aat3223igu-3.0-t1 Advanced Analogic Technologies, Inc., aat3223igu-3.0-t1 Datasheet - Page 14

• Manufacturer Part Number: aat3223igu-3.0-t1
• Description: Aat3223 | 250ma Nanopower Ldo Linear Regulator With Power Ok
• Manufacturer: Advanced Analogic Technologies, Inc.

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3.5
2.5
1.5
0.5
3.5
2.5
1.5
0.5
3.5
2.5
1.5
0.5
4
3
2
1
0
4
3
2
1
0
4
3
2
1
0
0
0
0
10
10
10
Device Duty Cycle vs. V
Device Duty Cycle vs. V
Device Duty Cycle vs. V
20
20
20
(V
30
30
30
(V
(V
DROP
250mA
DROP
DROP
Duty Cycle (%)
Duty Cycle (%)
Duty Cycle (%)
40
40
40
200mA
= 2.8V @ 50°C)
= 2.8V @ 85°C)
= 2.8V @ 25°C)
300mA
50
50
50
300mA
60
60
60
150mA
70
70
70
300mA
250mA
DROP
DROP
DROP
80
80
80
100mA
200mA
250mA
90
90
90
100
100
100
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 2.8V system using a AAT3223IGU-
2.8-T1 operates at a continuous 100mA load cur-
rent level and has short 250mA 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:
% 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 and then multi-
plied by the duty cycle to conclude the actual
power dissipation over time.
P
P
P
P
P
P
The power dissipation for a 100mA load occurring
for 91.8% of the duty cycle will be 207mW. Now
the power dissipation for the remaining 8.2% of the
duty cycle at the 150mA load can be calculated:
P
P
P
P
P
P
D(MAX)
D(100mA)
D(100mA)
D(91.8%D/C)
D(91.8%D/C)
D(91.8%D/C)
D(MAX)
D(250mA)
D(250mA)
D(8.2%D/C)
D(8.2%D/C)
D(8.2%D/C)
Linear Regulator with Power-OK
250mA NanoPower™ LDO
= (V
= (5.0V - 2.8V)250mA + (5.0V x 1.1µA)
= 550mW
= %DC x P
= 0.082 x 550mW
= 45.1mW
= (V
= (5.0V - 2.8V)100mA + (5.0V x 1.1µA)
= 225.5mW
= %DC x P
= 0.918 x 225.5mW
= 207mW
IN
IN
- V
- V
OUT
OUT
D(250mA)
D(100mA)
)I
)I
OUT
OUT
+ (V
+ (V
AAT3223
IN
IN
x I
x I
3223.2006.03.1.5
GND
GND
)
)