The integrated buck and asymmetrical half-bridge (IBAHB) converter is shown in
Figure 2. There are two elements,
and
, for reducing the energy of the leakage inductance, which enables the suppression of the peak voltage on the power switches, thus allowing power switches with lower
to be utilized; consequently, the reduced voltage stress improves its efficiency. A buck-type circuit is added to the primary side, which makes the voltage on the transformer unequal to
Therefore, the turns ratio of the transformer can be reduced to increase the coupling rate, which decreases the leakage inductance. The proposed topology uses four signals that are created by a pair of push–pull signals and a pair of complementary signals to control the power switches. The half-bridge ones are used for the main switches, and the complementary ones are used for synchronous rectification. The secondary side employs a dual-winding center-tapped rectifier circuit to double the frequency of the output inductor current, which can reduce the output current ripple. Therefore, the output inductor and the output capacitor can both be designed with a smaller volume.
2.1. Operating Principles
The equivalent circuit of the IBAHB is shown in
Figure 3. The
and
represent the energy-storing inductor and the output inductor, respectively. Switches
,
, and
are the main switches of this topology, and switches
and
are the synchronous rectifiers.
Diode
is a flywheel diode,
and
are the switching capacitors, and
is the filter capacitor. The primary side of the transformer is defined as
, and the secondary side is defined as
and
. The turns ratio is defined as
(
=
=
). The transient-state waveforms in CCM are shown in
Figure 4. There are eight transient states, which are depicted as follows.
Mode 1 [t0~t1]
In
Figure 5a, in this interval, the main switches
and
and the synchronous rectifier
are in the turned-on state. Voltage source
transfers the energy to the inductor
and capacitor
through the main switch
. At the same time, capacitor
discharges the energy stored by the previous cycle to
also through
. The transformer starts to send energy from the primary side to the secondary side. The current of the output inductor
maintains its freewheeling state while passing through
and the body diode of
. However, the current passing through
decreases, and the current of
increases. The inductor currents
,
,
and
are given, respectively, by
Mode 2 [t1~t2]
In
Figure 5b, in this interval, the main switches
and
and one of the synchronous rectifier switches,
, remain in the turned-on state. The voltage source
continues transferring the energy to the energy-storage inductor
and capacitor
, and the capacitor
is still discharging to
. The transformer continues transferring the energy to the secondary side, and output inductor
and output capacitor
are storing the energy provided by the transformer. The inductor currents
,
,
and
are given, respectively, by
Mode 3 [t2~t3]
In
Figure 5c, in this interval, the main switches
,
, and
are in the turned-off state, while the switches of the synchronous rectifier,
and
, are in the turned-on state. The parasitic body diode of switch
turns ON due to the freewheeling characteristic of leakage inductance. Additionally, the energy of leakage inductance can be retrieved by capacitor
. Inductor
releases energy through
to capacitor
by its freewheeling state; moreover, the output inductor
starts releasing the energy passing through
and
to provide the load
. The inductor currents
,
,
and
are given, respectively, by
Mode 4 [t3~t4]
In
Figure 5d, in this interval, the main switches
,
, and
remain in the turned-off state. Inductor
continues releasing energy to
through
. Output inductor
also keeps releasing energy to
, through
and
. The inductor currents
,
,
and
are given, respectively, by
Mode 5 [t4~t5]
In
Figure 5e, in this interval, the main switch
and one of the synchronous rectifiers
are in the turned-on state; the other switches are turned off. The inductor
keeps releasing energy to the capacitor
. When
turns on, the capacitor
starts to release the stored energy, which can transfer to
on the secondary side. Because the
releases energy, the current passing through
can decrease to zero, and the current passing through
can increase. The inductor currents
,
,
and
are given, respectively, by
Mode 6 [t5~t6]
In
Figure 5f, in this interval, switch
and the switch of the synchronous rectifier
remain in the turned-on state; the other switches are turned off. This mode is similar to mode 5. However, the current is no longer passing through switch
SR1. The inductor currents
,
,
and
are given, respectively, by
Mode 7 [t6~t7]
In
Figure 5g, in this interval, the main switches
,
, and
are in the turned-off state, while the switches of the synchronous rectifier
and
are in the turned-on state. The switching capacitor
can recover the leakage inductor energy. In this interval, the load energy is absorbed by the output inductor. The inductor currents
,
,
and
are given, respectively, by
Mode 8 [t7~t8]
In
Figure 5h, in this interval, the main switches
,
, and
are in the turned-off state and the switches of the synchronous rectifier
and
are in the turned-on state. Same as Mode 4, inductors
and
release the energy to the switching capacitor
and the load, respectively. The inductor currents
,
,
and
are given, respectively, by
2.2. Steady-State Analysis
In order to simplify the analysis, the proposed architecture is presumed to operate in continuous conduction mode (CCM), the method of control is shown in
Figure 6. The characteristics of the transient state over the circuit will be ignored, and the currents passing through all of the components will be considered in DC. In addition, there are some assumptions listed as follows:
- (1)
All components possess ideal characteristics.
