buck converter design

Power Electronics Technology June 2006 www.powerelectronics.com June 2006 www.powerelectronics.com46

Buck-Converter Design Demystifi ed

Though stepdown converters are extremely popular, the rules of thumb and calculations that speed their design can be hard to fi nd.

By Donald Schelle and Jorge Castorena, Technical Staff, Jorge Castorena, Technical Staff, Jorge Castorena,Maxim Integrated Products, Sunnyvale, Calif.

Stepdown (buck) switching converters are integral to modern electronics. They can convert a voltage source (typically 8 V to 25 V) into a lower regu-lated voltage (typically 0.5 V to 5 V). Stepdown converters transfer small packets of energy using a

switch, a diode, an inductor and several capacitors. Though substantially larger and noisier than their linear-regulator counterparts, buck converters offer higher effi ciency in most cases.

Despite their widespread use, buck-converter designs can pose challenges to both novice and intermediate power-supply designers because almost all of the rules of thumb and some of the calculations governing their design are hard to fi nd. And though some of the calculations are read-ily available in IC data sheets, even these calculations are occasionally reprinted with errors. In this article, all of the design information required to design a buck converter is conveniently collected in one place.

Buck-converter manufacturers often specify a typical application circuit to help engineers quickly design a working prototype, which in turn often specifi es component values and part numbers. What they rarely provide is a detailed description of how the components are selected. Suppose

a customer uses the exact circuit provided. When a critical component becomes obsolete or a cheaper substitute is needed, the customer is usually without a method for select-ing an equivalent component.

This article covers only one stepdown regulator topologyone with a fi xed switching frequency, pulse width modu-one with a fi xed switching frequency, pulse width modu-lation (PWM) and operation in the continuous-current mode (CCM). The principles discussed can be applied to other topologies, but the equations do not apply directly to other topologies. To highlight the intricacies of stepdown converter design, we present an example that includes a de-tailed analysis for calculating the various component values. Four design parameters are required: input-voltage range, regulated output voltage, maximum output current and the converter’s switching frequency. Fig. 1 lists these parameters, along with the circuit illustration and basic components required for a buck converter.

Inductor SelectionCalculating the inductor value is most critical in designing

a stepdown switching converter. First, assume the converter is in CCM, which is usually the case. CCM implies that the inductor does not fully discharge during the switch-off time. The following equations assume an ideal switch (zero on-resistance, infi nite off-resistance and zero switching time) and an ideal diode: (Eq. 1)

L VV

V f LIR IIN OUTOUVOUV T

INV fINV fSWV fSWV f OUTMAXMAXMA

MAX MSWX MSW OUX MOUTX MTMAX MMA AX

L V= −L V × ×× ×OU× ×OUT× ×T ×R I×R I

( )L V( )L V V( )VIN( )INL VINL V( )L VINL V OU( )OUVOUV( )VOUV T( )TMA( )

MAX( )

XMAXMA( )

MAXMA= −( )= −L V= −L V( )L V= −L V ,

1 1

where fSW

where fSW

where f is the buck-converter switching frequency and LIR is the inductor-current ratio expressed as a percentage of I

OUT (e.g., for a 300-mA

p-p (e.g., for a 300-mA

p-p (e.g., for a 300-mA ripple current with a 1-A output,

LIR = 0.3 A/1 A = 0.3 LIR).An LIR of 0.3 represents a good tradeoff between

effi ciency and load-transient response. Increasing the LIR constant—allowing more inductor ripple current—quickens Fig. 1. Basic stepdown converter circuit with operating parameters.

Controller

P1

VIN VOUT

L

COUTCIND

P MOSFET

VIN = 7 V 24 V VOUT = 2 V

IOUT fSW = 300 kHz MAX

= 7 A

+ +


BUCK-CONVERTER DESIGNS BUCK-CONVERTER DESIGNS

the load-transient response, and decreasing the LIR con-stant—thereby reducing the inductor ripple current—slows the load-transient response. Fig. 2 depicts transient response and inductor current for a given load current, for LIR con-stants ranging from 0.2 to 0.5.

