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SC4525DEVB 数据表(PDF) 14 Page - Semtech Corporation |
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SC4525DEVB 数据表(HTML) 14 Page - Semtech Corporation |
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14 / 21 page  © 2011 Semtech Corp. www.semtech.com SC4525D 14 Applications Information (Cont.) (2) Select the open loop crossover frequency, F C, between 10% and 20% of the switching frequency. At F C, find the required compensator gain, A C. In typical applications with ceramic output capacitors, the ESR zero is neglected and the required compensator gain at F C can be estimated by (9) (3) Place the compensator zero, F Z1, between 10% and 20% of the crossover frequency, F C. (4) Use the compensator pole, F P1, to cancel the ESR zero, F Z. (5) Then, the parameters of the compensation network can be calculated by (10) where g m=0.3mA/V is the EA gain of the SC4525D. Example: Determine the voltage compensator for an 350kHz, 12V to 3.3V/3A converter with 47uF ceramic output capacitor. Choose a loop gain crossover frequency of 35kHz, and place voltage compensator zero and pole at F Z1=7kHz (20% of F C), and FP1= 677kHz. From Equation (9), the required compensator gain at F C is Then the compensator parameters are CESAT D IN D O V V V V V D − + + = − = 1 V 0 . 1 V R R O 6 4 1 SW D O L L F ) D 1 ( ) V V ( I ⋅ − ⋅ + = D SW O D O 1 F I % 20 ) D 1 ( ) V V ( L ⋅ ⋅ − ⋅ + = ) D 1 ( D I I O CIN _ RMS − ⋅ ⋅ = ⋅ ⋅ + ⋅ D = D O SW L O C F 8 1 ESR I V SW IN O IN F V 4 I C ⋅ D ⋅ > , R G R G S CA PWM ⋅ ≈ ) / s Q / s 1 () / s 1 ( ) C R s 1 ( G V V 2 n 2 n p O ESR PWM c o ω + ω + ω + + = 7 1 Z 5 R F 2 1 C π = 7 1 P 8 R F 2 1 C π = , C R 1 O p ≈ ω , C R 1 O ESR Z = ω k 3 . 22 10 28 . 0 10 R 3 7 20 9 . 15 = ⋅ = − nF 45 . 0 10 1 . 22 10 16 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 12 10 1 . 22 10 600 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = ⋅ π ⋅ ⋅ − = O FB O C S CA C V V C F 2 1 R G 1 log 20 A dB 9 . 15 3 . 3 0 . 1 10 22 10 80 2 1 10 1 . 6 28 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − m 7 g 10 R 20 C A = CESAT D IN D O V V V V V D − + + = − = 1 V 0 . 1 V R R O 6 4 1 SW D O L L F ) D 1 ( ) V V ( I ⋅ − ⋅ + = D SW O D O 1 F I % 20 ) D 1 ( ) V V ( L ⋅ ⋅ − ⋅ + = ) D 1 ( D I I O CIN _ RMS − ⋅ ⋅ = ⋅ ⋅ + ⋅ D = D O SW L O C F 8 1 ESR I V SW IN O IN F V 4 I C ⋅ D ⋅ > , R G R G S CA PWM ⋅ ≈ ) / s Q / s 1 () / s 1 ( ) C R s 1 ( G V V 2 n 2 n p O ESR PWM c o ω + ω + ω + + = 7 1 Z 5 R F 2 1 C π = 7 1 P 8 R F 2 1 C π = , C R 1 O p ≈ ω , C R 1 O ESR Z = ω k 3 . 22 10 28 . 0 10 R 3 7 20 9 . 15 = ⋅ = − nF 45 . 0 10 1 . 22 10 16 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 12 10 1 . 22 10 600 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = ⋅ π ⋅ ⋅ − = O FB O C S CA C V V C F 2 1 R G 1 log 20 A dB 9 . 15 3 . 3 0 . 1 10 22 10 80 2 1 10 1 . 6 28 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − m 7 g 10 R 20 C A = CESAT D IN D O V V V V V D − + + = − = 1 V 0 . 1 V R R O 6 4 1 SW D O L L F ) D 1 ( ) V V ( I ⋅ − ⋅ + = D SW O D O 1 F I % 20 ) D 1 ( ) V V ( L ⋅ ⋅ − ⋅ + = ) D 1 ( D I I O CIN _ RMS − ⋅ ⋅ = ⋅ ⋅ + ⋅ D = D O SW L O C F 8 1 ESR I V SW IN O IN F V 4 I C ⋅ D ⋅ > , R G R G S CA PWM ⋅ ≈ ) / s Q / s 1 () / s 1 ( ) C R s 1 ( G V V 2 n 2 n p O ESR PWM c o ω + ω + ω + + = 7 1 Z 5 R F 2 1 C π = 7 1 P 8 R F 2 1 C π = , C R 1 O p ≈ ω , C R 1 O ESR Z = ω k 3 . 22 10 28 . 0 10 R 3 7 20 9 . 15 = ⋅ = − nF 45 . 0 10 1 . 