AAT3242
300mA/150mA Dual CMOS LDO Linear Regulator
The following is an example for a 2.5V output:
VOUTA = 2.5V
VOUTB = 1.5V
IOUTB
VIN
= 150mA
= 4.2V
IGND
= 125μA
909mW - (2 × 4.2V × 125μA) - (4.2 - 1.5) × 150mA
4.2 - 2.5
IOUTA(MAX)
=
IOUTA(MAX) = 296mA
From the discussion above, PD(MAX) was determined to equal 909mW at TA = 25°C.
Therefore, with Regulator B delivering 150mA at 1.5V, Regulator A can sustain a constant 2.5V output at a
296mA load current at an ambient temperature of 25°C. Higher input-to-output voltage differentials can be
obtained with the AAT3242, while maintaining device functions within the thermal safe operating area. To
accomplish this, the device thermal resistance must be reduced by increasing the heat sink area or by oper-
ating the LDO regulator in a duty-cycled mode.
For example, an application requires VIN = 4.2V while VOUT = 1.5V at a 500mA load and TA = 25°C. To main-
tain this high input voltage and output current level, the LDO regulator must be operated in a duty-cycled mode.
Refer to the following calculation for duty-cycle operation:
IGND = 125µA
IOUT = 500mA
VIN = 4.2V
VOUT = 1.5V
100(PD(MAX)
)
%DC =
%DC =
[(VIN - VOUTA)IOUTA + (VIN × IGND)] + [(VIN - VOUTB)IOUTB + (VIN × IGND)]
100(909mW)
[(4.2V - 1.5V)500mA + (4.2V × 125μA)] + [(4.2V - 1.5V)200mA + (4.2V × 125μA)]
%DC = 48.10%
PD(MAX) is assumed to be 909mW
For a 500mA output current and a 2.7V drop across the AAT3242 at an ambient temperature of 25°C, the max-
imum on-time duty cycle for the device would be 48.10%.
3242.2006.04.1.10
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AAT [ ADVANCED ANALOG TECHNOLOGY, INC. ]
