DOUBLE-COUPLED CURRENT-FED PUSH-PULL DC/DC CONVERTER:
ANALYSIS AND EXPERIMENTATION
Hugo R. E. Larico and Ivo Barbi IEEE, Senior Member
Power Electronics Institute – INEP
Federal University of Santa Catarina - UFSC
P. O. box 5119 – 88040-970 – Florianópolis – SC- Brazil
estofanero@inep.ufsc.br – ivobarbi@inep.ufsc.br
Abstract – This paper presents the double-coupled
current-fed push-pull dc/dc converter which operates
from 0 to 100% of duty cycle. One primary winding in
the transformer, low rms current in the output capacitor,
double switching frequency in the input and the output
filter, null ripple current in the input and in the output
for 50% of duty cycle, are some characteristics of this
new topology. These allow reductions of size and weight
of the components.
Qualitative and quantitative analysis are presented, in
which the simplified model of the inductor coupled and
the ideal model of the transformer are used. Finally,
experimental results of a prototype are shown for the
converter operating with a duty cycle higher or lower
than 50%.
The possible applications of this topology are
automotive, electric power generation such as fuel cells,
solar and wind.
Keywords – Buck-boost, dc/dc isolated, flyback and
push-pull converters.
I. INTRODUCTION
A comparison between conventional PWM converters:
push-pull, full-bridge and half-bridge topologies are shown
in [1], it reveals that the current-fed two-inductor boost
converter, proposed in [6], has important advantages over the
other topologies: conduction switching losses are the lowest
with the best utilization of the transformer, low input current
ripple and the transistors are referred to the ground. These
characteristics make this converter suited to low inputvoltage and high-current input applications. When the goal is
to process high-power and, consequently, the current stress in
the switching devices becomes a problem, the phase-shifted
parallel-input/series-output dual converter, proposed in [4], is
a possible solution. These topologies have in common the
output bridge rectifier; in applications when the output
voltage is higher there is no problem. However, if the output
voltage is lower, then the conduction and commutation losses
of the rectifier diodes are considerable.
Therefore, the dual inductor push-pull converter, which
has the same characteristics as the current-fed two-inductor
boost converter, was proposed in [2] and shown in the Fig. 1.
The difference between both is that, the first one has a halfbridge rectifier and the other one has a full-bridge rectifier in
the output. Consequently, two windings in the secondary side
of transformer are necessary. The high voltage stress on the
switching devices in the current-fed converters associated to
the high-current increase the switching losses. Some
solutions such as ZVS and ZCS are proposed in [3] and [5].
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They allow obtaining a higher efficiency of the converter.
These techniques improve the performance of the converter,
but they do not change the band of duty cycle operation
which is for more than 50%.
Fig. 1. Dual inductor push-pull converter.
A similar problem is presented by the current-fed pushpull converter [2], where, the switches can not be turned-off
simultaneously. A solution was proposed in [7] and is shown
in the Fig. 2. This technique consists of employing a coupled
inductor. Through the secondary winding of the inductor is
possible to transfer the energy-stored in this one to the input
or to the output [8], [9] and [10]. For multiple-output
applications, the transference to the input is an advantage; it
produces better tracking among the outputs of the converter.
If, however, it is desired to achieve non-pulsating output
current in single-output applications, the transference to the
output is preferable (Fig. 2). Therefore, the converter can
operate from 0 to 100% of duty cycle. If the duty cycle is less
than 50%, the diode D3 conducts when both switches are
turned-off. In the case, the duty cycle is 50% or more, the
diode D3 does not conduct all the time.
Fig. 2. Weinberg current-fed push-pull converter.
Soon after, this study proposes a new topology
denominated here double-coupled current-fed push-pull
305
dc/dc converter which employs two coupled inductors to
expand the duty cycle operation of the dual inductor pushpull converter. The paper will show that the characteristics of
this converter are suitable for low-voltages and high-current
applications as in fuel cells, solar and wind power
generation.
