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RT9602GS bảng dữ liệu(PDF) 7 Page - Richtek Technology Corporation |
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RT9602GS bảng dữ liệu(HTML) 7 Page - Richtek Technology Corporation |
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7 / 11 page ![]() RT9602 7 DS9602-08 March 2007 www.richtek.com According to the design of RT9602, before driving the gate of the high side MOSFET up to 12V (or 5V), the low side MOSFET has to be off; and the high side MOSFET is turned off before the low side is turned on. From Figure 1, the body diode "D 2" had been turned on before high side MOSFETs turned on Before the low side MOSFET is turned on, the Cgd2 have been charged to Vi. Thus, as Cgd2 reverses its polarity and g2 is charged up to 12V, the required current is g1 gs1 gs1 gs1 r1 g2 gs2 gs2 gs2 r2 dV C 12 l = C = (1) dt t dV C 12 l = C = (2) dt t × × gd2 gd2 gd2 r2 dV Vi+12V l = C = C (4) dt t gd1 g1 gd1 r1 dV 12V l = C = C (3) dt t -12 -9 gs1 1660 10 12 l = = 1.428 (A) (5) 14 10 ×× × -12 -9 gs2 2200 10 12 l = = 0.88 (A) (6) 30 10 ×× × It is helpful to calculate these currents in a typical case. Assume a synchronous rectified BUCK converter, input voltage Vi = 12V, Vg1 = Vg2 = 12V. The high side MOSFET is PHB83N03LT whose Ciss = 1660pF, Crss = 380pF,and tr = 14nS. The low side MOSFET is PHB95N03LT whose Ciss = 2200pF, Crss = 500pF, and tr = 30nS, from the equation (1) and (2) we can obtain Figure 2. Two- Phase Synchronous-Buck Converter Circuit BOOT1 UGATE1 PHASE1 LGATE1 VCC PVCC PWM1 PWM2 RT9602 11 14 5 1 2 4 13 12 UGATE2 PHASE2 LGATE2 BOOT2 GND PGND 10 1uF 12V 1uF PHB83N03LT PHB95N03LT 2uH 1uF 1000uF 1.2uH 12V 7 8 9 PHB83N03LT PHB95N03LT 2uH 1uF 1000uF 1uF V CORE 1500uF 1500uF 3 6 L1 C1 L2 C5 C6 Q3 Q4 L3 C3 C4 Q2 Q1 C2 Cb1 D1 D2 Cb2 R1 C7 PWM1 PWM2 10 Vin from equation. (3) and (4) the total current required from the gate driving source is By a similar calculation, we can also get the sink current required from the turned off MOSFET. Layout Consider Figure 2. shows the schematic circuit of a two-phase synchronous-buck converter to implement the RT9602. The converter operates for the input rang from 5V to 12V. -12 -9 gs1 380 10 12 l = = 0.326 (A) (7) 14 10 ×× × -12 -9 gs2 500 10 (12+12) l = = 0.4(A) (8) 30 10 ×× × g1 gs1 gd1 I = I +I = (1.428+0.326) = 1.745(A) (9) g2 gs2 gd2 I = I +I = (0.88+0.4) = 1.28(A) (10) |
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