The Use of Tapped Inductors and Autotransformers in DC/DC
Converters Shown for the SEPIC
FELIX A. HIMMELSTOSS, HELMUT L. VOTZI
Faculty of Electronic Engineering and Entrepreneurship
University of Applied Sciences Technikum Wien
Hoechstaedtplatz 6, 1200 Wien
AUSTRIA
Abstract: - Studying the function of DC/DC converters is an important topic when organizing a course in
Power Electronics. The use of tapped inductors or autotransformers changes the behavior of the converter and
also the stress of the semiconductors. In this paper, a simple way to teach these converters is shown and
demonstrated for two variants of the well-known SEPIC converter. Suggestive drawings and simple
calculations and simulations are used to demonstrate the operation of the converters. The large signal and the
small signal models for one variant are calculated.
Key-Words: - DC/DC converter, SEPIC, autotransformer, coupled coils, tapped inductor, large signal model,
small signal model, simulations.
Received: September 9, 2023. Revised: May 5, 2024. Accepted: June 11, 2024. Published: August 6, 2024.
1 Introduction
Replacing a coil of a DC/DC converter with a
tapped inductor or an autotransformer changes the
voltage transformation ratio, changes the stress of
the semiconductors, and gives an additional degree
of freedom for the design. The input and the output
voltages can be adapted in such a way that the duty
ratio of the active switch has no extreme values near
zero or near one. So the efficiency can be increased.
Here we discuss in a didactical way the function of a
Sepic converter, when the first coil is replaced by
two coupled windings. The best way to understand
the function of a converter is by sketching the
voltages across and the currents through the
components. So the comprehension of converters is
deepened. Both variants used here are done for the
first inductor, and in the first case, the coupled coils
behave like a transformer, and in the second variant
as a tapped inductor. First, a look at the literature
concerning tapped coils used in DC/DC converters,
especially the SEPIC converter is done.
A short paper describing basic bidirectional
converters with tapped inductors is [1]. In [2] tapped
inductor technology-based DC-DC converters are
shown. The modeling of the tapped inductor SEPIC
converter by the TIS-SFG approach is treated in [3].
A high step-up tapped inductor SEPIC converter
with a charge pump cell can be found in [4]. [5],
treats an analysis and design of a charge pump-
assisted high step-up tapped inductor SEPIC
converter with a regenerative snubber. The analysis
of a bidirectional SEPIC/Zeta converter with a
coupled inductor [6] uses the coupling of both
inductors of the original topologies. A family of
high step-up soft-switching integrated Sepic
converters with a Y-source coupled inductor is
discussed in [7]. Other applications with tapped
inductors are shown in [8] and [9]. It should be
mentioned that the original SEPIC topology is
treated in the textbooks on Power Electronic, e.g.
[10], [11], [12].
In its basic form, the Sepic converter consists of
an active (S) and a passive switch (D), two coils
(L1, L2), and two capacitors (C1, C2). The circuit
diagram is shown in Figure 1. During the steady
state the voltages across the inductors are zero in the
mean, so it is easy to see that the voltage across C1
must be equal to the input voltage U1. With the help
of the voltage-time balance, the voltage
transformation ratio is equal to a Buck-Boost
converter according to
d
d
U
U
M
1
1
2
(1)
Fig. 1: Circuit diagram of the Sepic converter
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The variants will be studied for ideal
components (no parasitic resistors, ideal switching),
in the continuous mode, in steady-state, and with
capacitors so large that the voltages across them are
constant during one period. In the continuous mode,
only two modes within a period follow each other in
a repeated way. In the first mode, M1 the active
switch S is on and the passive switch D is off, in the
second mode M2 the electronic switch S is off and
the diode D is conducting. When the converter is in
the discontinuous mode a third mode M3, when
both semiconductor components are off, occurs. The
drawings are done for a winding ratio of one to one
but lettered with the exact winding ratio.
2 Variant I
In the first variant, the first coil of the Sepic
converter is replaced by a tapped inductor which
acts as an autotransformer. Figure 2 shows the
circuit diagram.
2.1 Connection between the Voltages
When inspecting the circuit diagram one can
immediately see that in the steady state, the voltage
across C1 must be equal to the input voltage
11 UUC
. (2)
During M1 the input voltage U1 is across the
winding N11 and during M2 the input voltage minus
the output voltage minus the voltage across C2 is
across the complete winding of the autotransformer
(N11+N12). Only the corresponding part
121
1211
11
2,11 CMN UUU
NN
N
u
(3)
is across the first winding during M2. With an
arbitrarily chosen input voltage of two divisions,
one can now draw the voltage across N11 (Figure 3,
left).
