Choosing a buck inductor: ripple, saturation and heat

One equation gets you the value:

$L = \frac{V_{out}\left(1 - \dfrac{V_{out}}{V_{in}}\right)}{\Delta I \cdot f_{sw}}$

For 12 V down to 3.3 V at 2 A and 500 kHz, aiming at 30 % ripple ($\Delta I = 0.6$ A), that is 7.98 µH — fit 8.2 µH and the ripple comes out at 584 mA, or 29.2 %. The buck ripple calculator carries the same arithmetic through to the output voltage ripple that results. The buck inductor calculator runs this for any input range and checks the two ratings that follow.

The equation is the easy part. What sends inductors back is that a datasheet publishes two current ratings that mean different things, that the ripple you designed for exists at exactly one input voltage, and that the inductance you looked up is a small-signal value the part does not have at your operating current.

Where the equation comes from, and what it hides

While the switch is on, the inductor sees $V_{in} - V_{out}$ across it, so $di/dt = (V_{in}-V_{out})/L$ and the current ramps up linearly. While the switch is off it sees $-V_{out}$ and ramps back down. In steady state the two must cancel, which is what fixes the duty cycle at $D = V_{out}/V_{in}$ and gives the ripple above.

Two consequences fall straight out and both matter more than the equation itself.

The average is the load current and the peak is not. The core saturates on the instantaneous current, so the number to compare against a saturation rating is $I_{out} + \Delta I/2$ — 2.29 A in the worked case, not 2 A.

The RMS is almost exactly the load current. For a triangle riding on a DC level, $I_{rms} = \sqrt{I_{dc}^2 + \Delta I^2/12}$, and with 29 % ripple that is 2.0071 A against 2 A — a 0.35 % increase. Ripple at sane levels contributes essentially nothing to winding heating, which is why the two ratings behave so differently.

Why 20–30 %, and why it is a band

TI’s SNVA559 (Switching regulator fundamentals) states the convention plainly:

The peak-to-peak difference in the inductor current waveform is referred to as the inductor ripple current, and the inductor is typically selected large enough to keep this ripple current less than 20% to 30% of the rated DC current.

It is a band because it is a compromise between four things that do not all move the same way. Less ripple means more inductance — a larger, heavier, more expensive part with more turns and therefore more resistance — and it also means a converter that responds more slowly to a load step, because the same inductance that smooths the current resists changing it. More ripple means a bigger output capacitor for the same output ripple, more core loss, and a peak current that eats the saturation margin.

The ripple has a worst case, and it is not the nominal input

Substitute $D = V_{out}/V_{in}$ into the ripple expression and the input voltage appears only as $1 - V_{out}/V_{in}$, which increases monotonically towards 1. So:

$\Delta I \to \frac{V_{out}}{L f_{sw}} \quad \text{as } V_{in} \to \infty$

The ripple is largest at the highest input voltage, and approaches a ceiling of 805 mA for the worked case. Across a 5 V to 36 V input range the same 8.2 µH inductor gives:

Vin =  5 V     ΔI = 274 mA   (14 % of the load)
Vin = 12 V     ΔI = 584 mA   (29 %)
Vin = 24 V     ΔI = 694 mA   (35 %)
Vin = 36 V     ΔI = 731 mA   (37 %)

A design verified at 12 V has not been verified where its peak current is highest, nor where it drops out of continuous conduction earliest. Both checks belong at the top of the input range.

Two ratings, two failure modes

This is the part that most often goes wrong. An inductor datasheet gives a saturation current and a heating or RMS current, and they are not alternative names for the same thing.

• Saturation current is a magnetic limit. Beyond it the core’s permeability collapses and the part stops being the inductance you specified. It is compared against the peak current, and it is usually defined as the current at which the inductance has fallen by 20 % or 30 % — a definition, not a cliff edge.
• Heating current is a thermal limit, usually the DC current that produces a stated temperature rise (often 40 °C) in still air on a specified board. It is compared against the RMS current, and like every such figure it belongs to the test board rather than yours — the same problem θJA has.

