參數(shù)資料
型號(hào): AAT3223IGU-1.8-T1
廠商: Advanced Analogic Technologies, Inc.
元件分類(lèi): 基準(zhǔn)電壓源/電流源
英文描述: 3A Fixed Low Dropout Linear Regulator (LDO)
中文描述: 3A固定低壓差線性穩(wěn)壓器(LDO)
文件頁(yè)數(shù): 13/17頁(yè)
文件大小: 151K
代理商: AAT3223IGU-1.8-T1
AAT3223
250mA NanoPower LDO
Linear Regulator with Power-OK
3223.2005.04.1.4
13
The following is an example for an AAT3223 set for
a 2.8V output:
From the discussion above, P
D(MAX)
was deter-
mined to equal 667mW at T
A
= 25°C.
Thus, the AAT3223 can sustain a constant 2.8V
output at a 150mA load current as long as V
IN
is
9.11V at an ambient temperature of 25°C. 5.5V
is the maximum input operating voltage for the
AAT3223, thus at 25°C, the device would not
have any thermal concerns or operational V
IN(MAX)
limits.
This situation can be different at 85°C. The follow-
ing is an example for an AAT3223 set for a 2.8V
output at 85°C:
From the discussion above, P
D(MAX)
was deter-
mined to equal 267mW at T
A
= 85°C.
Higher input-to-output voltage differentials can be
obtained with the AAT3223, while maintaining
device functions in 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
= 5.0V
while V
OUT
= 2.8V at a 150mA load and T
A
= 85°C.
V
IN
is greater than 4.58V, which is the maximum
safe continuous input level for V
OUT
= 2.8V at
150mA for T
A
= 85°C. To maintain this high input
voltage and output current level, the LDO regula-
tor must be operated in a duty-cycled mode.
Refer to the following calculation for duty-cycle
operation:
P
D(MAX)
was assumed to be 267mW.
For a 150mAoutput current and a 2.2V drop across
the AAT3223 at an ambient temperature of 85°C,
the maximum on-time duty cycle for the device
would be 80.9%.
The following family of curves shows the safe
operating area for duty-cycled operation from
ambient room temperature to the maximum oper-
ating level.
%DC = 100
I
GND
= 1.1
μ
A
I
OUT
= 150mA
V
IN
= 5.0V
V
OUT
= 2.8V
%DC = 80.9%
P
D(MAX)
(V
IN
- V
OUT
)I
OUT
+ (V
IN
×
I
GND
)
%DC = 100
267mW
(5.0V - 2.8V)150mA + (5.0V
×
1.1
μ
A)
V
IN(MAX)
=
V
OUT
= 2.9V
I
OUT
= 150mA
I
GND
= 1.1
μ
A
V
IN(MAX)
= 4.58V
267mW + (2.8V
×
150mA)
150mA + 1.1
μ
A
V
IN(MAX)
=
V
OUT
= 2.9V
I
OUT
= 250mA
I
GND
= 1.1
μ
A
V
IN(MAX)
= 9.11V
667mW + (2.8V
×
150mA)
150mA + 1.1
μ
A
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