US3018
However if the inductor is too small , the output ripple
current and ripple voltage become too large. One solu-
tion to bring the ripple current down is to increase the
switching frequency , however that will be at the cost of
reduced efficiency and higher system cost. The follow-
ing set of formulas are derived to achieve the optimum
performance without many design iterations.
The maximum output inductance is calculated using the
following equation :
L = ESR *C *(Vi i -Vom ax )/(2*
∆I
)
nm n
Where :
Vinmin = Minimum input voltage
For Vo = 2.8 V ,
∆I
= 14.2 A
L =0.006 * 9000 * ( 4.75 - 2.8) / (2 * 14.2) = 3.7 uH
Assuming that the programmed switching frequency is
set at 200 KHZ , an inductor is designed using the
Micrometals’ powder iron core material. The summary
of the design is outlined below :
The selected core material is Powder Iron , the
selected core is T50-52D from Micro Metal wounded
with 8 Turns of # 16 AWG wire, resulting in 3 uH
inductance with
≈
3 mΩ of DC resistance.
Assuming L = 3 uH and the switching frequency ; Fsw =
200 KHZ , the inductor ripple current and the output
ripple voltage is calculated using the following set of
equations :
T = 1/Fsw
T
≡
Switching Period
D
≈
( Vo + Vsync ) / ( Vin - Vsw + Vsync )
D
≡
Duty Cycle
Ton = D * T
Vsw
≡
High side Mosfet ON Voltage = Io * Rds
Rds
≡
Mosfet On Resistance
Toff = T - Ton
Vsync
≡
Synchronous MOSFET ON Voltage=Io * Rds
∆Ir
= ( Vo + Vsync ) * Toff /L
∆Ir ≡
Inductor Ripple Current
∆Vo
=
∆Ir
* ESR
∆Vo ≡Output
Ripple Voltage
In our example for Vo = 2.8V and 14.2 A load , Assum-
ing IRL3103 MOSFET for both switches with maximum
on resistance of 19 mΩ, we have :
T = 1 / 200000 = 5 uSec
Vsw =Vsync= 14.2*0.019=0.27 V
D
≈
( 2.8 + 0.27 ) / ( 5 - 0.27 + 0.27 ) = 0.61
Ton = 0.61 * 5 = 3.1 uSec
Toff = 5 - 3.1 = 1.9 uSec
∆Ir
= ( 2.8 + 0.27 ) * 1.9 / 3 = 1.94 A
∆Vo
= 1.94 * .006 = .011 V = 11 mV
Power Component Selection
Assuming IRL3103 MOSFETs as power components,
we will calculate the maximum power dissipation as fol-
lows:
For high side switch the maximum power dissipation
happens at maximum Vo and maximum duty cycle.
Dmax
≈
( 2.8 + 0.27 ) / ( 4.75 - 0.27 + 0.27 ) = 0.65
Pdh = Dmax * Io^2*Rds(max)
= 0.65*14.2^2*0.029=3.8 W
Rds(max)=Maximum Rds-on of the MOSFET at 125°C
For synch MOSFET, maximum power dissipation hap-
pens at minimum Vo and minimum duty cycle.
Dmin
≈
( 2 + 0.27 ) / ( 5.25 - 0.27 + 0.27 ) = 0.43
Pds = (1-Dmin)*Io^2*Rds(max)
=(1 - 0.43) * 14.2^2 * 0.029 = 3.33 W
Heatsink Selection
Selection of the heat sink is based on the maximum
allowable junction temperature of the MOSFETS. Since
we previously selected the maximum Rds-on at 125°C,
then we must keep the junction below this temperature.
Selecting TO220 package gives
θjc=1.8°C/W
( From the
venders’ datasheet ) and assuming that the selected
heatsink is Black Anodized , the Heat sink to Case ther-
mal resistance is ;
θcs=0.05°C/W
, the maximum heat
sink temperature is then calculated as :
Ts = Tj - Pd * (θjc +
θcs)
Ts = 125 - 3.82 * (1.8 + 0.05) = 118
°C
With the maximum heat sink temperature calculated in
the previous step, the Heat Sink to Air thermal resis-
tance (θsa) is calculated as follows :
Assuming Ta=35
°C
∆T
= Ts - Ta = 118 - 35 = 83
°C
Temperature Rise
Above Ambient
θsa
=
∆T/Pd
θsa
= 83 / 3.82 = 22
°C/W
Next , a heat sink with lower
θsa
than the one calcu-
lated in the previous step must be selected. One way to
do this is to simply look at the graphs of the “Heat Sink
Temp Rise Above the Ambient” vs. the “Power Dissipa-
tion” given in the heatsink manufacturers’ catalog and
select a heat sink that results in lower temperature rise
than the one calculated in previous step. The following
heat sinks from AAVID and Thermaloy meet this crite-
ria.
Co.
Thermalloy
AAVID
Part #
6078B
577002
4-12
Rev. 1.4
12/8/00
UNISEM [ UNISEM ]