The validation of theoretical studies is made through the
experimental results obtained with a laboratory prototype.
II. THE PROPOSED TOPOLOGY
The proposed double-coupled current-fed push-pull dc/dc
converter is shown in Fig. 3. It comprises two coupled
inductors (Lf1, Lf2), one transformer (T), one capacitor (Co),
two switches (S1, S2) and four diodes (D1, D2, D3, D4).
- The coupled inductors are represented by the simplified
model, where only the magnetizing inductance was
considered;
- The coupled inductors are identical (Lf1m=Lf2m=Lfm);
- The transformer and semiconductors are ideal;
- The output capacitor is large enough so that the output
voltage is constant.
Due to magnetizing current and the duty cycle (D=ton /Ts),
four modes are presented:
- Continuous conduction mode for D < 0.5 ;
- Discontinuous conduction mode for D < 0.5 ;
- Continuous conduction mode for D > 0.5 ;
- Discontinuous conduction mode for D > 0.5 ;
According to the presented, if the duty cycle is 50% or
more all characteristics are preserved, therefore, this paper
only presents the operating principle for duty cycle less than
50%.
A. Continuous conduction mode for D < 0.50
For duty cycles less than 50%, the secondary windings of
the inductors transfer energy to the output when both
switches are turned-off. The topological operating stages are
shown in the Fig. 4 and the main waveforms in Fig. 5. The
inductances Lf1m and Lf2m represent the magnetizing
inductances of the coupled inductors. In this mode the
magnetizing current is continuous. Each stage is described
below.
Fig. 3. Double-coupled current-fed push-pull dc/dc converter.
This converter can operate from 0 to 100% of duty cycle.
If, the duty cycle is less than 50%, the secondary winding of
inductor Lf1 and Lf2 transfer energy to the load when both
switches are turned-off. In the other case, in which the duty
cycle is 50% or more, the diodes D3 and D4 must not conduct
any time, consequently, the relationship between the turns
ratio of the transformer (nT = Np/Ns) and the inductors (nL =
N1/N2) are selected to guarantee it. The most important
characteristics in this topology are listed:
- All characteristics presented in the dual inductor pushpull converter are preserved [2]. This occurs from 50% to
100% of duty cycle;
- Operation frequency of the filters is twice of the
switching frequency, which results in a reduction of the input
and output filter size;
- Non-pulsating input and output current for operation
with duty cycle equal to 50%, however, these currents
contain low ripple current at high frequency;
- Non-pulsating output current for relationship “ nT = 2nL ”,
however, the duty cycle is limited between 0 and 50%;
- Low peak current in the discontinuous conduction mode
for duty cycle less than 50%.
III. CIRCUIT OPERATION
The operating principle of the converter is made
considering the following:
- The converter is operating in steady state;
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1) 1st Stage –Switch S1 is turned-on in t0, diode D2 is on
and diodes D1, D3 and D4 are off. Inductor Lf1 is storing
energy and inductor Lf2 is transferring or storing energy
depending on the voltage over it. The variation rates of
magnetizing currents are given by (1) and (2). Both currents
of the inductors flow through to switch S1 and this is equal to
the input current (3). The output current is equal to the
magnetizing current of Lf2, referred to the secondary side of
transformer (4). The voltages over the semiconductors are
given by (5)-(8).
diLf 1m
E
(1)
= i
dt
L fm
diLf 2 m
dt
=
Ei − nT Vo
L fm
(2)
ii = iS 1 = iLf 1 p + iLf 2 p
(3)
io = iD 2
(4)
VS 2 = nT Vo
(5)
VD1 = −2Vo
(6)
⎛E
⎞
VD 3 =`− ⎜ i + Vo ⎟
⎝ nL
⎠
(7)
⎛E ⎛ n ⎞ ⎞
VD 4 = − ⎜⎜ i + ⎜ 1 − T ⎟ Vo ⎟⎟
(8)
⎝ nL ⎝ nL ⎠ ⎠
The condition to guarantee the reverse biased of diode D4
is obtained from (8) as it is shown in (9).
