1 boost converter design example m. t. thompson, 2008 power electronics notes 07c boost converter...

1Boost Converter Design Example M. T. Thompson, 2008

Power Electronics Notes 07CBoost Converter Design Example

© Marc Thompson, 2008

Marc T. Thompson, Ph.D.Thompson Consulting, Inc.

9 Jacob Gates RoadHarvard, MA 01451

Phone: (978) 456-7722 Fax: (888) 538-3824

Email: marctt@thompsonrd.comWeb: http://www.thompsonrd.com

2Boost Converter Design Example M. T. Thompson, 2008

Summary

• Design a boost converter with the following specifications:• Input voltage: 12V• Output: 24V @ 1A, 24 Watts• Continuous conduction mode• Inductor and capacitors: selected from following datasheets• Switching frequency 100 kHz• Output voltage ripple < 50 mV-pp

• Evaluate output ripple and estimate efficiency of converter

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Step-Up (Boost) DC-DC Converter

• Output voltage is higher than the input, without a phase inversion

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Boost Converter Waveforms• Continuous current conduction mode

Switch closed:

di

dt

V

LL CC

Switch open:

di

dt

V v

LL CC o

Inductor Volt-second balance:V DT

L

V V D T

L

VV

D

CC CC o

oCC

( )( )10

1

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Boost: Limits of Cont./Discont. Conduction

• The output voltage is held constant• For low load current, current conduction becomes discontinuous

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Boost Converter: Discont. Conduction

• Occurs at light loads

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Boost Converter: Effect of Parasitics

• The duty-ratio D is generally limited before the parasitic effects become significant

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Boost Converter Output Ripple

• ESR is assumed to be zero• Assume that all the ripple component of diode current flows through capacitor; DC component flows through resistor

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Boost Converter 1st-Cut Design --- Inductor

• D = 0.5• What is minimum inductor value to keep this converter in continuous conduction mode ? (I.e. this converter operates at the continuous/discontinuous conduction boundary)

• Average diode current: 0.5Ipk(1-D) = Io = 1A• Ipk = 4A• Lmin =(Vo – Vi)(1-D)T/Δi = (24-12)(0.5)(10-5)/4 = 15 µH

• For the diode, ID,rms = = 2.3A3

1pkI

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Inductor Datasheet

• Use 22 µH (ESR = 0.085 Ohms)• Note that series resonant frequency (SRF) is much higher than operating frequency• Note that IRMS rating of this inductor is 2.7A

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Boost Converter Current Waveforms

12Boost Converter Design Example M. T. Thompson, 2008

Boost Converter 1st-Cut Design --- Capacitor

• What is minimum capacitor value ?

Cf

D

R

VV

sw

oppo

FVRf

DVC

pposw

o 100)05.0)(24(10

)5.0)(24(5min

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Capacitor Datasheet

• Use 3 47 µF caps in parallel (35V, ESR = 0.9 Ohms)

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MOSFET Datasheet• This device is over-sized, but let’s use it anyway

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1st Cut Design

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Simulation Result --- Inductor Current

• Note that inductor ripple is about 3A peak to peak

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Simulation Result --- Output Ripple

• Why is output voltage ripple so large ?

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Simulation Result --- Analysis• The culprit is capacitor ESR. Ripple current is 3A pp, divided into 3 capacitors. Ripple voltage = ripple current x ESR• This is a problem with the boost converter --- large output ripple current makes sizing capacitor difficult

Ripple 1 V pp

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Mitigating Strategies• Parallel up more capacitors, or find capacitors with even lower ESR• Alternative strategy: use lower ESR caps with a post-filter

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2nd Cut Design

• Lower ESR capacitors and an LC post filter added

Lower ESR caps

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2nd Cut Design --- Simulation Results

Ripple 20 mV pp

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2nd Cut Design --- Efficiency Estimate

• Losses due to:• Inductor loss• Switch conduction loss• Switch switching loss• Diode loss• Capacitor ESR loss• Gate drive loss

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2nd Cut Design --- Efficiency EstimateBoost converter lecture exampleMTT 10-9-03fsw 1.00E+05L 2.20E-05Vi 12Vo 24Rinductor 0.085D 0.5Rsw 0.0825IL,avg 2IL,max 3.36E+00IL,min 6.36E-01tsw 1.50E-07IL,rms 2.15E+00Isw,RMS 1.52E+00Vd 0.8 Diode voltageIo 1 Output currentQg 1.00E-07

LOSSES CALCULATIONPdiode 0.80Pinductor 0.39Pswitch, conduction 0.19Pswitch, switching 0.61Gate drive loss 0.12Capacitor ESR loss 0.12

Total losses 2.23Output power 24Efficiency 91.5%

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2nd Cut Design --- Evaluation

• FATAL DESIGN FLAW• Note that ISAT rating of this inductor is 2.6A• Peak current in inductor is 3.4A• Therefore, this design will blow up

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3rd Cut Design --- Replace Inductor• Using next-size up Coilcraft inductor, Isat rating of 22 µH inductor is 7.0A, RMS rating is 3.5A, so this should be OK• Inductor loss will be lower due to lower DC resistance• This comes at the cost of a more expensive inductor, and more PC board space needed