Peak current through the inductor determines the inductor’s required saturation-current rating, which in turn dictates the approximate size of the inductor. Saturating the inductor core decreases the converter effi ciency, while increasing the temperatures of the inductor, the MOSFET and the diode. You can calculate the inductor’s peak operat-ing current as follows:

I II

I LIR I V

PEI IPEI IAKI IAKI IOUTINDUCTOR

OUI VOUI VT II VT II V

MATMAT XMAXMA

MAT IMAT IXT IXT IMAXMAT IMAT IXT IMAT I

= +I I= +I IOU= +OUT= +T

= ×I L= ×I LIR= ×IR I V=I V

∆

∆2

,

(I V(I VT I(T II VT II V(I VT II V

where

INI LINI LDUCTOI LDUCTOI LRI LRI L N ONN ON UT

OUT

IN SW

MAN OMAN OXN OXN OMAXMAN OMAN OXN OMAN O

MAXMAXMA

VN OVN O

VOUVOU

V fINV fIN SWV fSW L

− ×N O− ×N OUT− ×UT− ×N O− ×N OV− ×VN OVN O− ×N OVN O

× ×× ×

)− ×)− ×

.1 1

For the values listed in Fig. 1, these equations yield a cal-culated inductance of 2.91 µH (LIR = 0.3). Select an available value that is close to the calculated value, such as a 2.8 µH, and make sure that its saturation-current rating is higher than the calculated peak current (I

PEAK = 8.09 A).

Choose a saturation-current rating that’s large enough (10 A in this case) to compensate for circuit tolerances and the difference between actual and calculated component values. An acceptable margin for this purpose, while limiting the inductor’s physical size, is 20% above the calculated rating.

Inductors of this size and current rating typically have a maximum dc resistance range (DCR) of 5 m to 8 m. To minimize power loss, choose an inductor with the lowest possible DCR. Although data sheet specifi cations vary among vendors, always use the maximum DCR specifi cation for de-sign purposes rather than the typical value, because the maxi-mum is a guaranteed worst-case component specifi cation.

Output Capacitor SelectionOutput capacitance is required to minimize the voltage

overshoot and ripple present at the output of a stepdown converter. Large overshoots are caused by insuffi cient output capacitance, and large voltage ripple is caused by insuffi cient capacitance as well as a high equivalent-series resistance

(ESR) in the output capacitor. The maximum allowed output-voltage overshoot and ripple are usually specifi ed at the time of design. Thus, to meet the ripple specifi cation for a stepdown converter circuit, you must include an output capacitor with ample capacitance and low ESR.

The problem of overshoot, in which the output-voltage overshoots its regulated value when a full load is suddenly removed from the output, requires that the output capaci-tor be large enough to prevent stored inductor energy from launching the output above the specifi ed maximum output voltage. Output-voltage overshoot can be calculated using the following equation:

(Eq. 2)∆

∆

V VV VL I

I

CVOUV VOUV V T

OU

O

OUVOUV T

MAXMAXMA

V V=V V

V VV V

−2

2

2( )( )∆( )∆

L I( )L II( )I

OU( )OUT( )TIN( )INDUCTO( )DUCTOR( )R

MA( )

MATMAT( )TMAT X( )

XMAXMA( )

MAXMA+( )+

2( )

2 .

Rearranging Eq. 2 yields:

(Eq. 3)C

L II I

VO

OUTINDUCTOR

OUT OVT OV UT

MA MA TMAT X X MAXMA MA X MA =

+ L I

L I

L I

L IL I

L I

L I

L I

L I

L IL I

L I L I

L IL I

L IL I

L I

L I

L IL I

L I

L I

L I

+ −T O+ −T O

∆ ∆

2 2

2

2+ −2+ − 2( )V V( )V VOU( )OUV VOUV V( )V VOUV V T O( )T O+ −( )+ −V V+ −V V( )V V+ −V VOU+ −OU( )OU+ −OUV VOUV V+ −V VOUV V( )V VOUV V+ −V VOUV V T O+ −T O( )T O+ −T O∆( )∆,

where CO equals output capacitance and ∆V equals maxi-

mum output-voltage overshoot.Setting the maximum output-voltage overshoot to

100 mV and solving Eq. 3 yields a calculated output capaci-tance of 442 µF. Adding the typical capacitor-value tolerance (20%) gives a practical value for output capacitance of ap-proximately 530 µF. The closest standard value is 560 µF. Output ripple due to the capacitance alone is given by:

VC

V V

L

V

V fOUVOUV T

O

INV VINV VOUV VOUV V T OUVOUV T

INV fINV fSWV fSWV fCATCAT PACITOR

MAXMAXMA

MAXMAXMA

= ×= ×V V−V V

× ×× ×OU× ×OUT× ×T

× ×× ×

1

2

12

.

