Choosing Inductors
for Energy EfficientPower Applications
Energy efficiency can be as much about the inductors as the circuit topology

For battery powered applications, battery life is extended
by improving the efficiency of the entire power supply circuit, and inductor efficiency is often a major considerationin the design. Careful consideration of inductor efficiency
can mean the difference between having your battery workwhen you need it and having to stop in the middle of animportant task to plug it into a charger.
Inductor efficiency is highest when the combination ofcore and winding losses are the lowest. Therefore, thegoal of highest efficiency is met by selecting an inductor that provides sufficient inductance to smooth out theripple current while simultaneously minimizing losses. Theinductor must pass the current without saturating the coreor over-heating the winding.
Accurately predicting core and winding loss of an inductor
can be fairly complicated. Core loss depends on several
factors, such as peak-peak ripple current, ripple current
frequency, core material, core size, and turn count. The
required ripple current and ripple current frequency areapplication-dependent, while the core material, core size,
and turn count are inductor-dependent.
The most commonly-used equation to characterize coreloss is the Steinmetz equation:
Pcore = K(f)x(B)y
Where:
Pcore = power loss in the coreK, x, y = core material constants
F = frequency
B = flux density
This equation shows that core loss depends on frequency(f) and flux density (B). Flux density depends on ripple current, so both are application-dependent variables. It also
shows that the core loss is inductor-dependent, where thecore material determines the K, x, and y constants. Notethat flux density is also a function of the core area (Ae)and the number of turns (N), therefore core loss is both
application-dependent and inductor-dependent.
By comparison, DC winding loss is simple to calculate:
Pdc = Idc2 × DCR
WherePdc = DC power in Watts dissipated
Idc = Effective DC (rms) value of the inductor current.
DCR = DC resistance of the inductor winding
AC winding loss is more complicated and may include theeffects of increased resistance at higher frequency due toboth skin effect and proximity effect. ESR (effective seriesresistance) or ACR (AC resistance) curves may show
some of the increased resistance at higher frequency, however, these curves are typically made at very low current
levels, so they do not capture current-dependent (core)loss. They are also subject to possible misinterpretation.
For example, consider the ESR vs frequency curve shownin Figure 1
1000

Equivalent Series Resistance (Ohms)

In high frequency DC-DC converters, inductors filter outthe AC ripple current superimposed on the DC output.
Whether the converter steps the voltage down – buck –or steps the voltage up – boost – or both up and down
– SEPIC, the inductor smooths the ripple to provide apseudo-DC output.

100

10

1

0.1

10 µH

0.01

0.1

1

10Frequency (MHz)Figure 1. ESR vs Frequency
An initial observation indicates that the resistance looks
very high above 1 MHz. This would strongly suggest that
this part cannot or should not be used at that frequency
due to the expected very high loss due to the ESR. HowDocument 1400-1

Revised 07/16/20

ever, it has been observed that parts with curves like thishave performed very well in actual converters – muchbetter than would be suggested by these curves.

Let’s assume that the ripple current peak-peak is about
10% of the average current. From the previous example
this means:

Consider the following example:

Idc = 0.3 A
Ip-p = 0.03 A

Assume a converter is needed to provide an output of 5 V
at 0.3 A (1.5 Watts). We will use a 10 µH Coilcraft inductor
with a typical ESR vs frequency as shown in Figure 1. If
the converter operates at 250 kHz, we see from the graphthat the ESR, which includes both ac and dc resistance isapproximately 0.8 Ohms.
For a buck converter, the average inductor current equals
the load current, 0.3 A.
We can calculate the loss in the inductor:
I2R = (0.3 A)2 × (0.8Ω) = 0.072 W.
0.072 W ÷ 1.5 W = approx 5% of output power is lost in
the inductor.
However, if we were to run the same converter at 5 MHz,
we can see from the ESR curve that R is between
10 Ohms and 20 Ohms. If we even assume R = 10 Ohms,
then the power loss in the inductor should be:
I2R = (0.3 A)2 × (10Ω) = 0.9 W
0.9 W ÷ 1.5 W = 60% of the output power is lost in theinductor!
Based on this very simple example it would seem obviousthat a designer should not choose to use a component
like this.
It has been observed that converters, in fact, often achieve
better performance than the ESR curves predict. The following explanation illustrates why.
Figure 2 shows a very simplified version of a possible
buck converter waveform, with continuous conduction
and the ripple current is relatively small compared to theaverage current.
Buck Converter
Inductor Current
ton

