Application Note 1028 Basic Steps to Design a PSR Flyback Converter Using AP3706/08N Prepared by Su Qinghua and Dong Wenhui System Engineering Department Introduction Vdri is a simplified driving signal of primary transistor. IP is the primary side current. IS is the secondary side current. VS is the voltage of secondary side. AP3706/08N uses Pulse Frequency Modulation (PFM) method to realize Discontinuous Conduction Mode (DCM) operation for flyback power supplies; uses Primary Side Regulation (PSR) to provide Constant Voltage (CV) and Constant Current (CC) regulation. The principle of PFM is different from that of Pulse Width Modulation (PWM), so the design steps of the transformer for PFM operation is also different from that for PWM operation. In the other side, AP3706/08N uses PSR to provide CV/CC regulation without requiring opto-coupler and loop compensation, which makes the design of transformer more critical than that with PWM. The following design steps focus on the transformer design. Some design guides to select diodes and transistor are also included. Figure 2 shows the relatively ideal operation waveforms to illustrate some parameters used in the following design steps. The nomenclature of the parameters in Figure 2 are listed as below: TSW is the period of switching frequency Tonp is the time of primary side “ON” Tons is the time of secondary side “ON” Toff is the discontinuous time Ipk is the peak current of primary side Ipks is the peak current of secondary side VS equals the sum of VO and forward voltage of rectification diode Figure 1 is a simplified flyback converter controlled by AP3706 with a 3-winding transformer: Primary winding (NP), Secondary winding (NS) and Auxiliary winding (NAUX). The AP3706 senses the Auxiliary winding feedback voltage at FB pin and obtains power supply at VCC pin. In the circuit: The only difference between AP3706 and AP3708N is cable compensation. AP3708N has built-in it, while AP3706 has not. Their design steps are quite similar. Vg VS VIN + + CS OUT AP3706 BIAS NP NS LM LS Vd IS IO - + + VO - VAux Vdri VCC + NAux iP FB GND COMP VDD RCS Figure 1. Simplified Flyback Converter Using AP3706 Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 1 Application Note 1028 Figure 2. Ideal Operation Waveforms for a Simplified Flyback Converter Basic Design Step Tonp = Ipk × Step 1: Select a Reasonable Ipk of Flyback Converter Using AP3706/08N LM Vg where LM is the inductance of primary winding. 1-1. Calculate the Maximum Turn Ratio of the Transformer The maximum turn ration of the transformer should be calculated first to ensure the system working in DCM under all working conditions, especially at the minimum input voltage and full load. Vg is the rectified DC voltage of input. When Vg is at the minimum value, the maximum Tonp can be obtained. So, Tonp _ max = Ipk × If the system meets equation (1) at minimum input voltage and full load, the circuit can work in DCM in all working conditions. TSW ≥ Tonp + Tons LM Vg _ min ......(2) For the secondary side current, ......(1) Tons = Ipks × For the primary side current, LS VS ......(3) where Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 2 Application Note 1028 LS is the inductance of secondary winding. VS=VO+Vd, Vd is the forward voltage of secondary diode. PIN = At the full load, the system will work in the boundary of CC regulation. IO is determined by: In Flyback converter, the energy stored in the magnetizing inductance LM when the primary transistor turns ON. So the power transferring from the input to the output is given by: 1 Tons × Ipks IO = × 2 TSW Then, Ipks can be defined as: ......(4) ......(10) Ipks = k × I O then, TSW = ......(9) η where η is the system efficiency. For (3), in CV regulation, the VS is a constant voltage, so Tons is a constant value with different input voltage. 1 PI N = × LM × Ipk 2 × f SW 2 VO × I O In the design of AP3706/08N, 1 f SW = LM × Ipk 2 2 × PIN k= Substitute TSW, Tonp and Tons in equation (1) with (4), (2) and (3): 2 × TSW = 3 .5 Tons With (8), (9) and (10), it can be obtained: N PS ≤Vg _ min× ( LM × Ipk L LM ≥Ipks × S + Ipk × 2 × PIN Vs Vg _ min 2 N PS _ MAX = Vg _ min× ( Because the peak current and inductance of primary side and secondary side have the following relationship: ......(12) ......(6) Because above calculations are all based on ideal conditions without considering power loss, k is given an approximately value 4 instead of the real value 3.5. 1-2. Calculate the Peak Current of Primary Side and Current Sense Resistor Ipk can be calculated by the output current: ......(7) From (5), (6) and (7): Ipk 1 1 ≥ + 2 × PIN VS × N PS Vg 1 k ×η ) − 2 × VO VO + Vd (k ≈4) here, NPS=NP/NS is the turn ratio of primary to secondary sides, then L LS = M2 N PS ......(11) So, the maximum turn ration of primary and secondary side can be obtained: ......