V
(9)
1 + ( nL − nT ) o > 0
Ei
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2nd Stage – The switches are turned-on in t1, diodes D3
and D4 are on, and D1 and D2 are off. The two inductors
transfer energy to the load, the variation rates of the
magnetizing currents are equal in both inductors (10) and
(11). The input current is null (12) and the output current is
equal to the sum of the magnetizing currents, referred to
secondary side (13). The voltages over the semiconductors
are given in (14)-(15).
diLf 1m
nV
(10)
=− L o
dt
L fm
diLf 2 m
dt
=−
nLVo
L fm
ii = iS 1 = iS 2 = 0
io = iD 3 + iD 4
VS1 = VS 2 = Ei + nLVo
VD1 = VD 2 =`−Vo
(11)
(12)
(13)
(14)
(15)
2) 3rd Stage – Switch S2 is turned-on in t2, diode D1 is on
and diodes D2, D3 and D4 are off. Inductor Lf2 is storing
energy and inductor Lf1 is transferring or storing energy,
depending on the voltage over it, the variation rates of the
magnetizing currents are given by (16) and (17). Both
inductor currents flow through switch S2 and it is equal to the
input current (18). The output current is equal to the
magnetizing current of Lf2 referred to secondary side of the
transformer (19).
diLf 1m Ei − nT Vo
(16)
=
dt
L fm
diLf 2 m
dt
=
Ei
L fm
(17)
ii = iS 2 = iLf 1 p + iLf 2 p
(18)
io = iD1
(19)
Fig. 4. Topological stages in CCM for D<0.5.
Fig. 5. Main waveforms in CCM for D<0.5.
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307
3) 4th Stage – This stage is similar to the 1st one, therefore,
all equations are the same. The fourth stage finishes at t0+Ts
where the cycle begins again.
Im2 =
B. Discontinuous conduction mode for D < 0.50
In this mode, the magnetizing current of the inductors is
discontinuous. Because of that each inductor has four stages
(two energy storage and two energy transference stages or
one energy storage and three energy transference stages).
There are four cases of discontinuous mode, these cases are
described as follows:
Ei
ton
L fm
(23)
Ei − nT Vo
ton
L fm
(24)
I m1 =
1) 1st Case – The main waveforms of this case are shown
in Fig. 6. It shows that the converter has six stages. The
magnetizing current is null in the last energy transference
stage. The input current is discontinuous and the output
current is still continuous. The current through the
magnetizing inductance of inductor Lf1 in the last stages is
given by (20), (21) and (22).
E
(20)
I m1 = i ton
L fm
Im2 =
I m3
Ei + nLVo
nV T
ton − L o s
L fm
L fm 2
2E
nV T
V
= i ton − L o s + ( nL − nT ) o ton
L fm
L fm 2
L fm
(21)
Fig. 7. Main waveforms in the DCM for D<0.5: 2nd Case.
(22)
3) 3rd Case – The main waveforms of this case are shown
in Fig. 8. It shows that the converter has six stages. The
magnetizing current is null in the second energy transference
stage. The figure shows that the fourth stage is a transference
stage, since the voltage in the primary winding of the
transformer is higher than the input voltage. The input
current and the output current are discontinuous. The peak
current through the magnetizing inductance of the inductor
Lf1 is given by (20) and (21).
Fig. 6. Main waveforms in the DCM for D<0.5: 1st Case.
2) 2nd Case – The main waveforms of this case are shown
in Fig. 7. It shows that the converter has eight stages. The
magnetizing current is null in the first and in the last energy
transference stages. In this case, the voltage in the five stages
is positive, therefore, the voltage in the primary winding of
the transformer is less than the input voltage. The input
current and output current are discontinuous. The current
through the magnetizing inductance of the inductor Lf1 in the
last stages is given by (23) and (24).
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Fig. 8. Main waveforms in the DCM for D<0.5: 3rd Case.
