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AAT3221IJS-3.5-T1

更新时间： 2024-01-22 04:33:03
品牌 Logo 应用领域
ANALOGICTECH 稳压器调节器光电二极管输出元件
页数 文件大小 规格书
16页 222K
描述
150mA NanoPower™ LDO Linear Regulator

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AAT3221IJS-3.5-T1 技术参数

是否Rohs认证： 符合生命周期：Transferred
包装说明：TSOC, TSSOC8,.09,20Reach Compliance Code：unknown
风险等级：5.57可调性：FIXED
标称回动电压 1：0.18 V最大绝对输入电压：7 V
JESD-30 代码：R-PDSO-C8最大负载调整率：0.035%
湿度敏感等级：1输出次数：1
端子数量：8最高工作温度：85 °C
最低工作温度：-40 °C最大输出电流 1：0.1 A
标称输出电压 1：3.5 V封装主体材料：PLASTIC/EPOXY
封装代码：TSOC封装等效代码：TSSOC8,.09,20
封装形状：RECTANGULAR包装方法：TAPE AND REEL
认证状态：Not Qualified子类别：Other Regulators
表面贴装：YES技术：CMOS
端子形式：C BEND端子节距：0.5 mm
端子位置：DUAL最大电压容差：2%
Base Number Matches：1

AAT3221IJS-3.5-T1 数据手册

 浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第8页浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第9页浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第10页浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第12页浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第13页浏览型号AAT3221IJS-3.5-T1的Datasheet PDF文件第14页 
PRODUCT DATASHEET  
AAT3221/2  
PowerLinearTM  
150mA NanoPower™ LDO Linear Regulator  
maximum conditions are calculated at the maximum  
operating temperature where TA = 85°C, under normal  
ambient conditions TA = 25°C. Given TA = 85°C, the  
maximum package power dissipation is 267mW. At TA =  
25°C, the maximum package power dissipation is  
667mW.  
From the discussion above, PD(MAX) was determined to  
equal 667mW at TA = 25°C. Thus, the AAT3221/2 can  
sustain a constant 2.5V output at a 150mA load current  
as long as VIN is ≤6.95V at an ambient temperature of  
25°C. 5.5V is the maximum input operating voltage for  
the AAT3221/2, thus at 25°C the device would not have  
any thermal concerns or operational VIN(MAX) limits.  
The maximum continuous output current for the  
AAT3221/2 is a function of the package power dissipa-  
tion and the input-to-output voltage drop across the  
LDO regulator. Refer to the following simple equation:  
This situation can be different at 85°C. The following is  
an example for an AAT3221/2 set for a 2.5 volt output  
at 85°C:  
VOUT = 2.5 volts  
IOUT = 150mA  
IGND = 1.1μA  
PD(MAX)  
IOUT(MAX)  
=
(VIN - VOUT  
)
(267mW + [2.5V · 150mA])  
(150mA + 1.1µA)  
For example, if VIN = 5V, VOUT = 2.5V and TA = 25°C,  
IOUT(MAX) < 267mA. The output short-circuit protection  
threshold is set between 150mA and 300mA. If the out-  
put load current were to exceed 267mA or if the ambient  
temperature were to increase, the internal die tempera-  
ture would increase. If the condition remained constant  
and the short-circuit protection did not activate, there  
would be a potential damage hazard to the LDO regula-  
tor since the thermal protection circuit would only acti-  
vate after a short-circuit event occured on the LDO  
regulator output.  
VIN(MAX)  
=
VIN(MAX) = 4.28V  
From the discussion above, PD(MAX) was determined to  
equal 267mW at TA = 85°C.  
Higher input-to-output voltage differentials can be  
obtained with the AAT3221/2, while maintaining device  
functions in the thermal safe operating area. To accom-  
plish this, the device thermal resistance must be reduced  
by increasing the heat sink area or by operating the LDO  
regulator in a duty-cycled mode.  
To determine the maximum input voltage for a given  
load current, refer to the following equation. This calcu-  
lation accounts for the total power dissipation of the LDO  
regulator, including that caused by ground current.  
For example, an application requires VIN = 5.0V while  
VOUT = 2.5V at a 150mA load and TA = 85°C. VIN is  
greater than 4.28V, which is the maximum safe continu-  
ous input level for VOUT = 2.5V at 150mA for TA = 85°C.  
To maintain this high input voltage and output current  
level, the LDO regulator must be operated in a duty-  
cycled mode. Refer to the following calculation for duty-  
cycle operation:  
PD(MAX) = (VIN - VOUT)IOUT + (VIN · IGND  
)
This formula can be solved for VIN to determine the  
maximum input voltage.  
IGND = 1.1μA  
(PD(MAX) + [VOUT · IOUT])  
)
IOUT = 150mA  
VIN = 5.0 volts  
VOUT = 2.5 volts  
VIN(MAX)  
=
(IOUT + IGND  
The following is an example for an AAT3221/2 set for a  
2.5 volt output:  
PD(MAX)  
([VIN - VOUT]IOUT + [VIN · IGND])  
%DC = 100  
VOUT = 2.5 volts  
IOUT = 150mA  
IGND = 1.1μA  
267mW  
([5.0V - 2.5V]150mA + [5.0V · 1.1µA])  
%DC = 100  
(667mW + [2.5V · 150mA])  
(150mA + 1.1µA)  
%DC = 71.2%  
VIN(MAX)  
=
PD(MAX) is assumed to be 267mW.  
VIN(MAX) = 6.95V  
w w w . a n a l o g i c t e c h . c o m  
3221.2007.11.1.12  
11  
1...891011121314...16

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