Boost converter calculator
The inductor, the currents and the efficiency of a step-up power stage — built around the fact every other calculator hides behind 1/(1 − D): a boost converter has a maximum gain, it is set by resistance you already have, and at it the converter is exactly half efficient.
The ripple ratio is taken against the inductor current, not the load current — in a boost those are different numbers, and a calculator that confuses them hands you the wrong inductor. Gate charge is the total, driven at 5 V. Inductor core loss and reverse-recovery charge are not modelled, so treat the efficiency as an upper bound.
| Where the loss goes | Power (mW) | Share |
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| Output (V) | Ideal duty | Real duty | Inductor (A) | Efficiency | RHP zero (kHz) |
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The same stage asked for more and more output, at the load current you entered and with the inductance above. Watch the last two columns rather than the first: efficiency falls away slowly and then quickly, and the right-half-plane zero — which is what actually limits how fast the loop may respond — collapses as (1 − D)². The rows that report nothing are the ones past this stage's ceiling: not hard, not inefficient, impossible.
Why a boost converter has a maximum gain
A boost converter stores energy in an inductor while a switch holds one end at ground, then opens the switch so the inductor's current has nowhere to go but into the output, at whatever voltage it takes to get there. In the steady state the volt-seconds the inductor gains must equal the volt-seconds it loses, and with perfect parts that gives the formula every calculator prints:
Vout / Vin = 1 / (1 − D)
Which says that a duty cycle approaching 1 produces a voltage approaching infinity. It does not, and the reason is not subtle. The inductor carries the input current, and the input current is larger than the load current by exactly the gain. So as you ask for more gain, the current through the winding and the switch rises in proportion, and the voltage those resistances drop rises with it — until the drop eats the input.
Put the resistances in and solve the same volt-second balance. With x = 1 − D, a winding resistance RL, a switch resistance Rsw and a rectifier resistance Rrec, the balance is a quadratic rather than a division:
x² Vout − x [ Vin + Iout(Rsw − Rrec) ] + Iout(RL + Rsw) = 0
A quadratic has two roots, one root, or none, and each case means something physical. Two roots: the larger is the operating point, the smaller is a high-duty branch the converter does not sit on. One root: the two have met, and that is the ceiling. None: the stage cannot produce that output at that current, at any duty cycle at all.
In the terms this page works in — a load current you specify — the ceiling falls straight out of that quadratic having one root, and R is the resistance of the inductor-and-switch path only:
Vceil = [ Vin + Iout(Rsw − Rrec) ]² / (4 Iout R) at D = 1 − 2 Iout R / [ Vin + Iout(Rsw − Rrec) ]
The textbook version of the same fact holds a load resistance fixed instead, and it looks different:
Mmax = ½ √(Rload / R) at D = 1 − √(R / Rload)
Both are correct and they answer different questions, which is worth being careful about because they scale differently. Halve every resistance in the loop and a stage feeding a fixed current can reach twice the voltage; the same halving on a fixed resistor buys 41 % more gain, because the resistor draws less current as the voltage climbs and moves the target while you chase it. The square root in the familiar form is an artefact of which variable is being held still, not a property of the converter.
and the most useful consequence is what the efficiency does there. Conduction efficiency in a boost is exactly M·(1 − D), and substituting the two expressions above gives ½ — not approximately, not typically. At maximum gain a boost converter delivers exactly half the power it draws, the other half going into the resistance. So the ceiling is never a place to operate; it is a wall you find out about by getting slower and hotter on the way to it, which is why the efficiency curve on the chart above crosses 50 % at precisely the duty cycle where the gain curve peaks.
Either way the lever is the same one: milliohms, in the winding and in the switch. Note which resistances appear and which do not — the rectifier's barely matters, because at the duty cycles near the ceiling it conducts for almost none of the cycle, while the winding conducts for all of it.
The number that surprises people: the inductor carries the input current
Feeding the output through a switch rather than through an inductor changes which current is which. A buck converter's inductor carries the load current; a boost converter's inductor carries the input current, which is
IL = Iout / (1 − D) Ipeak = IL (1 + r/2)
and the ripple ratio r on this page is measured against IL, because that is the current the inductor actually sees. At the defaults — 5 V to 12 V at 1 A — the inductor's average current is 2.49 A and its peak is 2.84 A. A part chosen for a “1 A converter” saturates immediately.
Everything that scales with that current scales with it squared. The winding loss is IL²R, and since IL goes as 1/(1 − D), the copper loss in a boost rises with the square of the boost ratio: taking the same stage from 2× to 4× does not double its winding loss, it quadruples it. That is the mechanism behind the ceiling, seen from the other end. The same current has to reach the board, so the input copper is the wide trace here, not the output copper — the PCB trace width calculator takes IL, not the load current, and the same applies to the connector and the fuse.
If the source is a cell, the same factor decides runtime. A load drawing 1 A at 12 V from a 3.7 V cell is a 3.4 A drain on the cell before any losses, and the battery life calculator wants that figure — the converted one — not the load current.
Buck and boost swap which capacitor is hard
On a buck converter the input capacitor takes the chopped current and the output is fed smoothly through the inductor. A boost is the mirror image: the input is fed through the inductor, so the input capacitor sees only the ripple, while the output capacitor has to supply the entire load current for the whole of the on-time, every cycle, because during the on-time the output is disconnected from everything.
ICout(rms) ≈ Iout √(D / (1 − D))
At the defaults that is 1.23 A against the input capacitor's 0.20 A — six times as much — and it gets worse with the boost ratio without limit. Ripple-current rating, not capacitance, is what that part is chosen for; a ceramic bank is usually the honest answer, and a single electrolytic almost never is.
