Under the Hood of a Multiphase Synchronous Rectified Boost Converter

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Power Supply Design Seminar Topic 4 Presentation: Under the Hood of a Multiphase Synchronous Rectified Boost Converter Reproduced from 2014 Texas Instruments Power Supply Design

1 Power Supply Design Seminar Topic 4 Presentation: Under the Hood of a Multiphase Synchronous Rectified Boost Converter Reproduced from 2014 Texas Instruments Power Supply Design Seminar SEM2100, Topic 4 TI Literature Number: SLUP , 2015 Texas Instruments Incorporated Power Seminar topics and online power training modules are available at: ti.com/psds

2 Under the Hood of a Multiphase Synchronous Rectified Boost Converter David Baba

3 Agenda Synchronous boost introduction - Deciding how many phases to use Synchronous multiphase boost waveforms Design example single phase/two phase - Component selection - Loss calculations - Compensation Results Summary 4-2

4 Changing to Synchronous Rectification Losses = (1 D) I out VF Losses = I RMS 2 RDS on Losses = I out V F Losses = I RMS 2 RDS on 4-3

5 Boost Converter Basic Operation Boost During D ON Period Simple Boost Diagram Boost During D OFF Period 4-4

6 Determine Input Current per Phase Drawing Comparisons Between Buck and Boost P in_boost = P out_boost η P out_boost = V out_boost I out_boost V out_buck V in_boost I out_buck VIin_Boost in_boost Canonical Schematic 4-5

7 Interleaved Boost Basic Operation Vin Iin Ave + - I L1 V SW1 I D1 Iout I CinRMS I SW1 I CoutRMS + Vout Rout - Cout V SW2 Inductor Currents I L2 ISW2 I D2 V SW1 V SW2 Vout Vout Vin Vin I L1 Vin L D Fsw T IinAve n I L2 Vin L D Fsw T IinAve n Phase 1 Inductor Current T Phase 2 Inductor Current T 4-6

8 Interleaved Boost Basic Operation Boost Switch Currents V in L D F sw V in L D F sw Phase 1 Switch Current Phase 2 Switch Current 4-7

9 Interleaved Boost Basic Operation Rectifier Switch Currents V in L D F sw V in L D F sw I out n Iout n Phase 1 Rectifier Current Phase 2 Rectifier Current 4-8

10 Interleaved Boost Basic Operation V in L D F sw Input Capacitor Currents V in L D F sw DC content removed Ripple currents cancel at 50% 4-9

11 Interleaved Boost Basic Operation Output Capacitor Currents V in L D F sw I out n V in L D F sw I out n Ripple currents cancel at 50% 4-10

12 Design Example The Basic Boost Calculations Automotive trunk amplifier 14 V in 24 V 8 A Switching Frequency (F sw ) khz (single phase) khz (two phase) system switching frequency held constant Equation Single Phase Two Phase Comment V out V in = 1 1 D D = V out V in V out 250 khz 125 khz Per Phase F SW Transfer Function D = 0.42 D = 0.42 Rearranging for D P = P out in η P I = in in_ Avg V n in P in = 206 W P in = 206 W 7.3A 206W/14V W/14V 1 Efficiency Est. 93% n = No. Phase 4-11

13 Selecting the Right Inductor and Inductance Calculations Graph showing size factor as a function of ΔI L Slope reduces after ΔI L = I in Set ΔI L = 50% of I in_avg ΔI L I in 4-12

14 Boost Inductor Losses Equation Single Phase Two Phase Comment ΔI L = 0.5 I in _ Avg I L _ peak = ΔI L 2 + I in _ Avg L = V ind D ΔI L F SW 7.5 A 3.5 A Set ΔI L to 50% of l in_avg = 18.5 A = 8.4 A I sat to be set higher than I L_peak = 3.16 µh = 13.4 µh Boost inductor calculation 2 2 ΔI I L _ RMS = I in _ Avg + L 12 Selected Inductor =14.9 A = 7A Coilcraft XAL Coilcraft SER Inductor size 13.2, 14.1, , 13.5, 9 RMS current for DCR Loss 2 cores for the two phase Volume of inductor (in mm) DCR 3 mω 14 mω 2 DCR loss = I L _ RMS DCR = 0.6 W = 1.4 W Total DCR losses Coreloss = K 1 f x B y V E = 2.6 W =18 mw Total from online calculator 4-13

