Xác định một lịch trình để phân phối thư đồng ý cho bên thứ ba Trong những năm gần đây một số các phần khởi động đã cố gắng để crack không gian thị trường máy tính ở cấp độ chip với việc triển khai kỹ thuật số của thiết kế tương tự truyền thống của bộ điều biến PWM, | Buck Converters 61 Ay X 30 kHz X pH 3 mQ 226 Eq. 3-80 We know the ESR zero to be at 1 kHz fzESR 1 kHz Eq. 3-81 and fpLC i 2n LC f 1 X X Itr1 10000 27 370 Hz Eq. 3-82 Now from Eq. 3-78 Ay RqGịẶVQi X Ct 226 Eq. 3 83 and given V v 12 V VCT 1 V a 1 and Gyf 26 pA 26 mV 1 1 kQ we have Rc 226 X 1 kQ 12 19 kQ Eq. 3-84 Now from fzc U2nRcCc Eq. 3-85 setting fzc at 37 Hz lOx below the LC double pole we have Cc 1 X 19 kQ X 37 pF 220 nF Eq. 3-86 And so all the main parameters of the control loop are set. Input Filter The input current is chopped as indicated in Figure 3-29 so it will need some input filtering. Input Inductor LjN Assuming that at the input we need a current smoothed down to A ps with an input voltage ripple of V we have ưv Eq. 3-87 dỉ dt A ps 62 Chapter 3 Circuits Figure 3-29 Input filter. and from L n X dl dt dv Eq. 3-89 LlN V A ps 5 pH Eq. 3-90 Input Capacitor 1 20 A Eq. 3-91 dl dt o. A ps Eq. 3-92 hence the time to build up 20 A in the inductor is from Eq. 3-89 dT 20 A A ps 200 ps Eq. 3-93 From the formula CIN X dv dt I Knowing that dv dt V 200 ps v ms Eq. 3-95 c 20 A v ms 8 mF Eq. 3-96 This capacitor has to sustain an RMS current defined as a function of the DC and peak current as follows w AỠC - DC2 172 Eq. 3-97 from which 20 - 1 2 20 X 2 20 X 6 A Eq. 3-98 It is important to select the input capacitor capable of carrying the calculated RMS current. Buck Converters 63 Current Mode So far we have analyzed control schemes based on a single control loop the voltage control loop setting the output voltage. In any regulator when the output is low say at start-up the pass transistor will keep charging the output capacitor via the inductor until the output reaches final value. During this phase the voltage across the inductor is VIN - VOUT and the current is building in the inductor at a rate V N - VữUT L X t. If this phase lasts too long the current build up .