AAT4285 Datasheet, PDF (9/11 Pages) Advanced Analogic Technologies – 12V Slew Rate Controlled Load Switch

AAT4285
12V Slew Rate Controlled Load Switch
For maximum current, refer to the following equation:
I_OUT(MAX) <
P_D(MAX)
R_DS
For example, if V_IN = 12V, R_DS(MAX) = 307mΩ and
T_A = 25°C, I_OUT(MAX) = 1.53A. If the output load cur-
rent were to exceed 1.53A or if the ambient tem-
perature were to increase, the internal die temper-
ature would increase, and the device would be
damaged. Higher peak currents can be obtained
with the AAT4285. To accomplish this, the device
thermal resistance must be reduced by increasing
the heat sink area or by operating the load switch
in a duty cycled manner.
High Peak Output Current Applications
Some applications require the load switch to operate
at a continuous nominal current level with short
duration, high-current peaks. The duty cycle for both
output current levels must be taken into account. To
do so, first calculate the power dissipation at the
nominal continuous current level, and then add in
the additional power dissipation due to the short
duration, high-current peak scaled by the duty factor.
For example, a 12V system using an AAT4285
operates at a continuous 100mA load current level
and has short 2A current peaks.
The current peak occurs for 500µs out of a 5ms
period.
First, the current duty cycle is calculated:
⎛ x ⎞ ⎛ 500μs⎞
% Peak Duty Cycle = _{⎝ 100âŽ}  = _{⎝ 5.0msâŽ} 
% Peak Duty Cycle = 10%
The load current is 100mA for 90% of the 5ms peri-
od and 2A for 10% of the period.
De-rated for temperature:
240mΩ · (1 + 0.0028 · (125°C - 25°C)) = 307mΩ
4285.2007.04.1.0
The power dissipation for a 100mA load is calculat-
ed as follows:
P_D(MAX) = I_OUT2 · R_DS
P_D(100mA) = (100mA)² · 307mΩ
P_D(100mA) = 3.07mW
P_D(90%D/C) = %DC · P_D(100mA)
P_D(90%D/C) = 0.90 · 3.07mW
P_D(90%D/C) = 2.76mW
The power dissipation for 100mA load at 90% duty
cycle is 2.76mW. Now the power dissipation for the
remaining 10% of the duty cycle at 2A is calculated:
P_D(MAX) = I_OUT2 · R_DS
P_D(2A) = (2A)² · 307mΩ
P_D(2A) = 1.23W
P_D(10%D/C) = %DC · P_D(2A)
P_D(10%D/C) = 0.10 · 1.23mW
P_D(10%D/C) = 123mW
The power dissipation for 2A load at 10% duty
cycle is 123mW. Finally, the two power figures are
summed to determine the total true power dissipa-
tion under the varied load.
P_D(TOTAL) = P_D(100mA) + P_D(2A)
P_D(TOTAL) = 2.76mW + 123mW
P_D(TOTAL) = 125.76mW
The maximum power dissipation for the AAT4285
operating at an ambient temperature of 85°C is
286mW. The device in this example will have a total
power dissipation of 123mW. This is well within the
thermal limits for safe operation of the device; in fact,
at 85°C, the AAT4285 will handle a 2A pulse for up
to 23% duty cycle. At lower ambient temperatures,
the duty cycle can be further increased.
Printed Circuit Board Layout
Recommendations
For proper thermal management and to take
advantage of the low R_DS(ON) of the AAT4285, a few
circuit board layout rules should be followed: V_IN
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