15 AC Inductor Losses Coreloss = K 1 f x B y V E K 1 : Constant of the core material f: Switching frequency in khz - Higher frequencies results in higher losses B: Flux density in kguass - Lower flux density results in lower losses x: Frequency exponent for a specific core material y: is the flux exponent for a specific core material V E : Core volume - Larger volume results in more losses 4-14

16 Boost Convertor MOSFET Considerations VDS rating must be greater than output voltage - 25% margin is generally acceptable Calculate losses to determine suitability Losses ideally should be distributed evenly between conduction losses and switching losses - Higher RMS currents result in larger conduction losses - Higher gate charge results in higher switching losses 4-15

17 Control MOSFET Losses Equation Single Phase Two Phase Comment FET Selected CSD18531Q5A CSD18531Q5A I FET_RMS = D I in _ Avg FET Cond =I FET_ RMS 2 RDSon 9 A 0.3 W 0.8 W 4.47 A 0.16 W 0.4 W RDS on, 3 mω, hot 4 mω FET RMS current Total conduction losses Total transitional losses 4-16

18 Control MOSFET Transitional Losses Transitional Losses at Turn On SW TRANSLoss = Vout V in I in _ Avg T SLEW F SW Use triangular approximation - 1/2 base height - For worst case, the 1/2 drops out 4-17

19 Synchronous MOSFET Losses Equation Single Phase Two Phase Comment FET Selected CSD18531Q5A CSD18531Q5A RDS on, 3 mω, hot 4 mω I FET_RMS = 1 D I in _ Avg 11.2 A 5.3 A FET RMS current FETCond = I (FET_RMS) 2 RDS on 0.44 W 0.22 W Conduction losses Q OSSLoss = Q OSS 2 V out F SW 0.2 W 0.2 W Q OSS losses for both FETs Q RR_Loss = Q RR V out n F SW 0.6 W 0.6 W 100 nc of Q RR losses in boost FET I C_Loss = {( ) + I Q } V in n Q Gtot F SW 0.18 W 0.33 W Loss total in IC 4-18

20 Boost Converter FET Switching Control FET: Turn On V GATE V DS I D 4-19

21 Boost Converter FET Switching Control FET: Turn Off V GATE I D V DS 4-20

22 Boost Converter FET Switching Synchronous FET: Turn On V GATE V DS I D 4-21

23 Boost Converter FET Switching Synchronous FET: Turn Off V DS I D V GATE 4-22

24 Input/Output RMS Ripple Current Single Phase Two Phase Comment I C_in_RMS = ΔI L 12 = 2.1 A I C_in_RMS = ΔI L D 1 D = 0.9 A Two phase D < 0.5 I C_out_RMS I out D (1 D) = 6.7 A I C_out_RMS = 2.5 A I out 2 D (1 2D) (1 D) Two phase D < 0.5 ΔI C_out I in _ Avg =14.75 A ΔI C_out I in _ Avg = 6.8 A 2 x PCV1E391MCL2GS 1 x PCV1E391MCL2GS Pk-Pk ripple current in C OUT 390 µf electrolytic selected 4-23

25 Output Ripple Voltage Calculations Single Phase Two Phase Comment V C_out_Ripple = ΔI D C_out = 29 mv V F SW C C_out_Ripple = ΔI D C_out = 29 mv out C out F SW Ripple voltage due to charge C out ΔI Cout I out n (1 D) = 13.7 A ΔI Cout I out n (1 D) = 6.89 A V C_out_Ripple_ESR = ΔI C_out C out _ ESR V C_out_Ripple_ESR = ΔI C_out C out _ ESR Ripple voltage = 144 mv = 144 mv due to C outesr V out _ Ripple =! V C_out_Ripple 2 + V C_out_Ripple_ESR 2 = 147 mv! V out _ Ripple = V (C_out_Ripple) 2 + V (C_out_Ripple_ESR) 2 = 155 mv Total ripple voltage 4-24

