Successfully reported this slideshow.
Upcoming SlideShare
×

# Design of Buck Converter

A detailed step-by-step procedure for the design of a buck converter. Different active and passive components are selected as per the requirement specified in the design problem.

• Full Name
Comment goes here.

Are you sure you want to Yes No

### Design of Buck Converter

1. 1. Presented by, B. Naveen 207215001 Akhil S. 207215002 1
2. 2. Design a buck converter satisfying the following requirements:  Input Voltage, Vin=12V  Output Voltage, Vout=5V  Output Current, Io=2A  Switching frequency, fsw=400kHz  Ripple current, ΔIL=30% of IL (max.)  Ripple voltage, ΔVo=50mV (max.) 2Design of Buck converter Fig.1: Buck converter circuit arrangement
3. 3. Load resistance,RL = VO / ILOAD = 5/ 2 = 2.5X Ripplecurrent,IRIPPLE = 3 IL = 0.3 # ILOAD = 0.3 # 2 = 0.6A Output voltage, VO = DVIN ` Duty rati o, D = VIN VO = 12 5 = 0.4166 T = 1/ fsw = 400 * 10 3 1 = 2.5n sec TON = DT = 0.4166 * 2.5n = 1.04n sec TOFF = (1 - D) T = (1 - 0.4166) 2.5n = 1.45n sec 3Design of Buck converter
4. 4. 3 IL = fL D(1 - D) VS & L = f 3 IL D(1 - D) VS = 400 # 10 3 # 0.6 0.4166(1 - 0.4166) 12 = 12.152nH Also Vr ipple = 3 Vo= 8f 2 LC D(1- D)VS & C = 8f 2 L 3 Vo D(1 - D) VS = 8 # (400 # 10 3 ) 2 # 12.15 # 10 - 6 # 50 # 10 - 3 0.4166(1 - 0.4166) 12 = 3.75nF 4Design of Buck converter I max = IO + (3 IL / 2) = 2 + (0.6/ 2) = 2.3A I min = IO - (3 IL / 2) = 2 - (0.6/ 2) = 1.7A
5. 5. Critical Inductance (LC ) = 2f (1 - D) R = 2 # 400 # 10 3 (1 - 0.4166) 2.5 = 1.823nH Critical Capaci tance(CC ) = 16 # f 2 # Lc (1 - D) = 16 # (400 # 10 3 ) 2 # 1.823 # 10 - 3 (1 - 0.4166) = 0.125nF As the designed L & C values are greater than their corresponding critical values, the design is acceptable. 5Design of Buck converter Select L = 12nH,3A C = 10nF,16V Electrolytic capacitor
6. 6. Avg. current through diode, ID = (1 - D) # Iload = (1 - 0.416) # 2 = 1.17A Max. voltagea/ c diode, VD = Vin = 12V Peak current through diode,Ipeak = IL + 3 IL/ 2 = 2 + (0.6/ 2) = 2.3A ` Select1N5820,20V,3A Schottky diode 6Design of Buck converter
7. 7. Avg. current through MOSFET, ISW = D # Iload = 0.416 # 2 = 0.832 A Max. voltagea/ c MOSFET, VSW = Vin = 12V Peak current through MOSFET,Ipeak = IL + 3 IL/ 2 = 2 + (0.6/ 2) = 2.3A ` Select IRFZ1460V,10A MOSFET 7Design of Buck converter
8. 8. CIRCUIT DESIGN 8Design of Buck converter Fig.2: Final circuit design diagram of buck converter
9. 9. WAVEFORMS 9Design of Buck converter Fig.3: Simulation results of designed buck converter Simulation Results: ΔIL =2.3-1.7 =0.6A ΔVo=5.007-4.997=10mV
10. 10. Output power, Pout = V # I = 5 * 2 = 10W Capacitor loss = 0.01W (from ESR) MOSFET loss = 0.3W (from RDSon) Diodeloss = 0.47W (fromVDon) Inductor loss = 0.15W (coreloss) ` Total loss = 0.93W Efficiency = Pout + losses Pout = 10 + 0.93 10 = 91.5% Diodelossrepresentsone- half of thetotal losses. 10Design of Buck converter
11. 11. [1] Ned Mohan, Undeland and Robbin, ‘Power Electronics: Converters, Application and Design’, John Wiley and sons Inc., Newyork,2006. [2] Rashid M.H., ‘Power Electronics- Circuits, Device and Applications’, Prentice Hall India, New Delhi, 2009. [3] Datasheets  1N5820 http://www.farnell.com/datasheets/107972.pdf  IRFZ14 http://www.irf.com/product-info/datasheets/data/irfz14.pdf  Capacitor http://www.farnell.com/datasheets/1558295.pdf 11Design of Buck converter
12. 12. Design of Buck converter 12