ESR of the output capacitor dominates the output-voltage ripple. The amount can be calculated as follows:

V I ESR I ESROUV IOUV IT LV IT LV I C IR IC IR I NDUCTOR CESR CESRR CREST LEST LR RT LR RT L IPPLE OC IE OC I O

= ×V I= ×V IT L= ×T LV IT LV I= ×V IT LV I = ×R I= ×R IC I= ×C IR IC IR I= ×R IC IR I N= ×NDUCTO= ×DUCTOR C= ×R CR I= ×R IR IC IR I= ×R IC IR I∆R I∆R IR IC IR I∆R IC IR IR I= ×R I∆R I= ×R IR IC IR I= ×R IC IR I∆R IC IR I= ×R IC IR I .Be aware that choosing a capacitor with very low ESR may

cause the power converter to be unstable. The factors that affect stability vary from IC to IC, so when choosing an out-put capacitor, be sure to read the data sheet and pay special attention to sections dealing with converter stability.

LIR = 0.2 LIR = 0.3 LIR = 0.4 LIR = 0.5

Fig. 2. As LIR increases from 0.2 to 0.5, the load-transient response quickens. The top waveform is the ac-coupled output-voltage ripple, at 100 mV/div. The center waveform is the current load at 5 A/div. And the bottom waveform is the inductor current at 5 A/div. The time scale is 20 µs/div for all waveforms.

www.powerelectronics.com Power Electronics Technology June 2006www.powerelectronics.com Power Electronics Technology 49

Adding the output-voltage ripple due to capacitance value (the fi rst term in Eq. 4) and the output-capacitor ESR (the second term in Eq. 4) yields the total output-voltage ripple for the stepdown converter:

VC

V V

L

V

V f

I

OUVOUV T

O


INV fINV fSWV fSWV f

INDU

RIPPLTRIPPLT E

MAXMAXMA

MAXMAXMA

= ×= ×V V−V V

× ×× ×OU× ×OUT× ×T

× ×× ×

+1

2

12

∆ CTORCCTORC CESRO

× . (Eq. 4)

Rearranging Eq. 4 to solve for ESR yields:

ESRI

VC

V V

L

V

V

C

INDUCTOR

OUVOUV T

O


INVINV

O

RIPPLTRIPPLT E

MAXMAXMA

MAXMAXMA

= ×= ×

− ×− ×V V−V V

×

1

1

2

1

∆

fffSWfffSWfff

2

.

(Eq. 5)A decent stepdown converter usually achieves an output-

voltage ripple of less than 2% (40 mV in our case). For a 560-µF output capacitance, Eq. 5 yields 18.8 mΩ for the maximum calculated ESR. Therefore, choose a capacitor with ESR that’s lower than 18.8 mΩ and a capacitance that’s equal to or greater than 560 µF. To achieve an equivalent ESR value less than 18.8 mΩ, you can connect multiple low-ESR capacitors in parallel.

Fig. 3 presents output-ripple voltage versus output capaci-tance and ESR. Because our example uses tantalum capaci-tors, capacitor ESR dominates the output-voltage ripple.

Input Capacitor SelectionThe input capacitor’s ripple-current rating dictates its

value and physical size, and the following equation calculates the amount of ripple current the input capacitor must be able to handle:

I IV V

VC OI IC OI I UT

OUV VOUV VT IV VT IV V N OUT

INVINVIC OIC ORM

C ORM

C OS

C OS

C O MAUTMAUT XMAXMAI I=I I

( )V V( )V V V( )VT I( )T IV VT IV V( )V VT IV V N O( )N OVN OV( )VN OV UT( )UT−( )−.