26 C in RMS Ripple Current Rating Multiphase Boost Comparison of ripple current cancelation Boost convertor 1, 2, 3 and 4 phase approach Using a I L of 1 A peak to peak 4-25

27 Approximation C out RMS Ripple Current Rating Multiphase Boost Output ripple current cancelation for a boost convertor I out of 1 A using a 1, 2, 3 and 4 phase approach 4-26

28 C in RMS Ripple Current Condition 0 < D < 1 Condition 0 < D < < D < 1 Condition 0 < D < < D < < D < 1 ΔI L 12 Single Phase Two Phase ΔI L 12 ΔI L 12 ΔI L 12 2D 1 D Three Phase ΔI L D 1 D 1 3D 1 D (1 3D) (3D 2) 3D (1 D) ΔI L 12 3D 2 D 4-27

29 C in RMS Ripple Current Condition 0 < D < < D < < D < < D < 1 ΔI L 12 ΔI L 12 Four Phase ΔI L D 1 D (1 4D) (4D 2) 4D (1 D) (3 4D) (4D 2) 4D (1 D) ΔI L 12 4D 3 D 4-28

30 C out RMS Ripple Current* Condition Single Phase 0 < D < 1 I OUT D (1 D) Condition Two Phase 0 < D < 0.5 I OUT 2 D (1 2D) (1 D) 0.5 < D < 1 I OUT 2 2 (2D 1) 1 D Condition Three Phase 0 < D < < D < 0.66 I OUT 3 I OUT 3 D (1-3D) (1 D) (3D 2) (1 3D) (1 D) 0.66 < D < 1 I OUT 3 3D 2 1 D *Approxima5ons 4-29

31 C out RMS Ripple Current* Condition Four Phase 0 < D < 0.25 I OUT 2 D (1 4D) (1 D) 0.25 < D < 0.5 I OUT 2 (4D 2) (1 4D) 2 (1 D) 0.5 < D < 0.75 I OUT 2 (4D 2) (3 4D) 2 (1 D) 0.75 < D < 1 I OUT 2 4D 3 1 D *Approxima5ons 4-30

32 Loop Stability of a Current Mode Boost Current mode control modifies the complex conjugate double pole to two separate poles - The inductor pole pushes to a higher frequency Typically use current mode control (LM5122) Right Half Plane Zero (RHPZ) - RHPZ causes sudden decrease in the 1-D period due to control loop increasing D for sudden load step - Adds additional phase drop of negative 90 o phase shift Cross over frequency below RHPZ frequency to avoid additional phase shift For current mode control, duty cycles approaching 0.5 and beyond require modification to the current sense to avoid subharmonic oscillation 4-31

33 Loop Stability of a Dual Phase Current Mode Boost Adjustments to accommodate an interleaved configuration Divide down the output capacitor by number of phases - C out becomes 195 µf from 390 µf Multiply the output capacitor ESR by number of phases ESR - ESR becomes 40 mω, from 20 mω Multiply R out by number of phases - R out becomes 6 Ω from 3 Ω - All other elements stay the same 4-32

34 Current Mode Boost Power Stage Variable Equation V OUT V C AVC 1 S ω R 1+ S ω Z 1+ S 1+ S ω P ω L R I A CS R S A VC R out n (1 D) 2 R I A VC ω P 2π ω R 2π ω Z 2π ω L 2π ω P ω L K M ω Z = C out 2 R out = K R M I L C out V out V SLOPE 1 R ESR ω R = R out n (1 - D) 2 L V SLOPE (V out V in ) R i L f SW 4-33

35 Type II Error Amplifier A VM ω HF 2π ω ZEA 2π ω ZEA 2π = ω C 2π 10 ω HF 2π = ω R 2π 4-34

Type II Error Amplifier Variable Single Phase Two Phase Comment R FBT 10 kω 10 kω D MAX = 0.625 = 0.

