AAT3221/2
150mA NanoPower™ LDO Linear RegulatorPowerLinear
TM
PRODUCT DATASHEET
12 3221.2007.11.1.12
www.analogictech.com
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%.
The following family of curves shows the safe operating
area for duty-cycled operation from ambient room tem-
perature to the maximum operating level.
Device Duty Cycle vs. V
DROP
0
0.5
1
1.5
2
2.5
3
3.5
0 102030405060708090100
Duty Cycle (%)
Voltage Drop (V)
200mA
(V
OUT
= 2.5V @ 25
C)
Device Duty Cycle vs. V
DROP
(V
OUT
= 2.5V @ 50
C)
0
0.5
1
1.5
2
2.5
3
3.5
0 102030405060708090100
Duty Cycle (%)
Voltage Drop (V)
200mA
150mA
Device Duty Cycle vs. V
DROP
(V
OUT
= 2.5V @ 85
C)
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20304050607080 90100
Duty Cycle (%)
Voltage Drop (V)
200mA
150mA
100mA
High Peak Output Current Applications
Some applications require the LDO regulator to operate
at continuous nominal levels with short duration, high-
current peaks. The duty cycles for both output current
levels must be taken into account. To do so, one would
first need to calculate the power dissipation at the nom-
inal continuous level, then factor in the addition power
dissipation due to the short duration, high-current
peaks.
For example, a 2.5V system using an AAT3221/
2IGV-2.5-T1 operates at a continuous 100mA load cur-
rent level and has short 150mA current peaks. The cur-
rent peak occurs for 378μs out of a 4.61ms period. It
will be assumed the input voltage is 5.0V.
First, the current duty cycle percentage must be
calculated:
% Peak Duty Cycle: X/100 = 378ms/4.61ms
% Peak Duty Cycle = 8.2%
The LDO regulator will be under the 100mA load for
91.8% of the 4.61ms period and have 150mA peaks
occurring for 8.2% of the time. Next, the continuous
nominal power dissipation for the 100mA load should be
determined then multiplied by the duty cycle to conclude
the actual power dissipation over time.
P
D(MAX)
= (V
IN
- V
OUT
)I
OUT
+ (V
IN
· I
GND
)
P
D(100mA)
= (5.0V - 2.5V)100mA + (5.0V · 1.1mA)
P
D(100mA)
= 250mW
P
D(91.8%D/C)
= %DC · P
D(100mA)
P
D(91.8%D/C)
= 0.918 · 250mW
P
D(91.8%D/C)
= 229.5mW
The power dissipation for a 100mA load occurring for
91.8% of the duty cycle will be 229.5mW. Now the
power dissipation for the remaining 8.2% of the duty
cycle at the 150mA load can be calculated:
P
D(MAX)
= (V
IN
- V
OUT
)I
OUT
+ (V
IN
· I
GND
)
P
D(150mA)
= (5.0V - 2.5V)150mA + (5.0V · 1.1mA)
P
D(150mA)
= 375mW
P
D(8.2%D/C)
= %DC · P
D(150mA)
P
D(8.2%D/C)
= 0.082 · 375mW
P
D(8.2%D/C)
= 30.75mW
