18. Buck Converter Circuit Copyright (C) Bee Technologies Inc. 2011 14 Power Switches Filter & Load PWM Controller
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22. Step2: Set C1=1kF, C2=1fF, (always keep the default value) and R2= calculated value (Rupper//Rlower) as the initial values.
23. Step3: Select a crossover frequency (about 10kHz or fc < fosc/4). Then complete the table.
24. Step4: Read the Gain and Phase value at the crossover frequency (10kHz) from the Bode plot, Then put the values to the table
25. Step5: Select the phase margin at the fc ( > 45 ). Then change the K value until it gives the satisfied phase margin, for this example K=6 is chosen for Phase margin = 46.
26. Remark: If K-factor fail to gives the satisfied phase margin, Increase the output capacitor C then try Step1 to Step5 again.5 Load Transient Response Simulation 6
32. Error Amp. Gain is 100 (approximated) where VP is the sawtooth peak voltage. vFBH is maximum FB voltage where d = 0 vFBL is minimum FB voltage where d =1(100%) dMAX is maximum duty cycle, e.g. d = 0(0%) dMIN is minimum duty cycle, e.g. d =1(100%) Setting PWM Controller’s Parameters Copyright (C) Bee Technologies Inc. 2011 18 1 The PWM block is used to transfer the error voltage (between FB and REF) to be the duty cycle. If vFBH and vFBLare not provided, the default value, VP=2.5 could be used.
33. Copyright (C) Bee Technologies Inc. 2011 19 Setting PWM Controller’s Parameters (Example) 1 If the VP ( sawtooth signal amplitude ) does not informed by the datasheet, It can be approximated from the characteristics below. from VP= (Error Amp. Gain vFB )/d Error Amp. Gain = 100 (approximated) from the graph on the left, vFB= 25mV (15m - (-10m)) d = 1 – 0 = 1 VP ≈ ( 100 25mV )/1 ≈ 2.5V vFBH vFB = 25mV vFBL d = 1 (100%) dMIN dMAX LM2575: Feedback Voltage vs. Duty Cycle If vFBH and vFBLare not provided, the default value, VP=2.5 could be used.
39. Inductor Selection: L (Example) Copyright (C) Bee Technologies Inc. 2011 22 Inductor Value from Given: VI,max = 40V, VOUT = 5V IOUT,min = 0.2A RL,min = (VOUT /IOUT,min ) = 25 fosc = 52kHz Then: LCCM 210(uH), L = 330(uH) is selected 3
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41. Capacitor Selection: C, ESR (Example) Copyright (C) Bee Technologies Inc. 2011 24 Capacitor Value From and Given: VI, max = 40 V VOUT = 5 V L (H) = 330 Then: C 188 (F) In addition: ESR 100m 4
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44. Copyright (C) Bee Technologies Inc. 2011 27 Stabilizing the Converter (Example) 5 The element of the Type 2 compensator ( R2, C1, and C2 ), that stabilize the converter, can be extracted by using Type 2 Compensator Calculator (Excel sheet) and open-loop simulation with the Average Switch Models (ac models). Step2 Set C1=1kF, C2=1fF, and R2=calculated value (Rupper//Rlower) as the initial values. Step1 Open the loop with LoL=1kH and CoL=1kF then inject an AC signal to generate Bode plot. C1=1kF is AC shorted, and C2 1fF is AC opened (or Error-Amp without compensator).
45. Stabilizing the Converter (Example) Copyright (C) Bee Technologies Inc. 2011 28 5 Step3 Select a crossover frequency (about 10kHz or fc < fosc/4 ), for this example, 10kHz is selected. Then complete the table. values from 2 Calculated value of the Rupper//Rlower values from 1
46. Copyright (C) Bee Technologies Inc. 2011 29 Stabilizing the Converter (Example) 5 Gain: T(s) = H(s)GPWM Step4 Read the Gain and Phase value at the crossover frequency(10kHz) from the Bode plot, Then put the values to the table. Phase atfc Tip: To bring cursor to the fc = 10kHz type “ sfxv(10k) ” in Search Command. Cursor Search
47. Stabilizing the Converter (Example) Copyright (C) Bee Technologies Inc. 2011 30 5 Step5 Select the phase margin at fc (> 45 ). Then change the K value (start from K=2) until it gives the satisfied phase margin, for this example K=6 is chosen for Phase margin = 46. As the result; R2, C1, and C2 are calculated. Remark: If K-factor fail to gives the satisfied phase margin, Increase the output capacitor C then try Step1 to Step5 again. K Factor enable the circuit designer to choose a loop cross-over frequency and phase margin, and then determine the necessary component values to achieve these results. A very big K value (e.g. K > 100) acts like no compensator (C1 is shorted and C2 is opened).
48. Stabilizing the Converter (Example) Copyright (C) Bee Technologies Inc. 2011 31 5 The element of the Type 2 compensator ( R2, C1, and C2 ) extraction can be completed by Type 2 Compensator Calculator (Excel sheet) with the converter average models (ac models) and open-loop simulation. The calculated values of the type 2 elements are, R2=122.780k, C1=0.778nF, and C2=21.6pF. *Analysis directives: .AC DEC 100 0.1 10MEG
49. Copyright (C) Bee Technologies Inc. 2011 32 Stabilizing the Converter (Example) 5 Gain and Phase responses after stabilizing Gain: T(s) = H(s) G(s)GPWM Phase atfc Phase margin = 45.930 at the cross-over frequency - fc = 9.778kHz. Tip: To bring cursor to the cross-over point (gain = 0dB) type “ sfle(0) ” in Search Command. Cursor Search
50. Load Transient Response Simulation (Example) Copyright (C) Bee Technologies Inc. 2011 33 The converter, that have been stabilized, are connected with step-load to perform load transient response simulation. 5V/2.5 = 0.2A step to 0.2+0.8=1.0A load *Analysis directives: .TRAN 0 20ms 0 1u