Buck converter inductor calculator

The inductance a buck needs for a chosen ripple current, solved at the input voltage where the ripple is worst — then what a real part carries: the peak against its saturation rating, the RMS against its heating rating, and the copper loss its DCR costs.

Fig 1 — inductor current over one period at 12.0 V in, D = 0.27: 584 mA of ripple on 2.00 A of load, peaking at 2.29 A against a 3.60 A saturation rating.

Input voltage, minimum (V)

Input voltage, maximum (V)

Output voltage (V)

Load current (A)

Switching frequency (kHz)

Ripple target (% of load)

Inductance fitted (0 to use the solved value) (µH)

Saturation current (0 to skip) (A)

Heating current rating (0 to skip) (A)

DCR (0 to skip) (mΩ)

Inductance for 30 % ripple at 12.0 V in: 7.98 µH
Ripple with 8.20 µH fitted: 584 mA peak-to-peak at 12.0 V · 29 % of the load
Ripple at the lowest input · ceiling as V_in → ∞: 473 mA · 805 mA
Peak current · RMS current: 2.29 A · 2.01 A
Duty cycle, lowest to highest input: 0.412 to 0.275
Load below which conduction goes discontinuous: 292 mA
Load the part can carry: 3.31 A by saturation · 3.00 A by heating → heating binds
Saturation margin at the peak: 1.31 A
Copper loss I_rms²·R_DC: 201 mW · 3.05 % of the output power

How this is calculated

Standard: TI SNVA559 — Switching Regulator Fundamentals; TI SLUP123 — Magnetics Design for Switching Power Supplies (Dixon)

$D = \frac{V_{out}}{V_{in}}, \qquad L = \frac{V_{out}\,(1 - D)}{\Delta I \, f_{sw}}$: SNVA559: v = L·di/dt, and the CCM buck delivers V_in × D. During the off-time the inductor sees V_out for (1−D)/f_sw; that volt-second product over L is the ripple, and solving for L gives the value for a chosen ripple.
$\Delta I(V_{in}) = \frac{V_{out}\left(1 - \frac{V_{out}}{V_{in}}\right)}{L f_{sw}} \;\to\; \frac{V_{out}}{L f_{sw}}$: The same expression with the duty cycle substituted: ripple rises with input voltage towards a ceiling, so the inductor is sized and the peak is checked at the maximum input.
$I_{pk} = I_{out} + \frac{\Delta I}{2}, \qquad I_{rms} = \sqrt{I_{out}^2 + \frac{\Delta I^2}{12}}$: A triangle riding on a DC level. The peak is what the saturation rating is compared against; the RMS, almost exactly the load, is what the heating rating and the copper loss use. SLUP123: saturation follows the flux swing, which the peak sets.
$P_{Cu} = I_{rms}^2 \, R_{DC}$: Winding loss only. Core loss also rises with the flux swing and frequency but is not computable from a datasheet figure, so it is not computed here.

Assumptions

Continuous conduction and an ideal switch: duty is V_out/V_in. A real converter runs a few per cent higher to cover its losses, which slightly reduces the ripple against these figures. The tool flags a load below ΔI/2, where conduction goes discontinuous and the equations stop applying.
The inductance is taken as constant up to the saturation rating. Real parts lose inductance gradually as current rises, powdered iron more than gapped ferrite, so the ripple near the rating is higher than computed.
The heating rating is compared against the RMS current at the stated ambient of the datasheet's test board; on a different board, in an enclosure, the rating is lower.
Copper loss is the DC winding resistance times the RMS current. AC resistance from skin and proximity effect, and core loss, are not included.
SNVA559's 20 % to 30 % sizing band is a convention, not a limit; the calculator reports where the design sits against it and lets the ripple target be anything.

What sets a buck inductor's value

While the switch is on, the inductor sees V_in − V_outand its current ramps up; while the switch is off it sees V_outand the current ramps down. SNVA559 states the relation the whole page rests on, v = L·di/dt, and the consequence: for a buck in continuous conduction the output is V_in × D, so the off-time is (1 − V_out/V_in) of the period and the peak-to-peak ripple is V_out(1 − D)/(L f). Turn that round and the inductance for a chosen ripple falls out in one line.

The choice of ripple is the design decision. SNVA559 says the inductor "is typically selected large enough to keep this ripple current less than 20% to 30% of the rated DC current", and the band exists because both ends cost something: less ripple means a larger, slower, more expensive inductor and a smaller ramp for a current-mode controller; more ripple means a higher peak against the saturation rating, more core loss, and more output ripple across the capacitor.

Ripple grows with input voltage. Substituting the duty cycle back into the expression gives V_out(1 − V_out/V_in)/(L f), which rises towards a ceiling of V_out/(L f) as the input rises. The inductor is therefore sized at the highest input the converter will see, and that is where the calculator solves it; the ripple at the lowest input is reported for comparison and is always smaller.

Two datasheet ratings then decide whether the part survives, and they are compared against two different currents. The saturation current is a magnetic limit: SLUP123 puts it as the core's flux swing, and it is the peak current, I_out + ΔI/2, that reaches it. The heating current is a thermal limit, and it is the RMS current that reaches it, which for a triangle on a DC level is √(I_out² + ΔI²/12), a fraction of a per cent above the load. The lower of the two limits is the load the part can actually carry, and which one binds is a property of the part and the ripple, not a rule.

