www.fairchildsemi.com AN-4153 Designing Asymmetric PWM Half-Bridge Converters with a Current Doubler and Synchronous Rectifier using FSFA-Series Fairchild Power Switches (FPSTM) Introduction In general, high-frequency operation allows the use of smallsized passive components in switch-mode power supplies (SMPS), though it causes the switching losses to increase in a hard-switching mode. To reduce switching losses at high switching frequencies, many soft-switching techniques have been developed, including load-resonant and zero-voltagetransition techniques. Load-resonant techniques use a resonant feature of capacitors and inductors during the entire switching period to vary the switching frequency, depending on the input voltage and load current. The change of the switching frequency, i.e. pulse frequency modulation (PFM), makes it difficult to design an SMPS including input filters. Since there is no output inductor for filtering, the clamped voltage across output-rectifying diodes allows designers to select low-voltage-rating diodes. However, the absence of the output inductor burdens the output capacitors when the load current increases, making load-resonant techniques unsuitable for applications with high output current and low output voltage. On the other hand, zero-voltage-transition techniques use a resonant feature between parasitic components during turnon and/or turn-off transitions of the switching period. One of the advantages of these techniques is to use the parasitic components, such as the leakage inductance of the main transformer and the output capacitances of the switches, so there is no need to add more external components to achieve soft switching. In addition, these techniques take pulse-width modulation (PWM) up with fixed-switching frequency. Therefore, these are easier to understand, analyze, and design than load-resonant techniques. Due to its simple configuration and zero-voltage switching (ZVS) characteristic, an asymmetric PWM half-bridge converter is one of the most popular topologies using the zero-voltage-transition technique. In addition, the ripple component of the output current due to an output inductor becomes small enough to be handled by an appropriate output capacitor. Being easy to analyze and design and having an output inductor, it is generally used for applications with high output current and low output voltage © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 (e.g. game console power supplies). To handle the large output current, using a synchronous rectifier in the secondary side is popular to obtain the conduction losses as ohmic losses instead of diode losses. In addition, a current doubler increases the utilization of the main transformer when the output current is high. Fairchild’s FSFA-series of green power switches (FPS™) integrates a PWM controller and MOSFETs specifically designed for asymmetric-controlled topologies with minimal external components. Compared with discrete-PWMcontroller-and-MOSFETs solutions, FSFA-series switches can reduce total cost, bill of materials (BOM) list, size, and weight, while simultaneously increasing efficiency, productivity, and system reliability. This application note describes design considerations of an asymmetric PWM half-bridge converter with current doubler and synchronous rectifier employing FSFA-series switches. It includes a step-by-step design procedure as well as the general features and operational principles of the proposed topology. 1. Operational Principles of a Conventional Asymmetric PWM Half-Bridge Converter Figure 1 shows a conventional asymmetric PWM halfbridge converter with a center-tapped transformer. While the switch S1 operates with a duty D, depending on the input voltage and load current, the switch S2 operates with 1-D. During DTS, Vin-VCb is applied on the primary side of the transformer and the secondary diode D1 turns on. The primary current ipri increases since the magnetizing current im of the transformer (not illustrated) and the output inductor current iLo increase together. During (1-D)TS, VCb is applied on the transformer and D2 turns on. The capacitor Cb is not only a voltage source during (1-D)TS but also a DC-blocking capacitor to prevent transformer saturation. When the volt·sec balance for the magnetizing inductance of the transformer is applied, the following is obtained: www.fairchildsemi.com AN-4153 APPLICATION NOTE vCb n:1:1 Cb D1 iD1 iLo vT1 S2 DTS vGS Co Dloss2TS Dloss1TS LO (1-D)TS S1 S2 S1 Vo t Llk Vin ipri ipri S1 t D2 Lm/(Lm+Llk) *(Vin-VCb) iD2 vT1 Figure 1. Conventional Asymmetric PWM HalfBridge Converter with a Center-Tapped Transformer (Vin − VCb ) × D = VCb × (1 − D) ⇒ VCb = DVin Lm/(Lm+Llk) *(-VCb) t iD1 iD2 (1) The volt·sec balance for the output inductor yields: V ⎞ ⎛ ⎛ Vin − VCb ⎞ − VO ⎟ × D = ⎜VO − Cb ⎟ × (1 − D ) ⎜ n n ⎠ ⎝ ⎝ ⎠ t0 t1 (2) t t4 and scaling down vT1 by n. In addition, the applied voltage on the primary side of the transformer during powering modes (t1~t2 and t3~t4) is slightly less than Vin–VCb or –VCb due to the leakage inductance Llk as shown in Figure 3. Therefore, the output voltage equation could be obtained as: Combining Equations (1) and (2), the output voltage is obtained as: (3) ⎛ Lm VO = ⎜⎜ ⎝ Lm + Llk As can be seen in Figure 2 (the gain curve according to the duty cycle using Equation (3) ignoring turns ratio n), the gain is proportional to the duty cycle up to 50% and inversely proportional to it above 50%. Because of this