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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
Application ReportSLUA838–August 2017
Designing a Simple and Low-Cost Flybuck Solution Withthe TPS54308
RyanHu
ABSTRACTIn many applications, simple, low part-count, isolated power supplies working from an input voltage areneeded. A popular solution to these requirements is an isolated buck power supply. This application reportpresents a simple and low-cost flybuck solution using the TPS54308 device. The basic operatingprinciples of a flybuck converter are discussed, operating current and voltage waveforms are shown, andkey design equations are derived. Lastly, a step-by-step design example is presented.
Contents1 Introduction ................................................................................................................... 22 Flybuck Converter Device Overview ...................................................................................... 23 Design Flybuck With TPS54308 ........................................................................................... 64 Experimental Results ...................................................................................................... 105 Conclusion .................................................................................................................. 146 References .................................................................................................................. 14
List of Figures
1 Isolated Buck Converter With Two Outputs .............................................................................. 22 Simplified Isolated Buck Operating Waveforms ......................................................................... 33 Current Waveforms Affected by Leakage Inductance .................................................................. 44 TPS54308 Two Output Isolated Buck Solution .......................................................................... 65 Efficiency vs IOUT1 (IOUT1 = 0 A, IOUT2 = 0 A) ............................................................................... 106 Efficiency vs IOUT2 (IOUT1 = 0 A, IOUT3 = 0 A) ............................................................................... 107 Efficiency vs IOUT3 (IOUT1 = 0 A, IOUT2 = 0 A) ............................................................................... 108 VOUT2 Regulation vs IOUT2 (IOUT1 = 0 A, IOUT3 = 0 A) ....................................................................... 119 VOUT3 Regulation vs IOUT3 (IOUT1 = 0 A, IOUT2 = 0 A) ....................................................................... 1110 VOUT2 Regulation vs VIN (IOUT1 = 0 A, IOUT3 = 0 A) ........................................................................ 1111 VOUT3 Regulation vs VIN (IOUT1 = 0 A, IOUT2 = 0 A) ........................................................................ 1212 Steady State Waveforms (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A) ................................ 1213 Steady State Waveforms (VIN = 10 V, IOUT1 = 0 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A) ................................ 1214 Ripple Voltage Waveforms (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A) .............................. 1315 Load Step Response 50 mA to 200 mA (VIN = 10 V, IOUT1 = 0 A, and IOUT3 = 0 A) ................................. 1316 Line Step Response 10 V to 24 V (IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A).................................... 1317 Power Up (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)................................................... 14
List of Tables
1 Design Parameters .......................................................................................................... 6
TrademarksAll trademarks are the property of their respective owners.
CIN
VSWipri + Vpri -
D1
+Vd-
isec
N1
N2N = N2 : N1
COUT2
COUT1
IOUT2
IOUT1
R2
R1
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
1 IntroductionIsolated bias rails are common in many systems or subsystems, such as telecommunication equipment,medical equipment, and industrial factory automation. It is mandatory by safety standards to protect usersfrom the hazardous voltage of a power supply, or the isolation is installed to break the ground loopinterference for noise-sensitive applications.
In many cases, simple, low part count, isolated power supplies working from an input voltage are needed.This design solution is for when regulation may not be as important, but cost and board area are. Apopular solution to these requirements is an isolated buck power supply, which is a synchronous buckconverter, with the inductor replaced by a coupled inductor or flyback-type transformer. There is no needfor an optocoupler or auxiliary winding because the secondary output closely tracks the primary voltage,resulting in smaller solution size and cost. The flybuck can support a simple, small, and cost effectivepower solution making it suitable as a flyback alternative.
This application report presents the basic operating principles of a flybuck converter, shows some typicaloperating current and voltage waveforms, and the key design equations are derived. The design exampleshows a step-by-step procedure for designing one nonisolated and two isolated outputs with thesynchronous buck regulator TPS54308.
The TPS54308 device is a 4.5-V to 28-V input voltage range, 3-A switching regulator, that has twointegrated switching FETs, internal loop compensation, and 5-ms internal soft start to reduce componentcount. By integrating the MOSFETs and employing the SOT-23 package, the TPS54308 achieves high-power density and offers a small footprint on the PCB.
