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AAT3221/2

150mA NanoPower™ LDO Linear Regulator

PowerLinearTM

PRODUCT DATASHEET

3221.2007.11.1.12

www .analogictech.com

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

0.5

1.5

2.5

3.5

2030

405060

7080

100

Duty Cycle (%)

200mA

(VOUT = 2.5V @ 25

°C)

Device Duty Cycle vs. VDROP

(VOUT = 2.5V @ 50

°C)

0.5

1.5

2.5

3.5

2030

405060

7080

100

Duty Cycle (%)

200mA

150mA

Device Duty Cycle vs. VDROP

(VOUT = 2.5V @ 85

°C)

0.5

1.5

2.5

3.5

2030

405060

7080

100

Duty Cycle (%)

200mA

150mA

100mA

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

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

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