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Design of Buck Converter

Presented by,

B. Naveen 207215001

Akhil S. 207215002

Design of Buck converter 2

Design ProblemDesign 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.)

Fig.1: Buck converter circuit arrangement

Design of Buck converter 3

Loadresistance,RL =VO/ ILOAD=5/ 2=2.5X

Ripplecurrent,IRIPPLE =3 IL =0.3#ILOAD =0.3#2=0.6A

Outputvoltage,VO =DVIN

` Duty ratio, D =VINVO

=125

=0.4166

T =1/ fsw =400* 1031 =2.5n sec

TON =DT =0.4166* 2.5n =1.04n sec

TOFF =(1- D)T =(1- 0.4166)2.5n=1.45n sec

Initial Calculations

Design of Buck converter 4

3 IL = fLD(1- D)VS

& L = f 3 ILD(1- D)VS

= 400#103 #0.60.4166(1- 0.4166)12

=12.152nHAlso Vripple=3 Vo= 8f2LC

D(1- D)VS

& C= 8f2L 3 VoD(1- D)VS

=8#(400#103) 2 #12.15#10- 6 #50#10- 30.4166(1- 0.4166)12

=3.75nF

Inductor & Capacitor Selection

I max =IO +(3 IL/ 2)=2+(0.6/ 2) =2.3A

I min =IO - (3 IL/ 2)=2- (0.6/ 2) =1.7A

Design of Buck converter 5

Critical Inductance (LC) = 2f(1- D)R

= 2#400#103(1- 0.4166)2.5

=1.823nH

Critical Capaci tance(CC) =16#f2 #Lc(1- D)

=16#(400#103) 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.

Calculation of critical values

SelectL =12nH,3AC=10nF,16VElectrolytic capacitor

Design of Buck converter 6

Diode SelectionAvg. currentthroughdiode, ID =(1- D) #I load

=(1- 0.416) #2=1.17AMax. voltagea/cdiode, VD =Vin =12VPeakcurrentthroughdiode,Ipeak =IL +3 IL/ 2=2+(0.6/ 2)

=2.3A

` Select1N5820,20V,3ASchottkydiode

Design of Buck converter 7

MOSFET Selection

Avg. currentthroughMOSFET, ISW =D#I load=0.416#2=0.832A

Max.voltagea/c MOSFET, VSW =Vin =12VPeakcurrentthroughMOSFET,Ipeak =IL +3 IL/ 2=2+(0.6/2)

=2.3A

` SelectIRFZ1460V,10AMOSFET

Design of Buck converter 8

CIRCUIT DESIGN

Fig.2: Final circuit design diagram of buck converter

Design of Buck converter 9

WAVEFORMS

Fig.3: Simulation results of designed buck converter

Simulation Results:

ΔIL =2.3-1.7 =0.6AΔVo=5.007-4.997=10mV

Design of Buck converter 10

Efficiency CalculationOutput power, Pout=V#I =5* 2=10WCapacitor loss =0.01W (fromESR)MOSFET loss =0.3W (fromRDSon)Diodeloss =0.47W (fromVDon)Inductor loss =0.15W (coreloss)` Total loss =0.93WEfficiency =Pout+losses

Pout =10+0.9310

=91.5%Diodelossrepresentsone- half of thetotal losses.

Design of Buck converter 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

References

Design of Buck converter 12

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