Fig. 3. The output capacitor’s equivalent series resistance (ESR)dominates the output-voltage ripple.

Fig. 4 plots ripple current for the capacitor (shown as a multiple of the output current) against the input voltage of the buck converter (shown as a ratio of output voltage to input voltage). The worst case occurs when V

IN = 2V

OUT = 2V

OUT = 2V

(VOUT

(VOUT

(V /VIN

= 0.5), yielding IOUTMATMAT XMAXMA/ 2 for the worst-case

ripple-current rating.The input capacitance required for a stepdown converter

depends on the impedance of the input power source. For common laboratory power supplies, 10 µF to 22 µF of ca-pacitance per ampere of output current is usually suffi cient. Given the design parameters of Fig. 1, you can calculate the input-ripple current as 3.16 A. You then can start with 40 µF in total input capacitance and can adjust that value according to subsequent test results.

Tantalum capacitors are a poor choice for input capaci-tors. They usually fail “short,” meaning the failed capaci-tor creates a short circuit across its terminals and thereby raises the possibility of a fi re hazard. Ceramic or aluminum-electrolytic capacitors are preferred because they don’t have this failure mode.

Ceramic capacitors are the better choice when pc-board area or component height is limited, but ceramics may cause your circuit to produce an audible buzz. This high-pitched noise is caused by physical vibration of the ceramic capacitor against the pc board as a result of the capacitor’s ferroelectric properties and piezo phenomena reacting to the voltage ripple. Polymer capacitors can alleviate this problem.

BUCK-CONVERTER DESIGNS

ESRCO()

Outputcapacitance(F)

0.5

Outp

ut-v

olta

ge ri

pple

(V)

0.4

0.3

0.2

0.1

0

8-10-46-10-4

4-10-42-10-4

0

00.025

0.05 0.075 0.1

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Polymer capacitors also fail short, but they are much more robust than tantalums, and therefore are suitable as input capacitors.

Diode SelectionPower dissipation is the limiting factor when choosing

a diode. The worst-case average power can be calculated as follows:

(Eq. 6)P

V

VI VDIODPDIODP E

OUVOUV T

INVINV OUT D

MAXMAXMA

MAT DMAT DXT DXT DMAXMAT DMAT DXT DMAT D= −

= −= −

1 ,I V1 ,I VOU1 ,OUI VOUI V1 ,I VOUI VT D1 ,T DI VT DI V1 ,I VT DI VT DMAT D1 ,T DMAT D1 ,1 ,V

1 ,V

OU1 ,OUT1 ,TT DXT D1 ,T DXT DT DMAT DXT DMAT D1 ,T DMAT DXT DMAT D= −1 ,= −

1 ,

1 ,

1 ,

1 ,

1 ,

× ×1 ,× ×I V× ×I V1 ,I V× ×I VI VOUI V× ×I VOUI V1 ,I VOUI V× ×I VOUI VI VT DI V× ×I VT DI V1 ,I VT DI V× ×I VT DI V

where VD is the voltage drop across the diode at the given

output current IOUTMATMAT XMAXMA. (Typical values are 0.7 V for a silicon

diode and 0.3 V for a Schottky diode.) Ensure that the selected diode will be able to dissipate that much power. For reliable operation over the input-voltage range, you must also ensure that the reverse-repetitive maximum voltage is greater than the maximum input voltage (V

RRM VINVINV

MAXMAXMA). The diode’s

forward-current specifi cation must meet or exceed the maximum output current (i.e., IFAVFAVF IOUTMATMAT XMAXMA

).

MOSFET SelectionSelecting a MOSFET can be daunting, so engineers often

avoid that task by choosing a regulator IC with an internal MOSFET. Unfortu-nately, most manufacturers fi nd it cost prohibitive to integrate a large MOSFET with a dc-dc controller in the same pack-age, so power converters with integrated MOSFETs typically specify maximum output currents no greater than 3 A to 6 A. For larger output currents, the only alternative is usually an external MOSFET.

