Why Switching, Not Dissipating
Start here. Drop 100 W across a linear regulator, then do the same job with a switch and find the missing heat.
Skip to the animationPower dissipated in a device is V×I, so a semiconductor held part-way on wastes as much power as it delivers — while one that is only ever fully on or fully off wastes almost none, and the output is set by the ratio of the two times rather than by a resistance.
The problem, with numbers
Deliver 12 V at 10 A — 120 W — from a 24 V supply. Something must account for the other 12 V, and the whole subject is about what.
A linear regulator holds a transistor partly on so it drops the excess. That transistor has 12 V across it and 10 A through it, so it dissipates 12 × 10 = 120 W — exactly as much as the load receives. Efficiency is 50%, and 120 W in a small package is a soldering iron.
It gets worse as the ratio widens. Producing 5 V at 10 A from 24 V wastes 190 W to deliver 50 — 79% of the input, before the load has done anything useful.
The insight
| Device state | Voltage across | Current through | Power dissipated |
|---|---|---|---|
| Fully on | ~0.05 V | 10 A | 0.5 W |
| Fully off | 24 V | ~0 A | ~0 W |
| Half on | 12 V | 10 A | 120 W |
Power is a product. A closed switch has current but no voltage; an open switch has voltage but no current. In both states one factor is near zero, so the product is tiny. The only expensive place for a semiconductor to be is halfway — which is precisely where a linear regulator keeps it.
But a switch has only two outputs
A switch gives 24 V or 0 V, never 12 V. The resolution is to switch rapidly between them and let the average be the output: 50% duty cycle averages 12 V, 25% averages 6 V.
- 1Switch on: the inductor current rises, storing energy in its magnetic field.
- 2Switch off: the inductor opposes the change in current and keeps it flowing through the freewheeling diode.
- 3The capacitor holds the output voltage steady between pulses.
- 4The output is
D × V_infor a buck converter, where D is the duty cycle.
The control variable is now a ratio of times, not a resistance. Time costs no energy, which is the whole reason the efficiency is different in kind rather than just in degree.
What switching also makes possible
A linear regulator can only discard voltage; it has no mechanism to produce more than it was given. A switching converter stores energy in an inductor and can release it in series with the source, producing a higher output voltage than its input.
- Buck — steps down.
V_out = D·V_in. - Boost — steps up.
V_out = V_in/(1−D). - Buck-boost — either, with an inverted output.
- Inverter — arranges the duty cycle so the average traces a sinusoid. That is all PWM is.
- Isolated converters — a transformer in the switched path gives galvanic isolation and any ratio at all.
The price, honestly stated
A switch is only cheap when it is fully on or fully off. During a transition it is neither, and for tens of nanoseconds both voltage and current are substantial.
| Cost | Why it matters |
|---|---|
| Switching loss | Energy per transition × frequency. Dominant at high speed |
| EMI | Fast edges radiate; filtering and shielding are needed to be legal |
| Output ripple | The output is never perfectly flat, only filtered |
| Control complexity | A feedback loop that must remain stable across the load range |
| Magnetic components | Inductors and transformers are bulky, and often the largest part |
This sets up the central trade-off of converter design: a higher switching frequency allows a smaller filter, but multiplies the switching loss. Snubbers, soft switching and careful gate drive design all exist to buy back part of that loss.
When linear is still the right answer
- Low dropout — when input and output are close, the wasted power is small and switching complexity is not worth it.
- Low noise — a linear regulator produces no switching ripple, which matters for precision analogue and RF supplies.
- Low current — at milliamps, 50% of very little is still very little.
- Simplicity — three terminals and a capacitor, against a controller, a switch, a diode, an inductor and a compensation network.
A common design uses both: a switching converter to get close efficiently, then a small linear regulator to clean up the ripple. That is not a compromise but the correct use of each.
The numbers you will be asked for
- Dissipation in a device
P = V × I
Near zero when either factor is near zero — the entire argument.
- Buck converter output
V_out = D · V_in
D is the duty cycle, between 0 and 1.
- Boost converter output
V_out = V_in / (1 − D)
Always steps up; D → 1 is limited in practice by losses.
- Switching loss
P_sw = ½ · V · I · (t_on + t_off) · f_s
Linear in frequency, which is why speed is not free.
- Linear regulator efficiency
η = V_out / V_in
Independent of load, and set entirely by the voltage ratio.
Advantages and disadvantages
Advantages
- Efficiency of 90–98% instead of a voltage ratio.
- Can step up as well as down, which a linear regulator fundamentally cannot.
- Small heatsinks, so smaller and lighter equipment.
- The output is controlled by time, which costs nothing and is easy to control digitally.
Disadvantages
- Switching loss rises with frequency, capping how small the filter can get.
- Fast edges generate EMI that must be filtered and shielded.
- Output ripple is inherent — it is filtered, never eliminated.
- More components and a feedback loop that must be stable across the whole load range.
Watch it work
Check yourself
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