- (2)
The coupling coefficient of the transformer is unity.
- (3)
The duty cycle is low, under 50%.
- (4)
The subscript “pft” denotes the average current in the corresponding mode.
Mode 1 [0, DT]
The main switches
S1 and
S2 are in the turned-on state. The input voltage source transfers energy to the inductor
L1 and capacitor
CpT. At the same time, the capacitor
C1 also provides energy to
CpT through
S2. On the secondary side, the output inductor
L2 and the output capacitor
Co are charging from the transformer and the
Co supplies energy to load
RL. The equivalent circuit is shown in
Figure 7 and the formulas can be expresses as:
Mode 2 [DT, 0.5T]
The main switches
S1,
S2 and
S3 are in the turned-off state. Inductor
L1 changes into the freewheeling state to release energy to the capacitor
C1. Output inductor
Lo also turns into the freewheeling state to release energy to the capacitor
Co and load
RL. The equivalent circuit is shown in
Figure 8 and the formulas can be expressed as:
Mode 3 [0.5T, (0.5+D)T]
The main switch
S3 is in the turned-on state. Capacitor
CpT starts to release energy through the transformer to the secondary side. Inductor
L1 still keeps releasing energy to the capacitor
C1. The output inductor
L2 and the output capacitor
Co are charging using the transformer, and the
Co supplies energy to load
RL. The equivalent circuit is shown in
Figure 9 and the formulas can be expressed as:
Mode 4 [(0.5+D)T, T]
Mode 4 is similar to Mode 2. The main switches
S1,
S2 and
S3 are in the turned-off state. Inductor
L1 still keeps the freewheeling state to release energy to the capacitor
C1. Output inductor
Lo turns into the freewheeling state to release energy to the capacitor
Co and load
RL. The equivalent circuit is shown in
Figure 10 and the formulas can be expressed as:
2.4. Voltage Stress
The voltage stresses of semiconductor devices can be derived by the known voltage of all the capacitors. Therefore, when the main switches
S1 and
S2 are in the turned-on state, the voltage stress of switch
S3 will be equal to the voltage of the capacitor
C1, as shown in
Figure 7.
The voltage stress of the freewheeling diode
Dfw can be given by
Additionally, the voltage stress of the synchronous rectifier switch
SR2 will be equal to the sum of
VNs1 and
VNs2, which is also equal to
VC1 minus
VCpT and then multiplied two times by the turns ratio 2n.
when the switch
S3 is turned on, switches
S1 and
S2 are turned off. As shown in
Figure 9, the voltage stress of the
S1 is equal to the sum of the input voltage
Vi and the capacitor
C1. The voltage stress of
S2 is equal to the capacitor
C1 and the voltage stress can be given by
Because the secondary winding structure is symmetrical, the voltage stress of the synchronous rectifier
SR1 is the same as
SR2. The voltage stress can be given by
when the input voltage and the turns ratio n equal 380V and
, respectively, the relationship between the voltage stress and the duty cycle is shown in
Figure 12. It can be seen from Equations (53) and (57) that
S2 and
S3 have relatively lower voltage stress. Hence, low-voltage-stress power devices, such as MOSFETs with low R
DS(on), can be employed. Similar to switches
S2 and
S3,
SR1 and
SR2 can also adapt low-voltage-stress power devices with low R
DS(on) to reduce the loss of semiconductors to improve efficiency.
