Power supplies
The linear supply regulates by burning off whatever is left over. The switching supply doesn't regulate: it chops and averages, and that is why it reaches 90 % efficiency at a tenth of the size. All modern equipment lives off that difference.
01Linear versus switching
The linear supply was already covered with the LM317 regulator: mains transformer, rectifier, filter and a regulator that dissipates the difference between input and output.
A regulator delivering 5 V and 2 A from a 12 V input:
- Useful output: 5 × 2 = 10 W.
- The regulator has 12 − 5 = 7 V across it, with 2 A: it dissipates 7 × 2 = 14 W.
- Efficiency = 10/24 = 42 %, and a heat sink for 14 W is needed.
The same output from a switching supply at 90 % draws 11 W from the input and dissipates 1 W. The difference is not an improvement: it is another order of magnitude.
| Linear | Switching | |
|---|---|---|
| Efficiency | 30 to 60 % | 80 to 95 % |
| Size and weight | Large: 50 Hz transformer | Small: ferrite core at tens of kHz (transformers) |
| Electrical noise | Very low | High: it switches thousands of times per second |
| Response to load changes | Very fast | Depends on the control loop |
| Complexity | Low | High: feedback loop, protections, filters |
| Where linear supplies are still used | High-accuracy analog stages, reference-grade audio, power for noise-sensitive sensors | |
02The basic topologies
They all do the same thing: they switch at high frequency and use an inductor to store energy and deliver it at the right moment. What changes is the arrangement.
| Topology | Output | Isolated | Used in |
|---|---|---|---|
| Buck | Vo = D · Vi | No | Stepping voltage down efficiently: on-board regulators, chargers |
| Boost | Vo = Vi/(1 − D) | No | Stepping voltage up: LED backlights, power factor correction |
| Buck-boost | Inverts the polarity | No | When the input can be above or below the output |
| Flyback | Set by the turns ratio | Yes | Phone chargers, supplies up to ~100 W. The most widespread |
| Forward / half bridge / full bridge | Set by the turns ratio | Yes | PC power supplies, welders, industrial chargers of hundreds of watts |
24 V input, 12 V output at 3 A, switching at 100 kHz.
- Duty cycle: D = 12/24 = 0.5.
- Inductor current: its average value equals the output current, 3 A, with a typical ripple of 30 %: 0.9 A peak to peak.
- Inductance: L = (Vi − Vo)·D/(Δi·f) = 12 × 0.5/(0.9 × 100,000) = 66.7 µH. The commercial 68 µH value is chosen, checking that it can handle 3.5 A without saturating.
- Diode: it conducts 50 % of the time; a Schottky is chosen for its low drop.
- Output capacitor: chosen for the allowed ripple and, above all, for its ESR (impedance measurements): the ESR multiplied by the ripple current is most of the voltage ripple.
If the current through the inductor exceeds the value at which its core saturates, the inductance disappears: the current shoots up almost as in a short circuit and destroys the switch within one cycle. That is why an inductor is chosen by its saturation current, not its rated current, and why a supply that starts up fine and dies at the first load peak often has this problem.
03Controlled rectifiers
When power is taken directly from the mains and has to be regulated, thyristors are used instead of diodes. The firing angle sets the DC output voltage.
- Single-phase bridge on 220 V with α = 0: Vdc = 0.9 × 220 = 198 V. With α = 60°: 198 × 0.5 = 99 V.
- Three-phase bridge on 380 V with α = 0: 1.35 × 380 = 513 V. With α = 45°: 513 × 0.707 = 363 V.
- The three-phase bridge also delivers a much cleaner DC output: six pulses per cycle instead of two, so the ripple is smaller and at a higher frequency, and the filter ends up smaller.
At angles greater than 90°, the cosine becomes negative and so does the average voltage: if the load can deliver energy —a DC motor braking, for example— the circuit returns it to the mains. This mode is called inversion and it is the basis of regenerative braking in traction. The energy of a braking train goes back to the line instead of turning into heat.
04Inverters
They do the reverse: DC to AC. Four switches in a bridge chop the DC voltage and, by modulating the pulse width, build a sine wave.
| Output type | What it looks like | Where it is used |
|---|---|---|
| Square wave | Two levels. Many harmonics. | Very simple equipment. Harms motors and switching supplies. |
| Modified square wave | Three levels, with a step at zero. | Low-cost UPS units. Acceptable for resistive loads. |
| Sinusoidal PWM | Pulses of variable width plus a filter. | Quality UPS units, solar inverters, variable-speed drives. |
- Modulation index: the amplitude of the reference sine wave relative to that of the carrier. It controls the output voltage.
- Carrier frequency: typically 2 to 16 kHz. The higher it is, the easier to filter and the quieter the motor, but the greater the switching losses.
- The reference frequency is the output frequency: that is what makes it possible to vary the speed of a motor.
