Buck converter design

1. Power Electronics Technology June 2006 www.powerelectronics.comJune 2006 www.powerelectronics.com46
Buck-Converter
Design Demystified
Though stepdown converters are extremely
popular,therulesofthumbandcalculationsthat
speed their design can be hard to find.
By Donald Schelle and Jorge Castorena,Technical Staff,Jorge Castorena,Technical Staff,Jorge Castorena,
Maxim Integrated Products,Sunnyvale,Calif.
S
tepdown (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 efficiency 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 find. 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 specifies 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 topology
one with a fixed switching frequency, pulse width modu-one with a fixed 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 Selection
Calculating 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, infinite off-resistance and zero switching time)
and an ideal diode: (Eq. 1)
L V
V
V f LIR I
IN OUT
OUVOUV T
INV fINV fSWV fSWV f OUT
MAXMAXMA
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
IOUT
(e.g., for a 300-mAp-p
(e.g., for a 300-mAp-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
efficiency and load-transient response. Increasing the LIR
constant—allowing more inductor ripple current—quickensFig.1. Basic stepdown converter circuit with operating parameters.
Controller
P1
VIN
VOUT
L
COUT
CIN
D
P MOSFET
VIN
= 7 V � 24 V VOUT
= 2 V
IOUT
fSW
= 300 kHz
MAX
= 7 A
+ +

2. Power Electronics Technology June 2006 www.powerelectronics.comJune 2006 www.powerelectronics.com48
BUCK-CONVERTER DESIGNSBUCK-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 efficiency, 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 I
I
I LIR I V
PEI IPEI IAKI IAKI IOUT
INDUCTOR
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 (IPEAK
= 8.09 A).
Choose a saturation-current rating that’s large enough
(10Ainthiscase)tocompensateforcircuittolerancesandthe
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
possibleDCR.Althoughdatasheetspecificationsvaryamong
vendors,always use the maximum DCR specification for de-
sign purposes rather than the typical value,because the maxi-
mum is a guaranteed worst-case component specification.
Output Capacitor Selection
Output capacitance is required to minimize the voltage
overshoot and ripple present at the output of a stepdown
converter.Large overshoots are caused by insufficient output
capacitance,and large voltage ripple is caused by insufficient
capacitance as well as a high equivalent-series resistance
(ESR) in the output capacitor. The maximum allowed
output-voltage overshoot and ripple are usually specified at
the time of design. Thus, to meet the ripple specification 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 specified maximum output
voltage. Output-voltage overshoot can be calculated using
the following equation:
(Eq. 2)∆
∆
V VV V
L I
I
C
VOUV VOUV V T
OU
O
OUVOUV T
MAXMAXMA
V V=V V






V VV V







































−2
2
2
( )( )∆
( )∆
L I( )L I
I
( )I
OU( )OUT( )T
IN
( )INDUCTO
( )DUCTOR
( )R
MA
( )MATMAT( )TMAT X
( )XMAXMA
( )MAXMA
+( )+
2
( )
2 .
Rearranging Eq. 2 yields:
(Eq. 3)C
L I
II
V
O
OUT
INDUCTOR
OUT OVT OV UT
MAMATMAT XXMAXMAMAXMA
=
+

L I

L I

L I

L IL I

L I

L I

L I



L I

L IL I

L I

L I

L IL I

L IL I

L I

L I

L IL I

L I

L I

L I















+ −T O+ −T O
∆∆
22
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:
V
C
V V
L
V
V f
OUVOUV 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
1
2
.
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.

3. www.powerelectronics.com Power Electronics Technology June 2006www.powerelectronics.com Power Electronics Technology49
Adding the output-voltage ripple due to capacitance value
(the first 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:
V
C
V V
L
V
V f
I
OUVOUV T
O
INV VINV VOUV VOUV V T OUVOUV T
INV fINV fSWV fSWV f
INDU
RIPPLTRIPPLT E
MAXMAXMA
MAXMAXMA
= ×= ×
V V−V V
× ×× ×OU
× ×OUT
× ×T


