LT3083
16
3083fa
APPLICATIONS INFORMATION
The power in the drive circuit equals:
P
DRIVE
= (V
CONTROL
– V
OUT
)(I
CONTROL
)
where I
CONTROL
is equal to I
OUT
/60. I
CONTROL
is a function
of output current. A curve of I
CONTROL
vs I
OUT
can be found
in the Typical Performance Characteristics curves.
The total power equals:
P
TOTAL
= P
DRIVE
+ P
OUTPUT
The current delivered to the SET pin is negligible and can
be ignored.
V
CONTROL(MAX_CONTINUOUS)
= 3.630V (3.3V + 10%)
V
IN(MAX_CONTINUOUS)
= 1.575V (1.5V + 5%)
V
OUT
= 0.9V, I
OUT
= 3A, T
A
= 50°C
Power dissipation under these conditions is equal to:
P
DRIVE
= (V
CONTROL
– V
OUT
)(I
CONTROL
)
I
CONTROL
=
OUT
60
=
60
= 50mA
P
DRIVE
= (3.630V – 0.9V)(50mA) = 137mW
P
OUTPUT
= (V
IN
– V
OUT
)(I
OUT
)
P
OUTPUT
= (1.575V – 0.9V)(3A) = 2.03W
Total Power Dissipation = 2.16W
Junction Temperature will be equal to:
T
J
= T
A
+ P
TOTAL
• θ
JA
(using tables)
T
J
= 50°C + 2.16W • 16°C/W = 84.6°C
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 sacrific-
ing output current capability. Two techniques are available.
The first 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.
As an example, assume: V
IN
= V
CONTROL
= 5V, V
OUT
= 3.3V
and I
OUT(MAX)
= 2A. Use the formulas from the
Calculating
Junction Temperature
section previously discussed.
Without series resistor R
S
, power dissipation in the LT3083
equals:
P
TOTAL
= 5V − 3.3V
( )
•
2A
60
+ 5V − 3.3V
( )
• 2A
= 3.46W
If the voltage differential (V
DIFF
) across the NPN pass
transistor is chosen as 0.5V, then RS equals:
R
S
=
5V
3.3V
0.5V
= 0.6Ω
Power dissipation in the LT3083 now equals:
P
TOTAL
= 5V − 3.3V
( )
•
2A
60
+ 0.5V • 2A = 1.06W
The LT3083’s power dissipation is now only 30% compared
to no series resistor. R
S
dissipates 2.4W of power. Choose
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
+
–
LT3083
IN
V
CONTROL
OUT
V
OUT
V
IN
′
V
IN
C2
3083 F07
SET
R
SET
R
S
C1