Fig. 2: Variant I
The charge balance across N11
(4)
leads to the voltage transformation ratio
d
d
N
NN
U
U
M
1
11
1211
1
2
(5)
In the same way, one gets the voltage across the
second winding N12 of the autotransformer
(Figure 3, right).
Fig. 3: Voltages across the autotransformer: first
winding (left), second winding (right)
With a duty cycle of one-third the output
voltage is equal to the input voltage for N11=N12.
Figure 4 on the left side shows the input and the
output voltages, and the voltage across the first
capacitor.
The voltage across the coil L must be zero in the
mean. During mode M1 the sum of the transformed
voltage across N11 plus the voltage across C1
1
11
12
11, CML U
N
N
Uu
(6)
lies across the coil and during M2 the negative
output voltage (Figure 4 right).
.
Fig. 4: Input voltage and voltages across the
capacitors (left), voltage across the coil (right)
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Fig. 5: Voltages across the electronic switch S (left)
and the diode (right)
For the design of the converter the voltage
maxima across the semiconductors are important
(Figure 5). With the help of KVL one gets:
dU
US
1
1
1
max,
, (7)
dN
NN
U
UD
1
1
11
1211
1
max,
. (8)
2.2 Connection between the Currents
We start arbitrarily with a load current of two
divisions (Figure 6, left).
Fig. 6: Currents through the load (left) and through
the output capacitor (right)
The charge balance of the output capacitor C2
can easily be drawn (Figure 6, right) and leads to
dIIIdI LOAD
MNL
LOAD
1
2,12
__
. (9)
2,12
_
MN
I
is the mean value of the current through
the winding N12 referred to as the duration of
mode M2. The charge balance of the first capacitor
C1 is given by
dIdI MNL 1
2,12
__
. (10)
Solving
LOAD
L
MN I
I
I
dd
dd
0
1
1
11 2,12
_
(11)
leads to
d
d
I
I
LOAD
MN
1
2,12
_
, (12)
1
2
_
LOAD
L
I
I
. (13)
Figure 7, left shows the current through the
winding N12 which is equal to the current through
the capacitor C1. To distinguish the currents through
L and N12 the current ripple is chosen differently,
one division for the coil and half division for the
autotransformer winding. The current through coil is
shown in Figure 7, right.
Fig. 7: Current through the second winding (left)
and the current through the coil (right)
During M2 the transformer has no output
current, so the current through N12 is only the
magnetizing current imag. This current must also
flow through N11. During M1 the magnetizing
current flows only through the first winding N11
and must be therefore higher. The ampere windings
must be the same:
2,12111,11 )( MmagMmag iNNiN
(14)
therefore, the magnetizing current must be
2,
11
1211
1, MmagMmag i
N
NN
i
(15)
higher in mode M1. The current through N11 is in
M1 equal to this higher magnetizing current and the
current which is the transformed current through the
coil L. The current turns are:
11,1112 NLL iNiN
. (16)
So the current caused by the current through the
coil in N11 (Figure 8, left) can be obtained by:
LNL i
N
N
i
11
12
11,
(17)
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The magnetizing current is shown in Figure 8
on the right side.
Fig. 8: Current through the first winding (left),
magnetizing current (right)
Now one can construct the current through the
semiconductor devices. The current through the
electronic switch (Figure 9, left) is
111 CNS iii
(18)
and through the diode (Figure 9, right)
1CLD iii
(19)
Fig. 9: Currents through the semiconductors, active
switch (left), passive switch (right)
3 Variant II
Not only the first inductor is replaced by a tapped
inductor, also the connections of the electronic
switch S and of the intermediate capacitor C1 are
changed (Figure 10).
Fig. 10: Variant II
In this case, we can interpret the tapped inductor
as coupled coils.
3.1 Connection between the Voltages
During mode M1 (the active switch is turned on and
the diode blocks) the input voltage is across the
complete winding N1+N2. According to the
transformation law (the voltages behave like the
number of turns) only the part
1
1211
11
1,11 U
NN
N
u
N
MN
(20)
is across the winding N11. During M2 one gets the
sum of the input voltage minus the output voltage
and minus the voltage across C1 across N11
(Figure 11, left)
1212,11 CMN UUUu
. (21)
Inspecting the circuit of the converter, it is easy
to see that in the steady state, the voltage across C1
must be equal to the input voltage U1 (the inductive
components have zero voltage in the mean).
