AAT3221IJS-2.7-T1 데이터 시트보기 (PDF) - Analog Technology Inc
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AAT3221IJS-2.7-T1 Datasheet PDF : 16 Pages
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AAT3221/2
150mA NanoPower™ LDO Linear Regulator
Device Duty Cycle vs. VDROP
VOUT = 2.5V @ 25 degrees C
3.5
3
2.5
2
1.5
1
0.5
0
0
200mA
10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
Device Duty Cycle vs. VDROP
VOUT = 2.5V @ 50 degrees C
3.5
3
2.5
200mA
2
150mA
1.5
1
0.5
0
0
10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
Device Duty Cycle vs. VDROP
VOUT = 2.5V @ 85 degrees C
3.5
3
100mA
2.5
2
200mA
1.5
1
150mA
0.5
0
0 10 20 30 40 50 60 70 80 90 100
Duty Cycle (%)
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 a 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:
% 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.
PD(MAX) = (VIN - VOUT)IOUT + (VIN x IGND)
PD(100mA) = (5.0V - 2.5V)100mA + (5.0V x 1.1mA)
PD(100mA) = 250mW
PD(91.8%D/C) = %DC x PD(100mA)
PD(91.8%D/C) = 0.918 x 250mW
PD(91.8%D/C) = 229.5mW
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3221.2002.03.0.94
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