Application Note 1067 Application Notes for AP3770 System Solution Prepared by Su Qing Hua System Engineering Dept. Regulation (PSR). AP3770 has the special technique to suppress the audio noise, internal line compensation to reduce the number of system components, fixed cable compensation to compensate the voltage drop on different output cable for achieving good CV regulation. 1. Introduction The AP3770 uses Pulse Frequency Modulation (PFM) method to realize Discontinuous Conduction Mode (DCM) operation for FLYBACK power supplies. The principle of PFM is different with that of Pulse Width Modulation (PWM), so the design of transformer is also different. The AP3770 can achieve low standby power less than 30mW. The AP3770 can provide accurate constant voltage, constant current (CV/CC) regulation by using Primary Side Figure 1. 5V/1A Output for Battery Charger of Mobile Phone Figure 1 is AP3770 typical application circuit, which is a FLYBACK converter controlled by AP3770 with a 3-winding transformer---Primary winding (Np), Secondary winding (Ns) and Auxiliary winding (Na). The AP3770 senses the auxiliary winding feedback voltage at FB pin and obtains power supply at VCC pin. Vdri---A simplified driving signal of primary transistor Ip---The primary side current Is ---The secondary side current VSEC---The voltage of secondary tSW---The period of switching frequency tONP ---The time of primary side “ON” tONS ---The time of secondary side “ON” tOFF ---The discontinuous time tOFFS --- The time of secondary side “Off” IPK---Peak current of primary side IPKS---Peak current of secondary side VS---the sum of Vo and forward voltage of rectification diode Figure 2 is the typical operation waveforms of PFM controller. In this figure, a series of relative idea operation waveforms are given to illustrate some parameters used in following design steps. And the nomenclature of the parameters in Figure 2 is illustrated. Sep. 2011 Rev. 1. 0 BCD Semiconductor Manufacturing Limited 1 Application Note 1067 tONP Vdri tSW IPK IP IPKS IS tONS VS VSEC tOFFS tOFF Figure 2. Operation Waveforms Zero-Startup-Current technique, the startup up resistors R3+R4 should be higher to 10M to 14MΩ to further lower the power loss. The recommended value of dummy load resistor R13 is 4.7KΩ to 10KΩ for an model with 5V output voltage. The selection of dummy load resistor is a tradeoff between standby power and I-V Curve. 2. Five Aspects for System Design 1. 2. 3. 4. 5. Low Standby Power Design Switching Frequency Design Transformer and Power Devices Design Feedback Resistors Design Line Compensation Design 2.2 Switching Frequency Design 2.1 Low Standby Power Design In order to achieve low standby power, AP3770 decreases the minimum operating voltage. And due to proprietary VH=0.5V VCS_REF 1.4V VCPC fSW fSW 47.6kHz 20kHz IO 42%IO Figure 3. Relationship Between VCPC, fSW and IO with Constant Peak Current Sep. 2011 Rev. 1. 0 BCD Semiconductor Manufacturing Limited 2 Application Note 1067 Where, the fSW is the switching frequency. When the peak current IPK is constant, the output power depends on the switching frequency fSW. fSW is linearly increased with load increasing. When the constant peak current is adopted, the voltage of CPC pin is increased linearly with load increasing. The maximum value of VCPC is equal to t ONS 4 ⋅ VDD = ⋅ 3.5V = 1.4V t SW 10 (1) In AP3770, two-segmented peak current is used to realize audio noise suppression. The peak current is about 0.5V when IO>42%IOmax, and the peak current is about 0.5V/1.5 when IO<42%IOmax. The primary current ip(t) is sensed by a current sense resistor RCS as shown in Figure 2. The power transferring from input to output is given by: PO = 1 2 ⋅ LP ⋅ I pk ⋅ f SW 2 (2) VLOAD VCPC VH=0.5V VCS_REF VL=0.5V/1.5 0.42хIO_MAX IO_MAX fSW 55kHz 52kHz fSW 23.1kHz 20kHz 3.85kHz 0.42хIO_MAX IO_MAX ISOURCE 2/3*I SOURCE 0.42хIO_MAX IO_MAX Figure 4. Relationship Between VCPC, fSW and IO with Variable Peak Current So, the voltage of CPC pin and switching frequency has a mutation at about 42% of load. At the mutation point, if the peak current is changed from 0.5V( high IPK) to 0.33V(low IPK), the voltage of CPC pin at low IPK will be increased to 1.5 times of VCPC at high IPK and the switching frequency fSW at low IPK will be increased to 2.25 times of fSW at high IPK. So the range of load working in the audio frequency is suppressed. VCS_REF VH=0.5V VL=0.5V/1.5 39%IO 42%IO IO Figure 5. Hysteresis at Conversion Between Low IPK and High IPK Sep. 2011 Rev. 1. 