× ×× ×















+
1
2
1
2
∆ CTORCCTORC CESR O
× . (Eq. 4)
Rearranging Eq. 4 to solve for ESR yields:
ESR
I
V
C
V V
L
V
V
C
INDUCTOR
OUVOUV T
O
INV VINV VOUV VOUV V T OUVOUV T
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 presentsoutput-ripplevoltageversusoutputcapaci-
tance and ESR. Because our example uses tantalum capaci-
tors, capacitor ESR dominates the output-voltage ripple.
Input Capacitor Selection
The 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 I
V V
V
C OI IC OI I UT
OUV VOUV VT IV VT IV VN OUT
INVINVIC OIC ORM
C ORM
C OS
C OS
C O MAUTMAUT XMAXMA
I I=I I
( )V V( )V V V( )VT I( )T IV VT IV V( )V VT IV VN 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 VIN
= 2VOUT
= 2VOUT
= 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 sufficient.
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 fire 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
(�)
Output
capacitance(F)
0.5
Output-voltageripple(V)
0.4
0.3
0.2
0.1
0
8-10-4
6-10-4
4-10-4
2-10-4
0
0
0.025
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 Selection
Power dissipation is the limiting factor when choosing
a diode. The worst-case average power can be calculated as
follows:
(Eq. 6)
P
V
V
I 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
OU
1 ,OUT
1 ,T
T 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 0.3V 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 (VRRM
 VINVINV MAXMAXMA
). The diode’s
forward-current specification must meet or exceed the
maximum output current (i.e.,IFAVFAVF  IOUTMATMAT XMAXMA
).
MOSFET Selection
Selecting a MOSFET can be daunting, so engineers often
avoid that task by choosing a regulator
IC with an internal MOSFET. Unfortu-
nately, most manufacturers find it cost
prohibitive to integrate a large MOSFET
withadc-dccontrollerinthesamepack-
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 ( TJTJTMAXMAXMA
) and maximum ambient
temperature ( TATAT MAXMAXMA
) for the external
MOSFETmustbeknownbeforeyoucan
select a suitable device. TJTJTMAXMAXMA
should not
exceed 115o
C to 120o
C and TATAT MAXMAXMA
should
not exceed 60o
C. A 60o
C 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)
Insertingthevaluesmentionedabove
for TJTJTMAXMAXMA
and TATAT MAXMAXMA
into Eq. 7 yields a
maximum MOSFET temperature rise
of 55o
C. 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
JA
TO
(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 specified in the
data sheet,62o
C/W serves as a good esti-
mate for a standard SO-8 package (wire-
bond interconnect, without an exposed
BUCK-CONVERTER DESIGNSBUCK-CONVERTER DESIGNS
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BUCK-CONVERTER DESIGNSBUCK-CONVERTER DESIGNS
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 benefit of decreasing the JAJAJ
value quickly dwindles for
circuits that include more than 1 sq in. of pc-board copper.
Using JA
= 62JA
= 62JA
°
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 TJTJTHOT
. As a rule of thumb, a temperature
coefficient 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)
R
V
V
I T
P
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 T
DPDP
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
2
I T2
I 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 VN S
(Eq. 12)
N S
(Eq. 12)
f IN Sf IW Of IW Of I UT
GATE
SW
(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 VN 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 figure, you can calculate for the
MOSFET a maximum temperature rise of 101o
C, which is
within the acceptable temperature range.
Stepdown-Converter Efficiency
Minimizing 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 LCRRMR ORMR OS MR OS MR OUTS MUT AX
P 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 DS 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 D2
I D2
I DI D× ×I D2
I D× ×I D
Output capacitor ESR loss:Output capacitor ESR loss:
P I ESRC IP IC IP I NDUCTOR CESR CESRR CROC IOC IRM
C IRM
C IS
C IS
C I O
P 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 difficult 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 I2
R 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 efficiency:
=(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
h
l
b R = ρ x
b x h
l
b = l
R =
ρ = 1.72 x 10-8
x mΩ
h 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
IOUT
/2 = 0.5 when the variable input voltage equals twice the fixed
output voltage.
0
0.1
0.2
0.3
0.4
0.5
0 0.25 0.5 0.75 1
VOUT
/VIN
Ripplecurrent(multipleofIOUT
)

6. www.powerelectronics.com Power Electronics Technology June 2006www.powerelectronics.com Power Electronics Technology53
0.75 W, the efficiency for this converter is 69.5%. Replacing
the silicon diode with a Schottky diode increases the ef-
ficiency to 79.6%, and replacing the diode with a MOSFET
synchronous rectifier further increases the efficiency 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
efficiency 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 efficiency.
● Keep voltage- and current-sensing nodes and traces
away from switching nodes.
Verifying Performance
When 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 final electrical and
thermal specifications. For acceptable circuit operation, a
properpc-boardlayoutandjudiciouscomponentplacements
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 efficiency.
BUCK-CONVERTER DESIGNS