The DC-DC converter calculator above works the three basic non-isolated switching topologies — buck, boost and buck-boost — and reports the four numbers that actually constrain a design: duty cycle, inductor ripple current, peak switch current and output ripple voltage. It runs each topology with its own equations rather than fudging one formula to cover all three, because the current paths genuinely differ. In a buck converter the inductor carries the load current; in a boost converter it carries the input current, which is larger. Treating them the same is how inductors get undersized.
Arb Digital builds free calculators that state their assumptions instead of hiding them, and the assumption that matters most here is efficiency. Every figure on this page is derived under continuous conduction mode with a fixed efficiency you supply. That is a design starting point, not a simulation, and the section below on what the tool does not model is worth reading before you commit to component values.
What This DC-DC Converter Calculator Does
It takes the operating point you are designing for — input voltage, output voltage, load current and switching frequency — plus a candidate inductor and output capacitor, and tells you what the switching waveform looks like. The headline is the duty cycle: the fraction of each switching period for which the main switch is on. Everything else in a switching converter follows from that fraction.
The supporting grid gives the peak-to-peak ripple current in the inductor, the peak current the switch and inductor must survive, the average current drawn from the supply, and the peak-to-peak ripple appearing on the output. Those four numbers map directly onto component selection: ripple current sets the inductor, peak current sets the switch rating and the inductor saturation current, input current sets the supply and input capacitor, and output ripple decides whether the output capacitor is adequate.
The bar pair below the grid shows the on and off portions of the switching period as a visual split. It is there because duty cycle is easy to misread as a percentage of something else. It is a fraction of time, and the two bars always sum to the whole period.
How to Use It
- Pick the topology that matches your voltages. If the output is below the input, you want buck. Above it, boost. If you need a negative rail or the input can sit either side of the output, buck-boost. The tool flags an input and output combination that the chosen topology cannot produce.
- Enter the worst-case operating point, not the typical one. A converter has to survive the extremes of its input range. Run the calculation at minimum input voltage and again at maximum, because duty cycle, peak current and ripple all move as the input moves.
- Set the switching frequency your controller actually runs at. Frequency appears in the denominator of both the ripple current and the ripple voltage, so doubling it halves both. This is the main lever for shrinking magnetics.
- Try inductor values until the ripple ratio looks sensible. The note under the results reports ripple as a percentage of load current. Adjust the inductance until that percentage lands where you want it.
- Set efficiency honestly. Ninety per cent is a reasonable starting assumption for a modern synchronous buck at moderate voltages. A converter with a large step-up ratio or a Schottky diode instead of a synchronous switch will do worse, and the duty cycle it needs will be higher than the ideal figure.
The Formula: How Duty Cycle and Ripple Are Calculated
For a buck converter the duty cycle is D = Vout ÷ (Vin × η). For a boost converter it is D = 1 − (Vin × η) ÷ Vout. For an inverting buck-boost it is D = Vout ÷ (Vout + Vin × η), using the magnitude of the output. These forms, including the efficiency term, are the ones used in Texas Instruments application report SLVA477B, Basic Calculation of a Buck Converter's Power Stage, and its companion SLVA372D, Basic Calculation of a Boost Converter's Power Stage.
Ripple current is the volt-second product across the inductor during one part of the cycle divided by the inductance. For a buck converter the inductor sees Vin − Vout while the switch is on, so ΔIL = (Vin − Vout) × D ÷ (f × L). For a boost or buck-boost the inductor sees the full input voltage while the switch is on, giving ΔIL = Vin × D ÷ (f × L).
Take the default buck case and work it through. With 12 V in, 5 V out and ninety per cent efficiency, D = 5 ÷ (12 × 0.9) = 0.463, so the switch is on for 46.3 per cent of each period. Ripple current is (12 − 5) × 0.463 ÷ (500,000 × 0.00001) = 3.241 ÷ 5 = 0.648 A peak to peak, which is about 32 per cent of the 2 A load. The inductor's average current in a buck is the load current, so the peak is 2 + 0.324 = 2.324 A. Average input current comes from power balance: 5 × 2 ÷ (12 × 0.9) = 0.926 A. Output ripple, taking the capacitor as ideal, is ΔIL ÷ (8 × f × C) = 0.648 ÷ 400 = 1.62 mV. Those are exactly the figures the panel above reports.
Why the Inductor Current Is Not the Load Current in a Boost
This is the error that most often destroys a first prototype. In a buck converter the inductor sits in series with the output, so its average current is the load current and nothing more. In a boost converter the inductor sits in series with the input, and because a boost steps voltage up it must step current down at the output — which means the input current, and therefore the inductor current, is larger than the load current by roughly the voltage ratio.
Run the boost preset and watch it happen. Stepping 5 V up to 12 V at 1 A of load draws about 2.67 A from the supply once efficiency is accounted for. The inductor carries that 2.67 A average, plus half the ripple on top, so its peak is close to 3 A. An inductor chosen for a 1 A load would saturate, its inductance would collapse, the ripple current would rise sharply, and the converter would either shut down on overcurrent or fail. The same logic applies to a buck-boost, where the inductor average is the load current divided by one minus the duty cycle.
The practical consequence is that inductor saturation current, not the average rating, is the number to check. This tool reports the peak, and that peak is what the datasheet's saturation figure must exceed with margin.
