AAT3221/2
150mA NanoPower™ LDO Linear Regulator
culated at the maximum operating temperature
input operating voltage for the AAT3221/2, thus at
25°C the device would not have any thermal con-
cerns or operational VIN(MAX) limits.
where TA = 85°C, under normal ambient conditions
TA = 25°C. Given TA = 85°C, the maximum pack-
age 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 AAT3221/2 set for a 2.5 volt
output at 85°C:
The maximum continuous output current for the
AAT3221/2 is a function of the package power dis-
sipation and the input-to-output voltage drop
across the LDO regulator. Refer to the following
simple equation:
VOUT = 2.5 volts
IOUT = 150mA
IGND = 1.1µA
VIN(MAX)=(267mW+(2.5Vx150mA))/(150mA +1.1µA)
IOUT(MAX) < PD(MAX) / (VIN - VOUT
)
V
IN(MAX) = 4.28V
For example, if VIN = 5V, VOUT = 2.5V and TA = 25°C,
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 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 regulator since the thermal pro-
tection circuit would only activate after a short-circuit
event occured on the LDO regulator output.
From the discussion above, PD(MAX) was deter-
mined 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 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 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:
To determine the maximum input voltage for a
given load current, refer to the following equation.
This calculation accounts for the total power dissi-
pation of the LDO regulator, including that caused
by ground current.
PD(MAX) = (VIN - VOUT)IOUT + (VIN x IGND
)
This formula can be solved for VIN to determine the
maximum input voltage.
IGND = 1.1µA
IOUT = 150mA
VIN = 5.0 volts
VOUT = 2.5 volts
VIN(MAX) = (PD(MAX) + (VOUT x IOUT)) / (IOUT + IGND
)
The following is an example for an AAT3221/2 set
for a 2.5 volt output:
%DC = 100(PD(MAX) / ((VIN - VOUT)IOUT + (VIN x IGND))
%DC=100(267mW/((5.0V-2.5V)150mA+(5.0Vx1.1µA))
%DC = 71.2%
V
OUT = 2.5 volts
IOUT = 150mA
IGND = 1.1µA
PD(MAX) is assumed to be 267mW.
VIN(MAX)=(667mW+(2.5Vx150mA))/(150mA +1.1µA)
VIN(MAX) = 6.95V
For a 150mA output current and a 2.5 volt drop
across the AAT3221/2 at an ambient temperature
of 85°C, the maximum on-time duty cycle for the
device would be 71.2%.
From the discussion above, PD(MAX) was deter-
mined 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
The following family of curves shows the safe oper-
ating area for duty-cycled operation from ambient
room temperature to the maximum operating level.
12
3221.2005.12.1.11