PRODUCT DATASHEET
AAT3221/2
PowerLinearTM
150mA NanoPower™ LDO Linear Regulator
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%.
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.
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. VDROP
(VOUT = 2.5V @ 25°C)
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.
3.5
3
200mA
2.5
2
1.5
1
First, the current duty cycle percentage must be
calculated:
0.5
0
% Peak Duty Cycle: X/100 = 378ms/4.61ms
% Peak Duty Cycle = 8.2%
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
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.
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 50°C)
3.5
3
200mA
2.5
2
150mA
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND
)
1.5
1
PD(100mA) = (5.0V - 2.5V)100mA + (5.0V · 1.1mA)
PD(100mA) = 250mW
0.5
0
PD(91.8%D/C) = %DC · PD(100mA)
PD(91.8%D/C) = 0.918 · 250mW
PD(91.8%D/C) = 229.5mW
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
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:
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 85°C)
3.5
3
2.5
2
100mA
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND
PD(150mA) = (5.0V - 2.5V)150mA + (5.0V · 1.1mA)
PD(150mA) = 375mW
)
200mA
150mA
1.5
1
0.5
0
PD(8.2%D/C) = %DC · PD(150mA)
PD(8.2%D/C) = 0.082 · 375mW
PD(8.2%D/C) = 30.75mW
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
w w w . a n a l o g i c t e c h . c o m
12
3221.2007.11.1.12
ANALOGICTECH [ ADVANCED ANALOGIC TECHNOLOGIES ]
