AAT3223_06 Datasheet, PDF (14/17 Pages) Advanced Analogic Technologies – 250mA NanoPower™ LDO Linear Regulator with Power-OK

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Device Duty Cycle vs. V_DROP
(V_DROP = 2.8V @ 25°C)
4
3.5
3
2.5
2
1.5
1
0.5
0
0
250mA
300mA
10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
Device Duty Cycle vs. V_DROP
(V_DROP = 2.8V @ 50°C)
4
3.5
3
2.5
2
1.5
1
0.5
0
0
200mA
300mA
250mA
10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
Device Duty Cycle vs. V_DROP
(V_DROP = 2.8V @ 85°C)
4
3.5
200mA
100mA
3
150mA
2.5
2
1.5
250mA
1
0.5
300mA
0
0 10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
14
AAT3223
250mA NanoPower™ LDO
Linear Regulator with Power-OK
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_D(MAX)
P_D(100mA)
P_D(100mA)
= (V_IN - V_OUT)I_OUT + (V_IN x I_GND)
= (5.0V - 2.8V)100mA + (5.0V x 1.1µA)
= 225.5mW
P_D(91.8%D/C) = %DC x P_D(100mA)
P_D(91.8%D/C) = 0.918 x 225.5mW
P_D(91.8%D/C) = 207mW
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_D(MAX)
P_D(250mA)
P_D(250mA)
= (V_IN - V_OUT)I_OUT + (V_IN x I_GND)
= (5.0V - 2.8V)250mA + (5.0V x 1.1µA)
= 550mW
P_D(8.2%D/C) = %DC x P_D(250mA)
P_D(8.2%D/C) = 0.082 x 550mW
P_D(8.2%D/C) = 45.1mW
3223.2006.03.1.5