0 BCD Semiconductor Manufacturing Limited 3 Application Note 1067 in all working conditions. In order to avoid oscillation, a hysteresis is added at the conversion between low IPK and high IPK. Considering the relationship between audio noise and flux density of transformer, deltaB≤2500 gauss is better for audio noise suppression. t SW ≥ t ONS + t OFFS (7) For the primary side current, The low limitation of maximum switching frequency is given by audio noise suppression. And the upper limit of the AP3770 can be up to 120kHz. But this is only the limit of the IC; the finally designed maximum switching frequency is determined by the tradeoff between the efficiency, mechanical dimensions and thermal performance. t ONP = I pk ⋅ Lp (8) Vindc Where LP is the inductance of primary winding. Vindc is the rectified DC voltage of input. When Vindc is the minimum value, the maximum tONP can be obtained. So, 2.3 Transformer and Power Devices Design t ONP_MAX = I pk ⋅ Lp In Constant Current operation of AP3770, the CC loop control function of AP3770 will keep a fixed proportion between D1 (in Figure 1) on-time tONS and D1 off-time tOFFS (in Figure 2) by discharging or charging a capacitor embedded in the IC. The fixed proportion is For the secondary side current, t ONS 4 = t OFFS 6 In (10), LS is the inductance of secondary winding. t ONS = I pks ⋅ (3) t ONS 1 ⋅ I pks ⋅ 2 t ONS + t OFFS I pks (4) In FLYBACK converter, when the primary transistor turns ON, the energy stored in the magnetizing inductance Lp. So the power transferring from the input to the output is given by, t ONS 1 NP 1 N ⋅ ⋅ I pk ⋅ = ⋅ P ⋅ I pk 2 NS t ONS + t OFFS 5 N S Pin = Pin ⋅ηin = Vin ⋅ I in ⋅ηin (11) 1 2 ⋅ L p ⋅ I pk ⋅ f SW 2 (12) ' (5) ' Pin = Thus the output constant-current is given by: IO = (10) For (10), in CV regulation, the VS is a constant voltage, so tONS is a constant value with different input voltage. At the instant of D1 turn-on, the primary current transfers to the secondary at an amplitude of: N = P ⋅ I pk NS LS VS (9) VS = VO + Vd , Vd is the forward voltage of secondary diode. The relationship between the output constant-current and secondary peak current IPKS is given by: IO = Vindc _ min Here, Pin ' is input power of transformer, not including the all of the power loss at primary side (Rectifier, RCD snubber, BJT and so on). (6) Design Steps: ηin is definition to the input efficiency of system, which is about 0.9. Step 1, a reasonable IPK of FLYBACK with AP3770 should be designed Then, 1-1. Calculate the Max. turn ratio of XFMR t SW = The maximum turn ration of XFRM should be designed first, which is to ensure that the system should work in DCM in all working conditions, especially at the min. input voltage and full load. (13) 2 ⋅ Pin ⋅ η in tSW, tONP and tONS in (7) are replaced with (13), (9) and (10), 2 Lp ⋅ I pk 2 ⋅ Pin ⋅ ηin As we know, if the system can meet equation (7) at minimum input voltage and full load, it can work in DCM Sep. 2011 2 L p ⋅ I pk Rev. 1. 0 ≥ I pks ⋅ Lp Ls + I pk ⋅ Vs Vindc_min (14) BCD Semiconductor Manufacturing Limited 4 Application Note 1067 Because the peak current and inductance of primary side and secondary side have the following relationship, I pks = N ⋅ I pk ⋅ η i RCS = N2 (24) (15) So RCS can be obtained from (24) and selected with a real value from the standard resistor series. After RCS selected, IPK should be modified based on the selected RCS. (16) From now on, IPK and RCS have been designed. Lp Ls = 0.5V I pk Here, N is the turn ratio of primary and secondary sides. Step 2, Design Transformer With (14), (15) and (16), then, 2-1. Calculate the inductance of primary side---LP I pk 2 ⋅ Pin ⋅ηin ≥ ηi VS ⋅ N + 1 Vin (17) The primary side inductance LP is relative with the stored energy. LP should be big enough to store enough energy, so that PO_Max can be obtained from this system. Because, Pin = η VO ⋅ I O η From formula (18), the output power can be given by, (18) is the system efficiency. 