Now one can calculate the voltage
transformation ratio according to
d
d
NN
N
U
U
1
1211
11
1
2
. (22)
In Figure 11 on the right side the voltages across
the input, the capacitor C1, and the output are
depicted.
Fig. 11: Voltage across the first winding (left),
voltages across the input, the intermediate capacitor
and the output (right)
During mode M1 the voltage across the second
winding N12 of the first magnetic element is:
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1
1211
12
1,12 U
NN
N
uMN
, (23)
and the transformed voltage across N11 during M2
is:
121
11
12
2,12 CMN UUU
N
N
u
. (24)
The voltage across the second winding is drawn
in Figure 12 on the left side.
The third magnetic winding is the coil L. During
M2 the negative output voltage is across the coil and
during M1 the sum of the negative voltage across
N12 plus the voltage across C1 is there (Figure 12,
right).
Fig. 12: Voltage across the second winding (left),
voltage across the coil (right)
The maximum voltages across the
semiconductors (Figure 13) are
11
1211
21max, N
NN
UUUS
, (25)
2
1211
11
1max, U
NN
N
UUD
(26)
dNN
N
U
UD
1
1
1211
11
1
max,
(26a)
Fig. 13: Voltages across the electronic switch S
(left) and the diode (right)
3.2 Connection between the Currents
The connections between the currents are more
difficult to find. First one can write the charge
balance of the two capacitors (in the steady state the
mean current through a capacitor must be zero).
During mode M1 the output capacitor supplies the
load, and during mode M2 the diode transfers the
current through the coil L and the current through
the capacitor C1 to the output. The charge balance
for the output capacitor is therefore
dIIIdI LOAD
LMC
LOAD
1
_
2,1
_
(27)
where
2,1
_
MC
I
is the current through C1 during mode
M2 referred to the duration of mode M2.
2
_
L
I
represents the mean value of the current through the
coil L. The load current can be taken as constant
because the output capacitor is dimensioned so large
that the voltage across it stays nearly constant
during a period. When one looks at the first
capacitor one can see that the current through the
coil L must flow (discharging) through the capacitor
during M1, and during M2 a positive current flows
through it which must be equal to the one through
the first winding N1 (there is no current in the
winding N2).
2,11
_
2,1
_
MNMC II
. (28)
One can therefore write the charge balance
according to:
dIdI MCL 1
2,1
__
. (29)
One has to solve:
LOAD
L
MC I
I
I
dd
dd
0
1
1
11 2,1
_
(30)
leading to
d
d
I
I
LOAD
MC
1
2,1
_
, (31)
LOAD
LII
_
. (32)
We start to draw the signals with the load
current (arbitrarily 3.5 divisions, Figure 14, left)
The current through the capacitor C2 is the
negative load current during M1. The mean value of
the current during M2 must be half as large as the
load current (Figure 14, right).
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Fig. 14: Load current (left), current through the
output capacitor (right)
From (32) one knows that the mean value of the
current through L must be equal to the load current.
With an arbitrarily chosen current ripple one gets
Figure 15 on the left.
It is easy to draw the current through C1 (Figure
15, right). During M1 the current of the coil L must
flow in a negative direction through the first
capacitor. The mean value
2,1
_
2,1
_
CMC II
of the
current through C2 during M2 must be half as large.
Fig. 15: Current through the coil (left), current
through the intermediate capacitor (right)
In mode M2 no current flows through N12 and
the current through the first capacitor must flow
through N11. We can interpret this a magnetizing
current. The magnetizing current through N11
during M1 must be smaller because more windings
are available. The current turns (current linkage,
magnetomotive force) must be equal immediately
before
dT
and after the turn off
dT
of the
electronic switch
1111121111 )( NdTiNNdTiNN
(33)
Now one can draw the magnetizing current. For
a winding ratio of one to one we have half the
values of iN11, and M2 at the beginning and the end
of M2 to get the values of mode M1 (Figure 16).
Fig. 16: Magnetizing current
The current through the coil L flows to both
windings and compensates them (according to the
number of turns). The winding with a lower number
of turns must correspondingly take more current.
