PWM Step-down Converter(NJM2309). Concept kit provided by Bee Technologies.

1. Case Study:NJM2309 Application Circuit Design (PWM Step-down Converter) All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 1

4. VIN, MIN = 6 (V)

5. VOUT = 3.3 (V)

6. VOUT, Ripple = 1% ( 33mVP-P )

7. IOUT, MAX = 1.0 (A)

9. Switching Frequency – fosc = 105 (kHz)All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 3 NJM2309 Datasheet

10. NJM2309 Typical Application Circuit All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 4 Power Switches Filter & Load PWM Controller  Schematic is captured from NJM2309 datasheet page 4.

11. All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 5 TASK: Design and Evaluation of the Circuit 3? 2? 1 4? 5? NJM2309 Typical Application Circuit

13. Step2: Set C1=1kF, C2=1fF, (always keep the default value) and R2= calculated value (Rupper//Rlower) as the initial values.

14. Step3: Select a crossover frequency (about 10kHz or fc < fosc/4). Then complete the table.

15. Step4: Read the Gain and Phase value at the crossover frequency (10kHz) from the Bode plot, Then put the values to the table

16. 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.

17. 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

18. Buck Regulator Design Workflow All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 7 3 4 5 2 1

24. All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 13 Stabilizing the Converter (NJM2309) 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).

25. Stabilizing the Converter (NJM2309) All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 14 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

26. All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 15 Stabilizing the Converter (NJM2309) 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

27. Stabilizing the Converter (NJM2309) All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 16 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=3 is chosen for Phase margin = 48. 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).

29. C1=0.874nF ,

30. C2=97.07pF.*Analysis directives: .AC DEC 100 0.1 10MEG

31. All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 18 Stabilizing the Converter (NJM2309) 5 Gain and Phase responses after stabilizing Gain: T(s) = H(s) G(s)GPWM Phase  atfc Phase margin = 48.801 at the cross-over frequency - fc = 9.237kHz. Tip: To bring cursor to the cross-over point (gain = 0dB) type “ sfle(0) ” in Search Command. Cursor Search

32. Load Transient Response Simulation All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 19 The converter, that have been stabilized, are connected with step-load to perform load transient response simulation. 3 4 5 3.3V/16.5 = 0.2A step to 0.2+0.8=1.0A load 2 *Analysis directives: .TRAN 0 20ms 0 1u 1

35. A. Type 2 Compensation Calculation using Excel All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 23

36. All Rights Reserved Copyright (C) Bee Technologies Corporation 2011 24 B. Feedback Loop Compensators Type1 Compensator Type2 Compensator Type2a Compensator Type2b Compensator Type3 Compensator