Designing A Wide Input Range DCM Flyback Converter Using ...

1 AN730 Vishay SiliconixDocument Number: 7118008 Jul A wide Input Range DCM Flyback ConverterUsing the si9108 FEATURESDWide 10-V to 100-V Input voltage RangeDEnables Designs With Efficiency Above 80%D 12-V Outputs At 125 mADTotal 3-W Continuous PowerDDiscontinuous Conduction Mode Flyback dc-to-dcConverterD500-V Input /Output Isolation MinimumDNo Optocoupler Feedback NeededDLow Component Count, Low-cost DesignDComplete Solution Occupies 2 In2 Of Board Space InA Single-Sided FR4 BoardDMaximum Component Height Is 1/4 si9108 is a highly integrated, high-efficiencycurrent-mode regulator designed for telecom dc-to-dcconverters. Its wide operating duty cycle of up to issuitable for many power conversion applications, especiallythose with a wide operating Input voltage Range . Thisapplication note is intended to guide the user to design a verywide Input voltage Range , discontinuous conduction mode(DCM) Flyback use a Flyback Converter ?

2 Because the Flyback designcombines simplicity, a low parts count, and affordability forlow-power applications that require Input /output use discontinuous conduction mode? Here are some ofthe advantages:Dmaximizes energy storage in the magnetic component (asmaller core is needed for a given output power)Deliminates the output rectifying diode reverse recoveryproblem, hence there is no current shoot through in themain Input power switching deviceDhigher efficiency, especially at high Input voltageDno right half plane zero in the control loop, simplifyingfeedback compensation design, allowing a very stableand wide -bandwidth feedback DCM Flyback Design Has Some Limitations:Dthe peak primary current and output rectifying diodecurrent are large, although this is not a major concern forlow-power applicationsDduty cycle varies with both the Input voltage and outputload, thus wide duty cycle operation is usually design of the si9108 regulator helps to overcome thislimitation.

3 Figure 1 illustrates a simplified Flyback converterdesigned with the si9108 +iC1C1iC2C2+IO1D1V1 GND2V2 (NEG)D2IO2 FIGURE 1. Simplified Flyback Converter Using si9108 RegulatorAN730 Vishay Number: 7118008 Jul 02 Discontinuous Flyback Converter FundamentalsFigure 2 shows typical current waveforms of a DCM flybackconverter in simplify circuit analysis, we can convert the Flyback circuitto a basic buck-boost Converter configuration. The conversioncan be done in two steps:Step one: converting all outputs into a single output. Let VO =VO1, and PO = VO1 IO1 + VO2 IO2 IO = PO/VOStep two: reflecting the Input voltage to the output =Vi(Ns/Np)Where Ns is the secondary number of turns and Np is theprimary number of turns. Figure 3 shows the simplifiedbuck-boost pkiD1 Ipk10IO1iC1 Ipk1 IO10 IO1iD2 Ipk20IO2iC2 Ipk2 IO20 IO2 FIGURE 2. Typical Current Waveforms of DCM FlybackAN730 Vishay SiliconixDocument Number: 7118008 Jul +VDIOVO (neg)+ILLODIL0DD21/fQ+VSFIGURE 3.

4 Basic Buck-Boost ConverterQOFFQONDCM Buck-Boost Converter AnalysisThe inductor charging time interval is designated as t1. Whenthe power switch Q is on, VS is applied across the outputinductor, Lo. The current in the inductor starts to ramp up fromzero linearly, following the equation V = L(di/dt). By the end ofthis interval, when Q turns off, the inductor will have a peakcurrent of(1)Ipk VSDL ofwhere D is the duty cycle and f is the switching indicates that a certain amount of energy has been storedin the inductor. The inductor discharging time interval isdesignated as t2. After Q turns off, the current in Lo forcesdiode D to conduct. Lo sees the output voltage plus the diodeforward voltage drop across its terminal, but in the reversedirection. During this time, the inductor current decreases andthe energy in Lo is discharged to the output capacitor and DCM operation, all of the energy in Lo will dischargecompletely during this time interval.

5 An equation similar to t1results:(2)Ipk VOD2 Lofwhere D2 is the discharging cycle and Vo is the lump sum ofthe output voltage plus the rectifying diode forward this time interval, all of the energy stored in Lo has beendischarged to the output capacitor and load. The current in Lohas decreased to zero. The inductor current will remain at zerountil the next cycle every switching cycle, a package of energy is transferredto the output via Lo. The power associated with this energytransfer is:P 12 LoIp 2orIp 2 PLof (3)OrCombining the above equations and solving for D and D2:D 1Vs2 PLof andD2 1Vo2 PLof (4)(5)Equation (4) shows the duty cycle as a function of the outputpower and Input voltage , while equation (5) shows that D2 issolely a function of the output power. The maximum duty cycle,DM, occurs at maximum output power, PM, and minimum inputvoltage, Vsm. Maximum D2 occurs at maximum output Number: 7118008 Jul 02To maintain DCM operation, the inductor current mustdischarge completely before the next cycle start.