symmetry, the maximum duty cycle should be restricted up to 50% to regulate the output voltage. ⎞⎛ 2 D(1 − D )Vin 4 I O Llk ⎟⎟⎜ − 2 ⎜ n n TS ⎠⎝ ⎞ ⎟ − VF ⎟ ⎠ (4) where IO is the output load current and VF is the forward voltage drop of the secondary side-rectifying diodes. To design the transformer, the magnetizing current must be known. Assume that the magnetizing inductance and the output inductance are high enough for the current ripple on them to be ignored and the leakage inductance is low enough for the duty loss parts to be neglected. Then the current waveforms are simplified as shown in Figure 4. To meet the current·sec balance for Cb, the positive part of the primary current ipri is equivalent to the negative part such that the magnetizing current is obtained as: The loss parts of the duty cycle by the leakage inductance are not considered in Equation (3). Figure 3 shows the key waveforms of the conventional asymmetric PWM halfbridge converter illustrated in Figure 1. Since both secondary rectifying diodes conduct, the voltage across the primary side of the transformer becomes zero during Dloss1TS and Dloss2TS. As a result, the output voltage is not as high as in Equation (3), which is obtained by averaging, rectifying, Figure 2. Normalized Gain Curve © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 t3 Figure 3. Key Waveforms of the Conventional Asymmetric PWM Half-Bridge Converter where n is the turns ratio of the transformer. ⎛ 2 D(1 − D) ⎞ VO = ⎜ ⎟ × Vin n ⎝ ⎠ t2 Figure 4. Simplified Current Waveforms in the Primary Side www.fairchildsemi.com 2 AN-4153 I ⎞ I ⎞ ⎛ ⎛ ⎜ I m + O ⎟ × D = ⎜ − I m + O ⎟ × (1 − D) n n ⎠ ⎝ ⎠ ⎝ I ⇒ I m = (1 − 2 D) O n APPLICATION NOTE 2.1. Operational Principles Figure 7 shows the mode analysis for the asymmetric PWM half-bridge converter with the current doubler and corresponding key waveforms. Assume ZVS is achieved sufficiently with very short duration. The ZVS modes can be ignored in the mode analysis. The ZVS operation is discussed in detail in the next section. Other assumptions are: (5) where Im is the DC component of im. As can be seen in Equation (5), Im could be zero when the duty cycle is 50%. Generally im has a DC offset, so the core saturation has to be taken into account when the transformer is designed. (1) The DC-blocking capacitor Cb is large enough to neglect the voltage ripple on it, and 2. Operational Principles of an Asymmetric PWM Half-Bridge Converter with Current Doubler and Synchronous Rectifier (2) All elements in the circuit are ideal. For low-output-voltage and high-output-current applications, the current doubler is widely used. Figure 5 illustrates the asymmetric PWM half-bridge converter with the current doubler on the secondary side. The secondary winding is a single-ended configuration, while the output inductors are divided into two smaller inductors. To increase the total efficiency, a synchronous rectifier (SR) comprised of MOSFETs with low Rds(on) is used. The current doubler has several advantages compared to the conventional centertapped configuration. First, the DC component of the magnetizing current is lower than or equal to that of the center-tapped configuration, which makes it possible to use the smaller core for the transformer. The amount of the magnetizing current is the same as that of the center-tapped configuration when each output inductor carries half the load current. The amount of the magnetizing current is reduced when the output inductors carry the load current unevenly. Second, the root-mean-square (rms) value of the secondary winding current is smaller than that of the centertapped configuration, since almost half of the load current flows through each output inductor. As a result, the low current density for the secondary winding could be used with the same core and the same gauge of wire. Third, the winding itself is easier than the center-tapped configuration. This is notable especially for multi-output applications because of the limitation of the pin number of the bobbin of the transformer. Fourth, the gate signals for SR are obtained easily and effectively from the output inductors, as shown in Figure 6(b). An appropriate gate voltage (e.g., between 10 V and 20 V) could be easily obtained from the output inductors due to an enough number of turns, while the secondary side number of turns of the transformer is only a few. Additionally, the separated output inductors reduce the burden of the cost of the bigger core. These advantages make the current doubler one of the most popular topologies for high-output-current applications. Figure 5. Asymmetric PWM Half-Bridge Converter with the Current Doubler Figure 6. Methods for Producing the Gate Driver Signal Using; (a) the Transformer; (b) the Output Inductor © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 3 AN-4153 APPLICATION NOTE Let’s start with Mode 2, a powering mode. When S1 turns on, Vin-VCb is applied on the primary side of the transformer. The magnetizing current im increases with the slope of (VinVCb)/Lm. The slope of the current of LO1 is determined by subtracting the output voltage from (Vin-VCb)/n because SR2 turns off. On the other hand, the current of LO2 decreases with the slope of –VO/LO2, which is free-wheeling through SR1. While two output inductors share the load current, SR1 carries the whole load current. The secondary winding of the transformer handles only iLO1 so that iLO1/n is the reflected current to the primary side of the