AAT3242_08 Datasheet, PDF (12/15 Pages) Advanced Analogic Technologies – 300mA/150mA Dual CMOS LDO Linear Regulator

PowerLinear^TM
PRODUCT DATASHEET
AAT3242
300mA/150mA Dual CMOS LDO Linear Regulator
This formula can be solved for I_OUTA to determine the
maximum output current for LDOA:
I_OUTA(MAX) =
P_D(MAX) - (2×V_IN × I_GND) - (V_IN - V_OUTB) × I_OUTB
V_IN - V_OUTA
The following is an example for a 2.5V output in the
TSOPJW package:
V_OUTA = 2.5V
V_OUTB = 1.5V
I_OUTB = 150mA
V_IN
= 4.2V
I_GND
= 125μA
909mW - (2 × 4.2V × 125μA) - (4.2 - 1.5) × 150mA
I_OUTA(MAX) =
4.2 - 2.5
I_OUTA(MAX) = 296mA
From the discussion above, P_D(MAX) was determined to
equal 909mW at T_A = 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 operating
the LDO regulator in a duty-cycled mode.
For example, an application requires V_IN = 4.2V while
V_OUT = 1.5V at a 500mA load and T_A = 25°C. To maintain
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:
I_GND = 125μA
I_OUT = 500mA
V_IN = 4.2V
V_OUT = 1.5V
%DC
=
[(V_IN
-
V_OUTA)I_OUTA
+
100(P_D(MAX))
(V_IN × I_GND)] + [(V_IN
-
V_OUTB)I_OUTB
+
(V_IN
×
I_GND)]
100(909mW)
%DC = _{[(4.2V - 1.5V)500mA + (4.2V × 125μA)] + [(4.2V - 1.5V)200mA + (4.2V × 125μA)]}
%DC = 48.10%
P_D(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 maxi-
mum on-time duty cycle for the device would be
48.10%.
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3242.2008.08.1.11