AAT3223
250mA NanoPower™ LDO
Linear Regulator with Power-OK
High Peak Output Current Applications
Device Duty Cycle vs. VDROP
+
(VDROP = 2.8V @ 25°C)
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.
4
3.5
3
250mA
2.5
2
300mA
1.5
1
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.
0.5
0
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
First, the current duty cycle percentage must be
calculated:
Device Duty Cycle vs. VDROP
(VDROP = 2.8V @ 50°C)
% Peak Duty Cycle: X/100 = 378ms/4.61ms
% Peak Duty Cycle = 8.2%
4
3.5
3
200mA
2.5
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.
300mA
1.5
1
250mA
0.5
0
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
PD(MAX)
= (VIN - VOUT)IOUT + (VIN x IGND)
PD(100mA) = (5.0V - 2.8V)100mA + (5.0V x 1.1µA)
PD(100mA) = 225.5mW
Device Duty Cycle vs. VDROP
(VDROP = 2.8V @ 85°C)
PD(91.8%D/C) = %DC x PD(100mA)
PD(91.8%D/C) = 0.918 x 225.5mW
PD(91.8%D/C) = 207mW
4
3.5
3
200mA
100mA
150mA
2.5
2
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:
1.5
1
250mA
300mA
0.5
0
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
PD(MAX)
= (VIN - VOUT)IOUT + (VIN x IGND)
PD(250mA) = (5.0V - 2.8V)250mA + (5.0V x 1.1µA)
PD(250mA) = 550mW
PD(8.2%D/C) = %DC x PD(250mA)
PD(8.2%D/C) = 0.082 x 550mW
PD(8.2%D/C) = 45.1mW
14
3223.2006.03.1.5
AAT [ ADVANCED ANALOG TECHNOLOGY, INC. ]