With
12,11,NLNLL iii
, (34)
12,1211,11 NLNL iNiN
(35)
the currents through the windings caused by the
current through L during mode M1 can now be
calculated according to
1211
12
11,
NN
N
i
i
L
NL
(36)
1211
11
12,
NN
N
i
i
L
NL
(37)
The current through N11 (Figure 17, left) during
M1 is the magnetizing current minus the current
caused by L and the magnetizing current during M2.
The current through N12 (Figure 17, right) is the
sum of the magnetizing current and the current
caused by L during M1 and zero during M2.
Fig. 17: Currents through windings: first (left),
second (right)
The currents through the semiconductor devices
can now be found. The current through the
electronic switch is equal to the current through the
second winding N12. It is the sum of the
magnetizing current plus the current caused by the
coil L according to (37) (Figure 18, left). The
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current through the diode is the current through the
first winding N11 and the current through the coil
(Figure 18, right).
Fig. 18: Currents through the electronic switch (left)
and the diode (right)
4 Simulations
For better understanding, a few simulations with the
help of LTSpice are shown. The simulations show
excellent conformity with the theoretical
considerations in sections 2 and 3.
4.1 Variant I
Figure 19 shows the current through C1, the current
through C2, the current through the coil L, the
current through the load, the current through the
first winding N11, the current through the second
winding N12 of the autotransformer, the input
voltage, the output voltage, and the control signal.
Fig. 19: Variant I, up to down: current through C1
(black); current through C2 (grey); current through
the coil L (dark violet), current through the load
(brown); current through N11 (red), and N12
(violet) of the autotransformer; input voltage (blue),
output voltage (green), the control signal (turquoise)
4.2 Variant II
Figure 20 shows the voltage across the diode D, the
voltage across the active switch S, the voltage
across N12, the voltage across N11, the input
voltage, the control signal, and the output voltage.
Fig. 20: Variant II voltages across the components,
up to down: voltage across the diode D (dark green);
voltage across the active switch S (grey); voltage
across N12 (violet); voltage across N11 (red); input
voltage (blue), control signal (turquoise), output
voltage (green)
In Figure 21 the current through the output
capacitor C2, the current through the intermediate
capacitor C1, the current through L1, the current
through the winding N12, and the current through
winding N11 are depicted.
Fig. 21: Variant II currents through the components,
up to down: current through the output capacitor C2;
current through the intermediate capacitor C1;
current through L1; current through the winding
N12 (violet); current through winding N11 (red)
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All simulations were done with 40 µH for the
windings N11, and N12; 47 µH for L, and 330 µF
for the capacitors.
5 Modelling
In this section, the modeling of the converter when a
tapped inductor is used is sketched. Ideal
components and continuous mode are proposed.
Variant II (Figure 10) is taken as an example. A
tapped inductor with the windings N11, and N12
replaces the first coil. The first state variable is now
the flux. The basic equation is the induction law:
dt
d
N
dt
d
u
. (38)
A voltage is induced when the flux ψ is
changing. When discrete coils are employed, one
can use the flux per winding Φ and the number of
turns N.
During mode M1, when the active switch S is
turned on, the input voltage is across both windings
1211
1NN
u
dt
d
. (39)
With the help of KVL (Kirchhoff’s voltage law)
the voltage across the inductor L can be calculated.
The voltage across the second winding is given by
the transformer law (the voltages are proportional to
the number of turns) according to:
L
u
NN
N
u
dt
du C
L1
1211
12
1
. (40)
The changes in the voltages across the
capacitors are produced by the currents flowing
through them:
1
1C
i
dt
du L
C
, (41)
2
2
2CR
u
dt
du C
C
. (42)
The state equations can be combined with the
matrix equation
1
1211
12
1211
2
1
2
1
2
1
0
0
1
1
000
00
1
0
0
1
00
0000
u
LNN
NNN
u
u
i
RC
C
L
u
u
i
dt
d
C
C
L
C
C
L
. (43)
For mode M2, when the active switch S is off
and the diode D is conducting, the sum of the input
voltage and the negative voltages across the
capacitors C1 and C2 is across the first winding
leading to
11
121 N
uuu
dt
dCC
. (44)
The change of the current through L2 can be
found with the help of KVL according to
L
u
dt
du C
L2
. (45)
To obtain the current through the capacitor the
connection between flux and current is used. The
flux is directly proportional to the current which
produces it
Li
(46)
or using the flux per winding Φ
LiN
. (47)
The value of the inductor depends on the AL-
value (this value can be found in the datasheet of the
magnetic core which is used to build the coil; it
depends on the air gap) and the square of the
number of turns
2
NAL L
. (48)
The current can now be written as
2
NA
N
i
L
. (49)
The third state equation is therefore
111
1CNAdt
du
L
C
. (50)
The current through the output capacitor C2 is
the sum of the current through the first winding and
the current through the coil minus the load current
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L
C
L
L
CANNC
R
u
i
NA
dt
du
212
2
11
2
(51)
leading to (52)
1
11
2
1
22112
111
1111
2
1
0
0
0
1
1
0
11
000
1
1
000
11
00
u
N
u
u
i
RCCANC
ANC
L
NN
u
u
i
dt
d
C
C
L
L
L
C
C
L
.