4) 4th Case – The main waveforms of this case are shown
in Fig. 9. It shows that the converter has six stages. The
magnetizing current is null in the first energy transference
308
stage. In this case, only one inductor is transferring or storing
energy in each stage. The input current and the output current
are discontinuous. The peak current through the magnetizing
inductance of the inductor Lf1 is given by (20).
possible. For duty cycle lower than 50% the converter has a
buck-boost behavior, otherwise, it has a boost characteristic.
⎧2 D
⎪ n 1 − D , if, D < 0.5
⎪ T
(29)
a=⎨
⎪ 1 1 , if, D ≥ 0.5
⎪⎩ nT 1 − D
It is interesting to show that the situation where the
relationship is “nT = 2nL“. In this case, the operation can only
happen for duty cycle lower than 50% (30). Because of this
relationship, diodes D3 and D4 conduct for 50% or more of
duty cycle, which, according to (30), is prohibited. For the
first case of duty cycle range, the converter has buck
characteristic and the output current is non-pulsating.
⎧4
⎪ D, if, D < 0.5
(30)
a = ⎨ nT
⎪ prohibited , if, D ≥ 0.5
⎩
Fig. 9. Main waveforms in the DCM for D<0.5: 4th Case.
IV. MATHEMATICAL ANALYSIS
A. Static gain in CCM
In steady-state, the average voltage in each inductor in
CCM must be zero (25).
1 t0 +Ts
VLf 1med = ∫
vLf 1dt = 0
(25)
Ts t0
Replacing (1), (10), (16) and the intervals of time of Fig.
5 in (25), leads to (26).
1⎛
⎛ Ts
⎞⎞
(26)
⎜ ( 2 Ei − nT Vo ) ton − 2nLVo ⎜ − ton ⎟ ⎟ = 0
Ts ⎝
⎝2
⎠⎠
Therefore, the static gain in CCM is given by (27), where,
the static gain to 50% or more of duty cycle is transcribed
from [2].
D
⎧2
, if, D < 0.5
⎪n
⎛
nT ⎞
L
⎪
V
⎪ 1− ⎜ 2 − ⎟ D
nL ⎠
a= o =⎨
(27)
⎝
Ei ⎪
1 1
⎪
, if, D ≥ 0.5
⎪⎩ nT 1 − D
Replacing the static gain (27) in (9) one can obtain the
relationship between the turns ratio of the transformer and
the inductors (28). It shows that the relationship is the
function of the duty cycle, thus, for the operation in the
whole range of the duty cycle the turns ratio of the inductors
must be higher than the turns ratio of the transformer
(nL ≥ nT).
nL ≥ D ⋅ nT
(28)
Using the relationship “nL = nT“, the static gain is given by
(29). In this case, the operation in all range of duty cycle is
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B. Static gain in DCM
In the DCM the static gain is the function of the duty
cycle, the turns rate of the transformer and coupled inductors,
the magnetizing inductance, the switching frequency and the
output current. The static gain can be obtained by average
input current. In this section, it is obtained for duty cycle
lower than 50%, where four cases exist:
1) 1th Case – The expression (31), which evaluates the
average input current, can be obtained from Fig. 6.
1
I i = ( I m1 + I m 2 + I m3 ) ⋅ Δt10
(31)
Ts
Ideally, the relationship between the input current and the
output current is given by (32)
V
Ii = I o o
(32)
Ei
Replacing (20)-(22), (32) and the information shown in the
Fig. 6, result in (33), where, “γ“ is the parametric current
(34).
4D 2
(33)
a=
γ + ( nT − 2nL ) D 2 + nL D
γ =
I o L fm
Ts Ei
(34)
2) 2nd Case – In this case, the expression (35) is obtained
from Fig. 7.
1
I i = ( I m1 + I m 2 ) ⋅ Δt10
(35)
Ts
Replacing (23), (24), (34) and Fig. 7, gives (36).
2D 2
(36)
a=
γ + nT D 2
3) 3rd Case – In this case, the expression (37) is obtained
from Fig. 8.