Because the RMS is almost the DC current while the peak is meaningfully above it, the two ceilings are reached at different loads — and which is reached first is a property of the part and the ripple, not a rule. For a part rated 3.6 A saturation and 3.0 A heating, with 584 mA of ripple, the peak touches the saturation rating at 3.6 − 0.29 = 3.31 A of load, while the RMS touches the heating rating at 3.0 A. The heating rating binds first, by 0.31 A, and it is the thermal figure that sets the load this part can carry. Saturation takes over only when $I_{sat} - \Delta I/2$ falls below $I_{rms}$, which here needs more than $2 \times (3.6 - 3.0) = 1.2$ A of ripple — 60 % of the load — and at exactly 60 % the two only tie. A part whose two ratings sit closer together, or a design run with more ripple, reverses the order, which is why both comparisons are made every time: the lower of $I_{sat} - \Delta I/2$ and $I_{rms}$ is the load the inductor can actually carry.

Saturation is a curve, and its shape is not on the label

Gapped ferrite and powdered iron both saturate, and they do it very differently. TI’s magnetics seminar note SLUP123 (Magnetics Design for Switching Power Supplies) explains why the gap is there at all:

Since the magnetic core material itself is incapable of storing significant energy, energy storage is accomplished in a non-magnetic gap(s) in series with the core.

A gapped ferrite holds its inductance almost constant while the gap does the storing, then loses it abruptly when the ferrite itself saturates. A distributed-gap powdered-iron core has gap everywhere, so it rolls off gently from well below its rating. Both may be sold with the same “saturation current”, because both reach the same −30 % point.

The practical consequence: a soft-saturating part has already lost a useful fraction of its inductance at your operating peak — in the modelled curves of Fig 6, 17 % at the 2.29 A peak of the worked case, against 1 % for the gapped ferrite — which means more ripple than you designed for, which means a higher peak, which moves it further down the curve. The effect is mild and self-limiting, but the ripple in the finished supply is reliably worse than the calculation — and the fix is to look at the L-versus-I curve rather than the rating.

What hard saturation actually does

Integrating $di/dt = (V_{in}-V_{out})/L(i)$ across one on-time, with $L$ falling as the current rises, shows what saturation does to a single cycle. The slope is inversely proportional to the inductance, so the moment the inductance starts to go the current accelerates, which reduces the inductance further. Starting the on-time at 3.3 A with 12 V in, a constant-L model ends the 550 ns at 3.9 A; the modelled ferrite of Fig 6 ends it at 4.3 A, and the gap widens with every additional volt-second — at 36 V in the slope is nearly four times steeper.

Nothing in the control loop can intervene. The controller has already committed to this on-time; the only thing that can end it early is a cycle-by-cycle current limit, and only if that limit is both below the saturation current and fast enough to act inside a few hundred nanoseconds — the current a comparator delay lets through is $V \, t_{delay} / L$, and it grows in proportion as $L$ falls. A converter whose current limit sits above its inductor’s saturation current has no protection at all — the inductor becomes a wire and the switch takes the fault.

Losses: the winding you can compute, the core you cannot

Winding loss is straightforward. $P = I_{rms}^2 R_{DC}$, and since the RMS is essentially the load current, it is $I_{load}^2 R_{DC}$ to within a fraction of a per cent. At 2 A into a 50 mΩ part that is 201 mW against 6.6 W delivered — 3.05 % of the output, straight off the efficiency.

This is the real argument for a physically larger inductor. For the same inductance, a bigger part has thicker wire and a lower DCR, and the loss falls as the square of nothing — it falls in direct proportion to the resistance. Going from 100 mΩ to 20 mΩ saves 322 mW at this load — 403 mW against 81 mW on 6.6 W delivered — which is 4.9 points of efficiency.

Core loss is the term you cannot compute from a datasheet. It grows with frequency and with the flux swing, so it grows with ripple, and vendors publish it inconsistently or not at all. The practical approach is to keep the ripple in the 20–30 % band, prefer a part whose vendor publishes a loss curve, and measure the inductor’s temperature on the finished board.