Combined one gets ((43) is weighted by the
duty cycle d and (52) weighted by 1-d and added)
the large signal model of the converter (53)
)53(
0
0
1
1
0
11
00
1
11
00
11
00
1
1211
12
111211
2
1
22112
1111
22
1111
2
1
u
LNN
dN N
d
NN
d
u
u
i
RCC
d
ANC
dC
d
ANC
dL
d
L
dN
d
N
d
u
u
i
dt
d
C
C
L
L
L
C
C
L
Linearization leads to the small signal model
(54). With the help of it, the transfer functions can
be calculated and so a simple controller can be
designed.
d
u
C
I
CNA
C
I
CNA
LNN
UN
L
UU
LNN
ND NNN
UN
N
U
N
U
N
D
NN
D
u
u
i
RCC
D
ANC
DC
D
ANC
DL
D
L
DN
D
N
D
u
u
i
dt
d
L
L
L
L
CC
CC
C
C
L
L
L
C
C
L
1
2
0
211
0
1
0
111
0
1211
10122010
1211
120
121111
1012
11
20
11
10
11
0
1211
0
2
1
22
0
112
0
1
0
111
0
00
11
0
11
0
2
1
00
0
1
1
0
11
00
1
11
00
11
00
(54)
6 Conclusion
The application of two coupled coils instead of a
coil in a DC/DC converter changes the behavior of
the system. The voltage transformation ratio is
changed. In the first variant of the SEPIC treated
here, the voltage transformation ratio is increased,
and in the second variant, it is decreased by the
winding ratio. For the designer, this means an
additional degree of freedom for adapting the output
voltage to the input voltage. In the first variant, the
magnetic component can be interpreted as an
autotransformer, and in the second as a tapped
inductor. The method to describe the converters
shown here can be applied to other converters with a
tapped inductor or an autotransformer replacing an
inductor of the original topology. The text can be
utilized for a lecture note using tapped inductors in a
course concerning DC/DC converters.
References:
[1] F. A. Himmelstoss, and P.H.Wurm, Simple
Bi-Directional DC-to-DC Converter with
High Input to Output Voltage Ratio, 17th
European Photovoltaic Solar Energy
Conference and Exhibition, Munich,
Germany, 22-26 Oct. 2001, pp. 592-595,
published 2002, (ISBN: 3-936338-08-6).
[2] V. R. Tintu and M. George, Tapped inductor
technology based DC-DC converter, 2011
International Conference on Signal
Processing, Communication, Computing and
Networking Technologies, Thuckalay, India,
2011, pp. 747-753, doi:
10.1109/ICSCCN.2011.6024650.
[3] J. Yao, J. Zhao and A. Abramovitz, Modeling
of the Tapped Inductor SEPIC converter by
the TIS-SFG approach, IECON 2015 - 41st
Annual Conference of the IEEE Industrial
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Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
- Felix A Himmelstoss did the basic research, the
simulations and wrote the text,
- Helmut L Votzi did the proof reading and draw
the pictures Figures 3-9 and 11-19, and gave the
presentation at the conference.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
No funding was received for conducting this study.
Conflict of Interest
The authors have no conflicts of interest to declare.
Creative Commons Attribution License 4.0
(Attribution 4.0 International, CC BY 4.0)
This article is published under the terms of the
Creative Commons Attribution License 4.0
https://creativecommons.org/licenses/by/4.0/deed.en
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WSEAS TRANSACTIONS on CIRCUITS and SYSTEMS
DOI: 10.37394/23201.2024.23.10
Felix A. Himmelstoss, Helmut L. Votzi
E-ISSN: 2224-266X
113
Volume 23, 2024