transformer and it is superimposed on the magnetizing current, which constitutes the primary current ipri. In fact, vT2 is slightly lower than the value illustrated in Figure 7 due to leakage inductance (see Chapter 1.) It is ignored in this section to simplify analysis. body diode of SR2 turns on since its reverse-biased voltage is eliminated. Subsequently, both SRs turn on together during this mode. The body diode of S2 turns on after the S2 output capacitance is wholly discharged and that of S1 is entirely charged. Since both SRs turn on, iLO1 and iLO2 are free-wheeling with the slope of –VO/LO1 and –VO/LO2, respectively, and vT1 and vT2 are zero, VCb is applied only on the leakage inductance, causing the primary current’s polarity to change rapidly. When S2 turns on after the S2 body diode conducts, the S2 ZVS condition is achieved. The duration of this mode is obtained as: Dloss 2 = LO1 n:1 SR2 S2 Vin DTS Vo VGS SR1 Dloss2TS Dloss1TS Co Llk S1 (6) Mode 4, another powering mode, starts with the end of commutation between SRs. The applied voltage on the primary side of the transformer is –VCb so that the magnetizing current decreases with the slope of –VCb/Lm and When S1 turns off, Mode 3 begins. As the output capacitance of S2 is discharged, vT1 decreases. It becomes zero when the output capacitance voltage of S2 equals VCb. At this time, the Cb IO Llk × n DVin × TS (1-D)TS S1 S2 S1 t LO2 (a) Mode 1 (t0~t1) Cb LO1 n:1 im (Vin-VCb)/Lm ipri -VCb/Lm t SR2 S2 Co Llk Vin (Vin-VCb)/n Vo vT2 SR1 S1 t -VCb/n LO2 (b) Mode 2 (t1~t2) Cb LO1 n:1 -VO/LO1 iLo2 SR2 S2 -VO/LO2 Co Llk Vin Vo iLo1 ((Vin-VCb)/n-VO)/ LO1 SR1 S1 (VCb/n-VO)/LO2 t LO2 iSR1 (c) Mode 3 (t2~t3) Cb diLo1+diLo2 LO1 n:1 iSR2 diLo1+diLo2 SR2 S2 Co Llk Vin Vo S1 t0 t1 SR1 t2 t3 t4 t (e) Key waveforms LO2 (d) Mode 4 (t3~t4) Figure 7. Mode Analysis and Waveforms for Asymmetric PWM Half-Bridge Converter with the Current Doubler © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 4 AN-4153 APPLICATION NOTE Therefore, not only the energy in the leakage inductance, but also the energy of the load current, helps S1 be discharged from Vin to Vin-VCb. After vDS1 is reduced more than Vin-VCb, the dotted terminal of the transformer changes its polarity, which allows the body diode of SR1 to turn on. Therefore, the magnetizing inductance is short-circuited so that the switches are charged and discharged by the energy in the leakage inductance only from ta. Finally, vDS1 is fully discharged at tb, so the primary current flows through the body diode of S1, as can be seen in Figure 8(c). After tc, the primary current flows through both the channel and the body diode since the gate signal of S1 is applied. Mode 2 begins with the end of the commutation between SRs from t1. the slope of iLO2 is (VCb/n-VO)/LO2. The other inductor current is free-wheeling through SR2. As can be seen in Figure 7, the large ripple on each output inductor is cancelled because of the out-of-phase. Therefore, two smaller inductors can be used in the current doubler configurations compared with the center-tapped or bridge rectifying configurations. When S2 turns off, Mode 1 starts as another regenerating mode. The operating principle of Mode 1 is almost the same as Mode 3, except for a ZVS condition. In Mode 1, vT1 becomes zero at the instant when the output capacitance voltage of S1 is equivalent to Vin-VCb. Before this instant, the load current on the output inductor LO2 is reflected to the primary side of the transformer and helps to meet the ZVS condition of the switches. The energy stored in the leakage inductance only has to discharge and charge the output capacitance of the switches after this instant. Therefore, the ZVS condition for S1 is harder than S2 since Vin-VCb is higher than VCb in general. In all other respects, Mode 1 can be analyzed in the same way as Mode 3. The duration of Mode 1 is obtained as: Dloss1 = IO Llk × n (1 − D)Vin × TS For the ZVS operation of S1 there are three conditions in Figure 8, as follows: (1) The energy in the leakage inductance should be sufficient to discharge S1 from Vin-VCb to zero and charge S2 from VCb to Vin. (2) The instant tb must be earlier than when the primary current changes its polarity. Otherwise, the drain voltages of S1 and S2 are again charged and discharged, respectively. (7) (3) The gate signal of S1 must be applied before the primary current changes its polarity. The detailed output voltage is calculated with Equations (6) and (7) as: VO = Lm Lm + Llk ⎛ D (1 − D)Vin I O Llk ⎜ − 2 ⎜ n n TS ⎝ ⎞ ⎟ − VSR ⎟ ⎠ Figure 9 shows the detailed modes in Mode 3. While the detailed mode analysis is similar to the case of Figure 8, three conditions for the ZVS operation of S2 are different from those of Figure 8: (8) where VSR is the voltage across the MOSFET as an SR during powering modes. It is similar to Equation (4) except for the turns ratio, which is half that of the conventional converter. (1') Since the polarity of the transformer terminals changes when vDS2 reaches VCb, the portion of discharging S2 by the load current, t2~td, is much larger than t0~ta in Figure 8. Therefore, the remaining portion of discharging S2 by the energy in the leakage inductance only is shortened; the ZVS of S2 is easier to achieve compared with the ZVS of S1. Therefore, the energy in the leakage inductance should be sufficient to discharge S2 from VCb to zero and charge S1 from Vin-VCb to Vin. By modifying Equation (5), the DC and ripple components of im are obtained as: I m = (1 − D ) I I LO 2 − D LO1 n n Δim = ( DTS − Dloss1TS ) × (1 − D)Vin Lm + Llk (9) (2') The commutation between SRs begins with the change of polarity of the transformer terminals so that it takes longer from te to the instant when the currents in SRs are equal. In addition, the commutation slope is more sluggish than the case in Figure 8, since the applied voltage on the leakage inductance is reduced to VCb from Vin-VCb. (10) where ILO1 and ILO2 are the DC components of the output inductor currents. 