22 10 16 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 12 10 1 . 22 10 600 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = ⋅ π ⋅ ⋅ − = O FB O C S CA C V V C F 2 1 R G 1 log 20 A dB 9 . 15 3 . 3 0 . 1 10 22 10 80 2 1 10 1 . 6 28 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − m 7 g 10 R 20 C A = CESAT D IN D O V V V V V D − + + = − = 1 V 0 . 1 V R R O 6 4 1 SW D O L L F ) D 1 ( ) V V ( I ⋅ − ⋅ + = D SW O D O 1 F I % 20 ) D 1 ( ) V V ( L ⋅ ⋅ − ⋅ + = ) D 1 ( D I I O CIN _ RMS − ⋅ ⋅ = ⋅ ⋅ + ⋅ D = D O SW L O C F 8 1 ESR I V SW IN O IN F V 4 I C ⋅ D ⋅ > , R G R G S CA PWM ⋅ ≈ ) / s Q / s 1 () / s 1 ( ) C R s 1 ( G V V 2 n 2 n p O ESR PWM c o ω + ω + ω + + = 7 1 Z 5 R F 2 1 C π = 7 1 P 8 R F 2 1 C π = , C R 1 O p ≈ ω , C R 1 O ESR Z = ω k 3 . 22 10 28 . 0 10 R 3 7 20 9 . 15 = ⋅ = − nF 45 . 0 10 1 . 22 10 16 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 12 10 1 . 22 10 600 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = ⋅ π ⋅ ⋅ − = O FB O C S CA C V V C F 2 1 R G 1 log 20 A dB 9 . 15 3 . 3 0 . 1 10 22 10 80 2 1 10 1 . 6 28 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − m 7 g 10 R 20 C A = dB 7 3 . 3 0 . 1 10 47 10 35 2 1 10 53 . 3 18.5 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − dB 7 3 . 3 0 . 1 10 47 10 35 2 1 10 53 . 3 18.5 1 log 20 A 6 3 3 C = ⋅ ⋅ ⋅ ⋅ ⋅ π ⋅ ⋅ ⋅ ⋅ − = − − Select R 7=7.32k, C5=3.3nF, and C8= 33pF for the design. Compensator parameters for various typical applications are listed in Table 4. A MathCAD program is also available upon request for detailed calculation of the compensator parameters. Thermal Considerations For the power transistor inside the SC4525D, the conduction loss P C, the switching loss PSW, and bootstrap circuit loss P BST, can be estimated as follows: (11) whereV BST is the BST supply voltage and tS is the equivalent switching time of the NPN transistor (see Table 3). Table 3. Typical switching time In addition, the quiescent current loss is (12) The total power loss of the SC4525D is therefore k 4 . 7 10 3 . 0 10 R 3 7 20 7 = ⋅ = − nF 1 . 3 10 7.4 10 7 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 32 10 7.4 10 677 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = k 4 . 7 10 3 . 0 10 R 3 7 20 7 = ⋅ = − nF 1 . 3 10 7.4 10 7 2 1 C 3 3 5 = ⋅ ⋅ ⋅ ⋅ π = pF 32 10 7.4 10 677 2 1 C 3 3 8 = ⋅ ⋅ ⋅ ⋅ π = O C E S AT C I V D P ⋅ ⋅ = 40 I V D P O B S T B S T ⋅ ⋅ = DC 2 O IND R I ) 3 . 1 ~ 1 . 1 ( P ⋅ ⋅ = O D D I V ) D 1 ( P ⋅ ⋅ − = S W O IN S S W F I V t 2 1 P ⋅ ⋅ ⋅ ⋅ = Q B S T S W C T OT AL P P P P P + + + = mA 2 V P IN Q ⋅ = O C E S AT C I V D P ⋅ ⋅ = 40 I V D P O B S T B S T ⋅ ⋅ = DC 2 O IND R I ) 3 . 1 ~ 1 . 1 ( P ⋅ ⋅ = O D D I V ) D 1 ( P ⋅ ⋅ − = S W O IN S S W F I V t 2 1 P ⋅ ⋅ ⋅ ⋅ = Q B S T S W C T OT AL P P P P P + + + = mA 2 V P IN Q ⋅ = Table 3: Typical Switching Time 1A 2A 3A 5V 6.86ns 9.71ns 12.5ns 12V 12.5ns 15.3ns 18ns Load Current Input Voltage Table 3: Typical Switching Time 1A 2A 3A 5V 6.86ns 9.71ns 12.5ns 12V 12.5ns 15.3ns 18ns Load Current Input Voltage O C E S AT C I V D P ⋅ ⋅ = 40 I V D P O B S T B S T ⋅ ⋅ = DC 2 O IND R I ) 3 . 1 ~ 1 . 1 ( P ⋅ ⋅ = O D D I V ) D 1 ( P ⋅ ⋅ − = S W O IN S S W F I V t 2 1 P ⋅ ⋅ ⋅ ⋅ = Q B S T S W C T OT AL P P P P P + + + = mA 2 V P IN Q ⋅ = O C E S AT C I V D P ⋅ ⋅ = 40 I V D P O B S T B S T ⋅ ⋅ = DC 2 O IND R I ) 3 . 1 ~ 1 . 1 ( P ⋅ ⋅ = O D D I V ) D 1 ( P ⋅ ⋅ − = S W O IN S S W F I V t 2 1 P ⋅ ⋅ ⋅ ⋅ = Q B S T S W C T OT AL P P P P P + + + = mA 2 V P IN Q ⋅ = |
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