Selecting CoupledInductors for SEPIC
Applications
Application Design Example
The SEPIC (Single-Ended Primary Inductance Converter) topology is used in applications that require characteristics of both a buck and a boost regulator, specificallythe ability to step up and step down the input voltage.
Most often operated in CCM (Continuous ConductionMode), SEPIC provides a non-inverted output voltage.

The selection of one coupled inductor over two single
parts saves board space and can also save cost.

Typically, SEPIC is used in battery operated systems
and automotive applications. In these applications, thebattery input voltage, or bus line voltage, may be greateror less than that of the desired output voltage, dependingon the charge state of the battery. The SEPIC topology
can operate over more of the battery discharge cycle
because of the ability to regulate the output voltage overa wider input voltage range, including above and below
the output voltage.

Coupled Inductor Calculation

A simplified SEPIC schematic, using coupled inductors
is shown in Figure 1.
D1+L1
VIN

Q1+

C1L2C2+

Load–VOUT–Figure 1. Simplified SEPIC Circuit Diagram

The following examples demonstrate the calculations
necessary to identify the inductance, rms current andpeak current ratings.

Assume the following parameters for the circuit shownin Figure 1:
VIN = 2.7 V to 4.5 V
VOUT, IOUT = 3.3 V, 0.2 A
Fs (Switching Frequency) = 400 kHz
VD1 (Typical Voltage Drop) = 0.7 V
Target Efficiency = 90%
Step 1. Calculate minimum and maximum duty cycle(Dmin and Dmax)

Dmin =

VOUT + VD(VIN (max) + VOUT + VD )Dmin =

3.3 V + 0.7 V(4.5 V + 3.3 V + 0.7 V)

Dmin = 0.47

C1: Flying capacitor – Should be low ESR, most oftenceramic.

Dmax =VOUT + VD
(VIN (min) + VOUT + VD )C2: Output capacitor – Chosen based on output ripple
current tolerances, usually tantalum.

Dmax =

3.3 V + 0.7 V(2.7 V + 3.3 V + 0.7 V)

L1 and L2: Coupled inductor – Two inductors closely
coupled on a common core.

Dmax = 0.60

Inductors L1 and L2 do not have to be wound together
on a single magnetic core, however there are distinct advantages if they are. If L1 and L2 are closely coupled, theripple current is divided between them and the required
inductance is halved. For example, two 22 µH single
inductors can be replaced with one coupled inductor of
11 µH per winding, potentially allowing for the selectionof a smaller inductor, or one the same size that has lower
DCR and higher current handling.

Step 2. Calculate peak to peak ripple current
The allowed ripple current percentage must be chosen.
A good starting point is to limit the peak to peak ripple
current to 40% of the full load inductor Irms current.
This ensures a low output ripple, minimizes core loss
in the inductor and allows for light load continuous
conduction mode operation to current as low as 20%of full load.
Document 639-C

Revised 01/12/21

Allowing for 40% ripple, calculate the worst case peak
to peak ripple current at VIN(min) using the following
equation:
Iripple = IOUT ×VOUT× (% peak–peak ripple current)
VIN (min)

VOUT
× 0.4
Iripple = IOUT ×VIN (min)3.3 V
× 0.4Iripple = 0.2 A ×
2.7 V
Iripple = 0.098 A

Step 3. Calculate inductanceInductance is calculated by the fundamental equation:
itBecause the two windings of a coupled inductor share
the ripple current, the inductance value can be halved.

V=L×L1 = L2 = 0.5 ×

VIN (min)× DmaxIripple × FS
2.7 V× 0.60
0.098 A × 400 kHz

Step 6. Summarize inductor specifications
L1 = L2 = 22 µH
Irms (L1) = 0.27 A
Ipeak (L1) = 0.32 A
Irms (L2) = 0.20 A
Ipeak (L2) = 0.25 A
Step 7. Select the coupled inductor
Choose Coilcraft LPD4012-223ML.
Coilcraft LPD4012-223ML has 22 µH per winding, anda saturation rating of 0.70 A. This means that 0.35 A perwinding (or any combination that does not exceed 0.70 A)
can flow without saturation.
Coilcraft LPD4012-223ML has Irms rating of 0.34 A for
both windings and 0.48 A for a single winding. This meansthat for a 40°C temperature rise, up to 0.34 A can flowin each winding simultaneously or up to 0.48 A can flowin one winding.
For this example, Irms of L1 (0.27 A) and Irms of L2 (0.20 A)
are well below these limits. Refer to the LPD4012 data
sheet, and use the following formulas to calculate thetemperature rise (∆t):

Power loss = (IL12 + IL22 ) × DCR

135°C
W

L1 = L2 = 20.7 µH

Temperature rise (∆t) = Power loss ×
135°C∆t = power loss ×
WChoose standard value L = 22 µH

Power loss = (0.27 2 + 0.20 2) × 1.52 = 0.172 W

L1 = L2 = 0.5 ×

Step 4. Calculate required Irms

L1 Irms =VOUT × IOUT
VIN (min) × Efficiency

L1 Irms =3.3 V × 0.2 A
2.7 V × 0.9

L1 Irms = 0.27 A
L2 Irms = IOUT
L2 Irms = 0.2 A
Step 5. Calculate Ipeak

L1 Ipeak = Irms + (0.5 × Iripple)
L1 Ipeak = 0.27 A + (0.5 × 0.098 A)
L1 Ipeak = 0.32 A
L2 Ipeak = Irms + (0.5 × Iripple)
L2 Ipeak = 0.2 A + (0.5 × 0.098 A)
L2 Ipeak = 0.25 A∆t = 0.172 W ×135°C= 23°C
W

Step 8. Frequency Dependent Losses
In most cases the I2R loss calculated in Step 7 provides agood estimate of the inductor losses. However, frequency
dependent losses such as core loss and ac winding loss
should also be considered.
To estimate total inductor loss including ac winding
resistance and core loss refer to the Coilcraft Coupled
Inductor Finder and Analyzer tool at www.coilcraft.com/
en-us/tools/coupled-inductor-finder/#/search
Calculate the Total inductor loss for each winding. To
determine temperature rise, add the Total inductor loss
for each winding together, and multiply this number by
135 as shown in step 7.

References
Designing a SEPIC Converter, Wei Gu, National Semiconductor Application Note AN1484, June 2007
Versatile Low Power SEPIC Converter Accepts Wide
Input Voltage Range, Jack Palczynski, Unitrode Design
Note DN-48.
SEPIC Equations and Component Ratings, Maxim InDocument 639-C

Revised 01/12/21