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參數資料
| 型號： | IRU3005CW |
| 文件頁數： | 11/14頁 |
| 文件大?。?/td> | 83K |
| 代理商： | IRU3005CW |
IRU3004, IRU3005
4-13
Rev. 1.2
12/8/00
drooping during a load current step. However, if the in-
ductor is too small, the output ripple current and ripple
voltage become too large. One solution 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 following 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 * ( Vinmin - Vomax ) / ( 2*
I )
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
μ
H
Assuming that the programmed switching frequency is
set at 200KHZ, 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
μ
H inductance with
≈
3
m
of DC resistance.
Assuming L = 3
μ
H and the switching frequency; Fsw =
200KHZ , 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, assuming
IRL3103 MOSFET for both switches with maximum on-
resistance of 19m
, we have:
T = 1 / 200000
= 5
μ
Sec
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
μ
Sec
Toff = 5 - 3.1 = 1.9
μ
Sec
Ir = ( 2.8 + 0.27 ) * 1.9 / 3 = 1.94A
Vo = 1.94 * .006 = .011 V = 11mV
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)
Pdh= 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)
Pds=(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 TO-220 package gives
θ
jc=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;
θ
cs=0.05
°
C/W, the maximum heat
sink temperature is then calculated as:
Ts = Tj - Pd * (
θ
jc +
θ
cs)
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 resistance
(
θ
sa) is calculated as follows:
Assuming Ta=35
°
C
T = Ts - Ta = 118 - 35 = 83
°
C Temperature Rise
Above Ambient
θ
sa =
T/Pd
θ
sa = 83 / 3.82 = 22
°
C/W
Next, a heat sink with lower
θ
sa than the one calculated
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 Thermalloy meet this crite-
ria.
Co.
Thermalloy
AAVID
Part #
6078B
577002
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