(5) Ipks = N PS × Ipk 1 k ×η − ) 2 × VO VS Ipk = ......(8) Ipks N PS _ MAX = k × IO N PS _ MAX ......(13) here, k=4. In AP3706/08N, 0.5V is an internal reference voltage. If the sensed voltage VCS reaches to 0.5V, the Because Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 3 Application Note 1028 The turns of primary winding is: power transistor (APT13003 for this application) will be shut down and Tonp will be ended. So, RCS can be calculated as: RCS = 0.5V Ipk NP = ......(14) NS = N AUX = 2-1. Calculate the Inductance of Primary Side--LM The primary side inductance LM is relative with the stored energy. LM should be big enough to store enough energy, so that PO_max can be obtained from this system. ......(15) Vdr = VO + ......(16) Vg _ max× N S NP ......(21) Maximum reverse voltage of Auxiliary side is: Vdar = VAUX + 2-2. Re-calculate the Turn Ratio of Primary to Secondary Side---NPS From formula (13), the turn ratio of primary and secondary side NPS should be re-calculated. k × IO (k ≈4) Ipk ......(20) 3-1. Select Diodes of Secondary and Auxiliary Sides Maximum reverse voltage of secondary side is: To achieve good overall system performance, the optimum switching frequency fSW is recommended to be 50 to 60kHz under full load. N PS = N S × VAUX VS Step 3: Select Diode and Primary Transistor Then, LM is: 2 × PO Ipk × f SW ×η ......(19) here, VAUX can be set to its typical value 15V. Vs is equal to VO+Vd. Ae can be obtained automatically after core-type is selected. From equation (9), the output power can be obtained from: 2 NP N PS The turns of Auxiliary winding is: Step 2: Design Transformer LM = ......(18) The turns of secondary winding is: Then, select RCS with a real value from the standard resistor series. After RCS determined, Ipk should be modified according to the selected RCS. 1 PO = × LM × Ipk 2 × f SW ×η 2 LM × Ipk ×108 Ae × ∆B Vg _ max× N AUX NP Vdc _ max = Vdc _ spike + Vg _ max + ......(22) VS × N P NS ......(23) In (21) and (22), the maximum DC input voltage should be used. ......(17) 2-3. Calculate the Turns of Primary, Secondary and Auxiliary Windings First, select the reasonable core-type and ∆B . Then, decide the turns of 3-winding transformer respectively. 3-2. Select the Primary Side Transistor It should be noted that the value of Vdc_spike will be different in circuits with different snubber. Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 4 Application Note 1028 Design Example RCS = Design Specification 0.5V Ipk ......(14) RCS ≈ 2.1Ω Input Voltage 85 to 265 VAC Output Voltage VO=5.5V Re-calculate peak current of primary side, Output Current IO=0.5A Ipk = 238mA Efficiency 75% Step 2: Design Transformer Settings Selected by Users fSW=55kHz Switching Frequency Forward Diode Voltage of Secondary Vd=0.4V Forward Voltage of Auxiliary Diode Vda=1V Feedback Winding VAUX=15V Voltage 2-1. Calculate the Inductance of Primary Side--LM of Auxiliary Core_type EE16 (Ae=19.2mm2) ∆B ∆B=2850GS Vdc_spike 200V LM = ......(16) LM = 2.35mH 2-2. Re-calculate the Turn Ratio of Primary to Secondary Side---NPS N PS = Design Step k × IO (k ≈4) Ipk ......(17) N PS = 8.4 2-3. Calculate the Turns of Secondary and Auxiliary Windings Step 1: Select a Reasonable Ipk for the Flyback Converter Using AP3706 1-1. Calculate the Maximum Turn Ratio of the Transformer N PS _ MAX = Vg _ min× ( 2 × PO Ipk 2 × f SW ×η 1 k ×η − ) 2 × Vo Vo + Vd NP = LM × Ipk ×108 Ae × ∆B Primary, ......(18) N P = 102T ......(12) (k ≈4) NS = Vg _ min = VIN _ min× 2 − 40 NP N PS ......(19) N S = 12T N PS _ MAX = 8.259 N AUX = 1-2. Calculate the Peak Current of Primary Side and Current Sense Resistor N S × VAUX VS ......(20) N AUX = 31T Ipk = k × IO Ipks = N PS _ MAX N PS _ MAX ......(13) Ipk = 242mA Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 5 Application Note 1028 Step 3: Select Diode and Primary Transistor Maximum reverse voltage of secondary side. Vdar = V AUX + 3-1. Select Diodes of Secondary and Auxiliary Sides Maximum reverse voltage of secondary side. Vdr = VO + Vg _ max× N S NP Vg _ max × N AUX NP ......(22) Vdar = 129V 3-2. Select the Primary Side Transistor ......(21) Vdc _ max = Vdc _ spike + Vg _ max + Vg _ max = VIN _ max× 2 VS × N P NS ......(23) Vdc _ max = 625V Vdr = 50V Design Results Summary 1. Calculate the Maximum Peak Current of Primary Side and RCS Peak Current of Primary Side Ipk=238mA Current Sense Resistor RCS=2.1Ω 2. Design Transformer Inductance of Primary Side LM=2.35mH(+/-8%) Turn Ratio of Primary to Secondary Windings NPS=8.4 Turns of Primary Side NP=102T Turns of Secondary Side NS=12T Turns of Auxiliary Side NAUX=31T 3. Select Diode and Primary Transistor Maximum Reverse Voltage of Secondary Diode Vdr=50V Maximum Reverse Voltage of Auxiliary Diode Vdar=129V Voltage Stress of Primary Transistor Vdc_max=625V Sep. 2009 Rev. 1.1 BCD Semiconductor Manufacturing Limited 6

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