How to Select and Use Power Supplies and dc/dc Converters for Your ApplicationsContentsIntroduction to Power Supplies and dc/dc ConvertersIntroduction to Power Supplies and dc/dc Converters –cont.Power Supplies and dc/dc Converters –Types & TechnologiesAC-DC Power Supplies -Circuit Selection and DesignSlide 7Selecting the Right dc/dc ConverterSelecting the Right dc/dc Converter –cont.Slide 10Specs, Performance and ProtectionPower Losses and Thermal DesignPower Losses and Thermal Design --A more detailed thermal circuitSlide 14ExampleF. Z. Peng: Slide 1Feb. 15, 2006How to Select and Use Power Supplies and dc/dc Converters for Your ApplicationsFang Z. PengDept. of Electrical and Computer EngineeringMichigan State UniversityPhone: 517-336-4687, Fax: 517-353-1980Email: [email protected]. Z. Peng: Slide 2Feb. 15, 2006Contents•Introduction to Power Supplies and dc/dc Converters•Types & Technologies of Power Supplies and dc/dc Converters•Circuit Selection and Design•Circuit Performance and Protection Features•Thermal Requirements and Design IssuesF. Z. Peng: Slide 3Feb. 15, 2006Introduction to Power Supplies and dc/dc Converters•Available/Raw Power Sources–AC or DC (frequency)–Un-regulated (changes with load, prime source, etc.)–Voltage (different level, polarity, isolation)–Non-protected (against over load, fault, temp., etc.)•Load Demand–Different AC or DC (frequency)–Regulated (against load, prime source, etc.)–Voltage (different level, polarity, isolation)–Protected (against over load, fault, temp., etc.)F. Z. Peng: Slide 4Feb. 15, 2006Power SupplyIntroduction to Power Supplies and dc/dc Converters –cont.Power & ElectronicCircuitsRaw power inDesired power out(V, I, P, F)To loads:Electronic cktsMotorComputerEquipmentControlBatteryFuel CellAC OutletSolarF. Z. Peng: Slide 5Feb. 15, 2006Power Supplies and dc/dc Converters –Types & Technologies•AC-DC Power Supply (or AC Adapter)–Change ac power into regulated dc power, e.g., a typical AC Adapter takes 120 V ac input and converter it to regulated 5 Vdc.•Dc/dc Converters–Change dc at one voltage potential to a dc at a different voltage potential•DC-AC Power Supply (for example, UPS, 12Vdc-120Vac adapter)•AC-AC Power Supply/Regulator (for example, line regulator)F. Z. Peng: Slide 6Feb. 15, 2006AC-DC Power Supplies-Circuit Selection and Design•Using Linear Regulators•Using LDO Regulatorhttp://www.national.com/pf/LM/LM78M05.htmlStep-downXfmerRegulator120 VAC•For low power (several watts or below) applications.•Low efficiency, large size and weight (bulky step-down line transformer)•Low costF. Z. Peng: Slide 7Feb. 15, 2006AC-DC Power Supplies-Circuit Selection and Design•Using Switching-Mode•High efficiency•Small size and light weight•For high power (density) applicationshttp://www.linear.com/index.jsphttp://www.linear.com/3770http://www.electronicproducts.com/TI Power Supply Technologies PosterF. Z. Peng: Slide 8Feb. 15, 2006Selecting the Right dc/dc Converter•The Need for dc/dc Converters–E.g., a single AA alkaline battery produces 1.5 V when fully charged and its voltage drops to as low as 0.9 V when becoming depleted.•Dc/dc Converter Types–Buck–Boost–Buck-Boost•Dc/dc Converter Technologies–Linear Regulators–Switching Regulators–Charge PumpsThe MCP1703 LDO is one type of dc/dc linear regulatorF. Z. Peng: Slide 9Feb. 15, 2006Selecting the Right dc/dc Converter –cont. Dc/dc converter technology comparisonParameterLinear regulatorSwitching regulatorCharge pumpEfficiency Low High MediumEMI Noise Low High MediumOutput current Low to medium Low to High LowBoost (step-up) No Yes YesBuck (step-down) Yes Yes YesSolution size small Large MediumF. Z. Peng: Slide 10Feb. 15, 2006Selecting the Right dc/dc Converter –cont. VBAT = 3.7 V nom, BIN_BB = 1.2 VLoad Current = 600 mAPower delivered to load = 600 mA * 1.2 V = 720 mWPower converted to heat = 720 mW * ((3.7/1.2) 1) = 1,500 mWTotal power consumed = 720 mW + 1,500 mW = 2,200 mW32% goes to work, 68% goes to heating user hand and ear when using a Linear Regulator for a mobile deviceVBAT = 3.7 V nom; BIN_BB = 1.2 VLoad Current = 600 mAConverter efficiency = 90%Power delivered to load = 600 mA * 1.2 V = 720 mWPower converted to heat =720 mW * ((1/0.9) 1)=80 mWTotal power consumed = 720 mW + 80 mW = 800 mW90% goes to work, 10% goes to heating user hand and earWhen using a Switch-mode regulator for a mobile device. Linear regulators:•Inexpensive•small footprint•low part count•low noise•high ripple rejection Switching regulators:•a bigger footprint •higher part count, •more cost•prone to conducted and radiated EMI.F. Z. Peng: Slide 11Feb. 15, 2006Specs, Performance and Protection•Voltage ripple (+-50 mV, or 5%)•Isolation (e.g., 1,500 V ac for 1 min.)•Load regulation (e.g., 3%)•Dynamic response (transients, wake-up time, etc.)•Short circuit protection•OC protection•OV protection•OT protectionF. Z. Peng: Slide 12Feb. 15, 2006Power Losses and Thermal Design•For example, a 7815 linear regulator with input voltage of 20 V and output current of 1 A. The power loss is (20-15)Vx(1 A)=5 W.•From the chip to the ambient, Ti can be calculated according to the thermal circuit using Ohm’s law (R=V/I), where R is the thermal resistance, V is the temperature and I is the power dissipation.case ambientThcase ambientdissipationoutdissipation in out outopT TRPPP P P Ph--== - = -Where: Tcase is case Temp. Tambient is ambient Temp. Pdissipation is power loss Pin is input power Pout is output power op is efficiency under given operating conditionsF. Z. Peng: Slide 13Feb. 15, 2006Power Losses and Thermal Design--A more detailed thermal circuit•W : Device power loss•Tj : Junction temperature of device•Tc : Device case temperature•Tf : Temperature of heatsink•Ta : Ambient temperature•Rth(j-c) : Thermal resistance between junction and case, specified in datasheet•Rth(c-f) : Contact thermal resistance between case and heatsink, specified in datasheet•Rth(f-a) : Thermal resistance between heatsink and ambient air, specified by the heatsink manufacturerF. Z. Peng: Slide 14Feb. 15, 2006Power Losses and Thermal DesignTc=W×{Rth(c-f) + Rth(f-a)}+TaTj=W×Rth(j-c)+TcF. Z. Peng: Slide 15Feb. 15, 2006ExampleTc=5×(0.5 + 20)+25=127.5 °CTj=5×1+127.5=132.5 °CTj=82.5-25=107.5°C• Device : 7815 (Linear regulator)• Vin=20V, Vo=15V, Io=1A• W :
View Full Document
Unlocking...