In each leg of the bridge there are two switches in series between the positive and the negative rail. If for a moment both conduct, the result is a short circuit of the DC source that destroys them immediately. That is why the drive introduces a dead time —microseconds long— between turning one off and turning the other on. Every bridge driver IC includes it, and in a custom design it is the first thing to check on the oscilloscope.
05Firing and drive circuits
Synchronization with the mains, isolation by pulse transformer or optocoupler, and enough energy in the pulse. With inductive loads, a pulse train (automatic control systems).
They need a driver: the gate is capacitive and has to be charged and discharged with amperes of peak current for the switching to be fast. The high-side switches of the bridge are floating, so their drive is solved with a bootstrap circuit or an isolated supply.
If the driver is weak, the transition drags on: during that time the device has voltage and current at the same time, and the switching losses skyrocket. An IGBT that should switch in 200 ns and takes 2 µs dissipates ten times more. Many supplies that “heat up for no reason” have this problem, not a failed component.
06In the lab
Power the same load (12 V, 1 A) with a 7812 linear regulator from 20 V and with a switching buck module. Measure input voltage and current in both cases and calculate the efficiency. Touch both: the linear one burns, the switching one is lukewarm. Note the power dissipated in each.
On a buck module, use the oscilloscope to measure the voltage at the switching node —you see the square wave—, the inductor current with a current probe or a shunt resistor, and the output ripple with AC coupling. Check that D matches Vo/Vi and vary the load to see how the loop adjusts D.
With an isolation transformer and reduced voltage, build a controlled single-phase bridge and measure the average voltage for α = 0°, 45°, 90° and 135°. Compare with the formula. Repeat with an inductive load and observe the difference in the current waveform.
On a low-voltage H-bridge, use two channels to measure the drive signals of the upper and lower switch of one leg and check the dead time. Reduce it gradually and watch the peak of the supply current: it is the demonstration of why dead time exists, done without destroying anything.
07Common mistakes
| Symptom | Usual cause |
|---|---|
| The switching supply whistles | It is running in burst mode because of a light load, or the core vibrates from magnetostriction. Often normal with no load. |
| Dies as soon as it is asked for current | Inductor saturating, or undersized switch. |
| Lots of output ripple | Capacitor with high ESR, or the capacitor is far from the switching point. |
| Interferes with the radio and with other equipment | Input filter missing or badly placed, and large ground loops on the board (PCB). |
| Gets very hot for no apparent reason | Weak driver: slow transitions and high switching losses. |
| The bridge is destroyed at start-up | Insufficient or zero dead time: the two switches of a leg conduct at the same time. |
| The controlled rectifier voltage does not match the formula | The load is inductive and there is conduction during part of the negative half-cycle, or the freewheeling diode is missing. |
| The inverter drives the motor with noise and vibration | Carrier too low, or square-wave output instead of sinusoidal PWM. |
08Self-assessment
A linear regulator delivers 5 V and 1.5 A from 15 V. How much does it dissipate and what is its efficiency?
It dissipates (15 − 5) × 1.5 = 15 W and delivers 7.5 W: efficiency 7.5/22.5 = 33 %.
Why can a switching supply be so small?
Because it works at tens or hundreds of kHz instead of 50 Hz: the transformer core and the filters needed are much smaller, since the energy per cycle is far lower.
In a buck converter with 24 V input and D = 0.25, what is the output?
Vo = D · Vi = 0.25 × 24 = 6 V.
What happens if a converter's inductor saturates?
It stops behaving as an inductor: the current grows almost without limit in each cycle and destroys the switch. That is why the inductor is chosen by its saturation current.
Three-phase bridge on 380 V with α = 30°: what is the DC voltage?
V = 1.35 × 380 × cos 30° = 513 × 0.866 = 444 V.
What does it mean for a controlled rectifier to operate with α greater than 90°?
That its average voltage becomes negative and the circuit returns energy to the mains: it works as an inverter. This is what makes regenerative braking possible, if the load can deliver energy.
How is a sine wave built with PWM?
By comparing a reference sine wave with a triangular carrier of much higher frequency. The output switches at the carrier frequency with variable width, and its average value follows the sine wave. A filter lets only that component through.
What is dead time for in a bridge?
To ensure that the two switches of the same leg never conduct at the same time. Without it, the DC source is short-circuited through the two switches, which are destroyed immediately.
Why does an IGBT need a driver instead of being driven directly from a microcontroller?
Because its gate is capacitive —nanofarads— and to switch fast it has to be charged and discharged with peaks of several amperes. A microcontroller pin delivers 20 mA: the transition would be extremely slow and the losses huge.
Why does the three-phase bridge give a cleaner DC output than the single-phase one?
Because it delivers six pulses per cycle instead of two: the ripple is smaller and at a higher frequency (300 Hz), so the filter needed is much smaller.