2.2. ZVS Conditions (3') The gate signal of S2 must be applied before the primary current changes its polarity if Condition (3) is satisfied due to the same dead time and the reason mentioned in (2’). In the previous section, the duration for ZVS operation was omitted to simplify mode analysis. More detailed analysis for ZVS operation is given in this section to discover an exact ZVS condition for each switch. Figure 8 shows the detailed modes for Mode 1. With respect to both energy ((1) and (1’)) and timing ((2 & 3) and (2’ & 3’)), the ZVS condition of S1 is more difficult to achieve than that of S2. Therefore, the ZVS condition should be considered with S1 only. In general, the condition for timing is easily satisfied if the condition for energy is From t0, the primary current starts to charge and discharge the output capacitance of the switches. Before the drain voltage of S1, vDS1 reaches to Vin-VCb, the dotted terminal of the transformer is negative so that SR1 is still reverse-biased. © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 5 AN-4153 APPLICATION NOTE An easy way to achieve ZVS for both switches even at light load conditions is to increase Llk. However, the increased Llk increases duty loss parts by reducing the slope of the primary current in Modes 1 and 3. This results in the increase of conduction loss for the reduced effective duty cycle. Therefore, it is not recommended as a method to increase Llk for ZVS at very light load. According to Equation (9), as the load current decreases, the DC component of the magnetizing current decreases as well. If the DC component of im is less than half the ripple component of im, the ZVS operation is performed by the magnetizing inductance with the leakage inductance. In Equation (11), ignoring the second term in the denominator, satisfied. Therefore, the required leakage inductance for the ZVS of both switches at special load condition can be calculated as: Llk > 2COSS [(1 − D)Vin ]2 ⎧⎪ D (1 − D)Vin × TS I O, tar − ⎨ 2n ⎪⎩ 2( Lm + Llk ) (11) ⎛ Lm ⎜⎜1 − L m + Llk ⎝ ⎞ DI O, tar ⎫⎪ ⎟⎟ + ⎬ n ⎪⎭ ⎠ 2 where COSS is the output capacitance of the switch and IO,tar is the target load condition where a designer wants the system to operate in ZVS condition with the leakage inductance. Cb LO1 n:1 SR2 S2 Co Llk Vin Vo SR1 S1 Dloss1TS LO2 (a) Mode 1_1 (t0~ta) Cb LO1 n:1 (1-D)TS DTS S2 S1 t vDS1 SR2 S2 Co Vin-VCb Vo VCb vDS2 Llk Vin SR1 S1 iSR2 iSR1 t LO2 (b) Mode 1_2 (ta~tb) Cb LO1 n:1 t SR2 S2 Co Llk Vin Vo (Vin-VCb)/Llk SR1 S1 t ipri LO2 (c) Mode 1_3 (tb~tc) Cb t0 ta tb tc t1 LO1 n:1 (e) Key waveforms SR2 S2 Co Llk Vin Vo SR1 S1 LO2 (d) Mode 1_4 (tc~t1) Figure 8. Detailed Mode Analysis During Mode 1 © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 6 AN-4153 APPLICATION NOTE rearrangement for Lm yields: Lm + Llk < 2.3. Synchronous Rectifier D(1 − D)Vin × TS ⎧⎪ 2COSS ⎫⎪ DI 2×⎨ (1 − D)Vin − O, tar ⎬ n ⎪⎭ ⎪⎩ Llk It is more profitable that the conduction losses on the secondary rectifying stage are composed of ohmic losses instead of diode losses when the output current is high. Since most of load current flow through the channel, conduction losses can be reduced dramatically if synchronous MOSFETs with very low Rds(on) (less than several mΩ) turn on and off appropriately. In buck-derived topologies such as forward, half-bridge, and full-bridge converters, the gate signal for SR is easily obtained from the (12) To obtain appropriate Lm and Llk using Equations (11) and (12), iterations are necessary. An example of this is given in the next section. Cb LO1 n:1 SR2 S2 Co Llk Vin Vo SR1 S1 Dloss2TS LO2 DTS (a) Mode 3_1 (t2~td) Cb S1 LO1 n:1 (1-D)TS S2 t vDS2 SR2 S2 Co Vin-VCb Vo VCb vDS1 Llk Vin SR1 S1 iSR1 iSR2 t LO2 (b) Mode 3_2 (td~te) Cb LO1 n:1 t SR2 S2 ipri Co Llk Vin -VCb/Llk Vo SR1 S1 t LO2 (c) Mode 3_3 (te~tf) Cb t2 td te tf t3 LO1 n:1 (e) Key waveforms SR2 S2 Co Llk Vin Vo SR1 S1 LO2 (d) Mode 3_4 (tf~t3) Figure 9. Detailed Mode Analysis During Mode 3 © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 7 AN-4153 APPLICATION NOTE main transformer (as shown in Figure 6(a)). Unlike flyback or LLC converters, there is no need to add any other special functions, except for the driving circuit. Ease of construction of the SR driver is another reason to use the half-bridge topology for high-output-load-current applications. 