2 Flybuck Converter Device Overview
2.1 Operation DescriptionAn isolated buck converter is a synchronous buck converter with the inductor replaced by a coupledinductor or flyback-type transformer. The primary output is still regulated as in a sync buck. The secondaryoutput is generated by a diode rectifying the secondary winding. Figure 1 shows an isolated buckconverter with two outputs.
Figure 1. Isolated Buck Converter With Two Outputs
onOUT1 IN IN
on off
TV V D V
T T= = ´
+
TON TOFF
t
t
t
tVsw
VD
im
ipri
isec
tVpri
VIN-VOUT1
-VOUT1
VIN
¨im
t
IOUT2
IOUT1
VOUT2 + (VIN ± VOUT) N2
N1
IOUT1 + IOUT2 N2
N1
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Figure 2 shows typical isolated buck operating waveforms. During TON, the high-side MOSFET is on, andthe rectifier diode is turned off, because the reflected voltage across the secondary winding is negative.The isolated output capacitor COUT2 is discharged, supplying the load current, and the primary sidebehaves identical to a buck regulator. During TOFF, the low-side MOSFET is on, and the reflected voltageon the secondary winding turns positive, forcing the diode forward conducting. The current in the primarywinding splits into two parts: one part continues to supply the primary output (the magnetizing current, im,similar to a buck converter inductor current), the other part starts to flow to the secondary output. Thesecondary current waveform is determined by the load, leakage inductance, and output capacitance.
Figure 2. Simplified Isolated Buck Operating Waveforms
The primary output voltage is the same as a buck converter and is given by Equation 1:
(1)
sec OUT2I I=
pri OUT1I I=
t
t
ipri
isec
Ipri_pospk
Ipri_negpk
Isec_pk
TON TOFF
Higher leakage Lower leakageNormal leakage
tim
¨imIOUT1 + IOUT2 N2
N1
( )2D OUT2 IN OUT1
1
NV V V V
N= + -
2OUT2 OUT1 F
1
NV V V
N= ´ -
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
The secondary output voltage is given by Equation 2:
where• N1 and N2 are the turns of the primary winding and secondary winding• VF is the forward voltage drop of the secondary rectifier diode (2)
The blocking voltage of the rectifier during TON is given by Equation 3:
(3)
2.2 Equations for Maximum Output CurrentIn practice, the transformer has more or less leakage inductance, which determines the ramp rate of thecurrent in the secondary winding to charge the output capacitor. The simplified current waveforms inFigure 3 shows its relationship with leakage inductance. With lower leakage inductance, the current rampsup quickly to a high value, charging the output capacitor quickly. With the leakage inductance increasing,the current rises slowly, which can result in less energy being supplied to the output capacitance and lessoutput voltage. The higher leakage inductance is the higher peak charging current obtained in secondarywinding under the same output power rate. A high negative current is reflected in the primary winding atthe same time. Leakage inductance along with duty cycle impacts not only the output voltage regulation,but also limits the output power resulting from the minimum low-side sink current limit. Therefore, theleakage inductance must be minimized and the maximum duty cycle must be chosen carefully to mitigatetheir impacts.
Figure 3. Current Waveforms Affected by Leakage Inductance
The winding and output currents have a relationship as shown in Equation 4 and Equation 5 on one cycleaverage basis.
(4)
(5)
( )
( )
HS min
LSSOC min
pri_pospk LIM
pri_negpk LIM
i I
i I
£ìïí
£ïî
2 mpri_negpk OUT2 OUT1
1
N i2Di I I
N 1 D 2
æ ö D= - - +ç ÷
-è ø
2 2 mpri_negpk sec_pk pri_pospk m OUT2 OUT1
1 1
N N i1 Di i i i I I
N N 1 D 2
æ ö D+= - + - D = - - +ç ÷
-è ø
sec_pk OUT2
1 Di I
1 D
+=
-
sec_pk OUT2
2i I
1 D=
-
2 mSW _pospk pri_pospk OUT1 OUT2
1
N ii i I I
N 2
D= = + +
( ) ( )IN OUT1 IN OUT1OUT1m
pri SW IN pri SW
V V V VV Di
L f V L f
- -D = =
´
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
The magnetizing current in the transformer that combines the two windings current is identical to a buckconverter. Therefore, the magnetizing current ripple can be derived as in Equation 6.