6 In otherwords, D + D2 must be equal or less than 1 under all conditions:DM D2M 1(6)where DM is the maximum duty cycle and D2M is themaximum discharging cycle. Substituting (4) and (5) into theabove equation and solving for Lo, we have:Lo 12 PMf 1 Vsm 1Vo 2 Locrit 12 PMf 1 Vsm 1Vo 2(7)where PM is the maximum output power and Vsm is theminimum Input (7) imposes a maximum value for the output inductor,Locrit, to maintain discontinuous conduction mode whiledelivering the maximum output power at minimum inputvoltage for a buck-boost Converter . It is best to choose a Lovalue close to Locrit to maintain DCM while keeping theinductor peak current as low as to The Flyback ConverterIn a Flyback Converter , the Flyback transformer presentsdesigners with another choice, the secondary-to-primary turnsratio Ns/Np. The following equation was derived to assist thecalculation of the turns VoVim(1 DM)DM(8)where Vim is the minimum Input voltage .

7 Ns/Np determines amaximum duty cycle or in other words, allows designers tochoose a practical maximum duty cycle at the lowest operatinginput voltage and then calculate the required transformer turnsratio. It will be shown later that choosing DM is very importantto optimize the Converter following equations can be used to calculate the criticaloutput inductance once the maximum duty cycle is Vo2(1 DM)22 PMfOrLocrit Vo2(1 DM)22 PMf(9)Again, choose Lo = Locrit, sinceIp 2 PLof (10)Combining (9) and (10), we can now express the inductor peakcurrent in terms of the maximum duty cycle chosen:Ip 2 PMVo(1 DM)(11)As DM increases, the required inductance decreases, whilethe inductor peak current increases. Since the energy storagein the inductor is proportional to LIpk2 while the inductor coresize is proportional only to LIpk, doubling Ipk will reduce therequired inductance to 1/4. and reduce the required core sizeto 1/2. It is a good design practice to make D as large aspossible.

8 This minimizes inductance while keeping the peakcurrent to a manageable Side CalculationsThe primary inductance and peak current can be calculated byreflecting the output inductance and its peak current to theprimary side via the Flyback transformer turns ratio:Lp NpNs 2 LoIip NsNp2 PLof (12)where Lp is the primary inductance required and Iipk is theprimary peak current as a function of total output power. SinceVS = Vi(Ns/Np) and from (4):D NpNsVi2 PLof (13)As expected, the duty cycle D is a function of the inputvoltage,Vi, and the output power, (12) and (13), several equations can be derived tosupport design calculation:Iiave 12 Iip D PviIirms Iip D3 Icirms IiaveD26 2 D6D 3 (1 D) DVci 14 IiaveCif(2 D)2(14)whereIiave is the primary average currentIirms is the primary rms currentIcirms is the Input capacitor rms currentDVci is the Input voltage ripple, excluding ESR effectCi is the Input bypass capacitorOutput CalculationsAny output can be calculated referencing to the main outputusing the following equations:AN730 Vishay SiliconixDocument Number: 7118008 Jul VxVoNsLx NxNs 2 LoDx 1Vx2 PxLxf Ixp 2 PxLxf Ixrms Ixp Dx3 Icxrms IoxDX26 2 Dx6Dx 3 (1 Dx) DVcx 14 IoxCxf(2 Dx)2(15)where.

9 X denotes the particular output of interestVx is the output voltage plus the rectifier forward voltage dropNsx is the winding number of turnsLx is the winding inductanceDx is the discharge duty cyclePx is the output powerIxpk is the inductor, Lx, peak currentIxrms is the inductor rms currentIcxrms is the output capacitor rms currentIox is the output average load currentDVcx is the output voltage ripple, excluding capacitor ESR andESL effectsCx is the output capacitorDesign ExampleThis design example is based on the following specifications: Input voltage Range : Vi from 10 V to 100 VOutput voltages:+12 V and 12 V at 125 mA eachTotal output power is 3 WOutput power is reduced to 1 W forinput voltage <24 VSwitching frequency: 100 kHzCalculationLet us assume an efficiency of 75% and lump sum inductortolerance of 20%, plus 20% for power limit headroom. In thiscase, the total power processed by the Flyback Converter is:P (1 40%) W(16)FIGURE 4.

10 Normalized Lo Critical and Ipk vs. Maximum Duty CycleMaximum Duty Cycle CriticalFigure 4 above shows the normalized critical outputinductance and its peak current with respect to a maximumduty cycle varying from 20% to 90%. From the graph, we cansee that Lo decreases significantly for DM >40% and its peakcurrent decreases significantly for DM < 80%. This means thatit is desirable to choose DM anywhere in the 40% to 80% DM = 55% at full load and 24 V. Use Vo = 12 V asreference output plus V for the rectifier forward voltagedrop. We can now determine the required nominal outputinductance. Notice that the inductor tolerance has alreadybeen accounted for in the power calculation. The 55% DM waschosen at a 24-V Input so that at a 10-V Input and with 1-Woutput power, the duty cycle will be around 80% still anacceptable Vo2(1 DM)22 PMf (1 ) (17)Next, calculate the transformer turns ratio:NsNp VoVim(1 DM)DM (1 ) (18)Choose Ns = 40T and Np = 95T.