1 t I O = ⋅ ONS ⋅ I pks 2 t SW Then, LP can be gotten by, (19) In the design of AP3770, (21) N= The following can be obtained, N ≤ Vindc_min ⋅ ( k ⋅η ηi ) − 2 ⋅ VO ⋅ ηin ⋅ ηi VO + Vd (22) The turns of primary winding, Np = (23) LP ⋅ I PK Ae ⋅ ∆B (28) The turns of secondary winding, Here, k=5, η i = 0.9 , which is the efficiency of IPK and IPKS. NS = N is the calculated value of Nmax. NP N (29) The turns of auxiliary winding, In AP3770, 0.5V is an internal reference voltage. If the sensed voltage VCS reaches 0.5V, the power transistor will be shut down and tONP will be ended. Sep. 2011 (27) First, the reasonable core-type and ∆B should be selected. Then, the turns of 3-winding transformer can be obtained respectively. IPK can be calculated by the output current. k ⋅ IO = I pk ⋅η i = N N k ⋅ IO (k = 5 ) I pk ⋅η i 2-3. Calculate the turns of primary, secondary and auxiliary sides 1-2. Calculate the peak current of primary side and current sensed resistor I pks (26) 2 PK 2-2. Re-calculate the turn ratio of primary and secondary side---N From formula (24), the turn ratio of primary and secondary side N can be re-calculated. (20) 2 ⋅ t SW =5 t ONS 2 ⋅ PO ηin ⋅ I ⋅ f SW η LP = Then, IPKS can be defined, I pks = k ⋅ I O (25) Where fSW was set by the user based on definite requirement. At full load, the system will work in the boundary of CC regulation. IO can be given by, k= 1 η 2 ⋅ L p ⋅ I pk ⋅ f SW ⋅ 2 ηin PO = NA = Rev. 1. 0 N S ⋅ VA VS (30) BCD Semiconductor Manufacturing Limited 5 Application Note 1067 Ae can be gotten automatically after core-type is selected. Other setting by users: Switching frequency: fSW=54kHz (Should be equal to or higher than 54kHz) Forward voltage of secondary diode: Vd=0.4V Forward voltage of auxiliary diode: Vda=1.1V VCC voltage: VCC=12V Core_type: EE16 (Ae=19.2mm2) Set ∆B : ∆B <3000GS Vdc_spike=100V (with snubber circuit) 2-4. Check the maximum duty cycle of primary side Design Steps: After turn ratio of primary side and secondary side is designed, the maximum duty cycle of primary side at low line voltage can be calculated again. Step 1, a reasonable IPK of FLYBACK with AP3770 should be designed. Where VA=VCC+ Vda, VCC is the set IC supply voltage and Vda is the voltage drop of the auxiliary diode. For AP3770, the typical value of UVLO is decreased to 5.5V, so the supply voltage of IC，VCC can be set to a typical value---12V. VS is equal to VO+Vd. 1-1. Calculate the maximum turn ratio of XFMR Considering the Volt-second balance between magnetizing and de-magnetizing, the formula of duty cycle is D= k ⋅η ηi (35) − )(k = 5 ) 2 ⋅ VO ⋅ ηin ⋅ ηi VO + Vd = Vinac_min ⋅ 2 − 40 , η = 0.75 , ηin = 0.9 , ηi = 0.9 N MAX = Vindc_min ⋅ ( (VO + Vd ) ⋅ N ⋅ 0.4 Vindc (31) Vindc_min N MAX ≈ 22 The turn ratio is finally selected as: N = 18.5 Step 3, Select diode and primary transistor (36) 3-1. Select diodes of secondary and auxiliary sides 1-2. Calculate the peak current of primary side and current sensed resistor Maximum reverse voltage of secondary side， Vdr = VO + Vindc_max ⋅ N S NP (32) k ⋅ IO ( ηi = 0.9 , which is the transfer = N N efficiency of Ipk and Ipks) (33) I pk_max = 330 mA I pk ⋅η i = Maximum reverse voltage of auxiliary side, Vdar = V A + Vindc_max ⋅ N A NP RCS = 3-2. Select the primary side transistor VS ⋅ N P NS (37) Sensed current resistor, In (32) and (33), the maximum DC input voltage should be used. Vdc_max = Vdc_spike + Vindc_max + I pks VCS 0.5V = I pk _ max 359m (39) R CS ≈ 1.5Ω (34) (38) Re-calculate peak current of primary side, Be careful that the value of Vdc_spike will be different with different snubber circuit. (40) I pk_max = 333mA Design Example: Specification: Input voltage: 85VAC to 265VAC Output voltage: VO=5.3V (Considering compensation) Output current: IO=1.1A Efficiency: 80% the Step 2, Design Transformer 2-1. Calculate the inductance of primary side---LP cable LP = L p = 2.035 mH It is higher than the total efficiency because the loss in the input rectifier and the BJT are not included. Sep. 2011 2 ⋅ PO ηin ⋅ 2 ⋅ f SW η I PK Rev. 1. 