US3012/3012A

4-9

Rev. 1.0

5/6/98

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 *(V inm in -Vom ax )/(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 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 m

Ω, 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)

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 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.

Part #

Thermalloy

6078B

AAVID

577002