1
I i = ( I m1 ⋅ Δt20 + I m 2 ⋅ Δt43 )
(37)
Ts
309
Replacing (20), (21), (34) and Fig. 8, leads to (38).
4γ + 4 D ( ( nT + 2nL ) ⋅ D − nL )
a=
2
4γ nT − nL 2 (1 − 2 D )
(38)
4) 4th Case – Equation (39) shows the average current (see
Fig. 9).
1
I i = I m1 ⋅ Δt10
(39)
Ts
Replacing (20), (34) and the included information in
Fig. 9, gives (40).
D
a=
(40)
γ
capacitor is shown in Fig. 12. From the figure, the equations
for the ripple voltage (41) and the rms current (42) are
obtained; it is interesting to note that these parameters are
functions of the turns ratio of the transformer and the
inductors for duty cycle lower than 50%, though, they are not
functions otherwise.
ΔVo =
I Coef =
I Coef
Io
2nL − nT
⎧
(1 − 2 D ) D, if, D < 0.5
2ΔVo C ⎪
= ⎨ nL − ( 2nL − nT ) D
I oTs
⎪2 D − 1, D, if, D ≥ 0.5
⎩
⎧ 4n 2 + n ( n − 4n ) − 2 D ( n − 2 n ) 2
L
T
T
L
T
L
⎪
⎪
nL − ( nT − 2nL ) D
=⎨
⎪ 2D − 1
⎪ 2 1 − D , D ≥ 0.5
)
⎩ (
(41)
D
, D < 0.5
2
(42)
C. Output characteristics
With the equations (27), (34), (36), (38) and (40) one can
plot the output characteristics of the double-coupled currentfed push-pull dc/dc converter. Fig. 10 shows the graph for
the relationship “nT = nL = 1“, where, the four cases in DCM
are delimited. The graph for the relationship “nT = 2nL = 1” is
shown in Fig. 11.
Fig. 10. Output characteristics for relationship “nT = nL = 1“.
Fig. 12. Waveform of the current through filter capacitor Co.
The rms current through the capacitor for relationship
“nT = nL”, is plotted in Fig. 13, the result shows the existence
of one minimum point at 50% of the duty cycle, where,
ideally, the filter capacitor is not necessary. Replacing the
relationship “nT = 2nL” in (42), the result is zero rms current
to duty cycle lower than 50%, but, it should be remembered,
that operation with duty cycle 50% or higher is not admitted.
Fig. 11. Output characteristics for relationship “nT = 2nL = 1“.
D. Output filter capacitor
Parameters such as the ripple voltage and the rms current
of the filter capacitor are important to the design. However, if
it is considered, all details in the design process can be very
complex. On the other hand, assuming that the magnetizing
inductances of the coupled inductors were large enough that
its currents could be considered constant, the design process
becomes simple. The waveform of the current in the filter
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Fig. 13. Current through the capacitor to relationship “nT = nL“.
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E. Voltage stress on the semiconductors
The voltage stress on the switches and the diodes are
given by (43)-(45).
⎧ ⎛
nT ⎞
⎪ ⎜1+ n D ⎟
L
⎪ Ei ⎜
⎟ , if, D < 0.5
⎪ ⎜ nT ⎟
VS1mx = VS 2 mx = ⎨ ⎜ 1 − D ⎟
(43)
nL ⎠
⎪ ⎝
⎪
1
, if, D > 0.5
⎪ Ei
⎩ 1− D
VD1mx = VD 2 mx = −2Vo
(44)
VD 3 mx = VD 4 mx
⎧
⎪ E
⎪− i
⎪⎪ nL
=⎨
⎪
⎪ Ei
⎪−
⎪⎩ nL
nT
D
nL
, if, D < 0.5
⎛
nT ⎞
1− ⎜ 2 − ⎟ D
(45)
nL ⎠
⎝
1+
⎛ nL 1 ⎞
⎜1 +
⎟ , if, D > 0.5
⎝ nT 1 − D ⎠
V. EXPERIMENTAL RESULTS
The project parameters applied in the design of prototype
are listed in Table I.