(6)
The positive primary winding and switching peak current during TON is given by Equation 7.
(7)
Implementing the principle of charge balance to the secondary output capacitor, the secondary windingpeak current can be approximately derived as Equation 8 for a higher leakage case, and Equation 9 for anormal leakage case.
Considering the worst case, the following equations are derived based on having higher leakage.
(8)
(9)
Then the negative primary winding peak current can be derived as Equation 10 for a higher leakage case,and Equation 11 for a normal leakage case.
(10)
(11)
In fact, we must ensure that during TON the positive primary winding peak current does not exceed theminimum high-side source current limit, ILIMHS(min), and that during TOFF, the negative primary winding peakcurrent does not exceed the minimum low-side sink current limit, ILIMLSSOC(min). Therefore the positive andnegative primary winding peak current should meet that of Equation 12.
(12)
3OUT3 OUT1 F
1
NV V V 12 V
N= ´ - =
2OUT2 OUT1 F
1
NV V V 12 V
N= ´ - =
49.9R5
VIN =10V - 24V
510kR1
0
R3 C3
0.1uF
TP3
12
JP1VOUT1 = 5V, 1A
100kR6
GND1
SW2
VIN3
FB 4
EN5
BOOT6
U1
TPS54308DDCR
750343683
15µH
9
2
4
8
7T1
1
2
J2
1
2
J3
75pFC4
0R4
TP5
TP4
BOOT
SW
VSENSE
TP1
0.1µFC2
10µFC1
TP7
TP6
TP8
TP9
TP10
2200pFNC
C6
TP11
VOUT2 = 12V, 0.2A
VOUT3 = -12V, 0.2A
GND
GND_ISO
VIN
EN
1
2
J1
1
2
J4
N24:N78:N89=1:2.5:2.5 D2B270-13-F
D1B270-13-F
13.7kR7
10µFC8
10µFC9 10µF
NC
C11
10µFNC
C10
100R11
100pFC13
TP2
100R10
100pFC122.2kR8
2.2kR9
100NC
R12
100pFNC
C14
GND
22µFC5
22µFC768k
R2
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
3 Design Flybuck With TPS54308Figure 4 shows the design example with the typical flybuck circuit, using the synchronous buck regulatorTPS54308 from TI.
Figure 4. TPS54308 Two Output Isolated Buck Solution
Table 1 lists the design specification.
Table 1. Design Parameters
Design Parameter Example ValueInput voltage range (VIN) 10 V to 24 VPrimary output voltage (VOUT1) 5 VPositive isolated output voltage (VOUT2) 12 VNegative isolated output voltage (VOUT3) –12 VPrimary load current (IOUT1) 1 APositive isolated load current (IOUT2) 0.2 ANegative isolated load current (IOUT3) 0.2 ASwitching frequency (fsw) 350 kHz
In this example, one primary output and two isolated outputs are obtained. We will begin with the buckconverter component calculations and qualify the steps for the isolated configuration.
3.1 Primary Voltage and Turns RatioThe primary-side, nonisolated output is set to 5 V for several considerations. First, the setting is below theminimum input voltage, 10 V, and the theoretical duty cycle varies from 20 to 50 percent at the full VINrange, which is a balanced duty cycle during normal operation. Because the isolated outputs only have theoff-time window to transfer energy, for a duty cycle that is too high, the secondary winding current willhave a huge spike, which leads to poor regulation. Next, the turns ratio of the transformer is not too highto handle for the 5 V voltage level steps up to 12 V. Last, 5 V is one of the most common voltage levels inmany applications. In this design, a transformer turns ratio (N1:N2:N3) of 1:2.5:2.5 is selected (seeEquation 13 and Equation 14).
(13)
(14)
In this design, the VF is targeted at 0.5 V.