The maximum junction tempera-ture ( TJTJT

MAXMAXMA) and maximum ambient

temperature ( TATATMAXMAXMA

) for the external MOSFET must be known before you can select a suitable device. TJTJT

MAXMAXMAshould not

exceed 115oC to 120oC and TATATMAXMAXMA

should not exceed 60oC. A 60oC maximum ambient temperature may seem high, but stepdown converter circuits are typically housed in a chassis where such ambient temperatures are not unusual. You can calculate a maximum allow-able temperature rise for the MOSFET as follows:

T T TJ JT TJ JT T ATATRISEJ JRISEJ JMAX MAX MAMAX MMA AX

= −T T= −T T . (Eq. 7)Inserting the values mentioned above

for TJTJTMAXMAXMA

and TATATMAXMAXMA

into Eq. 7 yields a maximum MOSFET temperature rise of 55oC. The maximum power dissi-pated in the MOSFET can be calculated from the allowable maximum rise in MOSFET temperature:

(Eq. 8) (Eq. 8)P (Eq. 8)P (Eq. 8)T

(Eq. 8)D (Eq. 8) (Eq. 8)P (Eq. 8)D (Eq. 8)P (Eq. 8)J

(Eq. 8)J

(Eq. 8)TJT

JATO

(Eq. 8)TO

(Eq. 8)T

(Eq. 8)T

(Eq. 8)RISE

(Eq. 8)RISE

(Eq. 8) (Eq. 8)= (Eq. 8)Θ

(Eq. 8)Θ

(Eq. 8) (Eq. 8). (Eq. 8)

The type of MOSFET package and the amount of pc-board copper con-nected to it affect the MOSFET’s junc-tion-to-ambient thermal resistance (

JA). When

JA is not specifi ed in the

data sheet, 62oC/W serves as a good esti-mate for a standard SO-8 package (wire-bond interconnect, without an exposed


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paddle), mounted on 1 in.2 of 1-oz pc-board copper. There exists no inverse linear relationship between a

jA

value and the amount of copper connected to the device, and the benefi t of decreasing the

JAJAJ value quickly dwindles for

circuits that include more than 1 sq in. of pc-board copper. Using

JA = 62

JA = 62

JA°C/W in Eq. 8 yields a maximum allowable dis-

sipated power in the MOSFET of approximately 0.89 W.Power dissipation in the MOSFET is caused by on-

resistance and switching losses. On-resistance loss can be calculated as:

(Eq. 9)P (Eq. 9)P (Eq. 9)V (Eq. 9)V (Eq. 9)V

I R (Eq. 9)I R (Eq. 9)DPDP (Eq. 9)OU (Eq. 9) (Eq. 9)V (Eq. 9)OU (Eq. 9)V (Eq. 9) (Eq. 9)T (Eq. 9)

INVINV OUI ROUI RT DI RT DI R S O HOTRDS

MIN

MAT DMAT DXT DXT DMAXMAT DMAT DXT DMAT D= × (Eq. 9)= × (Eq. 9) (Eq. 9)= × (Eq. 9) (Eq. 9)OU (Eq. 9)= × (Eq. 9)OU (Eq. 9) (Eq. 9)T (Eq. 9)= × (Eq. 9)T (Eq. 9)I R×I R (Eq. 9)I R (Eq. 9)× (Eq. 9)I R (Eq. 9)I RT DI R×I RT DI R (Eq. 9)2 (Eq. 9) (Eq. 9)I R (Eq. 9)2 (Eq. 9)I R (Eq. 9)( )S O( )S ON( )N .

Because most data sheets specify the maximum on-

resistance only at 25°C, you may have to estimate the value of on-resistance at TJTJT

HOT. As a rule of thumb, a temperature

coeffi cient of 0.5%/°C provides a good indicator for maxi-mum on-resistance at any given temperature. Thus, the hot on-resistance is calculated as: (Eq. 10)

R T C RDSR TDSR THOT J DC RJ DC R S O CHOJ DHOJ DTJ DTJ D( )R T( )R TON( )ONR TONR T( )R TONR T ( )S O( )S ON( )N[ .R T[ .R T( )R T( )R T C R( )C RJ D( )J DR TJ DR T( )R TJ DR T C RJ DC R( )C RJ DC RJ DHOJ D( )J DHOJ DJ DTJ D( )J DTJ D] .C R] .C RJ D] .J DC RJ DC R] .C RJ DC R S O] .S O C

] .C( )] .( )S O( )S O] .S O( )S ON( )N] .N( )NR T= +R TR T[ .R T= +R T[ .R T( )− °( )

°] .