2.5. Current Stresses
Due to the law of conservation of energy, the output current
IRo is equal to the input current
Ii divided by the voltage gain, which can be expressed as
The main switches
S1 and
S2 are in the turned-on state and
S3 is in the turned-off state, which is shown in
Figure 7. The current through
S1 and
S2 can be expressed as
The main switches
S1,
S2 and
S3 are in the turned-off state, as shown in
Figure 8. During this period, the current stresses through other semiconductor devices can be expressed as
As shown in
Figure 9, the switch
S3 turns into the ON state, while the main switches
S1 and
S2 remain turned off. The current through the semiconductor devices can be expressed as
when the main switches
S1,
S2 and
S3 are in the turned-off state again, the current flowing through the semiconductor devices would be the same as the previous operation mode.
The highest current flowing through each semiconductor element is the criterion for selecting current stress. The current stress of all the semiconductor devices would be reorganized and expressed below
When the output current and the turns ratio n equal 40A and
, respectively, the relationship between the average current stress and the duty cycle is shown in
Figure 13.
2.6. Conduction Loss Analysis
The equivalent circuit for analyzing the conduction loss of inductors and semiconductor devices is shown in
Figure 14, in which
and
are the copper resistance of the inductors,
and
are the on-resistance and the forward voltage of the diode, respectively,
,
,
,
and
are the on-resistances of the switches.
Mode 1 [0, DT]
The main switches S1 and S2 are in turned-on state. The voltage source Vi transfers energy to the inductors L1 and the capacitor CpT receives energy from Vi and the capacitor C1.
Simultaneously, the output inductor
L2 and capacitor
Co are charging through the transformer, and then
Co supplies energy to the load
RL. The equivalent circuit is shown in
Figure 15 and the formulas are expressed as follows:
Mode 2 [DT, 0.5T]
The main switches
S1,
S2 and
S3 are in the turned-off state. In this interval, the inductor
L1 releases energy to the capacitor
C1, and the output inductor
L2 changes into the freewheeling state, releasing energy to the output load
RL. The equivalent circuit is shown in
Figure 16 and the formulas are expressed as follows:
Mode 3 [0.5T, (0.5+D) T]
In this interval, the main switch
S3 is in the turned-on state. The inductor
L1 keeps releasing energy to the capacitor
C1, and the capacitor
CpT sends the energy to the output inductor
L2 and output capacitor
Co through the transformer. Then,
Co provides energy to the output load
RL. The equivalent circuit is shown in
Figure 17 and the formulas are expressed as follows:
Mode 4 [(0.5+D) T, T]
Similar to the Mode 2 state, the main switches
S1,
S2 and
S3 are all in the turned-off state. The inductor
L1 keeps releasing energy to the capacitor
C1 and the output inductor
L2 changes into the freewheeling state, releasing energy to the output load
RL. The equivalent circuit is shown in
Figure 18 and the formulas are expressed as follows:
First, through volt-second balance, the equation of the inductor
L1 can be expressed as
By substituting Equations (80), (83), (85) and (88) into Equation (90), the voltage of the capacitor
C1 which is in the non-ideal state can be derived as
Second, the proposed circuit is an asymmetric topology, so the volt-second balance equation of the output inductor
L2 should be divided into two parts of derivations:
Substitute Equations (81) and (84) into (92) and simplify it. Then, replace
with
, which is shown below.
Substitute Equations (86) and (89) into (93) and simplify it. Then, replace
with
, which is shown below.
It is known that the summation and subtraction of Equations (94) and (95) are equal to zero, which are expressed as follows, respectively:
Substitute
of Equation (97) with Equation (91) and simplify it. Then, into Equation (95), as follows:
Then, transfer Equation (98) into an equivalent circuit module, as shown in
Figure 19.
Divide Equation (98) by
Vi to obtain the voltage gain of the non-ideal state, which is expressed as
Equation (98) multiplied by
is the efficiency equation of the non-ideal state, which is expressed as follows:
where
After calculating, the non-ideal voltage gain and the efficiency curve are shown below, as
Figure 20 and
Figure 21.