Datasheet > AAT3220IQY-30-T1

AAT3220IQY-30-T1

更新时间： 2022-11-25 13:29:50

品牌	Logo	应用领域
ANALOGICTECH		稳压器
页数	文件大小	规格书
16页	429K
描述
150mA NanoPower⑩ LDO Linear Regulator

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AAT3220

150mA NanoPower™ LDO Linear Regulator

High Peak Output Current Applications

Device Duty Cycle vs. V_DROP

V_OUT= 2.5V @ 25 degrees 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-

3.5

2.5

200mA

ous level, then factor in the addition power dissi-

pation due to the short duration high current peaks.

150mA

1.5

For example, a 3.0V system using a AAT3220IGV-

2.5-T1 operates at a continuous 100mA load cur-

rent level and has short 150mA current peaks. The

current peak occurs for 378µs out of a 4.61ms peri-

od. It will be assumed the input voltage is 5.0V.

0.5

100

Duty Cycle (%)

First the current duty cycle percentage must be

calculated:

ꢀ Peak Duty Cycle: X/100 = 378µs/4.61ms

ꢀ Peak Duty Cycle = 8.2ꢀ

Device Duty Cycle vs. V_DROP

V_OUT= 2.5V @ 50 degrees C

The LDO Regulator will be under the 100mA load for

91.8ꢀ of the 4.61ms period and have 150mA peaks

occurring for 8.2ꢀ 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.

3.5

2.5

200mA

150mA

1.5

P_D(MAX)= (V_IN- V_OUT)I_OUT+ (V_INx I_GND

P_D(100mA)= (4.2V - 3.0V)100mA + (4.2V x 1.1µA)

P_D(100mA)= 120mW

)

0.5

100

Duty Cycle (%)

P_{D(91.8ꢀD/C)}= ꢀDC x P_D(100mA)

P_{D(91.8ꢀD/C)}= 0.918 x 120mW

P_{D(91.8ꢀD/C)}= 110.2mW

The power dissipation for 100mA load occurring for

91.8ꢀ of the duty cycle will be 110.2mW. Now the

power dissipation for the remaining 8.2ꢀ of the

duty cycle at the 150mA load can be calculated:

Device Duty Cycle vs. V_DROP

V_OUT= 2.5V @ 85 degrees C

3.5

P_D(MAX)= (V_IN- V_OUT)I_OUT+ (V_INx I_GND

P_D(150mA)= (4.2V - 3.0V)150mA + (4.2V x 1.1µA)

P_D(150mA)= 180mW

)

100mA

2.5

P_D(8.2ꢀD/C)= ꢀDC x P_D(150mA)

P_D(8.2ꢀD/C)= 0.082 x 180mW

P_D(8.2ꢀD/C)= 14.8mW

1.5

200mA

150mA

0.5

The power dissipation for a 150mA load occurring

for 8.2ꢀ of the duty cycle will be 14.8mW. Finally,

the two power dissipation levels can be summed to

determine the total power dissipation under the

varied load.

100

Duty Cycle (%)

3220.2001.09.1.0

1...9 10 111213 14 15 16

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