參數(shù)資料
型號: AAT3218IJS-1.25-T1
廠商: Advanced Analog Technology,lnc.
英文描述: 150mA MicroPower⑩ High Performance LDO
中文描述: 150mA的微功耗高性能的LDO⑩
文件頁數(shù): 12/18頁
文件大?。?/td> 384K
代理商: AAT3218IJS-1.25-T1
AAT3218
150mA MicroPower High Performance LDO
12
3218.2006.04.1.8
This formula can be solved for V
IN
to determine the
maximum input voltage.
The following is an example for an AAT3218 set for
a 2.5V output:
From the discussion above, P
D(MAX)
was deter-
mined to equal 526mW at T
A
= 25°C.
Thus, the AAT3218 can sustain a constant 2.5V
output at a 150mA load current as long as V
IN
is
6.0V at an ambient temperature of 25°C. 6.0V is
the absolute maximum voltage where an AAT3218
would never be operated, thus at 25°C, the device
would not have any thermal concerns or opera-
tional V
IN(MAX)
limits.
This situation can be different at 85°C. The follow-
ing is an example for an AAT3218 set for a 2.5V
output at 85°C:
From the discussion above, P
D(MAX)
was deter-
mined to equal 211mW at T
A
= 85°C.
Higher input-to-output voltage differentials can be
obtained with the AAT3218, while maintaining
device functions within the thermal safe operating
area. To accomplish 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.
For example, an application requires V
IN
= 4.2V
while V
OUT
= 2.5V at a 150mA load and T
A
= 85°C.
V
IN
is greater than 3.90V, which is the maximum
safe continuous input level for V
OUT
= 2.5V at
150mA for T
A
= 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
(P
D(MAX)
is assumed to be 211mW):
For a 150mAoutput current and a 2.7V drop across
the AAT3218 at an ambient temperature of 85°C,
the maximum on-time duty cycle for the device
would be 85.54%.
The following family of curves show the safe oper-
ating area for duty-cycled operation from ambient
room temperature to the maximum operating level.
%DC = 100
I
GND
= 150
μ
A
I
OUT
= 150mA
V
IN
= 4.2V
V
OUT
= 2.5V
%DC = 85.54%
P
D(MAX)
(V
IN
- V
OUT
)I
OUT
+ (V
IN
×
I
GND
)
%DC = 100
211mW
(4.2V - 2.5V)150mA + (4.2V
×
150μA)
V
IN(MAX)
=
V
OUT
= 2.5V
I
OUT
= 150mA
I
GND
= 150μA
V
IN(MAX)
= 3.90V
211mW + (2.5V
×
150mA)
150mA + 150μA
V
IN(MAX)
=
V
OUT
= 2.5V
I
OUT
= 150mA
I
GND
= 150μA
V
IN(MAX)
= 6.00V
526mW + (2.5V
×
150mA)
150mA + 150μA
P
D(MAX)
+ (V
OUT
·
I
OUT
)
I
OUT
·
I
GND
V
IN(MAX)
=
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