( )( )( )
OUT1IN max OUT1pri
SW IN max
V V Vl 12.6 H
0.3 3 f V
-= = m
´ ´
( )( )( )
( )
OUT1IN max OUT1pri min
m SW IN max
V V Vl 1.79 H
i f V
-= = m
D ´
( )HS min
32m LIM OUT1 OUT2 OUT3
1 1
NNi 2 I I I I 4 A
N N
æ öæ öD = ´ - + + =ç ÷ç ÷ç ÷è øè ø
OUT3D2_pk
max
2 Ii 0.8 A
1 D
´
= =
-
OUT2D1_pk
max
2 Ii 0.8 A
1 D
´
= =
-
( )( )2D2 OUT1 OUT3IN max
1
NV V V V 59.5 V
N= ´ - + =
( )( )2D1 OUT1 OUT2IN max
1
NV V V V 59.5 V
N= ´ - + =
6 FB7
OUT1 FB
R VR 13.53 k
V V
´= = W
-
6OUT1 FB
7
RV V 1
R
æ ö= ´ +ç ÷
è ø
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
3.2 Feedback ResistorWith the expected primary voltage set point at 5 V, and VFB = 0.596 V (typical), VOUT1 is calculated to be asin Equation 15:
(15)
Start with 100 kΩ for the upper resistor divider, R6 = 100 kΩ (see Equation 16).
(16)
Next choose R7 = 13.7 kΩ. The 49.9-Ω resistor, R5, is provided as a convenient location to break thecontrol loop for stability testing.
3.3 Rectifier DiodeThe rectifier diode D1 and D2, must meet the blocking voltage and maximum current requirements (seeEquation 17, Equation 18, Equation 19, and Equation 20).
(17)
(18)
(19)
(20)
Considering the voltage spike and keeping some margin, as well as the less forward voltage drop, aschottky diode of 70 V or higher reverse voltage rating is required. A 70-V, 2-A diode is selected for eachrectifier diode in this design.
3.4 Primary InductanceFor the TPS54308, the minimum high-side current is 4 A, and therefore the maximum magnetizing currentripple that can be tolerated is given by Equation 21:
(21)
Using Equation 6 for the maximum magnetizing current ripple calculation, the minimum primaryinductance is given by Equation 22.
(22)
Too much high Δim may not be good for the efficiency and output voltage ripple. Assuming the 30% rippleof the rate output current of the device, then the optimized minimum primary inductance is given byEquation 23:
(23)
( )32
OUT1 OUT2 OUT3
1 1IN
SW IN
NNI I I
1 2.5 0.2 2.5 0.2 AN NC 3.6 F
8 f V 8 350 kHz 0.2 V
+ ++ ´ + ´
³ = = m´ ´ D ´ ´
( )pri SW _pospk max 1
1max e
L iN
B A
´=
´
( )( )m min3 max2
OUT2 OUT3SW _negpk max 21 1 max
iN 2DNi I I 2.24 A
N N 1 D 2
Dæ öæ ö= - + - = -ç ÷ç ÷
-è ø è ø
( )( )m min3 max2
OUT2 OUT3SW _negpk max 21 1 max
iN 1 DNi I I 3.24 A
N N 1 D 2
Dæ öæ ö += - + - = -ç ÷ç ÷
-è ø è ø
( )( )m max32
OUT2 OUT3SW _pospk max 21 1
iNNi I I 1.13 A
N N 2
D= + + =
( )( )m min3 max2
OUT2 OUT3 OUT1SW _negpk max 11 1 max
iN 2DNi I I I 1.24 A
N N 1 D 2
Dæ öæ ö= - + - + = -ç ÷ç ÷
-è ø è ø
( )( )m min3 max2
OUT2 OUT3 OUT1SW _negpk max 11 1 max
iN 1 DNi I I I 2.24 A
N N 1 D 2
Dæ öæ ö += - + - + = -ç ÷ç ÷
-è ø è ø
( )( )m max32
OUT1 OUT2 OUT3SW _pospk max 11 1
iNNi I I I 2.13 A
N N 2
D= + + + =
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
We chose 15 μH for the primary inductor, therefore Δim(max) = 0.75 A and Δim(min)= 0.48 A. Use Equation 7,Equation 10, and Equation 11 to check the positive and negative primary winding peak current (seeEquation 24, Equation 25, and Equation 26.
(24)
(25)
(26)
The primary output side is not always loaded, and without the primary load, the negative current is moreobvious. So consider the worst case, IOUT1 = 0 A, check the positive and negative primary winding peakcurrent again, as given by Equation 27, Equation 28, and Equation 29.