TABLE I
Project parameters
Parameter
Input voltage (Ei)
Output voltage (Vo)
Output power (Po)
Switching frequency (fs)
Turns ratio ( nL = nT )
Fig. 15. Channel 1: input current – 5A/div., Channel 2: output
current – 2A/div, Channel 3: output voltage – 50V/div., time scale
10μs/div. Duty cycle D=0.55.
The current and voltage in one switch are shown in Fig.
16 and Fig. 17. For 36% of duty cycle the switch conducts all
current of the input. The voltage, when both switches are
turned-off, is 96V and when the other switch is turned-on, it
is 48V. In case the duty cycle is 55%, when both switches
conduct the current, a half part of the input current flows
through each switch; the voltage over the switch is 48V.
Value
48-24V
48V
140W
30kHz
1
The experimental duty cycle for 48V and 24V of voltage
input was 36% and 55%, respectively. The efficiency
obtained in the prototype was 88%.
Fig. 14 shows the input current and output current for 36%
of duty cycle; in these waveforms can be seen that the input
current is zero when both switches are turned-on and the
output current is the sum of the magnetizing currents of the
inductors. The result for 55% is shown in Fig. 15; here only
the transformer transfers energy to the load, this
demonstrates that the characteristics of the original converter
ware preserved.
Fig. 16. Channel 1: switch current – 2A/div., Channel 2: switch
voltage – 50A/div, time scale 10μs/div. Duty cycle D=0.36.
Fig. 17. Channel 1: switch current – 2A/div., Channel 2: switch
voltage – 50A/div, time scale 10μs/div. Duty cycle D=0.55.
Fig. 14. Channel 1: input current – 5A/div., Channel 2: output
current – 2A/div, Channel 3: output voltage – 50V/div., time scale
10μs/div. Duty cycle D=0.36.
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In Fig. 18 and Fig. 19 are shown the magnetizing currents
through the inductors, this is done through the sum of
primary winding current plus the secondary winding current
of each inductor. For 36% of duty cycle, it is possible to see
that the current presents four stages, a storage one, two for
transference and one neutral stage; in this last one the voltage
in the inductor is zero, because of the primary voltage in the
311
transformer is equal to the input voltage. In the other case,
for 55% of duty cycle, two stages are presented in the
current, one for storage and one for transference. The figures
show the magnetizing currents are almost equal, which
proofs the distribution of the input current.
operation), but, the converter can only operate with duty
cycle lower than 50%. Some possible applications of this
topology are solar and wind power generation, fuel cells, and
aircraft.
ZVS and ZCS techniques could also be applied to improve
the efficiency of the converter. The association of various
converters in order to increase the power is also possible.
REFERENCES
Fig. 18. Channel 1: magnetizing current in Lf1 – 2A/div., Channel
2: magnetizing current in Lf2 – 2A/div, Channel 3: primary voltage
on Lf1 – 50V/div., time scale 10μs/div. Duty cycle D=0.36.
Fig. 19. Channel 1: magnetizing current in Lf1 – 2A/div., Channel
2: magnetizing current in Lf2 – 2A/div, Channel 3: primary voltage
on Lf1 – 50V/div., time scale 10μs/div. Duty cycle D=0.55.
VI. CONCLUSION
The qualitative and quantitative analysis of the doublecoupled current-fed push-pull dc/dc converter was presented
in this paper, subsequently the theoretical study is verified
through the experimental results for duty cycles lower or
higher than 50%.
The study has demonstrated that through the use of
coupled inductor, the duty cycle operation can be extended
from 0 to 100%. The new converter preserved all
characteristics of the original one and presented some
improvements, such as low rms current in the input and in
the output, low size of the filter capacitor, low voltage stress
on the semiconductors. The operation with the relationship
“nT =2nL” gives non-pulsating output current (buck-like
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