AAT3221IJS-1.8-T1 View Datasheet(PDF) - Advanced Analogic Technologies
Part Name
Description
MFG CO.
'AAT3221IJS-1.8-T1' PDF : 16 Pages View PDF
PowerLinearTM
PRODUCT DATASHEET
AAT3221/2
150mA NanoPower™ LDO Linear Regulator
For a 150mA output current and a 2.5 volt drop across
the AAT3221/2 at an ambient temperature of 85°C, the
maximum on-time duty cycle for the device would be
71.2%.
The following family of curves shows the safe operating
area for duty-cycled operation from ambient room tem-
perature to the maximum operating level.
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 25°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°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 (%)
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 calculate the power dissipation at the nom-
inal continuous level, then factor in the addition power
dissipation 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 cur-
rent level and has short 150mA current peaks. The cur-
rent 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 dissipation over time.
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND)
PD(100mA) = (5.0V - 2.5V)100mA + (5.0V · 1.1mA)
PD(100mA) = 250mW
PD(91.8%D/C) = %DC · PD(100mA)
PD(91.8%D/C) = 0.918 · 250mW
PD(91.8%D/C) = 229.5mW
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 85°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 (%)
The power dissipation for a 100mA load occurring for
91.8% of the duty cycle will be 229.5mW. Now the
power dissipation for the remaining 8.2% of the duty
cycle at the 150mA load can be calculated:
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND)
PD(150mA) = (5.0V - 2.5V)150mA + (5.0V · 1.1mA)
PD(150mA) = 375mW
PD(8.2%D/C) = %DC · PD(150mA)
PD(8.2%D/C) = 0.082 · 375mW
PD(8.2%D/C) = 30.75mW
12
www.analogictech.com
3221.2007.11.1.12
Share Link: 