0 (41) (42) BCD Semiconductor Manufacturing Limited 6 Application Note 1067 2-4. Check the maximum duty cycle of primary side The maximum duty cycle of primary side is calculated as following, 2-2. Re-calculate the turn ratio of primary and secondary side---N N= k ⋅ IO (k = 5 ) I pk ⋅η i D= (43) (44) N = 18.3 D |Vindc _ min = 2-3. Calculate the turns of primary, secondary and auxiliary sides The turns of primary winding, Np = (VO + Vd ) ⋅ N ⋅ 0.4 Vindc LP ⋅ I PK Ae ⋅ ∆B N P > 118T (VO + Vd ) ⋅ N ⋅ 0.4 (5.3 + 0.4) ⋅ 18.3 ⋅ 0.4 = = 0.528 Vindc _ min 80 (45) Step 3, Select diode and primary transistor 3-1. Select diodes of secondary and auxiliary sides Maximum reverse voltage of secondary side， (46) Vdr = VO + NP (55) Vindc_m ax = 265V ⋅ 2 N NS = P N (47) N S ≈ 7T Vdr = 5.4 + (48) Vdar = V A + (49) N P ≈ 128T (50) Vdar The turns of auxiliary winding, N S ⋅V A Vo + Vd N A ≈ 15T (56) 375 ≈ 28V 17 (57) Maximum reverse voltage of auxiliary side, Recalculate the primary winding, N P = NS ⋅ N (54) Vindc_max ⋅ N S The turns of secondary winding, NA = (53) Vindc_max ⋅ N A NP (58) 375 ⋅ 15 = 12 + ≈ 56V 128 (59) 3-2. Select primary side transistor VS ⋅ N P NS (51) Vdc_max = Vdc_spike + Vindc_max + (52) Vdc _ max = 100 + 375 + 5.4 * 17 ≈ 567V (60) (61) Design Results Summary: 1.Calculate the maximum peak current of primary side and RCS IPK= 333 mA Peak current of primary side RCS= 1.5 Current sensed resistor Ω 2.Design transformer LP= 2 mH(+/-8%) Inductance of primary side N= 17 Turn ratio of primary and secondary NP= 128 T Turns of primary side NS= 7 T Turns of secondary side N A= 15 T Turns of auxiliary side DMAX 0.52 Maximum duty cycle of primary side at VINDC=80V 3. Select diode and primary transistor Vdr= 28 V Maximum reverse voltage of secondary diode Vdar= 56 V Maximum reverse voltage of auxiliary diode VdcMax= 567 V Voltage stress of primary transistor Sep. 2011 Rev. 1. 0 BCD Semiconductor Manufacturing Limited 7 Application Note 1067 2.4 Feedback Resistors Design Figure 6. Feedback Resistors Circuit From above Figure 6, (R + RFB 2 ) N S Vo = VFB ⋅ FB1 ⋅ − VD RFB 2 NA 2.5 Line Compensation Design The internal line compensation function in AP3770 is shown in Figure 7. S1 is closed when the primary switch is “ON”. The line voltage can be detected from the FB pin. The detected voltage internally compensates the peak current. So the line compensation is determined by RLINE. In different application, the value of RLINE is different. (62) Through adjusting RFB1 and R FB2, a suitable output voltage can be achieved. The recommended values of RFB1 and R FB2 are within 5kΩ to 50kΩ. Figure 7. Line Compensation Circuit Sep. 2011 Rev. 1. 0 BCD Semiconductor Manufacturing Limited 8 Application Note 1067 tSW FB tONP t VN tOFF Figure 8. Waveform of FB Pin So, RLINE can be adjusted to achieve excellent line regulation of output current. The negative voltage VN of FB pin (in Figure 8) is linear to line voltage. The AP3770 samples VN to realize the line compensation. N RFB 2 VN = ⋅ a ⋅ Vindc RFB1 + RFB 2 N p 3. Summary In order to get good performance of AP3770, it’s important to design transformer, line compensation and feedback resistance correctly. This application only gives a preliminary design guideline about these aspects and considers ideal conditions, so some parameters need to be adjusted slightly on the basis of the calculated results. (63) The compensated voltage of line compensation (VCS_LINE) can be calculated by the following formula, 1 ⋅ VN 670k N RFB 2 1 = Rline ⋅ 0.8 ⋅ ⋅ ⋅ a ⋅ Vindc 670k RFB1 + RFB 2 N p Vcs _ line = Rline ⋅ K ⋅ Sep. 2011 (64) Rev. 1. 0 BCD Semiconductor Manufacturing Limited 9

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