°] .[ .1 0[ .R T[ .R T1 0R T[ .R TR T[ .R T= +R T[ .R T1 0R T[ .R T= +R T[ .R T005R T005R T( )25( )J D( )J D25J D( )J D( )− °( )25( )− °( )

25] .

25] .

Assuming the on-resistance loss is approximately 60% of the total MOSFET losses, you can substitute in Eq. 10 and rearrange to yield Eq. 11, the maximum allowable on-resistance at 25°C:

(Eq. 11)

RV

V

I TP

DS C

INVINV

OU (Eq. 11)

OU (Eq. 11)

VOUV T (Eq. 11)

T (Eq. 11)

OUI TOUI TT JI TT JI TT JI TT JI TDPDP

MIN

MAT JMAT JX HT JX HT JMAX HMAT JMAT JX HT JMAT J OT

TO

( )ON( )ON

[ .I T[ .I TT J[ .T JI TT JI T[ .I TT JI TT JX HT J[ .T JX HT J( )I T( )I T C( )CT J( )T JI TT JI T( )I TT JI TX H

( )X HT JX HT J( )T JX HT J OT

( )OT

]

25

2I T2I T

1 (Eq. 11)

1 (Eq. 11)

I T[ .I T1 0I T[ .I TT J[ .T J1 0T J[ .T JI TT JI T[ .I TT JI T1 0I TT JI T[ .I TT JI TI T005I TT J005T J( )25( )

°= ×= ×MI= ×MIN= ×N

I T+ ×I TI TT JI T+ ×I TT JI TI T+ ×I TI TT JI T+ ×I TT JI TI T[ .I T+ ×I T[ .I TI TT JI T[ .I TT JI T+ ×I TT JI T[ .I TT JI TI T[ .I T1 0I T[ .I T+ ×I T[ .I T1 0I T[ .I TI TT JI T[ .I TT JI T1 0I TT JI T[ .I TT JI T+ ×I TT JI T[ .I TT JI T1 0I TT JI T[ .I TT JI TI T005I T+ ×I T005I TI TT JI T005I TT JI T+ ×I TT JI T005I TT JI T( )− °( )( )25( )− °( )25( ) TTT× 60%.

Switching losses constitute a smaller portion of the MOSFET’s power dissipation, but they still must be taken into account. The following switching-loss calculation provides only a rough estimate, and therefore is no substi-tute for evaluation in the lab, preferably a test that includes

a thermocouple mounted on P1 as a sanity check.

(Eq. 12) (Eq. 12)P (Eq. 12)P (Eq. 12)C V f I

I (Eq. 12)

I (Eq. 12) (Eq. 12)D (Eq. 12) (Eq. 12)P (Eq. 12)D (Eq. 12)P (Eq. 12)

RS (Eq. 12)

RS (Eq. 12)

C VRSC VS I (Eq. 12)

S I (Eq. 12)

C VS IC V N S (Eq. 12)

N S (Eq. 12)

f IN Sf IW Of IW Of I UT

GATESW

(Eq. 12)SW

(Eq. 12) (Eq. 12)MA (Eq. 12)N SMAN S

(Eq. 12)N S

(Eq. 12)MA (Eq. 12)N S

(Eq. 12)X M (Eq. 12)X M (Eq. 12)N SX MN S

(Eq. 12)N S

(Eq. 12)X M (Eq. 12)N S

(Eq. 12)W OX MW O

(Eq. 12)W O

(Eq. 12)X M (Eq. 12)W O

(Eq. 12)UTX MUT

(Eq. 12)UT

(Eq. 12)X M (Eq. 12)UT

(Eq. 12) (Eq. 12)MA (Eq. 12)X M (Eq. 12)MA (Eq. 12)N SMAN SX MN SMAN S

(Eq. 12)N S

(Eq. 12)MA (Eq. 12)N S

(Eq. 12)X M (Eq. 12)N S

(Eq. 12)MA (Eq. 12)N S

(Eq. 12)AX (Eq. 12)AX (Eq. 12) (Eq. 12)= (Eq. 12)× ×C V× ×C VS I× ×S IC VS IC V× ×C VS IC V N S× ×N Sf I×f If IW Of I×f IW Of I2× ×2× ×

(Eq. 12), (Eq. 12)

where CRSS

is the reverse-transfer capacitance of P1, IGATE

is the peak gate-drive source/sink current of the controller and P1 is the high-side MOSFET.