The Efficiency Assumption and What It Hides
An ideal converter with no losses would need a buck duty cycle of exactly Vout ÷ Vin. A real one needs more, because some of the input power never reaches the load. Putting efficiency into the duty-cycle denominator is a standard first-order way to account for that, and it is the approach the application notes above take. It is not a loss model. It does not tell you where the power went, and it does not predict temperature rise.
Losses in a switching converter come from several places at once: conduction loss in the switch and the inductor's winding resistance, switching loss that scales with frequency, core loss in the magnetics, forward drop in a diode if the design is not synchronous, and quiescent current in the controller. Their balance shifts with load. A converter that reaches ninety-four per cent at half load may fall below eighty-five at very light load, because fixed losses dominate when the output power is small. Entering a single efficiency figure means you are designing for one point on that curve, so pick the point you care about.
Efficiency also gets worse as the conversion ratio gets more extreme. A boost from 3 V to 24 V runs at a high duty cycle, which means high peak currents and a short window for the output to be recharged. Assuming ninety per cent there is optimistic, and the duty cycle the calculator returns will be lower than the real one. If the tool tells you the duty cycle is above about ninety per cent, treat that as a signal that the topology is being pushed rather than as a working design.
What This Calculator Deliberately Does Not Model
It assumes continuous conduction mode, meaning the inductor current never falls to zero within a cycle. At light load a converter naturally enters discontinuous conduction, where the duty cycle for a given output voltage changes and these equations no longer apply. The boundary is where half the ripple current equals the average inductor current, which is easy to spot in the numbers above: if the ripple approaches twice the average, you are near or past that boundary.
The output ripple figure treats the capacitor as pure capacitance. Real capacitors have equivalent series resistance, and for many electrolytic and tantalum parts the ripple caused by that resistance dominates the capacitive term completely. Multiply the ripple current by the capacitor's ESR and compare it with the figure reported here; whichever is larger is the one that governs. Ceramic capacitors have very low ESR but lose a large fraction of their nominal capacitance under DC bias, which is a separate trap the datasheet curve will show you.
Finally, nothing here addresses control-loop stability, layout, EMI or thermal design, and those are where working converters are actually won or lost. Component selection is the first step, not the whole job. For the surrounding circuit arithmetic, the Ohm's law calculator, the voltage divider calculator for the feedback network, and the capacitance calculator all help, while the electrical power calculator covers the power balance side.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Sizing a boost inductor for the load current — the inductor carries the input current, which is larger. Use the peak figure reported here against the part's saturation rating.
- Designing only at nominal input — duty cycle, ripple and peak current all shift across the input range. Check both ends of it.
- Ignoring capacitor ESR — for many capacitor types the resistive ripple term is far larger than the capacitive one, and the tool reports only the capacitive part.
- Assuming the ceramic capacitor keeps its rated value — class II ceramics can lose more than half their capacitance at their rated DC voltage, which multiplies the output ripple.
- Pushing the duty cycle past about ninety per cent — the equations still return a number, but the peak currents and the shrinking off-time make such a design impractical. Change topology instead.
Related Free Tools From Arb Digital
The feedback divider that sets the output voltage is handled by the voltage divider calculator, and its resistor values by the resistor color code calculator. For basic circuit relationships use the Ohm's law calculator and the electrical power calculator. Capacitor combinations are covered by the capacitance calculator, and the input rectification stage by the bridge rectifier calculator. If your power figures are quoted logarithmically, the dBm to watts converter brings them back to linear units. The full free online tools hub lists everything Arb Digital publishes.
Frequently Asked Questions
Whatever the voltage ratio demands. Duty cycle is not a design choice; it is set by the input voltage, the output voltage and the losses. For a twelve volt input and a five volt output at ninety per cent efficiency it comes out near forty-six per cent.
A common design convention is twenty to forty per cent of the maximum load current. Lower ripple needs a larger inductor and gives cleaner output; higher ripple allows a smaller part but raises peak currents and core loss. The convention is a starting point, not a rule.
Because the inductor current has a triangular ripple riding on its average value, and in a boost or buck-boost the average itself is larger than the load current. The peak is what the switch and the inductor must survive, so it is the figure to check against saturation ratings.
It reduces both ripple current and ripple voltage in proportion, which allows smaller magnetics and smaller capacitors. It also increases switching losses and makes layout and electromagnetic interference harder, so the gain is a trade rather than a free improvement.
Almost always because of capacitor equivalent series resistance, which this figure excludes. Multiply the ripple current by the capacitor's ESR and compare. For electrolytic and tantalum parts that term usually dominates the capacitive ripple entirely.
It means the inductor current never reaches zero during a switching cycle. All the equations on this page assume it. At light load a converter drops into discontinuous conduction, where the relationship between duty cycle and output voltage changes and these results no longer describe the circuit.
No. A buck topology can only step down, because the duty cycle cannot exceed one. If you need a higher output, use a boost. If the input can be either above or below the output, a buck-boost or a four-switch topology is required.
This tool is provided for educational and engineering study use. It performs first-order continuous-conduction design arithmetic and does not model losses, control-loop stability, thermal behaviour or layout, so verify any design against the controller datasheet and on the bench.
