US3007
For example, if the desired current limit point is set to
be 22A for the synchronous and 16A for the non syn-
chronous , and from our previous selection, the maxi-
mum MOSFET Rds-on =19mW, then the current sense
resistor Rcs is calculated as :
The same heat sink as the one selected for the switcher
MOSFETs is also suitable for the 1.5V regulator.
2.5V, Clock Supply
Vcore
The US3007 provides a complete 2.5V regulator with a
minimum of 200mA current capability. The internal regu-
lator has short circuit protection with internal thermal
shutdown.
Vcs=IcL*Rds=22*0.019=0.418V
Rcs=Vcs/Ib=(0.418V)/(200uA)=2.1kW
Where: Ib=200uA is the internal current setting of the
US3007
3.3V supply
1.5V and 2.5V Supply Resistor Divider Selection
Vcs=IcL*Rds=16*0.019=0.3V
Rcs=Vcs/Ib=(0.3V)/(200uA)=1.50kW
Since the internal voltage reference for the linear regula-
tors is set at 1.26V for US3007, there is a need to use
external resistor dividers to step up the voltage. The re-
sistor dividers are selected using the following equations:
Vo=(1+Rt/Rb)*Vref
Where:
Rt=Top resistor divider
Rb=Bottom resistor divider
Vref=1.26V typical
1.5V, GTL+ Supply LDO Power MOSFET Selection
The first step in selectiong the power MOSFET for the
1.5V linear regulator is to select its maximum Rds-on of
the pass transistor based on the input to output Dropout
voltage and the maximum load current.
Rds(max)=(Vin - Vo)/IL
For Vo=1.5V, and Vin=3.3V , IL=2A
Rds-max=(3.3 - 1.5)/2= 0.9W
For 1.5V supply :
Assuming Rb=1kW
Note that since the MOSFETs Rds-on increases with
temperature, this number must be divided by » 1.5,
inorder to find the Rds-on max at room temperature. The
Motorola MTP3055VL has a maximum of 0.18W Rds-on
at room temperature, which meets our requirement.
To select the heatsink for the LDO Mosfet the first step
is to calculate the maximum power dissipation of the
device and then follow the same procedure as for the
switcher.
Rt=Rb*[(Vo/Vref) - 1]
Rt=1*[(1.5/1.26) - 1]=191W
For 2.5V supply :
Assuming Rb=1.02kW
Rt=Rb*[(Vo/Vref) - 1]
Rt=1.02*[(2.5/1.26) - 1]=1kW
Switcher Output Voltage Adjust
Vcore
Pd = ( Vin - Vo ) * IL
Where :
Pd = Power Dissipation of the Linear Regulator
IL = Linear Regulator Load Current
For the 1.5V and 2A load:
Pd = (3.3 - 1.5)*2=3.6 W
Assuming Tj-max=125°C
Ts = Tj - Pd * (qjc + qcs)
Ts = 125 - 3.6 * (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 (qsa) is calculated as follows :
Assuming Ta=35 °C
DT = Ts - Ta = 118 - 35 = 83 °C Temperature Rise
Above Ambient
qsa = DT/Pd
As it was discussed earlier,the trace resistance from
the output of the switching regulator to the Slot 1 can be
used to the circuit advantage and possibly reduce the
number of output capacitors, by level shifting the DC
regulation point when transitioninig from light load to full
load and vice versa. To account for the DC drop, the
output of the regulator is typically set about half the DC
drop that results from light load to full load. For example,
if the total resistance from the output capacitors to the
Slot 1 and back to the GND pin of the 3007 is 5mW and
if the total DI, the change from light load to full load is
14A, then the output voltage measured at the top of the
resistor divider which is also connected to the output
capacitors in this case, must be set at half of the 70 mV
or 35mV higher than the DAC voltage setting. To do this,
the top resistor of the resistor divider(R17 in the applica-
tion circuit) is set at 100W, and the R19 is calculated.
For example, if DAC voltage setting is for 2.8V and the
desired output under light load is 2.835V, then R19 is
calculated using the following formula :
qsa = 83 / 3.6 = 23 °C/W
Rev. 1.8
12/8/00
4-13
UNISEM [ UNISEM ]