Assuming a gate drive of 1 A (obtained from the gate driver/ controller data sheet) and a reverse-transfer capaci-tance of 300 pF (obtained from the MOSFET data sheet), Eq. 11 yields a maximum RDS C( )ON( )ON 25°

of approximately 26.2 m. Recalculating and summing the on-resistance losses and the switching losses yields a net dissipated power of 0.676 W. Using this fi gure, you can calculate for the MOSFET a maximum temperature rise of 101oC, which is within the acceptable temperature range.

Stepdown-Converter Effi ciencyMinimizing power loss throughout the converter will

extend battery life and reduce heat dissipation. The follow-ing equations calculate power loss in each section of the converter.

Input capacitor ESR loss: P I ESRC CP IC CP I CIC CIC CRM

C CRM

C CS

C CS

C CIRMS I= ×P I= ×P IC C= ×C CP IC CP I= ×P IC CP I 2= ×2= × .

Refer to Eqs. 6, 9 and 12 for losses due to the diode, the RM

Refer to Eqs. 6, 9 and 12 for losses due to the diode, the RMS

Refer to Eqs. 6, 9 and 12 for losses due to the diode, the S RM

Refer to Eqs. 6, 9 and 12 for losses due to the diode, the RMS

Refer to Eqs. 6, 9 and 12 for losses due to the diode, the S

MOSFET on-resistance and the MOSFET switching loss.Inductor DCR loss:P I I DCRDCP IDCP IR OP IR OP I INDUCTOR LI DR LI DCRR LCR

RMR ORMR OS MR OS MR OUTS MUT AXP I= +P IP IR OP I= +P IR OP I I D× ×I DI DR LI D× ×I DR LI D( )( )P I( )P I I D( )I DI D( )I DR O( )R OUT( )UT IN( )INI DINI D( )I DINI DDUCTO( )DUCTOI DDUCTOI D( )I DDUCTOI DR L( )R LR L( )R LI DR LI D( )I DR LI DI DR LI D( )I DR LI D

S M( )

S MR OS MR O( )R OS MR OUTS MUT( )UTS MUT AX( )

AX= +( )= +P I= +P I( )P I= +P IR O= +R O( )R O= +R OP IR OP I= +P IR OP I( )P IR OP I= +P IR OP I UT= +UT( )UT= +UT I D× ×I D( )I D× ×I DI DR LI D× ×I DR LI D( )I DR LI D× ×I DR LI DI D× ×I D( )I D× ×I DI D× ×I D( )I D× ×I DI DR LI D× ×I DR LI D( )I DR LI D× ×I DR LI DI DR LI D× ×I DR LI D( )I DR LI D× ×I DR LI D .( )∆( )I D( )I D2I D( )I DI DR LI D( )I DR LI D2I DR LI D( )I DR LI DI D× ×I D( )I D× ×I D2I D× ×I D( )I D× ×I DI DR LI D× ×I DR LI D( )I DR LI D× ×I DR LI D2I DR LI D× ×I DR LI D( )I DR LI D× ×I DR LI D2I D2I DI D× ×I D2I D× ×I D

Output capacitor ESR loss:Output capacitor ESR loss:P I ESRC IP IC IP I NDUCTOR CESR CESRR CR