(27)
(28)
(29)
For the TPS54308, the minimum high-side source current limit, ILIMHS(min), is 4 A and the minimum low-side sink current limit, ILIMLSSOC(min), is 2.6 A. Therefore the positive and negative primary winding peakcurrent can meet the current requirements with the normal leakage of a transformer.
3.5 Primary TurnsA coupled inductor or a flyback-type transformer is required for flybuck topology. Calculate the minimumprimary turns for a flyback-type transformer using Equation 30.
where• Bmax is the maximum flux density.• Ae is the effective core area of the chosen transformer. (30)
In this design, the 750343683 inductor from Würth is selected for the transformer. Würth offers many otherflyback-type transformers and ready-made coupled inductors, in a variety of standard values, saturationcurrents, and sizes.
3.6 Input and Output CapacitorThe input capacitor must be large enough to limit the input voltage ripple, see Equation 31.
(31)
Choosing ΔVIN= 0.2 V gives a minimum of CIN= 3.6 μF. Considering derating, one standard value 10-μF,35-V capacitor is selected for better input ripple performance. Another 0.1μF capacitor has been added asa bypass capacitor to clear high-frequency noise.
Choosing ΔVOUT1= 0.05 V gives a minimum of COUT1= 28.5 μF. Considering derating, two standard value22-μF, 16-V capacitors are selected to maintain a small voltage ripple.
Energy is transferred from primary to secondary when the synchronous switch of the buck converter is on.The primary output capacitance is given by Equation 32:
OUT2 maxOUT2
SW OUT2
I D 0.2 0.5C 2.9 F
f V 350 kHz 0.1V
´ ´= = = m
´ D ´
( )32
OUT2 OUT3 max
1 1
OUT1
SW OUT1
NNI I D
2.5 0.2 2.5 0.2 0.5N NC 28.5 F
f V 350 kHz 0.05 V
æ ö+ ´ç ÷ ´ + ´ ´è ø³ = = m
´ D ´
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
(32)
The secondary output current is sourced by COUT2 during TON. Ignoring the current transitions time in thesecondary winding, the value of the secondary output capacitor can be calculated using Equation 33:
(33)
Choosing ΔVOUT1= 0.1 V gives a minimum of COUT2= 2.9 μF. Considering derating, a standard value 10-μF,25-V capacitor is selected, the same as the COUT3.
3.7 Pre-LoadPre-load is necessary to prevent VOUT2 and VOUT3 from going too high at light load. The required pre-loadcurrent is usually set around 5 mA as a starting point, but it should be adjusted based on the circuit testand application requirements. In this design, the pre-load resistor is selected as 2.2 kΩ for ±12-V outputs.Users can use a Zener diode instead of the pre-load resistor to increase the transfer efficiency.
3.8 Factors Affecting Voltage RegulationA flybuck converter regulates the primary output voltage with a closed-loop controller. The isolated outputis achieved by rectifying the secondary winding of the coupled inductor when the low side MOSFET is on.Therefore the isolated output regulation is passive, affected by the winding leakage inductance, windingresistances, diode drop, and low-side MOSFET, RDSON.
3.9 Avoiding Low-Side Sink Current LimitIn a flybuck application, the isolated output power rate may be limited by the low-side sink current limit.Therefore, users must select design parameters elaborately to promote the isolated power rate. FromEquation 25 and Equation 26, we know that the key factors affecting the negative primary winding peakcurrent are Dmax, N2/N1, N3/N1, and the output current. The following tips are for trying to keep the margin:
1. Select a reasonable range of duty cycle. TI recommends a duty cycle ranging from 20% to 50% inmost cases. Users can reduce Dmax by setting a lower primary voltage or increasing the minimum inputvoltage to increase the negative current margin.
2. Minimize the leakage inductance. Leakage inductance is a crucial factor determining the ramp rate ofthe current in the secondary winding which charges the output capacitor. A nominal amount of leakageis within 1% of primary inductance.
3. Pick the correct turns ratio. A lower turns ratio (isolated side to non-isolated side) results in a lowerreflected current in the primary winding.
4. Raise the non-isolated output power, which will reduce the negative winding peak current.5. Reduce the isolated output power. This is the most direct way to lower the level of the negative
current.
