lt3083idf Linear Technology Corporation, lt3083idf Datasheet - Page 17

• Manufacturer Part Number: lt3083idf
• Description: Lt3083 - Adjustable 3a Single Resistor Low Dropout Regulator
• Manufacturer: Linear Technology Corporation

APPLICATIONS INFORMATION
The power in the drive circuit equals:
where I
of output current. A curve of I
in the Typical Performance Characteristics curves.
The total power equals:
The current delivered to the SET pin is negligible and can
be ignored.
Power dissipation under these conditions is equal to:
Junction Temperature will be equal to:
In this case, the junction temperature is below the maxi-
mum rating, ensuring reliable operation.
Reducing Power Dissipation
In some applications it may be necessary to reduce the
power dissipation in the LT3083 package without sacrifi c-
ing output current capability. Two techniques are available.
The fi rst technique, illustrated in Figure 7, employs a
resistor in series with the regulator’s input. The voltage
drop across RS decreases the LT3083’s input-to-output
differential voltage and correspondingly decreases the
LT3083’s power dissipation.
P
P
V
V
V
P
P
P
P
Total Power Dissipation = 2.16W
T
T
I
CONTROL
J
J
DRIVE
TOTAL
CONTROL(MAX_CONTINUOUS)
IN(MAX_CONTINUOUS)
OUT
DRIVE
DRIVE
OUTPUT
OUTPUT
= T
= 50°C + 2.16W • 16°C/W = 84.6°C
CONTROL
= 0.9V, I
A
= (V
= (V
= (3.630V – 0.9V)(50mA) = 137mW
= P
+ P
= (V
= (1.575V – 0.9V)(3A) = 2.03W
=
DRIVE
TOTAL
CONTROL
CONTROL
I
OUT
60
is equal to I
IN
OUT
– V
+ P
=
• θ
= 3A, T
OUT
3A
60
OUTPUT
JA
– V
– V
= 1.575V (1.5V + 5%)
)(I
(using tables)
= 50mA
OUT
OUT
OUT
OUT
A
CONTROL
= 50°C
)(I
)(I
= 3.630V (3.3V + 10%)
/60. I
)
CONTROL
CONTROL
CONTROL
vs I
OUT
)
)
can be found
is a function
As an example, assume: V
and I
Junction Temperature section previously discussed.
Without series resistor R
equals:
If the voltage differential (V
transistor is chosen as 0.5V, then RS equals:
Power dissipation in the LT3083 now equals:
The LT3083’s power dissipation is now only 30% compared
to no series resistor. R
appropriate wattage resistors or use multiple resistors in
parallel to handle and dissipate the power properly.
Figure 7. Reducing Power Dissipation Using a Series Resistor
P
R
P
TOTAL
TOTAL
S
OUT(MAX)
=
5V − 3.3V − 0.5V
= 3.46W
= 5V − 3.3V
= 5V − 3.3V
(
(
C1
= 2A. Use the formulas from the Calculating
2A
V
SET
R
CONTROL
SET
S
LT3083
+
–
dissipates 2.4W of power. Choose
S
)
)
, power dissipation in the LT3083
•
•
IN
⎛
⎜
⎝
⎛
⎜
⎝
= 0.6Ω
= V
2A
2A
60
60
DIFF
3083 F07
CONTROL
⎞
⎟ + 0.5V • 2A = 1.06W
⎠
⎞
⎟ + 5V − 3.3V
⎠
) across the NPN pass
OUT
IN
(
R
= 5V, V
S
C2
LT3083
V
V
V
IN
IN
OUT
)
OUT
• 2A
17
= 3.3V
3083f