OC IOC IRM

C IRM

C IS

C IS

C I OP I= ×P IP IC IP I= ×P IC IP I ×R C×R C( )( )P I( )P IC I( )C IP IC IP I( )P IC IP I N( )NDUCTO( )DUCTOR C( )R CR C( )R C= ×( )= ×P I= ×P I( )P I= ×P IC I= ×C I( )C I= ×C IP IC IP I= ×P IC IP I( )P IC IP I= ×P IC IP I N= ×N( )N= ×NDUCTO= ×DUCTO( )DUCTO= ×DUCTOR C= ×R C( )R C= ×R C .( )∆( )P I( )P I∆P I( )P IP IC IP I( )P IC IP I∆P IC IP I( )P IC IP IP I= ×P I( )P I= ×P I∆P I= ×P I( )P I= ×P IP IC IP I= ×P IC IP I( )P IC IP I= ×P IC IP I∆P IC IP I= ×P IC IP I( )P IC IP I= ×P IC IP I( )3( )R C( )R C3R C( )R C

2

Pc-board copper Loss: Pc-board copper loss is diffi cult to calculate accurately, but Fig. 5 provides a rough estimate of the amount of resistance per square area of pc-board cop-per. With Fig. 5, you can use a simple I2R power dissipation equation to calculate the power loss.

The following equation sums all of the power losses throughout the converter, and accounts for those losses in the expression for converter effi ciency:

=(VOUT

=(VOUT

=(V IOUT

)/(VOUT

)/(VOUT

)/(V IOUT

+ P P PC CP PC CP P DCPDCP RIC CIC CRM

C CRM

C CS

C CS

C CORMS RMS+ +P P+ +P PC C+ +C CP PC CP P+ +P PC CP P +

P P P PD DP PD DP P DIODP PDIODP PE CP PE CP P URDD DRDD DS SD DS SD D W+ +P P+ +P PD D+ +D DP PD DP P+ +P PD DP P + ×P P+ ×P PE C+ ×E CP PE CP P+ ×P PE CP P U+ ×U ) %+ ×) %+ × .) %100) %

Assuming a reasonable net copper loss of approximately

hl

b R = ρ x b x h

l

b = l

R = ρ = 1.72 x 10-8 x mΩh 1.4 mil

R = 0.484 mΩ =~ 0.5 mΩ

Fig. 5. The resistance of one square of 1-oz copper is approximately 0.5 m.

Fig. 4. Ripple current for the input capacitors reaches a worst case of I

OUT /2 = 0.5 when the variable input voltage equals twice the fi xed

output voltage.

0

0.1

0.2

0.3

0.4

0.5

0 0.25 0.5 0.75 1

VOUT/VIN

Ripp

le c

urre

nt (m

ultip

le o

f IOU

T)

www.powerelectronics.com Power Electronics Technology June 2006www.powerelectronics.com Power Electronics Technology 53

0.75 W, the effi ciency for this converter is 69.5%. Replacing the silicon diode with a Schottky diode increases the ef-fi ciency to 79.6%, and replacing the diode with a MOSFET synchronous rectifi er further increases the effi ciency to 85% at full load.

Fig. 6 depicts a breakdown of power losses in the con-verter. Doubling the copper weight to 2 oz or tripling it to 3 oz minimizes the copper loss and thereby increases the effi ciency to approximately 86% to 87%.

Careful pc-board layout is critical in achieving low switch-ing losses and stable operation for a stepdown converter. Use the following guidelines as a starting point:

Keep the high-current paths short, especially at the ground terminals.

Minimize connection lengths to the inductor, MOSFET and diode/synchronous MOSFET.

Keep power traces and load connections short and wide. This practice is essential for high effi ciency.

Keep voltage- and current-sensing nodes and traces away from switching nodes.

Verifying PerformanceWhen designing or modifying a stepdown switching-

regulator circuit (one that operates in CCM, using PWM), you can use the equations in this article to calculate values for the critical components and characteristics required. You should always lab-test the circuit to verify fi nal electrical and thermal specifi cations. For acceptable circuit operation, a proper pc-board layout and judicious component placements are as critical as choosing the right components. PETech

Diode 63%

MOSFET switching 5%

MOSFET on-resistance 6%

Inductor dc resistance 8%

0utput capacitor ESR 7%

Copper 8%

Input capacitor ESR 3%

Fig. 6. Power loss caused by the freewheeling diode should be eliminated to increase the converter’s effi ciency.

BUCK-CONVERTER DESIGNS

buck converter design

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