AAT3221/2
150mA NanoPower™ LDO Linear Regulator
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 calcu-
late the power dissipation at the nominal continu-
ous level, then factor in the addition power dissipa-
tion 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
current level and has short 150mA current peaks.
The current 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:
Device Duty Cycle vs. V
DROP
(V
OUT
= 2.5V @ 50°C)
Device Duty Cycle vs. V
DROP
(V
OUT
= 2.5V @ 25°C)
3.5
Voltage Drop (V)
3
2.5
2
1.5
1
0.5
0
0
10
20
30
40
50
60
70
80
90
100
200mA
Duty Cycle (%)
% 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 dis-
sipation over time.
100
3.5
Voltage Drop (V)
3
2.5
2
1.5
1
0.5
0
0
10
20
30
40
50
60
70
80
90
200mA
150mA
Duty Cycle (%)
P
D(MAX)
= (V
IN
- V
OUT
)I
OUT
+ (V
IN
x I
GND
)
P
D(100mA)
= (5.0V - 2.5V)100mA + (5.0V x 1.1mA)
P
D(100mA)
= 250mW
P
D(91.8%D/C)
= %DC x P
D(100mA)
P
D(91.8%D/C)
= 0.918 x 250mW
P
D(91.8%D/C)
= 229.5mW
Device Duty Cycle vs. V
DROP
(V
OUT
= 2.5V @ 85°C)
3.5
Voltage Drop (V)
3
2.5
2
1.5
1
0.5
0
0
10
20
30
40
50
60
100mA
200mA
150mA
70
80
90
100
Duty Cycle (%)
3221.2005.12.1.11
13
AAT [ ADVANCED ANALOG TECHNOLOGY, INC. ]