3. Design Procedure and Example In this chapter, a design procedure is shown using the design reference illustrated in Figure 11. The target system for this example is a game console power-supply unit with 12 V of output voltage and 30 A of output load current. To handle the large output-load current, the current doubler with the synchronous rectifier discussed in the previous chapter is used. Since the input comes from a power factor correction (PFC) circuit, the input-voltage range is not wide. Moreover, when using a current doubler, it is more efficient to get the gate signal from the output inductors than from the main transformer (as shown in Figure 6(b)). Figure 10 illustrates the SR gate-signal waveforms from the main transformer and the output inductors. It is difficult to tune up the turns ratio na and nb (where na=ns/n1 and nb=ns/n2) to make the sufficient gate voltage (as shown in Figure 10(a)). This is because both positive and negative parts are dependent not only on the turns ratios na and nb but also on VCb. Additionally, the power loss by the negative part of the gate signal is determined by the turns ratio, VCb, and Vin. On the other hand, the power loss by the negative part of the gate signal does not depend on the load condition (as shown in Figure 10(b), where nc=n3/n1 and nd=n4/n2). In addition, during the duty loss part, Dloss1TS and Dloss2TS, the gate signals change their polarity to a negative value so that the SRs turn off rapidly and definitely. This helps to reduce the turn-off loss of the synchronous MOSFETs. Dloss1TS [STEP 1] System Specifications The first step in designing is to define the system specifications. Generally, a PFC circuit is used for mediumor high-power applications such as LCD/PDP TV systems, game console power supplies, and beam projectors to meet international harmonic regulations. Thus, the input voltage range for the main power stage (i.e. the output voltage of PFC stage) is almost fixed (e.g., 370~410 Vdc). However, the input voltage range may be widened to meet special requirements. In this chapter, the target specifications are: ▪ Nominal input voltage: 390 Vdc ▪ Input voltage range: 370~410 Vdc ▪ Output voltage: 12 V ▪ Output current: 30 A ▪ Switching frequency: 100 kHz Dloss2TS (Vin-VCb)/n vT2 -VCb/n t [STEP 2] Turns Ratio and Duty Cycle (Vin-VCb)/n/na The output voltage equation (Equation 8) is used to determine turns ratio n. However, the output voltage equation contains the leakage and magnetizing inductance, which are not yet determined. Therefore, a designer should make assumptions for the following: vgate_SR1 -VCb/n/na vgate_SR2 t VCb/n/nb t ▪ ▪ ▪ -(Vin-VCb)/n/nb t0 t1 t2 t3 t4 (a) Dloss1TS vgate_SR1 ▪ Same shape as vLO1 except for nc Dloss2TS According to Equation (8), the turns ratio is obtained as: ((Vin-VCb)/nVO)/nc -VO/nc t Dn (1 − Dn )Vin , n + ( Dn (1 − Dn )Vin, n ) 2 − 4(VO + VSR ) Same shape as vLO2 except for nd vgate_SR2 n= (VCb/n-VO)/nd -VO/nd t0 t1 VSR considering Rds(on) of used MOSFETs as an SR; α, the ratio between Lm and Lm+Llk; The leakage inductance that will be changed later by iterations of Equations (11) and (12); The nominal duty cycle at the nominal input voltage. α t t2 t3 2(VO + VSR ) where Vin,n and Dn are the nominal input voltage and the nominal duty cycle at Vin,n, respectively. t4 (b) For turns ratio n, the duty cycle at an input voltage and a load current is calculated as: Figure 10. Gate Signals for Synchronous Rectifier; (a) Transformer Coupled; (b) Output Inductor Coupled ⎛ n(VO + VSR ) I O Llk 1 − 1 − 4⎜⎜ + αVin nVinTS ⎝ D= 2 © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 I O Llk αTS (13) ⎞ ⎟⎟ ⎠ (14) www.fairchildsemi.com 8 AN-4153 APPLICATION NOTE C102 Dvcc Vcc Cvcc #7 FUSE1 R107 #9 C107 #2 VFB Cfb R102 L202 Csn1 Cout2 N4 Rdr1 Vo = 12V/30A Rdr2 Qdr2 Rsn2 Cout1 Csn2 L201 Q102 #10 #4 CS Qdr1 Ns Chvcc Control IC U2-2 Clink Np H-Vcc #3 RT Vin PFC output Rsn1 Q101 #1 VDL L-Vcc R105 Ddr1 N1 Rvcc VCTR N2 Ddr2 N3 U1 FSFA2100 #5 SG #6 C103 PG R101 Figure 11. Reference Design Schematic Design Example For the example, the following values are assumed: ▪ ▪ ▪ D@ 30% VSR = 0.3V α is 0.95 The initial leakage inductance is 20 µH. This may be increased after checking the ZVS condition. Taking the core size for 360W into account, if Llk is less than this value, the productivity is not good. The nominal duty cycle at 390 Vdc is 0.4. ▪ Since COSS of the FSFA2100 MOSFETs is 150 pF, the required leakage inductance is obtained with [email protected]%=0.305 as: n= 2COSS [(1 − D )Vin ] 2 Llk > Using these values, the turns ratio is obtained by Equation (13) as: 0.4 ⋅ 0.6 ⋅ 390 + (0.4 ⋅ 0.6 ⋅ 390) 2 − 4(12 + 0.3) ⎛ 6.5 ⋅ (12 + 0.3) ⎞ 9 ⋅ 20 μ 1 − 1 − 4⎜⎜ + ⎟⎟ μ 0 . 95 ⋅ 410 6 . 5 ⋅ 410 ⋅ 10 ⎝ ⎠ = 2 = 0.305. ⎧ D (1 − D )Vin × TS I O ,tar ⎛ Lm ⎞ DI O ,tar ⎫ − ⎨ ⎜⎜1 − L + L ⎟⎟ + n ⎬ L L n 2 ( ) 2 + m lk m lk ⎠ ⎝ ⎭ ⎩ 2 2 ⋅150 p ⋅ [(1 − 0.305) ⋅ 410] 2 = 30 ⋅ 20μ 0.95 ⋅ 10μ ⎧ 0.305 ⋅ (1 − 0.305) ⋅ 410 × 10 μ ⎞ 0.305 ⋅ 9 ⎫ 9 ⎛ 400 μ − ⎨ ⎜⎜1 − 400 μ + 20μ ⎟⎟ + 6.5 ⎬ + × 2 ( 400 μ 20 μ ) 2 6 . 5 ⎝ ⎠ ⎩ ⎭ = 12.0 μH . 2(12 + 0.3) 0.95 = 6.52 2 The required leakage inductance is 12.0 µH, which is too small to control in a mass production. If the obtained value is larger than the assumed value, the obtained value is used, and the SMPS designer must repeat Step 2 to check if the turns ratio is still valid. However, in this design example, the designer chooses the initial value for productivity, and there is no iteration needed. yielding a turns ratio is 6.5. The nominal duty cycle at the nominal input voltage is recalculated by Equation (14) as: ⎛ 6.5 ⋅ (12 + 0.3) 30 ⋅ 20 μ ⎞ + 1 − 1 − 4⎜⎜ ⎟ 6.5 ⋅ 390 ⋅ 10 μ ⎟⎠ ⎝ 0.95 ⋅ 390 Dn = = 0.397 2 [STEP 3] Magnetizing and Leakage Inductance Using the turns ratio obtained in Step 2, the ZVS condition could be checked with Equations (11) and (12). Design Example This example is designed to achieve the ZVS operation from full- to 30%-load condition using the leakage inductance and the magnetizing inductance. The duty cycle at 30% load condition and the maximum input voltage is obtained by Equation (14) as: © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 Figure 12. Primary Current Waveform www.fairchildsemi.com 9 AN-4153 APPLICATION NOTE addition, it is better to choose a wire with multi-strands of thinner wire, such as Litz wire, to minimize the skin effect. The magnetizing inductance can be determined using Equation (12) as: D(1 − D)Vin × TS Lm + Llk < When the output inductor current ripple is ignored, Figure 12 shows the primary-current waveform. The rms value of this waveform is given as: ⎧ 2COSS ⎫ DI 2× ⎨ (1 − D)Vin − O,tar ⎬ n ⎭ ⎩ Llk 0.305 ⋅ (1 − 0.305) ⋅ 410 ⋅10μ = ⎧ 2 ⋅150 p 0.305 ⋅ 9 ⎫ 2× ⎨ (1 − 0.305) ⋅ 410 − ⎬ 6.5 ⎭ ⎩ 20μ = 638μH . ( I P12 + I P1I P 2 + I P 2 2 ) ( I 2 + I P3 I P 4 + I P 4 2 ) D + P3 (1 − D) 3 3 iP rms = Therefore, Lm is selected as 600 µH. I P1 = I LO1 Δi + Im − m 2 n (20) I P2 = I LO1 Δi + Im + m 2 n (21) [STEP 4] Transformer Using Equations (9) and (10), the peak magnetizing current is obtained as: I Δi I I m + m = (1 − D) LO 2 − D LO1 2 n n (1 − D)Vin + ( DTS − Dloss1TS ) × 2( Lm + Llk ) (15) IO 2n Lmim max Ae Bmax I LO 2 Δi + Im − m 2 n (23) im max = IO 30 = = 2.31A. 2 n 2 ⋅ 6 .5 The given core is EER4042 (Ae=158 mm2). The minimum turns number for the transformer primary side is calculated as: (17) NP min Lmim 600 μ ⋅ 2.31 = = 38.14 Ae Bmax 158μ ⋅ 0.23 max = When NP is selected as 39, the secondary turns number is obtained as 6. Use Equations (19)-(23) to get the rms value of the transformer primary side current. Assume each output inductor carries the output load current evenly at the nominal condition (Dn=0.397). (18) where NP is larger than NPmin in Equation (17). The diameter of the wire is selected based on the current density, whose range is generally 4~10 A/mm2. It is recommended to select the current density as low as possible to reduce conduction losses on the wire. However, try to reduce the winding layers at the same time. The more winding layers, the more circulating current caused by the proximity effect. Sometimes trying to reduce conduction losses by increasing the wire diameter makes conduction losses increase by increasing the circulating current. In © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 I P4 = − When the duty cycle is zero, the maximum im is obtained as: (16) The number of turns for the transformer secondary side is obtained as: NP n (22) Design Example where Ae is the effective cross-sectional area of the used core in mm2, and Bmax is the maximum flux density in Tesla. Bmax = 0.2~0.25 T is recommended if there is no reference data. NS = I LO 2 Δi + Im + m 2 n For the secondary-side winding, half the load current is the rms value when it is assumed that each output inductor carries the load current evenly and the ripple on the output inductor is small enough to be ignored. The minimum number of turns for the transformer primary side is given as: N P min = I P3 = − where Im and Δim are defined in Equations (9) and (10). The maximum value of the peak magnetizing current occurs when each output inductor carries half the load current for the worst case and the duty cycle is zero during startup or transient instant. Therefore, the maximum im is: im max = (19) I P1 = 15 1.357 + 0.475 − = 2.10 A 6 .5 2 I P2 = 15 1.357 + 0.475 + = 3.46 A 6.5 2 I P3 = − 15 1.357 + 0.475 + = −1.15 A 2 6.5 I P4 = − 15 1.357 + 0.475 − = −2.51A 2 6 .5 www.fairchildsemi.com 10 AN-4153 APPLICATION NOTE Therefore, the rms value of the transformer primary side current is obtained by Equation (19) as: i P rms = fS = ( 2.12 + 2.1 ⋅ 3.46 + 3.46 2 ) 0.397 3 For the design example, the frequency setting resistor R105 is selected as 27 kΩ for 100 kHz operation. [STEP 7] DC-Blocking Capacitance The rms value of the transformer secondary side current is half the load current, so iSrms=15 A. It has been assumed that the DC-blocking capacitor is large enough to neglect the voltage ripple on it. However, too large a DC-blocking capacitor leads to slow dynamic response. Therefore, it is recommended to make the voltage ripple on the DC-blocking capacitor around 10% of the input voltage. The voltage ripple on the DC-blocking capacitor is obtained as: Since the diameter of the wire becomes too thin, it is not easy to wind 39 turns for the primary side of the transformer in two layers. Choose the biggest wire that can be wound 13 turns in one layer of the bobbin for EER4042. Due to consideration of the skin effect, Litz wire of 100 strands with AWG38 (American wire gauge) is selected as the primary wire. In this case, the current density is around 2.9 A/mm2. For the secondary side, 250-strand Litz wire with AWG36 is chosen where the current density is around 4.7 A/mm2. ΔvCb [STEP 5] Output Inductance (V + VSR )(1 − D + Dloss1 )TS = O (24) LO1 (VO + VSR )( D + Dloss 2 )TS ⎛ Dloss1TS × I P1 Dloss 2TS × I P 2 ⎞ + ⎜ ⎟ 1 2 2 ⎟ C102 ≅ ×⎜ 2ΔvC102 ⎜ ( D − Dloss1 )TS × ( I P1 + I P 2 ) ⎟ ⎜+ ⎟ 2 ⎝ ⎠ ⎛ 0.039 × 10μ × 2.1 0.06 × 10μ × 3.47 ⎞ + ⎜ ⎟ 1 2 2 ⎜ ⎟ = × 2 × 30 ⎜ (0.397 − 0.039) × 10μ × (2.1 + 3.47) ⎟ ⎜+ ⎟ 2 ⎝ ⎠ = 190nF (25) LO 2 Design Example In the design example, the ripple on each output inductor is selected to be less than 20% of the rated output load current. The inductances are calculated as: Therefore, 220 nF is selected as the DC-blocking capacitor. (V + VSR )(1 − D + Dloss1 )TS = O [STEP 8] Sensing Resistor ΔiLO1 = L202 = = (12 + 0.3)(1 − 0.397 + 0.039) × 10 μ = 13.2μH 6 The pulse-by-pulse current limit of the FSFA-series switches can be adjusted by changing R101 in Figure 11. It is determined