AAT3221IGV-1.7-T1

更新时间： 2024-02-02 18:24:03

品牌	Logo	应用领域
ANALOGICTECH		稳压器
页数	文件大小	规格书
16页	222K
描述
150mA NanoPower™ LDO Linear Regulator

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PRODUCT DATASHEET

AAT3221/2

PowerLinear^TM

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%.

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.

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. V_DROP

(V_OUT= 2.5V @ 25°C)

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.

3.5

200mA

2.5

1.5

First, the current duty cycle percentage must be

calculated:

0.5

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

% Peak Duty Cycle = 8.2%

100

Duty Cycle (%)

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.

Device Duty Cycle vs. V_DROP

(V_OUT= 2.5V @ 50°C)

3.5

200mA

2.5

150mA

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

)

1.5

P_D(100mA)= (5.0V - 2.5V)100mA + (5.0V · 1.1mA)

P_D(100mA)= 250mW

0.5

P_D(91.8%D/C)= %DC · P_D(100mA)

P_D(91.8%D/C)= 0.918 · 250mW

P_D(91.8%D/C)= 229.5mW

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:

Device Duty Cycle vs. V_DROP

(V_OUT= 2.5V @ 85°C)

3.5

2.5

100mA

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

P_D(150mA)= (5.0V - 2.5V)150mA + (5.0V · 1.1mA)

P_D(150mA)= 375mW

)

200mA

150mA

1.5

0.5

P_D(8.2%D/C)= %DC · P_D(150mA)

P_D(8.2%D/C)= 0.082 · 375mW

P_D(8.2%D/C)= 30.75mW

100

Duty Cycle (%)

w w w . a n a l o g i c t e c h . c o m

3221.2007.11.1.12

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