_2=22.1\ \text{A}
A 24% reduction.
Cable loss saving. Loss goes as I^2:
1-\left(\frac{22.1}{29.2}\right)^2=1-0.573=42.7\%\ \text{less loss in the supply cable}
Tariff saving:
254 kvar x 8 euro x 12 months = 24,384 euro per year.
Capacitor bank cost: roughly €25,000 for 254 kvar at medium voltage.
Payback: about 13 months.
Three practical cautions:
Do not overcorrect. A leading power factor is penalised as heavily as a lagging one, and it can cause voltage rise. Automatic switched banks track the load in steps, rather than a single fixed bank sized for peak.
Harmonics. With non-linear loads present — variable-speed drives, rectifiers — the capacitors can resonate with the supply inductance at a harmonic frequency, amplifying it enormously and destroying the capacitors. Detuned reactors in series with each step shift the resonance below the lowest significant harmonic, and they are standard practice in any modern installation.
Switching transients. Energising a capacitor bank produces a large inrush. Point-on-wave switching or damping reactors limit it.
Problem 14 — Diagnose a failing supply
A 500 W switching power supply that has run for four years now trips its overcurrent protection on start-up, and its output ripple has risen from 50 mV to 400 mV. What has failed?
Solution
The two symptoms together point to one component.
Rising ripple is the diagnostic. Chapter 9.5 established that output ripple has two parts:
\Delta V=\frac{\Delta I}{8fC}+\Delta I\times ESR
An eightfold ripple increase with unchanged load and frequency means either C has fallen or ESR has risen.
And the failure mechanism of an electrolytic capacitor does both at once. The electrolyte gradually evaporates through the rubber seal, which reduces capacitance and raises ESR — typically ESR rises much faster.
The rate doubles for every 10 °C, and the standard model is
L=L_0\times2^{(T_0-T)/10}
Worked example. A capacitor rated 2000 hours at 105 °C, running at 75 °C:
L=2000\times2^{3}=16{,}000\ \text{hours}=1.8\ \text{years continuous}
Four years of operation is well past it — the diagnosis fits the age exactly.
Why the overcurrent trip. A capacitor with high ESR fails to hold the output steady during the start-up surge, so the feedback loop demands more current, and the peak exceeds the protection threshold. The two symptoms are the same failure seen from two directions.
Confirming it, without guessing:
- Visual inspection. Bulging tops, split vent scores, or dried electrolyte residue around the base. Often visible immediately, and this failure is common enough that it has a name — the capacitor plague of the early 2000s, from a stolen and incomplete electrolyte formulation.
- ESR meter, in circuit. A capacitor whose ESR is more than twice the specification is failed even if it measures the right capacitance.
- Oscilloscope on the output, looking at the ripple's shape. A sharp spike at each switching edge indicates ESR; a smooth triangular ramp indicates low capacitance.
The other candidates, and why they fit less well:
- A failing MOSFET would cause overheating and probably a hard failure, not gradually rising ripple.
- A degraded feedback opto-isolator — their current transfer ratio does fall with age — would cause output voltage drift, not ripple.
- A worn fan raising the internal temperature is a genuine possibility, and it would accelerate the capacitor failure rather than replacing it as the explanation. Check it too, because replacing the capacitors without fixing the cooling means doing it again in two years.
The repair: replace all electrolytics in the output stage with 105 °C low-ESR parts of the same or higher capacitance and voltage. Replace all of them, not just the visibly bad ones — they have all had the same thermal history, and the ones that look fine are close behind.
The design lesson. Specify 105 °C parts rather than 85 °C, give them a longer rated life, keep them away from heatsinks and transformers, and provide airflow across them specifically. A 10 °C reduction in capacitor temperature doubles the supply's service life, which is a remarkably large return for a layout decision.
Part 10 assembles everything in this volume into the objects that contain it.