36 Type II Error Amplifier Variable Single Phase Two Phase Comment R FBT 10 kω 10 kω D MAX = = Choose value between 2 kω 100 kω = V out V in_ Min V out,v in_ Min = 9V R I = 40 mω = 80 mω A CS R S G M_Mod = = = 1 D max R I 4-35

37 Boost Compensation Approach Variable Single Phase Two Phase Comment RHPZ 52 khz 21 khz = R n (1 D) out L 2π F C 12.5 khz 5 khz ω C 12.5 khz 2π 5 khz 2π A VM = 4.4 = 1 = = RHPZ 4 C out n G M _ Mod ω C = ω R 4 R COMP = 44 kω = 10 kω = A VM R FBT C COMP 2.8 nf 27 nf C COMP = R COMP 1 ω ZEA C HF 68 pf 720 pf C HF = R COMP 1 ω HF 4-36

38 Asymptotic Power Stage Current Mode Boost A VC ω P 2π ω R 2π ω Z 2π ω L 2π ,000 10, ,000 1,000, Type II Error Amplifier ω ZEA 2π A VM ω HF 2π ,000 10, ,000 1,000,000 Control Loop ω C 2π ,000 10, ,000 1,000,

39 Compensation Results Single Phase Simulation results A crossover frequency of ~13 k and a PM of ~50 degrees MathCAD results F C of ~13 k and a PM of ~75 degrees 4-38

40 Dual Phase Compensation Results Simulation results (Simplis) Shows an F C of ~5 khz and a PM of ~56 degrees Mathcad results Mathcad result correlate well to simulation showing an Fc of ~5 khz and a PM of ~60 degrees 4-39

41 Parameter Summary of Results Per phase switching frequency 250 khz Inductance value 3 µh I sat 15 A Energy 1/2 x L x I µj Inductor DCR losses 0.6 W Inductor core losses 2.6 W 4 mω R sense R sense losses Boost FET conduction losses Boost FET transitional losses FET Q OSS losses Q RR losses Synchronous FET conduction losses I C losses Total losses Calculated efficiency C in RMS ripple current rating C out RMS ripple current rating C in C out F C Single Phase 0.9 W 0.3 W 0.8 W 0.2 W 0.6 W 0.44 W W ~6.097 W ~97% 2.1 A 6.7 A 22 µf 780 µf 12.5 khz Dual Phase 125 khz 15 µh 9 A mj 1.4 W Total W Total 8 mω 0.8 W Total 0.16 W Total 0.4 W Total 0.2 W Total 0.6 W 0.22 W W ~3.734W ~98% 0.9 A 2.5 A 22 µf 390 µf 5 khz 4-40

42 Summary of Results Component Count Comparison Part Number Single Phase Part Number Dual Phase MOSFETs CSD18531Q5A 2 CSD18531Q5A 4 C in 25 V Ceramic 1 25 V Ceramic 1 C out PCV1E391MCL2GS 2 PCV1E391MCL2GS 1 Inductor XAL SER IC LM LM R sense 2 W Current Sense 1 2 W Current Sense 2 Total

43 Bench Test Results: Single Phase (PMP9385) Efficiency and Thermals Comparison F C ~15 khz; PM ~

44 Bench Test Results: Dual Phase (PMP9386) Efficiency and Thermals Comparison F C ~7.5 khz, PM ~

45 Conclusion Using equations and step-by-step approach provided herein enables designer to adjust design for optimizing efficiency or size Both size, cost and performance can be modified by using multiphase boost approach Thermal performance improved using two phase approach - Thermal stress on FETs significantly reduced with multiphase approach For single phase boost - Increasing switching frequency in an attempt to reduce size will result in exceeding FET thermal limits For two phase boost - Increasing switching frequency is feasible without thermal stress on FETs - Significant reduction in size can be further gained 4-44

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