Buck inductor chart: inductance for common conversions

The inductance that gives 30 % ripple at 1 A, computed by the calculator above for the conversions a board actually has. Inductance is inversely proportional to the load current, so halve the cell for 2 A and take a tenth of it for 10 A; and every doubling of the switching frequency halves it again.

Conversion	250 kHz	500 kHz	1 MHz	2 MHz
5 V → 3.3 V	15.0 µH	7.48 µH	3.74 µH	1.87 µH
5 V → 1.8 V	15.4 µH	7.68 µH	3.84 µH	1.92 µH
12 V → 5 V	38.9 µH	19.4 µH	9.72 µH	4.86 µH
12 V → 3.3 V	31.9 µH	16.0 µH	7.98 µH	3.99 µH
24 V → 12 V	80.0 µH	40.0 µH	20.0 µH	10.0 µH
24 V → 5 V	52.8 µH	26.4 µH	13.2 µH	6.60 µH
48 V → 12 V	120 µH	60.0 µH	30.0 µH	15.0 µH
48 V → 5 V	59.7 µH	29.9 µH	14.9 µH	7.47 µH

Read down a column and the 48 V to 5 V row is only half again the 12 V to 5 V row, though the input is four times higher, because the ceiling V_out/(L f) has nearly been reached: once the duty cycle is small the input voltage stops mattering to the ripple. Read across and the 2 MHz column is an eighth of the 250 kHz one, which is the whole case for high switching frequency, and the case against it is the switching loss the page on gate resistors accounts for.

Worked example: 12 V to 3.3 V at 2 A and 500 kHz

The defaults: 8 V to 12 V in, 3.3 V out at 2 A, 500 kHz, aiming at 30 % ripple, then fitting 8.2 µH with a part rated 3.6 A saturation, 3.0 A heating and 50 mΩ.

D at 12 V       = 3.3 / 12                            = 0.275
L for 30 %      = 3.3 × (1 − 0.275) / (0.6 A × 500 kHz) = 7.98 µH   →  fit 8.2 µH
ΔI at 12 V      = 3.3 × 0.725 / (8.2 µH × 500 kHz)     = 584 mA    (29 % of the load)
ΔI at 8 V       = 3.3 × 0.5875 / 4.1                   = 473 mA
ceiling         = 3.3 / (8.2 µH × 500 kHz)             = 805 mA
I_pk            = 2 + 0.292                            = 2.29 A
I_rms           = √(2² + 0.584² / 12)                  = 2.007 A
load by saturation   3.6 − 0.292                       = 3.31 A
load by heating      √(3.0² − 0.584²/12)              = 2.99 A     →  heating binds
copper loss     = 2.007² × 50 mΩ                       = 201 mW    (3.05 % of 6.6 W)

The part is fine at 2 A with a full amp of headroom on both ratings, and the thermal rating, not saturation, is what would give out first as the load rose. The 201 mW of copper loss is three per cent of the output, which is the real argument for the physically larger part with the lower DCR: at 20 mΩ the same current costs 81 mW.

Where the ripple equation stops being valid

Discontinuous conduction. SNVA559 defines continuous mode as the inductor current never dropping to zero. Below a load of ΔI/2 it does, the duty cycle rises above V_out/V_in, and the ripple equation no longer describes the waveform. The calculator reports that boundary and flags a load below it; the efficiency and ripple measured at full load say nothing about idle.

Saturation is a curve. The rating is the current at which the inductance has fallen by a stated fraction, and gapped ferrite and powdered iron get there by different paths, one sharply and one gradually. The peak compared here is against the number on the label; a part already 20 % down at that number is delivering more ripple than this predicts, and the effect feeds on itself.

Core loss is not computed. It rises with frequency and with the flux swing, which is the ripple, and vendors publish it inconsistently. The copper loss here is the winding only. Keeping the ripple in the band, preferring a part with a published loss curve, and measuring the inductor's temperature on the finished board is what remains.

The ideal duty cycle. A real converter runs slightly above V_out/V_in to cover its losses, which lengthens the on-time and slightly reduces the ripple against these figures. The error is a few per cent at ordinary efficiencies and on the conservative side.

Common buck inductor mistakes

Sizing at the nominal input instead of the maximum. Ripple is largest at the highest input; a part sized at 12 V for a 9 V to 36 V range is carrying 25 % more ripple than intended at the top of the range.
Comparing the load current against the saturation rating. The peak is the load plus half the ripple; at 30 % ripple that is 15 % more than the load, and it is the peak that saturates the core.
Comparing the peak against the heating rating. The RMS is what heats the winding, and it is almost exactly the load; using the peak overstates the loss and picks a larger part than needed.
Chasing minimum ripple. A very large inductor is slow, expensive and gives a current-mode controller a small ramp to work with; the 20 % to 30 % band is a design point, not a compromise.
Ignoring DCR. Copper loss goes directly as the winding resistance, and at a few amps the difference between a 100 mΩ part and a 20 mΩ part is several points of efficiency.