AAT3201
150mA OmniPower™ LDO Linear Regulator
TA = 25°C. Given TA = 85°, the maximum package
at 25°C, the device would not have any thermal con-
cerns or operational VIN(MAX) limits.
power dissipation is 267mW. At TA = 25°C°, the
maximum package power dissipation is 667mW.
This situation can be different at 85°C. The follow-
ing is an example for an AAT3201 set for a 2.5 volt
output at 85°C:
The maximum continuous output current for the
AAT3201 is a function of the package power dissi-
pation and the input to output voltage drop across
the LDO regulator. Refer to the following simple
equation:
From the discussion above, PD(MAX) was deter-
mined to equal 267mW at TA = 85°C.
VOUT = 2.5 volts
IOUT = 150mA
IGND = 20µA
IOUT(MAX) < PD(MAX) / (VIN - VOUT
)
For example, if VIN = 5V, VOUT = 2.5V and TA = 25°,
IOUT(MAX) < 267mA. The output short circuit protec-
tion threshold is set between 150mA and 300mA. If
the output load current were to exceed 267mA or if
the ambient temperature were to increase, the inter-
nal die temperature will increase. If the condition
remained constant and the short circuit protection
did not activate, there would be a potential damage
hazard to LDO regulator since the thermal protection
circuit will only activate after a short circuit event
occurs on the LDO regulator output.
VIN(MAX)=(267mW+(2.5Vx150mA))/(150mA+20µA)
VIN(MAX) = 4.28V
Higher input to output voltage differentials can be
obtained with the AAT3201, while maintaining
device functions in 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.
To figure what the maximum input voltage would be
for a given load current refer to the following equa-
tion. This calculation 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 continuous 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:
P
D(MAX) = (VIN - VOUT)IOUT + (VIN x IGND)
This formula can be solved for VIN to determine
the maximum input voltage.
PD(MAX) is assumed to be 267mW
VIN(MAX) = (PD(MAX) + (VOUT x IOUT)) / (IOUT + IGND
)
IGND = 20µA
IOUT = 150mA
VIN = 5.0 volts
VOUT = 2.5 volts
The following is an example for an AAT3201 set for
a 2.5 volt output:
From the discussion above, PD(MAX) was deter-
mined to equal 667mW at TA = 25°C.
%DC = 100(PD(MAX) / ((VIN - VOUT)IOUT + (VIN x IGND))
%DC=100(267mW/((5.0V-2.5V)150mA+(5.0Vx20µA))
%DC = 71.2%
VOUT = 2.5 volts
IOUT = 150mA
IGND = 20µA
For a 150mA output current and a 2.5 volt drop
across the AAT3201 at an ambient temperature of
85°C, the maximum on time duty cycle for the
device would be 71.2%.
VIN(MAX)=(667mW+(2.5Vx150mA))/(150mA +20µA)
VIN(MAX) = 6.95V
Thus, the AAT3201 can sustain a constant 2.5V out-
put at a 150mA load current as long as VIN is ≤ 6.95V
at an ambient temperature of 25°C. 5.5V is the max-
imum input operating voltage for the AAT3201, thus
The following family of curves shows the safe oper-
ating area for duty cycled operation from ambient
room temperature to the maximum operating level.
3201.2002.3.0.91
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