AAT3216 ETC, AAT3216 Datasheet - Page 13
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AAT3216
Manufacturer Part Number
AAT3216
Description
150mA MicroPower LDO with PowerOK
Manufacturer
ETC
Datasheet
1.AAT3216.pdf
(16 pages)
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Applications Information
High Peak Output Current Applications
Some applications require the LDO regulator to
operate at continuous nominal level 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 calcu-
late the power dissipation at the nominal continu-
ous level, then factor in the additional power dissi-
pation due to the short duration high current peaks.
For example, a 2.5V system using a AAT3216IGV-
2.5-T1 operates at a continuous 100mA load cur-
rent level and has short 150mA current peaks. The
current peak occurs for 378µs out of a 4.61ms peri-
od. It will be assumed the input voltage is 4.2V.
First the current duty cycle in percent must be cal-
culated:
% Peak Duty Cycle: X/100 = 378µs/4.61ms
% Peak Duty Cycle = 8.2%
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 dis-
sipation over time.
3216.2004.01.0.94
150mA MicroPower™ LDO with PowerOK
P
P
P
P
P
P
The power dissipation for 100mA load occurring for
91.8% of the duty cycle will be 156.6mW. Now the
power dissipation for the remaining 8.2% of the
duty cycle at the 150mA load can be calculated:
P
P
P
P
P
P
The power dissipation for 150mA load occurring for
8.2% of the duty cycle will be 21mW. Finally, the
two power dissipation levels can summed to deter-
mine the total true power dissipation under the var-
ied load.
P
P
P
The maximum power dissipation for the AAT3216
operating at an ambient temperature of 85°C is
211mW. The device in this example will have a total
power dissipation of 177.6mW. This is well within
the thermal limits for safe operation of the device.
D(MAX)
D(100mA)
D(100mA)
D(91.8%D/C)
D(91.8%D/C)
D(91.8%D/C)
D(MAX)
D(150mA)
D(150mA)
D(8.2%D/C)
D(8.2%D/C)
D(8.2%D/C)
D(total)
D(total)
D(total)
= P
= 156.6mW + 21mW
= 177.6mW
= (V
= (V
= (4.2V - 2.5V)100mA + (4.2V x 150µA)
= 170.6mW
= (4.2V - 2.5V)150mA + (4.2V x 150mA)
= 255.6mW
= %DC x P
= 0.082 x 255.6mW
= 21mW
D(100mA)
= %DC x P
= 0.918 x 170.6mW
= 156.6mW
IN
IN
- V
- V
OUT
OUT
+ P
)I
)I
OUT
OUT
D(150mA)
D(150mA)
D(100mA)
+ (V
+ (V
IN
IN
x I
x I
AAT3216
GND
GND
)
)
13