The light-load end

Continuous conduction ends when the valley of the ripple reaches zero — at a load of $\Delta I / 2$. For the worked case that is 292 mA at 12 V in and 366 mA at 36 V in. A device that idles at a few milliamps spends almost its whole life below that line.

Three things change there. The duty cycle stops following $V_{out}/V_{in}$, so the switching frequency or the on-time varies with load depending on the control scheme. The output ripple changes shape and often gets worse before it gets better. And in a synchronous converter the inductor current will go negative: during the low-side on-time it reverses, flowing from the output capacitor back through the inductor and the low-side switch to ground, and when the high-side switch next turns on that reversed current is returned to the input — unless the controller detects the zero crossing and turns the low-side switch off. That is the difference between the two light-load modes a controller offers: diode emulation (or pulse skipping) stops the reversal and lets the converter go discontinuous; forced continuous mode keeps the low-side switch on, accepts the circulating current and its loss, and in return holds the switching frequency and the ripple shape fixed.

None of this is a fault, but it means the ripple and efficiency measured at full load say nothing about the ripple and efficiency at idle, which for a battery product is where the device lives.

Frequency buys size, and sells it back elsewhere

The inductance for a given ripple falls as $1/f_{sw}$, so the switching frequency is the strongest lever on inductor size there is:

100 kHz    39.9 µH
500 kHz     8.0 µH
  2 MHz     2.0 µH

That is the whole reason modern point-of-load converters run at megahertz. The bill arrives in switching loss, in core loss, and in an EMI spectrum whose fundamental has moved into the band the conducted-emissions input filter has to attenuate — and the faster edges that come with a higher frequency worsen the switch-node ringing at the same time.

The margin is smaller than the nameplate suggests

Fitting a 3.6 A part in a 2 A design looks like 1.8× of headroom. Work through what actually consumes it:

load current                              2.00 A
steady-state peak, +ΔI/2                  2.29 A
with −20 % inductance tolerance           2.36 A
plus a 25 % load-step overshoot           2.86 A

The real margin against a 3.6 A saturation rating is 1.26×, not 1.8×. And the converter’s own current limit has to fit underneath all of it. A cycle-by-cycle limit acts on the instantaneous current, so a limit set at 3.3 A caps the peak at 3.3 A plus whatever the current gains during the comparator’s response delay — with (12 − 3.3) V across 8.2 µH the ramp is 1.06 A/µs, so 100 ns of delay adds 0.1 A; at 36 V in the ramp is 3.99 A/µs and the same delay adds 0.4 A, past the rating. What the 3.3 A setting leaves under a 3.6 A rating is 0.3 A less that overshoot, and that is the whole margin between a fault and a saturated core. A limit referenced to the average or the valley of the ripple rather than the peak lets the peak sit ΔI/2 above the setting — 3.59 A at 12 V in and 3.67 A at 36 V in, the second of them past the rating — so the number that matters is the peak the limit lets through, not the setting.

Add temperature to that. Ferrite’s saturation flux density falls with temperature, so the saturation current of the finished assembly at 85 °C is below the room-temperature figure on the datasheet — by an amount most vendors publish as a derating curve and some do not publish at all.

The order that finds the problems

1. Pick the ripple — 20–30 % of the rated load current, from SNVA559. Lower if the output ripple is critical, higher if size is.
2. Compute the inductance at the highest input voltage, because that is where the ripple peaks. Round up to a standard value and recompute the actual ripple.
3. Two separate checks. Peak current, $I_{out} + \Delta I/2$, against the saturation rating. RMS current against the heating rating. Then confirm the converter’s current limit is below the saturation rating, not above it.
4. Check both ends of the load range. The light-load DCM boundary, and the peak during a load step. Then repeat the peak check with the inductance 20 % low, because the tolerance moves the ripple the wrong way.

Then look at the part’s actual L-versus-I curve rather than its saturation number, and if the vendor does not publish one, treat that as information about the part. Work the ripple and the resulting output voltage ripple out in the buck ripple calculator, which uses the exact waveform rather than a linear approximation and reports where the two models diverge — and read where the buck ripple comes from for what happens to it after it leaves the inductor.