by the peak of the primary current obtained using Equation (21) when the input voltage is maximized. Due to the ripple current of the magnetizing inductance, the maximum peak of the primary current happens when the input voltage is maximized. (VO + VSR )( D + Dloss 2 )TS ΔiLO 2 (27) When the voltage ripple on the capacitor is 30 V, the DCblocking capacitance is calculated using Equation (27) as: In general, the current ripple on the output inductor is set to 10-20% of the rated output load current. L201 ⎛ Dloss1TS × I P1 Dloss 2TS × I P 2 ⎞ + ⎜ ⎟ 1 ⎜ 2 2 ⎟ ≅ × 2Cb ⎜ ( D − Dloss1 )TS × ( I P1 + I P 2 ) ⎟ ⎜+ ⎟ 2 ⎝ ⎠ Design Example The output inductor current ripple is given as: ΔiLO 2 = (26) Design Example ((−1.15) 2 + ( −1.15)(−2.51) + ( −2.51) 2 ) (1 − 0.397) 3 = 2.29 A. + ΔiLO1 27 kΩ × 100 [kHz] R105 . (12 + 0.3)(0.397 + 0.060) × 10μ = 9.4 μH 6 Design Example To increase productivity, both output inductors are selected as the same value, 15µH. Continuing with the example, calculate the duty cycle at the maximum input voltage and full-load conditions using Equation (14). [STEP 6] Operating frequency In Figure 11, the operating frequency fS is obtained by using the following equation when FSFA-series is used. © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 11 AN-4153 APPLICATION NOTE D@ 410V ,100% ⎛ n(VO + VSR ) I O Llk + 1 − 1 − 4⎜⎜ nVinTS ⎝ αVin = 2 An N-channel power MOSFET with 8mΩ of Rds(on) and 100V of the voltage rating, HUF75652G3, is selected for both SRs with consideration of the voltage ringing and overshoot. ⎞ ⎟⎟ ⎠ ⎛ ⎞ ⎜ 6.5 ⋅ (12 + 0.3) 30 × 20 μ ⎟ + 1 − 1 − 4⎜ ⎟ ⎜ 600 μ 620 μ × 410 6.5 × 410 × 10 μ ⎟ ⎝ ⎠ = 2 = 0.338. The voltages across the output inductors during powering modes are: VL 201min = − VO n (1 − 0.458) ⋅ 370 = − 12 = 19 6 .5 Then, the peak of the primary current is obtained combining Equations (7), (9), (10), and (21) as: I P2 (1 − D@ 370V ,100% ) ⋅ 370 Δi I = LO1 + I m + m 2 n I LO1 I I = + (1 − D) LO 2 − D LO1 n n n 1⎛ I O Llk ⎞ (1 − D)Vin ⎟× + ⎜⎜ DTS − 2⎝ n(1 − D)Vin ⎟⎠ Lm + Llk VL 201max = (1 − 0) ⋅ 410 410 − VO = − 12 = 51 n 6.5 VL 202 min = 0 × Vin − VO = −12 n VL 202 max = 15 15 15 = + (1 − 0.338) − 0.338 ⋅ 6.5 6.5 6.5 ⎞ (1 − 0.338) ⋅ 410 1⎛ 30 × 20 μ ⎟× + ⎜⎜ 0.338 × 10μ − 6.5 ⋅ (1 − 0.338) ⋅ 410 ⎟⎠ 600μ + 20 μ 2⎝ D@ 370V ,100% × 370 n − VO = 0.458 × 370 − 12 = 14 6.5 To protect the SRs, the gate signal has to be restricted ±20 V. The turns ratios between the output inductors and the windings for the gate drivers are: N3 =3 N1 = 3.72 A Since the internal threshold voltage for the pulse-by-pulse current limit is -0.58 V, 0.1 Ω is selected as the sensing resistor, R101. N4 =1 N2 [STEP 9] Synchronous Rectifier [STEP 10] External Soft Start The voltage stresses on the SRs are calculated as: At startup, the duty cycle starts increasing slowly to establish the correct working conditions for transformers, inductors, and capacitors. The voltage on the output capacitors is progressively increased to smoothly establish the required output voltage. For the FSFA-series, the softstart time is internally implemented for 15ms when the operating frequency is set to 100 kHz. In addition, to help the soft-start operation, a capacitor and a resistor are connected on the RT pin externally, as shown in Figure 11. Before the power supply is powered on, the capacitor C107 remains fully discharged. After power-on, C107 becomes charged gradually by the current through the RT pin, which determines the operating frequency. The current through the RT pin is inversely proportional to the total impedance of the connected resistors. The total impedance during startup is lower than that of the normal operation because R107 is added on R105 in parallel, which means the operating frequency decreases continuously from higher to nominal. Eventually, C107 is fully charged to the RT pin voltage and the operating frequency is determined by R105 only. VSR1 = DVin n (28) VSR 2 = (1 − D)Vin n (29) For windings to drive the gate of the SRs, during powering modes the voltages across the output inductors are: VLO1 = (1 − D)Vin − VO n (30) VLO 2 = DVin − VO n (31) Design Example Considering the worst case for each SR, the voltage stresses on them are: VQ101 = DVin 0.5 × 410 = = 32 n 6 .5 VQ102 = (1 − D)Vin (1 − 0) × 410 = = 64 n 6 .5 © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 During C107 charging time, the operating frequency is higher than during normal operation. In asymmetric PWM halfbridge converters, a switching period contains powering and commutation periods. The energy cannot be transferred to the output side during the commutation period. Since the DC www.fairchildsemi.com 12 AN-4153 APPLICATION NOTE link voltage applied to the VDL pin and the leakage inductance of the main transformer are fixed, the powering period over the switching period is shorter in high switching frequencies. As C107 is charged, the switching frequency decreases so that the powering period over the switching period increases. It is helpful to start SMPS with the internal soft-start time together. 