PRODUCT DATASHEET
AAT3221/2
PowerLinearTM
150mA NanoPower™ LDO Linear Regulator
maximum conditions are calculated at the maximum
operating temperature where TA = 85°C, under normal
ambient conditions TA = 25°C. Given TA = 85°C, the
maximum package power dissipation is 267mW. At TA =
25°C, the maximum package power dissipation is
667mW.
From the discussion above, PD(MAX) was determined to
equal 667mW at TA = 25°C. Thus, the AAT3221/2 can
sustain a constant 2.5V output at a 150mA load current
as long as VIN is ≤6.95V at an ambient temperature of
25°C. 5.5V is the maximum input operating voltage for
the AAT3221/2, thus at 25°C the device would not have
any thermal concerns or operational VIN(MAX) limits.
The maximum continuous output current for the
AAT3221/2 is a function of the package power dissipa-
tion and the input-to-output voltage drop across the
LDO regulator. Refer to the following simple equation:
This situation can be different at 85°C. The following is
an example for an AAT3221/2 set for a 2.5 volt output
at 85°C:
VOUT = 2.5 volts
IOUT = 150mA
IGND = 1.1μA
PD(MAX)
IOUT(MAX)
=
(VIN - VOUT
)
(267mW + [2.5V · 150mA])
(150mA + 1.1µA)
For example, if VIN = 5V, VOUT = 2.5V and TA = 25°C,
IOUT(MAX) < 267mA. The output short-circuit protection
threshold is set between 150mA and 300mA. If the out-
put load current were to exceed 267mA or if the ambient
temperature were to increase, the internal die tempera-
ture would increase. If the condition remained constant
and the short-circuit protection did not activate, there
would be a potential damage hazard to the LDO regula-
tor since the thermal protection circuit would only acti-
vate after a short-circuit event occured on the LDO
regulator output.
VIN(MAX)
=
VIN(MAX) = 4.28V
From the discussion above, PD(MAX) was determined to
equal 267mW at TA = 85°C.
Higher input-to-output voltage differentials can be
obtained with the AAT3221/2, while maintaining device
functions in the thermal safe operating area. To accom-
plish 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.
To determine the maximum input voltage for a given
load current, refer to the following equation. This calcu-
lation accounts for the total power dissipation of the LDO
regulator, including that caused by ground current.
For example, an application requires VIN = 5.0V while
VOUT = 2.5V at a 150mA load and TA = 85°C. VIN is
greater than 4.28V, which is the maximum safe continu-
ous input level for VOUT = 2.5V at 150mA for TA = 85°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:
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND
)
This formula can be solved for VIN to determine the
maximum input voltage.
IGND = 1.1μA
(PD(MAX) + [VOUT · IOUT])
)
IOUT = 150mA
VIN = 5.0 volts
VOUT = 2.5 volts
VIN(MAX)
=
(IOUT + IGND
The following is an example for an AAT3221/2 set for a
2.5 volt output:
PD(MAX)
([VIN - VOUT]IOUT + [VIN · IGND])
%DC = 100
VOUT = 2.5 volts
IOUT = 150mA
IGND = 1.1μA
267mW
([5.0V - 2.5V]150mA + [5.0V · 1.1µA])
%DC = 100
(667mW + [2.5V · 150mA])
(150mA + 1.1µA)
%DC = 71.2%
VIN(MAX)
=
PD(MAX) is assumed to be 267mW.
VIN(MAX) = 6.95V
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