Buck, Boost & Linear Picker
A boost draws more current than it delivers — and at microamp loads the "inefficient" linear regulator beats the switcher outright.
Linear against switcher, across the load range
A boost draws more current than it delivers
Power in equals power out over efficiency, and the input voltage is the lower one — so input current is output current times Vout/(Vin·η). A 3.7 V cell boosted to 12 V at 1 A out pulls 3.8 A in. People size the cell, the fuse, the switch and the wiring for the output current, then wonder why a protection circuit trips or the cell sags. A buck does the opposite — 12 V to 5 V at 2 A out draws only 0.93 A — which is why stepping down feels safe and stepping up quietly isn't.
A linear regulator's efficiency is exactly Vout/Vin, with no design freedom. 12 V to 5 V is 41.7% whatever part you buy, dropping the difference across itself as heat — 7 W at 1 A, which is a heatsink rather than a package. That's arithmetic, not quality, and it's the entire reason switching converters exist.
But at microamp loads the linear wins, which is exactly backwards from the rule of thumb. Quiescent current dominates once the load is small enough: a 2 µA LDO against a 50 µA switcher crosses over at 94 µA of load. Below that the "inefficient" regulator draws less total power — at 1 µA of load it's 16.8× better, and at 10 µA still 4.6×. For a sensor node that sleeps most of the time, that's the number that decides battery life.
And the crossover moves with the switcher's quiescent current, not its efficiency rating: a 5 µA buck crosses at about 6 µA of load, a 200 µA one at 388. That's the spec to compare for anything battery-powered and mostly idle — the 90%-efficiency figure on the front of the datasheet is measured at full load and says nothing about the sleeping case.
- If the input range crosses the output, neither half works. A 1S LiPo runs 4.2 V down to 3.0, so a 3.3 V rail starts above it and ends below — the classic trap. It needs a buck-boost or SEPIC, and that costs efficiency and parts.
- A boost can't current-limit a short on its output. There's a diode path straight through the inductor, so the converter has no way to disconnect — a shorted output is a shorted battery. Protection has to live elsewhere.
- Efficiency curves matter more than peak efficiency. A converter at 92% at full load may be at 60% at a tenth of it, and the shape differs far more between parts than the headline number does.
- Check the input at its low end. That's where a boost draws its worst current and where a linear runs out of dropout — the nominal voltage tells you about neither.
How to use
- Check the input range at its low end, not the nominal voltage.
- Size the input side of a boost for the amplified current.
- Compare quiescent current, not peak efficiency, for sleepy loads.
- Use a buck-boost if the input range crosses the output.
Frequently asked questions
Does a boost converter draw more current than it delivers?
Yes, in proportion to the voltage ratio divided by efficiency. Power in equals power out over efficiency, and the input voltage is the lower one, so a 3.7 volt cell boosted to 12 volts at 1 amp out pulls about 3.8 amps in. People size the cell, the fuse, the switch and the wiring for the output current, which is the single commonest mistake with boost converters.
Why does my battery protection trip when the boost converter starts?
Because the input current is several times the output current, and the inrush at startup is higher still. A protection circuit chosen for the output rating will trip on the real input draw. Work out the input current at the lowest input voltage in the range — the cell sags under load, so the bottom end is worse than the datasheet cut-off suggests.
What efficiency does a linear regulator have?
Exactly the output voltage divided by the input voltage, with no design freedom at all. Twelve volts to five is 41.7 per cent, and twenty-four to five is 20.8, whatever part you buy. The regulator drops the difference across itself as heat, so the wasted power is the voltage difference times the current — seven watts at one amp for a 12 to 5 volt conversion.
Is a switching regulator always more efficient than a linear one?
No, and the exception matters for battery-powered designs. At small enough loads the quiescent current dominates, and a modern LDO idling at two microamps beats a switcher idling at fifty despite the far worse ratio efficiency. With those figures the crossover is around 94 microamps of load — below that the linear draws less total power, and at one microamp it is nearly seventeen times better.
What is quiescent current and why does it matter?
The current a regulator draws to run itself, independent of the load. It is irrelevant at an amp and decisive at a microamp, which is exactly the regime a sleeping sensor node lives in. For anything battery-powered and mostly idle it is the specification to compare — the efficiency figure on the front of a datasheet is measured at full load and says nothing about the sleeping case.
Should I use an LDO or a buck converter for a battery sensor node?
It depends on the duty cycle rather than on which is more efficient in general. If the node sleeps at a few microamps and wakes briefly, the LDO usually wins on total consumption because the switcher burns its quiescent current continuously. If it draws milliamps most of the time, the switcher wins comfortably. The crossover moves with the switcher own quiescent current: about six microamps of load for a good one and nearly four hundred for a cheap one.
Why does a 1S LiPo to 3.3V need a buck-boost?
Because the cell runs from about 4.2 volts down to 3.0, so a 3.3 volt rail starts below the input and ends above it. A buck stops regulating partway through the discharge and a boost cannot start at the top. It is the classic topology trap and almost every single-cell battery causes it, which is why buck-boost and SEPIC parts exist despite costing efficiency.
Can a boost converter current-limit a short on its output?
No, and this surprises people the first time. There is a diode path straight from input to output through the inductor and the catch diode, so the converter has no way to disconnect the two — a shorted output is a shorted battery regardless of what the control chip does. Any protection has to sit somewhere else in the circuit.
How much heat does a linear regulator produce?
The voltage difference times the current, all of it. Twelve volts to five at one amp is seven watts, which in a small package with sixty degrees per watt of thermal resistance reaches four hundred degrees of junction temperature — far past any limit. It will thermally shut down and restart repeatedly, which presents as an intermittent fault rather than a dead one.
What efficiency should I assume for a switching converter?
Eighty-five to ninety per cent is a reasonable planning figure, but the curve matters more than the peak. A converter rated at ninety-two per cent at full load may be at sixty at a tenth of it, and the shape of that curve varies far more between parts than the headline number does. If your load swings over a wide range, compare curves rather than peaks.
Should I check the input voltage at its high or low end?
Both, for different reasons. The low end is where a boost draws its worst input current and where a linear runs out of dropout headroom. The high end is where a linear wastes the most heat and where the converter sees its highest stress. The nominal voltage in the middle tells you about neither, and it is the number people design to.
What is a SEPIC converter?
A topology that can step voltage both up and down without inverting it, using two inductors and a coupling capacitor. It solves the same problem as a buck-boost — an input range that crosses the output — with different trade-offs in part count, noise and efficiency. Both cost something against a plain buck or boost, and that cost is the price of not having a dead zone mid-discharge.
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