The ripple voltage follows the same asymmetry. The capacitance term is not the buck's tidy ΔI/(8fC) — it is the load current draining the capacitor for the on-time:
ΔVC = Iout D / (f C) ΔVESR = Ipeak · ESR
and the ESR term is driven by the peak inductor current, not by the ripple, because the capacitor's current steps by the whole inductor current the instant the rectifier starts conducting. Both terms are therefore several times what the same parts would give on a buck: at the defaults this stage has 40 mV of output ripple where the buck page's example, with an identical 100 µF at 10 mΩ and the same frequency, has 7 mV. If the ripple matters, the answer is usually lower ESR or a second-stage filter — an ordinary LC section — rather than more capacitance.
The right-half-plane zero, and why your loop is slow
This is the constraint nobody puts on a calculator, and it is the one that decides whether a design is buildable. Suppose the load suddenly takes more current and the controller responds the way it must, by increasing the duty cycle. Increasing the duty cycle means the rectifier conducts for a shorter fraction of each cycle. The inductor current takes several cycles to build, so for those first cycles less charge reaches the output, not more: the converter's immediate response to a demand for more output is less output.
fRHPZ = Vout (1 − D)² / (2π L Iout)
A right-half-plane zero adds phase lag while adding gain, which no amount of compensation removes. The standard remedy is to cross the loop over well below it — a third of fRHPZ at the most aggressive, a fifth for comfort — and this page reports both the zero and that fifth.
The numbers are humbling. At the defaults the zero sits at 38 kHz and a comfortable crossover is about 7.5 kHz, which is 66 times below the 500 kHz switching frequency. Raising the switching frequency does not help, because fRHPZ does not contain it. What does help is a smaller inductor, a lower boost ratio, or a lighter load — the zero moves as (1 − D)², so doubling the load current halves it and every extra volt of output pushes it down twice over. A boost that is fine on the bench and oscillates under load usually met its zero rather than its phase margin.
Where the loss goes
Six terms, and their shapes differ from a buck's in two places worth knowing. The switch and the winding both conduct IL, which is the input current, so their losses carry the 1/(1 − D)² factor above. And the switch node swings to the output voltage rather than the input, so the transition loss is ½VoutIL(tr+tf)f — a term that grows on both of its factors as the boost ratio rises.
A Schottky diode in the rectifier position carries the load current on average, exactly, so it dissipates Vf·Iout and nothing else. That is easy to reason about and easy to underestimate: at the defaults a 0.45 V part dissipates 450 mW, most of a watt once its reverse recovery is counted, and it wants a real thermal path — a heatsink question with a real answer rather than a shrug. A synchronous rectifier replaces that fixed drop with I²R and usually wins outright, which is why modern boost controllers have one.
Two things the board decides rather than the maths. The loop carrying the chopped rectifier current — switch node, rectifier, output capacitor — is the fast loop here and should be physically tiny; on a buck it is the input loop instead, so a layout habit carried over from one to the other optimises the wrong rectangle. And the feedback network is an ordinary voltage divider from an output that is now the noisiest node on the board, so its source resistance and its routing matter more than they would on a step-down stage.
FAQ
Why is my duty cycle higher than 1 − Vin/Vout?
Because the converter has to make up what it drops. Every resistance in the loop subtracts from the volt-seconds the inductor collects, so it has to collect them for longer. The gap widens with load current and with boost ratio, and it is the beginning of the ceiling described above — at the defaults the ideal duty is 58.3 % and the real one is 59.8 %, but push the same stage towards its limit and the two diverge completely.
Why can't a boost converter output less than its input?
Because there is a permanent conducting path from input to output through the inductor and the rectifier, whatever the switch is doing. Even with the converter shut down, the output sits at roughly the input minus a diode drop. A boost cannot regulate below its input, cannot disconnect its load, and cannot current-limit a short at its output — if you need any of those, you need a different topology (SEPIC, buck-boost, or a buck in front) or a series switch.
What inductance should I use?
Whatever gives the ripple ratio you want against the inductor current: L = VonD/(r·IL·f). Between 20 % and 40 % is ordinary. The boost-specific pressure is that a larger inductor lowers the right-half-plane zero, so unlike a buck — where more inductance is simply bigger and quieter — here it costs you loop bandwidth, and the sensible ratio is often higher than it would be on a step-down stage.
What is the right-half-plane zero in plain terms?
The converter briefly does the opposite of what it was asked. To deliver more current it must first spend more of each cycle charging the inductor, which means spending less of each cycle delivering — so output current dips before it rises. A control loop fast enough to notice that dip will respond to it, make it worse, and oscillate. The fix is to be slow, and this page tells you how slow.
Does this handle discontinuous conduction?
Yes. Below a load current of about (1 − D)·ΔI/2 the inductor current reaches zero each cycle, and the page switches to the discontinuous model — different duty cycle, different peak current, different everything. It reports the load at which that happens and marks it in the warnings. The discontinuous branch uses the ideal duty cycle, because carrying the resistive drops through it would make the expression implicit for a sub-percent correction, so the two models meet with a small step rather than exactly.
Can I boost from one cell?
Only if the resistance allows it, and it often does not. Try 1.2 V in, 5 V out, 0.5 A, with 300 mΩ between the winding and the switch: this page will tell you the stage has no operating point, and that its highest possible output at that current is 2.4 V. Nothing about the controller changes that. Single-cell boost designs live or die on milliohms, which is why their datasheets specify the inductor so tightly.
Why is the efficiency worse than the datasheet's?
Core loss and reverse recovery are not modelled here, and both are worse in a boost than in a comparable buck — the core swings a larger current and the rectifier commutates the full inductor current into a hard-switched node. The figures here are an upper bound from conduction, switching, gate and quiescent losses. If the measured number is more than a few points below, suspect the inductor's core material first and the rectifier second.