AAT3215
150mA CMOS High Performance LDO
Thus, the AAT3215 can sustain a constant 2.5V out-
put at a 150mA load current as long as VIN is ≤ 6.00V
at an ambient temperature of 25°C. 6.0V is the
absolute maximum voltage where an AAT3215
would never be operated, thus at 25°C, the device
would not have any thermal concerns or operational
VIN(MAX) limits.
For a 150mA output current and a 2.7V drop
across the AAT3215 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.
This situation can be different at 85°C. The follow-
ing is an example for an AAT3215 set for a 2.5V
output at 85°C:
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 25°C)
3.5
3
VOUT = 2.5V
IOUT
IGND
= 150mA
= 150µA
2.5
200mA
2
1.5
1
211mW + (2.5V
=
· 150mA)
150mA + 150μA
VIN(MAX)
0.5
0
VIN(MAX) = 3.90V
0
10
20
30
40
50
60
70
80
90
100
From the discussion above, PD(MAX) was deter-
mined to equal 211mW at TA = 85°C.
Duty Cycle (%)
Higher input-to-output voltage differentials can be
obtained with the AAT3215, 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.
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 50°C)
3.5
3
For example, an application requires VIN = 4.2V
while VOUT = 2.5V at a 150mA load and TA = 85°C.
VIN is greater than 3.90V, which is the maximum
safe continuous 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:
2.5
2
200mA
1.5
1
150mA
0.5
0
0
10
20
30
40
50
60
70
80
90
100
Duty Cycle (%)
IGND = 150µA
IOUT = 150mA
VIN = 4.2V
Device Duty Cycle vs. VDROP
(VOUT = 2.5V @ 85°C)
VOUT = 2.5V
3.5
3
100mA
PD(MAX)
(VIN - VOUT)IOUT + (VIN · IGND)
%DC = 100
2.5
2
200mA
150mA
211mW
(4.2V - 2.5V)150mA + (4.2V
1.5
1
%DC = 100
·
150μA)
0.5
0
%DC = 85.54%
0
10
20
30
40
50
60
70
80
90
100
PD(MAX) was assumed to be 211mW.
Duty Cycle (%)
12
3215.2006.05.1.6
ANALOGICTECH [ ADVANCED ANALOGIC TECHNOLOGIES ]