IOUT3 (mA)
Effi
cien
cy (�
)
0 20 40 60 80 100 120 140 160 180 20040
45
50
55
60
65
70
75
80
85
90
D003
VIN = 10 VVIN = 12 VVIN = 16 VVIN = 24 V
IOUT2 (mA)
Effi
cien
cy (�
)
0 20 40 60 80 100 120 140 160 180 20040
45
50
55
60
65
70
75
80
85
90
D002
VIN = 10 VVIN = 12 VVIN = 16 VVIN = 24 V
IOUT1 (A)
Effi
cien
cy (�
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.255
60
65
70
75
80
85
90
95
100
D001
VIN = 10 VVIN = 12 VVIN = 16 VVIN = 24 V
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
4 Experimental ResultsFigure 5 through Figure 17 show the experimental test results of Figure 4.
Figure 5. Efficiency vs IOUT1 (IOUT1 = 0 A, IOUT2 = 0 A)
Figure 6. Efficiency vs IOUT2 (IOUT1 = 0 A, IOUT3 = 0 A)
Figure 7. Efficiency vs IOUT3 (IOUT1 = 0 A, IOUT2 = 0 A)
VIN (V)
VO
UT
2 (V
)
10 11 12 13 14 15 16 17 18 19 20 21 22 23 2411.5
12
12.5
13
13.5
14
14.5
15
D006
IOUT2 = 0 AIOUT2 = 10 mAIOUT2 = 80 mAIOUT2 = 150 mA
IOUT3 (mA)
VO
UT
3 (V
)
0 20 40 60 80 100 120 140 160 180 20011.5
12
12.5
13
13.5
14
14.5
15
D005
VIN = 10 VVIN = 12 VVIN = 16 VVIN = 24 V
IOUT2 (mA)
VO
UT
2 (V
)
0 20 40 60 80 100 120 140 160 180 20011.5
12
12.5
13
13.5
14
14.5
15
D004
VIN = 10 VVIN = 12 VVIN = 16 VVIN = 24 V
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
Figure 8. VOUT2 Regulation vs IOUT2 (IOUT1 = 0 A, IOUT3 = 0 A)
Figure 9. VOUT3 Regulation vs IOUT3 (IOUT1 = 0 A, IOUT2 = 0 A)
Figure 10. VOUT2 Regulation vs VIN (IOUT1 = 0 A, IOUT3 = 0 A)
VIN (V)
VO
UT
2 (V
)
10 11 12 13 14 15 16 17 18 19 20 21 22 23 2411.5
12
12.5
13
13.5
14
14.5
15
D007
IOUT2 = 0 AIOUT2 = 10 mAIOUT2 = 80 mAIOUT2 = 150 mA
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
Figure 11. VOUT3 Regulation vs VIN (IOUT1 = 0 A, IOUT2 = 0 A)
Figure 12. Steady State Waveforms (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)
Figure 13. Steady State Waveforms (VIN = 10 V, IOUT1 = 0 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
Figure 14. Ripple Voltage Waveforms (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)
Figure 15. Load Step Response 50 mA to 200 mA (VIN = 10 V, IOUT1 = 0 A, and IOUT3 = 0 A)
Figure 16. Line Step Response 10 V to 24 V (IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)
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Designing a Simple and Low-Cost Flybuck Solution With the TPS54308
Figure 17. Power Up (VIN = 10 V, IOUT1 = 1 A, IOUT2 = 0.2 A, and IOUT3 = 0.2 A)
5 ConclusionThis application report presented the operating principles of a flybuck converter, showed typical operatingcurrent and voltage waveforms, and provided a detailed, step-by-step design example which explainedhow to select the external components. Data and waveforms tested based on the example design showthat the flybuck can support a simple, small, and cost effective power solution, making it suitable as aflyback alternative, and the TPS54308 makes the advantage more obvious.
6 References• Texas Instruments, TPS54308: 4.5-V to 28-V Input, 3-A Output, Synchronous 350-kHz FCCM Step-
Down Converter in SOT23 Package , data sheet• Texas Instruments, Designing an Isolated Buck (Flybuck) Converter, application report• Texas Instruments, Design a Flybuck Solution With Optocoupler to Improve Regulation Performance
Create an Inverting Power Supply, application report• Texas Instruments, Creating a Split-Rail Power Supply With a Wide Input Voltage Buck Regulator,
application report
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