Ip-p

Idc
ContinuousCurrent
Mode

toff

Figure 2. Ideal Converter Waveform with
Small Ripple Current

In order to predict the inductor losses correctly, this must
be separated into two components. For the low frequency
or dc loss, we use the low frequency resistance (effectivelyDCR), which we can see from the graph is 0.7 Ohms. Thecurrent is the rms value of the load current plus the ripple
current. In this case the ripple current is small, so the value
is approximately equal to the dc load current.
Low frequency loss = Idc2R = (0.3 A)2 × (0.7Ω) = 0.063 W
To get the total loss, we must add that to the high frequency
loss, which is I2R. In this case the R is the ESR and the Iis the rms value of the ripple current only.
Approximate rms ripple current:
Ip-p ÷ 2    3 ) = 0.03 A ÷ 3.464 = 0.0087 AAt 250 kHz the ac loss would be:(0.0087 A)2 × (0.8 Ω) = 0.00006 W.
Therefore, at 250 kHz, we predict the total inductor loss
is 0.063 W + 0.00006 W = 0.06306 W.
We see that operating at 250 kHz predicts only slightly
more loss (less than 1%) than predicted simply by theDCR.
Now, let’s look at the same example at 5 MHz. The low
frequency loss is still the same 0.063 W.
The ac loss calculation must use the ESR, which waspreviously estimated at 10 Ohms:(0.0087 A)2 × (10 Ω) = 0.00076 W.
So, the total inductor loss at 5 MHz: 0.063 W + 0.00076
W = 0.06376 W.
This loss is more significant, with a predicted loss of about
1.2% greater than DCR loss, but is not nearly the 0.9 W
originally predicted by multiplying the ESR by the entire
load current. Also, this example is not exactly fair, because
we wouldn’t use the same inductor value at 5 MHz as wewould at 250 kHz. We would use a much smaller L andtherefore we would get a much smaller DCR.
In summary, the inductor loss must be calculated by a
combination of the DCR and ESR, and for a continuous current mode converter in which the ripple current
is small compared to the load current, the losses will bereasonable.
In typical applications, ripple current is kept to approximately 40% of the load current or less. Regardless of ripple
content, ESR curves do not capture current-dependent

Document 1400-2

Revised 07/16/20

core loss at higher current, and total inductor loss determines the overall inductor efficiency.
Therefore, inductor manufacturers optimize inductor efficiency by selecting low loss materials and designing
inductors for minimal total loss. The use of rectangular“flat” wire may provide the lowest DCR in a given size tominimize DC loss. Improvements in core materials have led
to inductors with very low AC core loss at high frequency
resulting in higher inductor efficiency.
For example, Coilcraft’s industry-leading XGL Family ofmolded power inductors are optimized for high frequency,
high peak current applications. These offer soft-saturation,
while also providing the lowest AC loss at frequencies of
2 MHz and higher. They also have extremely low DCR
for their size.
Figure 3 shows the inductance vs current characteristicsof the 2.2 μH value in the XGL, XEL, XAL, and XFL Series.
The XGL, XEL, and XAL series are clearly the best choicefor holding inductance at around 3 A or higher current.
Table 1 compares the DCR and Isat of these inductors.
Table 1. Comparing XAL, XEL and XFL
L nom

DCR typ

Isat (30%)

XGL4020-222

2.2 µH

19.5 mOhms

5.9 A

XEL4020-222

2.2 µH35.2 mOhms

5.9 A

XAL4020-222

2.2 µH35.2 mOhms

5.6 A

XFL4020-222

2.2 µH

21.4 mOhms

3.7 A

Figure 3. Comparing L vs I of XAL, XEL XFL
and XGL 2.2 μH inductors
Figure 4 compares the AC loss and total loss of thesame inductors at 2 MHz. The XGL utilizes an innovative
construction that exceeds all previous designs, resulting
in a combination of the lowest DC and AC losses. This
makes the XGL Family the best choice for high frequency
power converter applications that must withstand highpeak current with lowest DC and AC losses.
To speed up the design process for engineers selectinginductors, Coilcraft has developed tools that calculate
measurement-based core and winding loss for each possible application condition. The results from these tools

Figure 4. Comparing AC Losses and Total Lossesof XAL, XEL, XFL, and XGL at 2 MHz
include current-dependent and frequency-dependent core
and winding loss, eliminating the need to request proprietary inductor design information, such as core material,
Ae, and number of turns, and the need to perform handcalculations.
If your application is a DC-DC converter, the Coilcraft
DC-DC Optimizer Tool calculates the inductance value,
peak current, and peak-peak current requirements basedon your operating conditions and amount of AC ripple
current you choose. It then feeds this information into ourPower Inductor Finder tool to display a list of inductorsthat may meet these requirements. The list includes theinductance at peak current, current rating, total losses,
and resulting part temperature for each inductor listed.
If you already know the inductance value and current
ratings required for your application, you can enter thisinformation directly into the Power Inductor Finder. The
results include core and winding (total) loss and saturation current ratings for each inductor, to verify that theinductance will remain close to the design requirement at
the peak current condition for your application.
The tool may also be used to graph the inductance vscurrent behavior to compare traditional hard-saturating
inductors to soft saturation types. To select the highest
efficiency inductor, the results can be first sorted by total
loss. Multiple sorts allow selection by multiple parameters.
Document 1400-3

Revised 07/16/20

Inductor loss is closely related to core size and wire size. In
many cases, lowest loss corresponds to larger part size, or
it corresponds to using a hard-saturation core material. Aswith any design, there may be compromises that require
analyzing trade-offs in size or inductance at peak current
vs efficiency. Having all of the inductor information in acomplete list that allows multiple sorting facilitates suchan analysis.

References:

Conclusion

Coilcraft Power Inductor Finder and Analyzer Tool,
Coilcraft website, https://www.coilcraft.com/en-us/tools/
power-inductor-finder/#/search

Designing for highest efficiency performance requires selection of inductors with the lowest total loss at application
conditions. Calculating total loss can be complicated, butthese calculations are built into the Coilcraft power magnetics tools, making selection, comparison, and analysis
as simple as possible.

XGL, XEL, XAL, or XFL? – Which Molded Power
Inductor is Right for You?, Coilcraft web page:https://www.coilcraft.com/en-us/other/xal-or-xfl-or-xgl/
Coilcraft DC-DC Optimizer, Coilcraft website,
https://www.coilcraft.com/en-us/tools/dc-dc-optimizer/#/
searchDocument 1400-4

Revised 07/16/20