4. Design Summary Figures 13 and 14 show the full schematic of the reference design and its transformer configuration. Table 1 shows the detailed wire information of the transformer. The electrical features of the transformer are described in Table 2. Design Example In the design example, 2.2 µF and 12 kΩ are selected as C107 and R107, respectively. Figure 13. Full Schematic with Components Values of the Reference Design 1 16 2 15 3 14 4 13 NP NS 5 12 6 11 7 10 3 mm 3 mm 8 1 27 28 29 30 31 32 33 34 35 36 37 38 39 26 25 24 23 22 21 20 19 18 17 16 15 14 1 2 3 4 5 6 7 8 9 10 11 12 13 5 4 9 6 9 8 3 2 1 16 Figure 14. Transformer Construction of the Reference Design © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 13 AN-4153 APPLICATION NOTE Table 1. Transformer Winding Specifications Pin No. Winding (start → end) Wire Turns Winding Method 39 T Solenoid 6T Solenoid Insulation Tape (25 um) 1T Litz wire (AWG38×100 strands) (1) 1Æ8 NP 2 Insulation Tape (25 um) 1T NS 1 16 Æ 9 Litz wire (AWG36×250 strands) NOTE: 1. Insulation tape (25 µm, 1T each) should be inserted between the layers. Table 2. Transformer Electrical Characteristics Pin Spec. Remark Magnetizing Inductance (Lm) 1－8 600 µH (typical) (600 µH ± 5%) Leakage Inductance (Llk) 1－8 20 µH ± 10% 100 kHz, 1 V All other pins open 100 kHz, 1 V All other pins shorted illustrated in Figure 6(a). As can be seen in Figure 17, since the winding voltage decreases to the negative at the turn-off transition in the upper waveforms, the SRs are turned off more rapidly and definitely than in the transformer coupling case. In Figure 17, the smaller negative parts in the inductor coupling case are shown compared to the transformer coupling case. The smaller negative parts allow the power losses on the gate driver circuit for SRs to be reduced. 5. Experimental Results Figures 15 and 16 show the experimental waveforms of the converter designed in the previous chapter at the nominal input and the full-load condition. The gate signal of S1, the primary- and secondary-side voltages across the main transformer, and the primary current are shown in Figure 15. These waveforms are consistent with the theoretical analysis, including the ZVS operation. The output inductor currents and the SRs’ currents are shown in Figure 16. The output inductor currents are unbalanced due to the duty cycle and the parasitic components, which means the averaged magnetizing current is smaller than that of the center-tapped configuration.[1] The ZVS operations at various load conditions are shown in Figure 18. The drain voltage and the gate signal of the lower side switch are displayed. As designed in the previous chapter, the converter shows ZVS operation downs to 30% load condition. The efficiency of the converter is shown in Figure 19. The measured efficiencies are 93.7%, 94.6%, and 93.1% at 20%, 50%, and 100% of the rated load condition, respectively. It shows a marginal performance so that the 85 PLUS program can be achieved with well-designed PFC and DC-DC stages. Figure 17 shows the winding voltages for the gate driver circuits of SRs at the full-load condition. The upper waveforms are for the inductor coupling illustrated in Figure 6(b), while the lower ones are for the transformer coupling Figure 16. Waveforms for the Secondary Side Figure 15. Experimental Waveforms © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 14 AN-4153 APPLICATION NOTE 96 Efficiency [%] 95 93.7% 94 94.6% 93 93.1% 92 91 90 89 88 10 20 30 40 50 60 70 80 90 100 Load [%] Figure 19. Measured Efficiency of the Designed Converter 6. References [1] Hong Mao, Songquan Deng, Yangyang wen, and Issa Batarseh, “Unified Steady-State Model and DC Analysis of Half-Bridge DC-DC Converters with Current Doubler Rectifier,” APEC '04. Nineteenth Annual IEEE, Vol. 2, 2004, pp. 786-791. Figure 17. Gate Signals for SR; Upper Waveforms Use Output Inductors; Lower Waveforms Use the Transformer [2] Yu-Chieh Hung, Fu-San Shyu, Chih Jung Lin, and YenShin Lai, “Design and Implementation of Symmetrical HalfBridge DC-DC Converter”, The Fifth International Conference on PEDS 2003. Vol. 1, Nov. 2003 pp. 338-342. vgs_S1 (20V/div.) [3] Panov, Y. and Jovanovic, M.M., “Design and Performance Evaluation of Low-Voltage / High-Current DC/DC On-board Modules,” IEEE Transactions on Power Electronics, Vol. 16, Issue 1, Jan. 2001 pp. 26-33. vds_S1 (200V/div.) (a) ZVS fail begins vgs_S1 (20V/div.) vds_S1 (200V/div.) (b) Figure 18. ZVS Verification; (a) at 40% Load; (b) at 30% Load © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 www.fairchildsemi.com 15 AN-4153 APPLICATION NOTE DISCLAIMER FAIRCHILD SEMICONDUCTOR RESERVES THE RIGHT TO MAKE CHANGES WITHOUT FURTHER NOTICE TO ANY PRODUCTS HEREIN TO IMPROVE RELIABILITY, FUNCTION, OR DESIGN. FAIRCHILD DOES NOT ASSUME ANY LIABILITY ARISING OUT OF THE APPLICATION OR USE OF ANY PRODUCT OR CIRCUIT DESCRIBED HEREIN; NEITHER DOES IT CONVEY ANY LICENSE UNDER ITS PATENT RIGHTS, NOR THE RIGHTS OF OTHERS. LIFE SUPPORT POLICY FAIRCHILD’S PRODUCTS ARE NOT AUTHORIZED FOR USE AS CRITICAL COMPONENTS IN LIFE SUPPORT DEVICES OR SYSTEMS WITHOUT THE EXPRESS WRITTEN APPROVAL OF THE PRESIDENT OF FAIRCHILD SEMICONDUCTOR CORPORATION. As used herein: 1. Life support devices or systems are devices or systems which, (a) are intended for surgical implant into the body, or (b) support or sustain life, or (c) whose failure to perform when properly used in accordance with instructions for use provided in the labeling, can be reasonably expected to result in significant injury to the user. © 2008 Fairchild Semiconductor Corporation Rev. 1.0.0 • 12/9/08 2. A critical component is any component of a life support device or system whose failure to perform can be reasonably expected to cause the failure of the life support device or system, or to affect its safety or effectiveness. www.fairchildsemi.com 16

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